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Montana Pete
09-19-2008, 06:59 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?

Bear in mind that a rifle is a machine that produces useful work. Work is measured in foot/pounds. That's a simple calculation and can be found in most reloading manuals.

Now . . . horsepower is a measure of work plotted vs. time. In other words, the power of this machine that is the rifle. A machine rated at 10 horsepower can perform more useful work in a given unit of time than a machine rated at 5 horsepower.

Now, take any rifle and any load you want, and see if you can determine the horsepower produced by said rifle.

I would try this, but I'm an English major, unfortunately, and incapable of the analysis.

I've often wondered about this . . . hope someone takes me up on it.

Montana Pete

4Mesh
09-19-2008, 07:13 PM
Horsepower is weight over time. You'd have to define the time. You can calculate the energy generated because that's a finite amount. Horsepower is something that is assumed to be continous.

You could calculate how many horsepower is generated during the acceleration phase of the bullets path down the barrel, but would have to have an accruate time of acceleration to do so. Of course the "horsepower" ends when the powder is burnt.

Energy is easiser. Measure how fast the bullets going, how much it weighs and whalla.

Montana Pete
09-19-2008, 08:16 PM
4Mesh--

I have thought about this. You can only consider the "machine" (rifle) to be operating when it is actually accelerating the bullet down the barrel.

You need to come up with an estimate of the duration the "machine" operates. That is, what is the time element between ignition of the primer and departure of the bullet from the muzzle? If we do not have an actual acceleration curve, we would need to average the rest speed at ignition against the muzzle velocity-- which contains a time factor.

Once you get the duration that the machine operates -- accelerates one round from rest to the muzzle -- then you have the information needed to assess the horsepower.

I'm the English major, you are the engineer or math teacher -- whatever.

You should be able to do this. Just be prepared for a large number.

I read in the Speer Reloading Manual a situation involving the .223, I believe-- not sure -- where a 1/7 twist barrel would put the spin on a 3500 fps bullet of -- get this -- 330,000 rpm. In this instance, most varmint bullets would literally blow up out of the muzzle due to incredible angular rotation.

The numbers are huge, but that's no reason not to do the numbers.

Hope you can give it a try--

jackie schmidt
09-19-2008, 08:22 PM
Then there are Rifles that sacrifice ALL other aspects of Rifle performance for one thing. The ability of that Rifle to stack one bullet on top of another in a Competitive Arena......jackie

4Mesh
09-19-2008, 08:35 PM
Stolen from elsewhere


Power is the measurement of how much work can be done per unit of time. It can be expressed in either watts or horsepower, depending on whether metric (International System) or English (Imperial) units of measure are used to calculate it. Horsepower is a term first coined by James Watt, who calculated that a single horse can do 33,000 foot-pounds of work in 1 minute.


Step 1
Calculate how much work is done by the device. Work is the measure of force times distance. For instance, raising an object weighing 1 Newton 1 meter results in 1 Newton-meter (or 1 Joule) of work. In English units, raising an object weighing 1 pound 1 foot results in 1 foot-pound of work.

Step 2
Understand the definition of "weight." Weight is a measurement of mass multiplied the acceleration of gravity. In SI units, kilogram is a measurement of mass, although it is commonly used to refer to weight. 1 kilogram of mass multiplied by the acceleration of gravity is 1 Newton of weight. Similarly, in English units, a pound is a measurement of weight, and slugs is a measurement of mass, making a pound equal to 1 slug multiplied by the acceleration of gravity.

Step 3
Learn how to define "work." To do work, you must move an object in the same direction as the force that was applied to it. For instance, if you try to lift a box, but can't, you haven't done any work even though the attempt my tire you. If you lift the box and carry it across the room, only the distance that you lifted the box counts as work.

Step 4
Determine how much time was required to do the work. If it took 1 second to raise the 1-Newton weight 1 meter, the power of the device that raised the weight is 1 Joule per second or 1 watt. In English units, the device that raised a 1-pound weight 1 foot in 1 second is 1 foot-pound per second.

Step 5
Convert foot-pounds per second into horsepower by dividing by 550. You can also convert watts to horsepower if you know that 746 watts equals 1 horsepower.

There you go. I have to load ammo and all I'm concerned with is how well they group and score. HP don't win, and this ain't a race!

longshooter
09-19-2008, 09:30 PM
It seems to me if the horsepower were determinable as associated with a rifle/cartridge combination, that there would be two horsepowers that you would be looking at. The first would be the indicated or calculated amount within the barrel (I guess you could say as related to internal ballistics), and would have to include in the equation all of the losses associated with that process. The other horsepower would have to do with the bullet at the moment it exits the muzzle. In my way of thinking (i know, that could be very bad:confused:), the bullet would have to be immediately arrested, and it's energy available put into a device for measuring brake horsepower. Perhaps we might call this external horsepower. It too would have losses that would have to be taken into consideration.

So, we would be talking about the produced h.p., or the delivered h.p.

The internal portion would seem to come from an elaborate laboratory test and associated with calculation.
The external portion might be pulled off if there were a device suitable for that determination.

Perhaps a rocket scientist could calculate the whole thing in a few moments.

My $.02

Con Cross
09-19-2008, 09:43 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?

Bear in mind that a rifle is a machine that produces useful work.
Montana Pete


Mr. Montana Pete would you believe, that it can be calculated.

Mr. Montana Pete in your own words please explain to us, why do you feel to know how many horsepower a particular cartridge will have?

Mr. Montana Pete would you believe, that to any non-scientist shooter who can't already flawlesly calculate the horsepower, this information is totally useless.

Con

J. Valentine
09-19-2008, 09:54 PM
550 foot pounds per second is 1 horsepower.
How it can be related to a bullet I have no idea and I don't think the math exists as it has never been required as far as I know .
Very loosely if you have 1100 foot pounds of muzzel energy you could loosely describe the the gun as 2 HP at the muzzel and a half dead Donkey at 1000 yards .:D

SBH
09-20-2008, 11:13 AM
I could be waaay off but taking a quick stab at it I came up with this based on a 300 win mag load.

3633 ft/lbs energy at muzzle = .001835 hp/hour
Time from ignition to bullet exit from muzzle according to quick load .0012 seconds

Number of .0012 time slices per hour = 3,000,000

.001835 hp/hr x 3,000,000 = 5505 hp

As I said above I could be way off but maybe it will get someone smarter thinking.

Stephen Hall

alinwa
09-20-2008, 06:34 PM
I could be waaay off but taking a quick stab at it I came up with this based on a 300 win mag load.

3633 ft/lbs energy at muzzle = .001835 hp/hour
Time from ignition to bullet exit from muzzle according to quick load .0012 seconds

Number of .0012 time slices per hour = 3,000,000

.001835 hp/hr x 3,000,000 = 5505 hp

As I said above I could be way off but maybe it will get someone smarter thinking.

Stephen Hall


Yup Steve, that's WAYYY!!! off :)

There's really nothing to "think about" here....... this idea that the discharge of a rifle is super-duper powerful or "mega-horsepower" is what's way off.

The act of firing a rifle is no more "powerful" nor mystical than the controlled deflagration or "explosion" that goes on inside of an internal combustion motor. The difference is that the rifle only does it ONCE.......perty wimpy really......EVERY DAY when you drive your car you fire hundreds of thousands of "rifle shot equivalents".


BUT..... with a huge difference. ONE firing stroke of your 350Chev has more energy that a shot from a rifle, and it fires OVER and OVER and OVER.....

Now a Gatling gun or another sort of machine gun makes "horsepower" over time BUT, to any of you service guys who've actually held the power....... just how much "power" does that little thing have if you were trying to power your APC or HumVee with it? Think about it.....


Now, if one were to actually take the firing cycle of a submachine gun over time......say 3000rds firedX cyclic rate..... one could then quite easily express this in "horsepower". Meaningless but easy.


The problem here is in the terminology,the definition of horsepower VS the thing that a rifle does. It doesn't take a "rocket scientist" nor an Engineer nor even any sort of a math whiz to figure this out, just a basic understanding of "work" and "horsepower".......... given a proper definition the "horsepower" can easily be computed.expressed with any unite from watts to ergs to joules to BTU's if you'd like....

al

longshooter
09-20-2008, 07:59 PM
if i were trying to decide the best method, i would chose SBH'S method first, because he mentioned 300 win mag, and we all know that that is the finest cartridge in north america, and alot of the world.

second best method would go to 4MESH, because he obviously has an honest handle on it. he would have been first, but he didn't mention 300 win mag.

third best method would go to jackie schmidt, because he mentioned accuracy, which is what we are all after.

next best method would go to alinwa, because, well, as my father used to say, "that's so far out, it's almost back in again!"

Con Cross
09-20-2008, 08:14 PM
if i were trying to decide the best method, i would chose SBH'S method first, because he mentioned 300 win mag, and we all know that that is the finest cartridge in north america, and alot of the world.




Gentleman, what do you think?

Con

Bill Wynne
09-20-2008, 08:43 PM
During my freshman year in collage I took Physics 1 and 2 and got very interested in Newton's laws of motion. I actually built a ballistic pendulum to test the velocities of various rifles.

The ballistic pendulum was a 3" pipe filled with sand and paper and capped with a piece of inter-tube rubber. I suspended this thing from thin wire and when I shot into the pendulum it moved forward and upward. The total upward movement would tell you by calculation how fast the pendulum was moving. The pendulum weighed 14 pounds or (14 x 7000) grains. Going to Newton's laws you can determine the weight of the bullet times its velocity= the weight of the pendulum times its velocity.

To make a long story brief, I tested a .222 and several other rifles including a full blown 300 H & H Ackley improved. The 300 Ackley improved with a 150 grain bullet at about 3500 feet per second lifted the pendulum less than 1 foot.

I am not getting into the math again, but that is not much horse power.

Concho Bill

4Mesh
09-20-2008, 08:56 PM
On the 300 Win Mag, I'll give another try.

210 grain Bullet. = .03 Lbs.
Launched at 3000 FPS

Muzzle pointed vertically and plumb. Firing straight up.

.03x3000=90 Ft Lbs per second of energy created. (though the energy is created in far less time than 1 second.)

Time already factored by the muzzle velocity. Power is only created for the hypothetical 1 or 2 ms.

90/550 = 0.1636HP

Now, lets say the estimate of ~1ms bullet in barrel is accurate for our discussion. You could then say the gun produces 163.6HP for one thousanth of a second, or you could say it produces 81.8HP for 2 thousanths of a second (milliseconds, ms).

Now, I can hang on to a gun that's 163.6 HP for no time, but would not want to hang onto the shaft of a 1HP motor when someone turns it on if that motor is to be on for any amount of time. The HP is pretty slight if you spread it out over 1 second.

So, if you were to fire continously you could make about enough horsepower to actually do something, but, to get to the numbers we are discussing here, the gun would have to fire at a cyclic rate of 60,000 rounds per minute to even produce 163hp. Unlikely?

Now, take a top fuel dragster that truly does have 5000hp (and then a bunch), can light the candles at the line, blow up or break instantly, and still cover the traps at pretty significant speed. Now THAT'S horsepower.

If the gun made thousands of HP, it would propel us pretty violently back into the cleaning area when shooting.

If someone sees an error in my math, please advise. Keep in mind, we are only talking about the production of HP during the acceleration phase, not some hypothetical HP of millions of shots being fired in succession for a greater timespan.

Montana Pete
09-20-2008, 09:02 PM
I should stay out, because I'm that English major again without the math or engineering savvy.

Who am I to comment? So don't take me too seriously.

However, I am for Steve. The time the "machine" is operating is very brief. Steve comes up with .0012 seconds. This seems about right.

During that time the rifle does measurable work, in feet/seconds. It is NOT necessary that the "machine" be operating continually over a period of time. Horsepower is force plotted against time. Probably Steve's estimate of 5500 hp. for the rifle is about right.

---------

I think about the example I gave earlier in the thread. Taking a .223 with a 1/7 twist. A bullet exiting the muzzle at 3300 fps will be spinning at 330,000 rpm and is likely to blow up within ten feet of the muzzle. (From the new Speer Manual #14.)

Big numbers and strange physics are actually rather common with high-power rifles.

I wish we could get one of the ballistics engineers from Speer or Sierra to comment here. They would have this totally nailed down, and they would be SURE what they were saying.

A good high-school physics teacher could probably handle this one also.

Not meaning to put down anyone. I honestly do not know for sure what the answers are.

Louis Boyd
09-20-2008, 09:22 PM
Horrsepower is a rate of energy release, not unlike Watts in the metric system.
1 horsepower = 745.67 Watts
foot pounds (ft*lbf) is a unit of energy, not unlike Joules in the metric system
1 foot pound(f) = 0.73756 Joules
1 horsepower = 550 ft*lbf/sec (foot pounds force per second.
1 watt = 1 joule/second

note, lbf is a force not a weight.

So in J Valentines example if you fired a 300 Win Mag with 1100 ft*lbf muzzle energy repeatedly once per second the average energy rate would be two horsepower.

You could also describe the rate of enegy release for an individual shot such as the rate the rifle pumps kinetic energy into the bullet as its traveling down the bore. if that 1100 ft*lbf was put into the bullet in two milliseconds of bore time the horsepower would be 1100 ft*lbf / .002 sec / 550 sec/ft*lbf = 1000 horsepower for 2 milliseconds.

You could also apply horsepower to describe the rate of energy release when an inch long bullet from that 300 WIn Mag strikes and imbeds in a heavy thick armor plate at close range. (1100 ft*lbf energy * (3000 fps * 12 in/ft) / 550 sec/ft*lb = 72,000 horsepower for 27 microseconds.

longshooter
09-20-2008, 09:25 PM
i like 4mesh's. let's see; energy, distance, time, 300 wm, yep, i think he's onto it.

4Mesh
09-20-2008, 09:33 PM
Louis,

I'm not sure which of us is incorrect, but one of us is.

Our numbers are off by approximately one decimal place.

Mine is, 210gn = .03 pounds. 100 of these weigh 3 Lbs.

Launched at 3000 Feet Per Second, = .03 x 3000 or 90 ft Lbs / Sec.

Why is our energy number in such disparity? Am I doing something wrong with that calculation?

EDIT -------------

Possibly answering my own question. Mine is Ft Lbs/Sec where you are talking about Ft Lbs.

I see Sierra Infinity puts my bullet at 4195Ft Lbs, and if you divide by 90, you get 46.6 accelerations per second if a continous stream of bullets is sent. That's saying a millisecond and a quarter per bullet, roughly what we've accepted. Perhaps this is part of it. If so, I like my 163hp example better.

Louis Boyd
09-20-2008, 11:07 PM
I'm 63 and my 300 win mag (a Rem 700 Sendero) has been fired under 100 shots. That says it's delivered about one ten-millionth of a horsepower on average during the life of this owner. I doubt i'll ever put a total of 200 though it.

The only thing in question about appling the concept of horsepower to a rifle is what time period should be used. For a machine gun using the cyclic rate would be appropriate.

Roger T
09-20-2008, 11:54 PM
The 30mm cannon on a A-10 Thunderbolt with sustained fire would overcome the combined thrust of Both it's turbo-fan engines and push the plane backwards. That's at approx 4000 rounds per min. useing depleted uranium rounds.Check the Mil. specs for Vel. and proj.-powder weights.

Bill Wynne
09-21-2008, 07:26 AM
The 300 Ackley improved with a 150 grain bullet at about 3500 feet per second lifted the pendulum less than 1 foot.

I am not getting into the math again, but that is not much horse power.

Concho Bill

Power enough to lift a 14 pound object less than one foot is not very much.

We can say that it takes less than one horsepower to kill a horse.

Concho Bill

Jetmugg
09-21-2008, 08:11 AM
The concept and measure of horsepower was developed to measure the ability to do actual work over a macroscopic period of time (not 1 millisecond). Trying to apply horsepower measurements to a rifle shot is not an appropriate use of this unit of measure (HP). It's a distortion of the intent and usefulness of the horsepower concept.

There are certainly not thousands of usable horspower generated by a single rifle shot. Try to lift a weight or move an object by firing a rifle at it and it should be obvious to anyone who has a grasp of horsepower that you are not generating much. On the other hand, there are thousands of horsepower generated by a top fuel dragster. Your shoulder can tolerate the recoil from even a .300 Win mag. However, try to use your body to keep a top fuel dragster from moving off the starting line and you will find out what horsepower really means.

SteveM.

Vibe
09-21-2008, 05:46 PM
Power is a measure of energy per unit time. In the case of a rifle, we only have the muzzle energy numbers to go by - E=1/2 M * V^2, or 1/2 the Mass times the Velocity squared. Divide this by the time in the barrel and you will have a close approximation of HP. Keeping the units straight from metric to SAE is the main obstacle.

Another formula is Force times Velocity over 33000, but that involves knowing some instantanious values that we do not really know - we may know peak pressure, and thus force values, but usually not the Vel of the bullet at that instant, we also kniow muzzle Vel, but usually not the force at that instant. But we could calculate the average force F=MA needed to achieve the measured MV and use that F and the measured MV to calculate the value.
A=(V2=V1)/t [and V1=0 here].

alinwa
09-21-2008, 07:04 PM
Lynn,

I've got a standard computer, you've got my email....... alinwa1@hotmail.com

send it over

al

alinwa
09-21-2008, 07:53 PM
Ask and ye SHALL receive :D

Here's a copy of HBC's email to Lynn, she's a Honker eh!!!






Lynn,



Yes that is a lot of acceleration for
a machine carrying a human and is probably exceeded only by some Air
Force fighters and some manned space flight velichles at maximum g
levels.



But here are a few interesting facts about, say a
30 cal. 1000 yard load that shoots a bullet having a maximum acceleration
that makes the acceleration of the fuel dragster pale by an order of many
magnitudes:



First let me give you the true definition of
acceleration, "a": a = dv/dt, where dv in an infestimally small change in
veloctiy and dt is an infestimally small change in time corresponding to
the change in velocity and occurs at a point in time. I
realize that is not too colorful but you can use that equation to estimate the
average acceleration of the dragester. That is a = (333
mph*88f/s/60 mph)/4.441 sec = 109.97 f/sec2 or 109.97 ft/sec2/32.17405 ft/sec2
= 3.41 g's. Acceleration defined in words is the rate of change in
velocity with respect to the accompanying change in time.



Now for the 30 cal mag. load shooting a 220 grain
MK at a muzzle velocity of 3002 f/s with a barrel time of 0.001711
seconds:



The average bullet acceleration in the
barrel is 54,532 g's (That is the fuel dragester's average
acceleration increased by a multiple of 15,991 times.)

The maximum acceleration, which occurs near max.
chamber pressure, is 126,849 g's (Under that amount of acceleration, a
free standing column of pure lead taller than 0.077" will began to collapse upon
itself.)

Power is the rate that energy is transmitted.
One Horsepower is defined as delivering or transmitting 550 ft-lbf in one
second. Thus one can calculate the average rate that energy is delivered
to the 220 grain VLD thus:



(3002)^2*220/450436.7/0.001711sec/(550
ft-lbf/sec/HP) = 4677 Horspower That is a lot of Horsepower for
such a small bore but true, although the delivery time is very short for
the single stroke rifle "engine" and the peak Horsepower would likely
be much higher but I would have to modify my internal ballistics program to
estimate the peak HP. Peak energy delivery rate to the bullet, or
peak HP, would likely occur well past maximum chamber pressure.



If the 300 Mag. were fired, such that the bullet
was exiting the cartridge case and entering the barrel bore as the
fule dragster was about 80 yards from the finish line, the bullet would reach
the finish line first.







Henry





Subject:
FW: Definition of Acceleration
Date: Fri, 14 Mar 2008 08:16:49 -0700







'DEFINITION
OF ACCELERATION'

One top fuel dragster 500 cubic inch Hemi engine
makes more horsepower than the first 4 rows of stock cars at the Daytona
500.

Under full throttle, a dragster engine consumes 1-1/2 gallons of
nitro methane per second; a fully loaded 747 consumes jet fuel at the
same rate with 25% less energy being produced.

A stock Dodge Hemi V8
engine cannot produce enough power to drive the dragster's
supercharger.

With 3,000 CFM of air being rammed in by the supercharger
on overdrive, the fuel mixture is compressed into a near-solid form before
ignition.

Cylinders run on the verge of hydraulic lock at full
throttle.

At the stoichiometric (stoichiometry: methodology and
technology by which quantities of reactants and products in chemical reactions
are determined) 1.7:1 air/fuel mixture of nitro methane, the flame front
temperature measures 7,050 deg F.

Nitro methane burns yellow. The
spectacular white flame seen above the stacks at night is raw burning
hydrogen, dissociated from atmospheric water vapor by the searing
exhaust gases.

Dual magnetos supply 44 amps to each spark plug. This is
essentially the output of an arc welder in each cylinder.

Spark plug
electrodes are totally consumed during a pass. After halfway, the engine is
dieseling from compression, plus the glow of exhaust valves at 1,400 deg
F. The engine can only be shut down by cutting the fuel flow.

If spark
momentarily fails early in the run, unburned nitro builds up in the affected
cylinders and then explodes with sufficient force to blow cylinder heads
off the block in pieces or split the block in half.

In order to exceed
300 mph in 4.5 seconds, dragsters must accelerate an average of over 4G's. In
order to reach 200 mph (well before half-track), the launch acceleration
approaches 8G's.

Dragsters reach over 300 miles per hour before you
have completed reading this sentence.

Top fuel engines turn
approximately 540 revolutions from light to light! Including the burnout, the
engine must only survive 900 revolutions under load.

The redline is
actually quite high at 9,500 rpm.

Assuming all the equipment is paid
off, the crew worked for free, and for once NOTHING BLOWS UP, each run costs
an estimate $1,000.00 per second.

The current top fuel dragster elapsed
time record is 4.441 seconds for the quarter mile (10/05/03, Tony Schumacher).
The top speed record is 333.00 mph (533 km/h) as measured over the last 66' of
the run (09/28/03 Doug Kalitta).

You are driving the average $140,000
Lingenfelter 'twin-turbo' powered Corvette Z06. Over a mile up the road, a top
fuel dragster is staged and ready to launch down a quarter mile strip as you
pass. You have the advantage of a flying start. You run the 'Vette hard up
through the gears and blast across the starting line and pass the dragster at
an honest 200 mph. The 'tree' goes green for both of you at that moment.
The dragster launches and starts after you.

You keep your foot down
hard, but you hear an incredibly brutal whine that sears your eardrums and
within 3 seconds, the dragster catches and passes you. He beats you to
the finish line, a quarter mile away from where you just passed him. Think
about it; from a standing start, the dragster had spotted you 200 mph and not
only caught you, but nearly blasted you off the road when he passed you within
a mere 1,320 foot long race course.









That folks is
acceleration...

alinwa
09-21-2008, 07:55 PM
BTW Lynn,

please don't advise Henry to read this thread..... it'd make him very frustrated :)


al

Bill Wynne
09-21-2008, 08:43 PM
"(3002)^2*220/450436.7/0.001711sec/(550
ft-lbf/sec/HP) = 4677 Horspower That is a lot of Horsepower for
such a small bore but true, although the delivery time is very short for
the single stroke rifle "engine" and the peak Horsepower would likely
be much higher but I would have to modify my internal ballistics program to
estimate the peak HP. Peak energy delivery rate to the bullet, or
peak HP, would likely occur well past maximum chamber pressure."

Your formula is flawed. The bullet does not weight 220 pounds but 220 grains. There are 7000 grains in a pound so divide by 7000, maybe. If the rest of the formula is right that would be .668 Hp. I am not sure about the rest of the formula either.

If I have time, I will work on it.

Concho Bill

4Mesh
09-22-2008, 08:24 AM
The 30mm cannon on a A-10 Thunderbolt with sustained fire would overcome the combined thrust of Both it's turbo-fan engines and push the plane backwards. That's at approx 4000 rounds per min. useing depleted uranium rounds.Check the Mil. specs for Vel. and proj.-powder weights.
I had also thought about this before you mentioned it, but the A10 does not fire 210 grain bullets, those 4000/minute from multiple cannons, are firing projectiles that weigh well in the pounds, not grains. I do remember hearing what these projectiles weigh and it's significantly more than a comparable lead core bullet, and at 30mm, they're a good bit bigger. I can't recall the exact numbers and won't guess here.

For those using Ft Lbs of energy and then calculating HP, this is incorrect. You are using the wrong units. You can't just take an arbitrary energy number and then calculate HP by applying a hypothetical time. The unit to use is Ft Lbs/Sec. Now you know how much force and for how long representing total work. Not, how much force is generated over a timespan, and then multiply by that timespan as I see being done here.

Vibe
09-22-2008, 11:33 AM
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

Working backwards
.668 Hp = .668 * 550 ft-lb/sec = 367.4 ft-lbs/sec

367.4 ft-lbs/sec * 0.001711 sec =0.6286214 ft-lbs muzzle energy = pretty anemic.

Bill Wynne
09-22-2008, 11:42 AM
On the 300 Win Mag, I'll give another try.

210 grain Bullet. = .03 Lbs.
Launched at 3000 FPS

Muzzle pointed vertically and plumb. Firing straight up.

.03x3000=90 Ft Lbs per second of energy created. (though the energy is created in far less time than 1 second.)

Time already factored by the muzzle velocity. Power is only created for the hypothetical 1 or 2 ms.

90/550 = 0.1636HP

4 Mesh is right, all others are wrong.

The answer is exactly .16363636 horse power. (more or less)

I came up with this using an independent method and then I found 4mesh's post.

Concho Bill

Vibe
09-22-2008, 01:08 PM
I suppose the only thing left to add is to repeat the signature line I use on many other forums.

The opinion of 10,000 men is of no value if none of them know anything about the subject.
- Marcus Aurelius -
:D :D :D :D :D

4Mesh
09-22-2008, 02:26 PM
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

Working backwards
.668 Hp = .668 * 550 ft-lb/sec = 367.4 ft-lbs/sec

367.4 ft-lbs/sec * 0.001711 sec =0.6286214 ft-lbs muzzle energy = pretty anemic.

Vibe,

in your example above, you are taking this TOTAL generated energy value which is generated over the 1.7ms, and you are saying that that much work is being done for the entire time which it is not and that is incorrect.

An analogy would be you have 60 1 pound weights on the floor. You need to lift all of these exactly 1 foot each, and do one per second for one minute. Then when you are done you say you were doing 60 ft lbs of work per second. No, you were doing 1 ft lb of work per second and did a total of 60 ft lbs of work.

And Gee, thanks for the vote of confidence there Bill!

Vibe
09-22-2008, 02:42 PM
Vibe,

in your example above, you are taking this TOTAL generated energy value which is generated over the 1.7ms, and you are saying that that much work is being done for the entire time which it is not and that is incorrect.

An analogy would be you have 60 1 pound weights on the floor. You need to lift all of these exactly 1 foot each, and do one per second for one minute. Then when you are done you say you were doing 60 ft lbs of work per second. No, you were doing 1 ft lb of work per second and did a total of 60 ft lbs of work.

And Gee, thanks for the vote of confidence there Bill!
Nope. I took the total amount of energy generated, divided by the total amount of time to do it. In your example of 60 weights that would be 60ft-lbs in one minute or 1ft-lb/sec. And the 4458HP is the average HP generated from chamber to muzzle, I'm sure the instantaneous peak HP would be a bit higher.

Take the same 60 lbs of weights and put them 1ft higher up...but do it in 1/1000 sec. Same 60 ft-lbs of work was done, but the HP required to do it is much higher.

Think of HP as a RATE of energy change...it really has not so much to do with how much work was actually done, or how much energy was generated, so much as how FAST it got there.

4Mesh
09-22-2008, 03:15 PM
Take the same 60 lbs of weights and put them 1ft higher up...but do it in 1/1000 sec. Same 60 ft-lbs of work was done, but the HP required to do it is much higher.

Think of HP as a RATE of energy change...it really has not so much to do with how much work was actually done, or how much energy was generated, so much as how FAST it got there.

Work is work. There's a mass being lifted. It's not much mass. There's a time to lift and it's not much time. In the early post I have the description of horsepower and how it is calculated. You are taking total generated energy numbers and using them as if they are rates of force, they are not. Do not work backward from a ft-lbs number and say we have this many foot lbs for x amount of time. That is not how it works. Work in from the other direction with values you know and units that are not confusing or ambiguous.

Here's a better way to say it. You take a 3000 # car up to speed and then smash head on into a concrete wall. The car stops in .0017 seconds, and you then attempt to calculate how many HP that car is from how hard it hit. no no no no no. All you have is an energy number to work with, not a force of acceleration.

Vibe
09-22-2008, 03:27 PM
Whatever you think 4Mesh.
I got my BS in Engineering in 1985. And worked as an engineer and machine designer for around 25 years. I have a fair handle on the differences between power, energy and work. I have successfully built and used a ballistic pendulum - not an exercise to sneeze at, simple as it may sound.

But think what you like. Or you can re-read reply #32.
:D

4Mesh
09-22-2008, 03:32 PM
Whatever you think 4Mesh.

I got my BS in Engineering in 1985. And worked as an engineer and machine designer for around 25 years. I have a fair handle on the differences between power, energy and work. I have successfully built and used a ballistic pendulum - not an exercise to sneeze at, simple as it may sound.

But think what you like. Or you can re-read reply #32.
Vibe, I'm not really concerned with what degree you have and that does not automatically mean you are correct. Degrees do not make one incapable of mistake.

In any case, the energy number you are using in YOUR calculations included within it a SQUARE of the velocity. HP does not include any such squaring of that quantity.

You believe what YOU want. There isn't 4000 horsepower in a gun.

Vibe
09-22-2008, 03:39 PM
In any case, the energy number you are using in YOUR calculations included within it a SQUARE of the velocity. HP does not include any such squaring of that quantity.

LOL. Actually in the complete derivation of the term....it does. :D
(Hint - it's in the g portion of M*g*delta-H/t)
And equals 32ft/sec^2, or 9.8m/sec^2 if you prefer metric.

4Mesh
09-22-2008, 03:42 PM
LOL. Actually in the complete derivation of the term....it does. :D
(Hint - it's in the g portion of M*g*delta-H/t)
And equals 32ft/sec^2, or 9.8m/sec^2 if you prefer metric.
Ah, yes that is correct. Unfortunately it is also already included in the 550 ft-lb/sec and thus, you are applying the square TWICE.

Vibe
09-22-2008, 03:56 PM
Ah, yes that is correct. Unfortunately it is also already included in the 550 ft-lb/sec and thus, you are applying the square TWICE.
Wrong again. The terms for HP are actually dimensionless, which you can only achieve by dividing ft-lb/sec by something else measured in ft-lb/sec. So not only are we not applying the square twice, we MUST use it again in order to eliminate it as a term in the result.

By the way...A ballistic pendulum works because of conservation of MOMENTUM (Mass times Velocity), and not energy, or work.

4Mesh
09-22-2008, 04:03 PM
Taken from my post #5 quite some time ago.

You need only read steps 4 and 5.

Now, how much does the bullet weigh.

How far can we lift that bullet in one second.

This just ain't that tough.

In your defense, my calculation is not spot on correct as we are not truly lifting the bullet 3000 ft up in the one second. It will be lifted less than that so you could say, I also overestimated the HP by whatever amount the bullet does not travel. Take the 27' off of the 3000' due to gravity, and now you can move on. So, the bullet will not make it 3000' high in one second.

However, for the purpose of this discussion, I think the numbers are acceptable. My example is easy enough to divide the figure by whatever muzzle time one wishes to use to find the "horsepower".


Power is the measurement of how much work can be done per unit of time. It can be expressed in either watts or horsepower, depending on whether metric (International System) or English (Imperial) units of measure are used to calculate it. Horsepower is a term first coined by James Watt, who calculated that a single horse can do 33,000 foot-pounds of work in 1 minute.


Step 1
Calculate how much work is done by the device. Work is the measure of force times distance. For instance, raising an object weighing 1 Newton 1 meter results in 1 Newton-meter (or 1 Joule) of work. In English units, raising an object weighing 1 pound 1 foot results in 1 foot-pound of work.

Step 2
Understand the definition of "weight." Weight is a measurement of mass multiplied the acceleration of gravity. In SI units, kilogram is a measurement of mass, although it is commonly used to refer to weight. 1 kilogram of mass multiplied by the acceleration of gravity is 1 Newton of weight. Similarly, in English units, a pound is a measurement of weight, and slugs is a measurement of mass, making a pound equal to 1 slug multiplied by the acceleration of gravity.

Step 3
Learn how to define "work." To do work, you must move an object in the same direction as the force that was applied to it. For instance, if you try to lift a box, but can't, you haven't done any work even though the attempt my tire you. If you lift the box and carry it across the room, only the distance that you lifted the box counts as work.

Step 4
Determine how much time was required to do the work. If it took 1 second to raise the 1-Newton weight 1 meter, the power of the device that raised the weight is 1 Joule per second or 1 watt. In English units, the device that raised a 1-pound weight 1 foot in 1 second is 1 foot-pound per second.

Step 5
Convert foot-pounds per second into horsepower by dividing by 550. You can also convert watts to horsepower if you know that 746 watts equals 1 horsepower.
Edited to remove statement about velocity loss due to drag... Not applicable.

Vibe
09-22-2008, 04:27 PM
LOL. If this were not so tiresome it might be fun.
Lift a 1 lb weight 1 ft and you have changed it's (potential) energy by 1 ft-lb
do it in 1 second and you have demonstrated 1 ft-lb/sec of power

Fire a bullet straight up and you have changed it's energy (in this case kinetic) from 0 to 4195 ft-lb, and it took 0.001711 sec. barrel time to do this.
That is still a 2451782.583 ft-lb/sec rate of change in energy. Divide this value by the energy rate of change defined as ONE HP or 550 ft-lbs/sec and the result is STILL 4458 HP.

Just because you stole the rather well written description, is no indication that you understand what it says.
Determine how much time was required to do the work

This whole discussion reminds be of a comedy skit.

Hippy: Oh I'm a happy hippy, takin' a little trippy.....
Cop: Pull over boy, I got you clocked at better'n 80 miles an hour.
Hippy: Man, that's impossible. I haven't even been out an hour.

It's 4 pm here and I have to be at work at 11pm. No more replies from me until tomorrow.

4Mesh
09-22-2008, 04:47 PM
LOL. If this were not so tiresome it might be fun.
Lift a 1 lb weight 1 ft and you have changed it's (potential) energy by 1 ft-lb
do it in 1 second and you have demonstrated 1 ft-lb/sec of power

Fire a bullet straight up and you have changed it's energy (in this case kinetic) from 0 to 4195 ft-lb, and it took 0.001711 sec. barrel time to do this.
That is still a 2451782.583 ft-lb/sec rate of change in energy. Divide this value by the energy rate of change defined as ONE HP or 550 ft-lbs/sec and the result is STILL 4458 HP.

Just because you stole the rather well written description, is no indication that you understand what it says.

This whole discussion reminds be of a comedy skit.

Hippy: Oh I'm a happy hippy, takin' a little trippy.....
Cop: Pull over boy, I got you clocked at better'n 80 miles an hour.
Hippy: Man, that's impossible. I haven't even been out an hour.

I'll tell you what's funny. You now are telling me the bullet weighs TWO TONS and doing calculations, telling me you have a BS in engineering, and can't do basic math.

Where the h___ did you get an energy number to apply into the front side of the horsepower equasion? How many times need I say you are SQUARING the velocity. Your projectile YOU are measuring is traveling 9 million feet straight up, NOT 3000 feet.

Let me spell it out REAL CLEAR for you since all you have is a BS (and probably should not have received it). In your first example, you are WRONG.

Lift a 1 lb weight 1 ft and you have changed it's (potential) energy by 1 ft-lb
do it in 1 second and you have demonstrated 1 ft-lb/sec of power

POWER has nothing to do with it. You did not "demonstrate power" you demonstrated "WORK". Power is how much energy potential is available, work is how much has been done in a period of time.

We do not care about remaining energy. We lifted that weight 1 foot in one second. It was LIFTED 1 foot. Do NOT square the velocity now and say that is the input to the HP equasion.

Now, take a .03 pound weight, lift it 3000 feet in ONE SECOND and you have 90 FOOT POUNDS PER SECOND of WORK. GET IT, I DON'T CARE ABOUT RESIDUAL ENERGY. WORK IS WORK.


Fire a bullet straight up and you have changed it's energy (in this case kinetic) from 0 to 4195 ft-lb, and it took 0.001711 sec. barrel time to do this.
That is still a 2451782.583 ft-lb/sec rate of change in energy. Divide this value by the energy rate of change defined as ONE HP or 550 ft-lbs/sec and the result is STILL 4458 HP.

Just because you stole the rather well written description, is no indication that you understand what it says.
Just because you have a BS does not mean you weren't doing too much "LDS as Spock would say". You are unbelieveable.

I have a great excersize for you. Why don't you take your muzzle energy value and try to tell me how much that bullet weighs? Lol. Hell, you'll probably come up with 210 grains again doing the reverse math wrong too! :D

That second line is simply so far out in left field, it's not fit to discuss. I sure hope you didn't engineer anything I use on a daily basis.

All in fun man.

Let's break this down into really little pieces so you can see how this works. Just pretend we're all back in school.

How much does the bullet weigh.

How far did we move the bullet in one second.

Now, just stop there and tell me how much work was done. Can you do that?

Con Cross
09-22-2008, 04:55 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?

I've often wondered about this . . . hope someone takes me up on it.

Montana Pete


Gentleman, would you believe, that Mr. Montana Pete actually specified the type of people he thought would be correct.

Con

Bill Wynne
09-22-2008, 05:19 PM
Whatever you think 4Mesh.
I got my BS in Engineering in 1985. And worked as an engineer and machine designer for around 25 years. I have a fair handle on the differences between power, energy and work. I have successfully built and used a ballistic pendulum - not an exercise to sneeze at, simple as it may sound.
:D

Vibe,

You and I may be the only two nuts who have built a ballistic pendulum. After I made a shot, there was about one hour of math to do without a calculator or a computer. On this horsepower thing, I seem to remember a picture in my physics book of a horse pulling a rope over a pulley lifting a 550 pound weight one foot in one second (550 foot pounds per second) and that represented one horsepower. It had nothing to do with acceleration.

I believe that it is just that simple. Study this thing a little longer and let us know what you think. It has been an interesting little problem and it doesn't really mean anything. I don't believe that anyone will describe their rifle in horse power.

Concho Bill

4Mesh
09-22-2008, 06:02 PM
Now addressing the edit you did on the post 2 above...

Breaking this statement down, let's see how we arrive at these numbers.

Fire a bullet straight up and you have changed it's energy (in this case kinetic) from 0 to 4195 ft-lb, and it took 0.001711 sec. barrel time to do this.
That is still a 2451782.583 ft-lb/sec rate of change in energy. Divide this value by the energy rate of change defined as ONE HP or 550 ft-lbs/sec and the result is STILL 4458 HP.

Part one.

Fire a bullet straight up and you have changed it's energy (in this case kinetic) from 0 to 4195 ft-lb

Yup, this is how we calculate energy, has nothing to do with work.

Energy is calculated by taking the weight of the projectile in grains and multiplying it by the SQUARE of it's VELOCITY in feet per second, then dividing by 450240. IE, 210 grain bullet multiply by 3000fps^2 then divide by 450240 and you get ENERGY in foot lbs. NOT WORK. You are saying the bullet moved NINE MILLION feet and it did not. Work is simply how much weight you lift adn how far, not how far squared.

210 x 9,000,000 / 450240 = 4198 Ft Lbs ENERGY ( why sierra infinity said 4195 ???)

Here's a site with a great example. http://www.pyramydair.com/site/articles/formulas/

Now, you are taking that value and using it as if it was work, and it is not. It is energy. Then you are taking the time in the barrel which has not one thing to do with this quantity, and dividing by that infintesimally small time in seconds to come up with an astronomicaly wrong value for work done, then somehow converting it to horsepower.

In other words, you are saying 4198 ft lbs of work was done in .0017 seconds, (or .0012 or whatever value we are using) and that is so far off base I can't tell ya.

WORK is calculated by taking the weight in LBS, and multiplying by the distance you lift the mass in feet, represented by foot lbs. Please re-read the example I gave on how to calculate horsepower AND DO THAT. If I need to make that simpler we are at an impass because I can't do it. To arrive at work per second you must then take how many seconds it took to do that work and divide to get foot lbs per second.

How much work was done is completely independant of the time it took.

How much power it took to do the work is directly related to how long it took.

You are taking the velocity number for our mass, squaring it and trying to call it the distance moved. BZZZZZZZZ Wrong.

Please work through that energy calculation MANUALLY and tell me how it is applicable or comparable to the calculation of work described above several times. How do you arbitrarily define the ending energy units as work? Would you please explain that in detail?

One last time. WORK = Weight * Distance. THAT'S IT. To convert to HP you need to know how long it took to do that work. NOW, you have HP. End of story. You do NOT square the amount of distance in feet and then call it work. Then backtrack time and call that HP.

All I can say is, PLEASE give a 100% full example showing ALL of your calculations in parts, so as to prove your point if you still feel it is correct. Also, please correlate those examples completely with MY formulas and show where my formula is flawed. I am no longer interested in hearing how your example is right, please show me what part of mine does not work in the schoolbook example written very simply above...

HP = feet distance * lbs moved / seconds elapsed / 550

Bill Wynne
09-22-2008, 09:05 PM
HP = feet distance * lbs moved / seconds elapsed / 550

Check out that last formula. By distance moved do you mean dragging a weight is as as much effort as lifting it?:)

Concho Bill

Vibe
09-22-2008, 10:53 PM
Vibe,

You and I may be the only two nuts who have built a ballistic pendulum. After I made a shot, there was about one hour of math to do without a calculator or a computer.
I set up an Excell spreatsheet so that part took a bit less time. But I designed my pendulum on the front end to where the difference between a 12" swing and a 13" swing was 100ft/sec. That way I could come close to knowing the velocity of each shot right away. And I was "only" dealing with 35 grain projectiles at between 1500 adn 2000 fps. I was developing the Cricket and did not have a chronograph of my own.




On this horsepower thing, I seem to remember a picture in my physics book of a horse pulling a rope over a pulley lifting a 550 pound weight one foot in one second (550 foot pounds per second) and that represented one horsepower. It had nothing to do with acceleration. LOL. I must have had the same book, or that picture was prionted in all of them. Yes, even in that picture there was still acceleration...but it was "just" gravity.


I believe that it is just that simple. Study this thing a little longer and let us know what you think. It has been an interesting little problem and it doesn't really mean anything. I don't believe that anyone will describe their rifle in horse power.
LOL. I know I'm not going to be one of them if they do.
I figure if anything is "off" it's the time value.

Concho Bill[/QUOTE]

alinwa
09-22-2008, 11:33 PM
In this world there's gener'ly two sorts of people, guys that FIGGER and guys that DO. Also most gener'ly, the guys that DO hire the guys that FIGGER whenever they need them....... kinda' like hiring a taxi to get you from the airport to the meeting.


RARELY comes a guy that can FIGGER AND DO, guys like Harold Vaughn, Thomas Edison, Leonardo Da Vinci................even some folks here on this board, and some who USEta' hang out on this board until they just got frustrated.


Now, if I have a thing to be done and a choice between a "FIGGERER" and a DOER to do it, I'll pick the DOER every time since't he ain't hampered by all the figgers, he's got a job to do. :D figgerers tend to get lost in the elegance of their theory...... to these folks, reality means NOTHING, to them Schroedinger's cat was an actual experiment......


reality CHECK!


LOL


al

alinwa
09-23-2008, 12:48 AM
Ohhh yeahhh, Henry's numbers are good :)

Henry is most definitely both a figgerer and a doer. ;)

Henry's post is about acceleration. The question is about horsepower.

al

alinwa
09-23-2008, 01:02 AM
OOOPS......

Henry's post IS about horsepower :o

(wiping egg off my face)

I coulda' just deleted the post but this'll do my pride some good if I'm wrong....Maybe we DO need to involve heem. :D

LOL


al

alinwa
09-23-2008, 01:14 AM
I finally took the time to look at how Henry expressed himself.... (I'm surprised that he used the term horsepower at all :) it ain't LIKE Henry ) We don't need to bring him in...... the kicker lies in the time interval. The work being done is still only a fraction of "a horsepower." BECAUSE horsepower isn't a rate, it's a defined unit of measurement.

Ahhh, Henry will be PISSED if we drag him into this one :D

LOL


al

Bill Wynne
09-23-2008, 06:34 AM
In this world there's gener'ly two sorts of people, guys that FIGGER and guys that DO. Also most gener'ly, the guys that DO hire the guys that FIGGER whenever they need them.

Now, if I have a thing to be done and a choice between a "FIGGERER" and a DOER to do it, I'll pick the DOER every time.

al

The figgerer is the guy who considers what a thing is going to look like or what a thing is going to do or not do before he starts doing. In the Boy Scouts we had a term for this, "Be prepared".

Al, I figger I am a doer and a figgerer if that term means thinking and planning before doing and I kinda think you are too.

You wouldn't set up some forms that you intend to pour concrete into without figgering what the concrete thing is going to look like after you remove the forms, would you?

Would you call a figgerer to tell you how to brace the forms?

And I don't think that you would need to rely on a figgerer to tell you how much concrete to order to fill your forms.

Of course not! It would delay the project by several days.

I figger that this horsepower thing as applied to a gun is like a crossword puzzle to some of us. Nothing more and nothing less. Just a mind game, that's all.

By the way, me and old 4Mesh are still right.:)

Concho Bill

4Mesh
09-23-2008, 09:07 AM
Vibe, Lynn, those who are still disputing this... mother of gawd guys, you have to come away from this acceleration thing.

Now, I want you to read this message real close like, and at each line, determine if my math is correct or incorrect. Now, please quote this entire message, and in the middle where I make my mistake, I want you to put a big bold red message that says, 4Mesh is WRONG HERE. Ok?

Now.

Formula for HP is --------- WORK / TIME = HP

Are we still ok to this point?

Work is defined as

Step 1
Calculate how much work is done by the device. Work is the measure of force times distance. For instance, raising an object weighing 1 Newton 1 meter results in 1 Newton-meter (or 1 Joule) of work. In English units, raising an object weighing 1 pound 1 foot results in 1 foot-pound of work.

I do not see any velocity in that calculation, do you? Now, above Vibe, you were kind enough to mention that I am unable to understand this text I've provided. Would you like for me to quote your "degree in BS" statement about that? It looks to me like you multiply one value times the other. WEIGHT in POUNDS times DISTANCE LIFTED = WORK PERFORMED.

Are we ok so far folks? Problems here?


Step 2
Understand the definition of "weight." Weight is a measurement of mass multiplied the acceleration of gravity. In SI units, kilogram is a measurement of mass, although it is commonly used to refer to weight. 1 kilogram of mass multiplied by the acceleration of gravity is 1 Newton of weight. Similarly, in English units, a pound is a measurement of weight, and slugs is a measurement of mass, making a pound equal to 1 slug multiplied by the acceleration of gravity.

Ok, so weight is expressed in POUNDS and it already has gravity included into it. We don't need to include it again do we?

Are we OK to here? We know what a pound is?

1 Pound = 1 Slug * gravity acceleration. ? Still Ok? Should be, there's no calculations to do if we weigh the object we are propelling. Nature does that math for us! Whoopeee.


Step 3
Learn how to define "work." To do work, you must move an object in the same direction as the force that was applied to it. For instance, if you try to lift a box, but can't, you haven't done any work even though the attempt my tire you. If you lift the box and carry it across the room, only the distance that you lifted the box counts as work.

Only the distance you LIFT the box counts as work.

For our purpose, let's just say, ONLY THE DISTANCE YOU LIFT THE BULLET COUNTS AS WORK.

Any troubles to this point Vibe?

Step 4
Determine how much time was required to do the work. If it took 1 second to raise the 1-Newton weight 1 meter, the power of the device that raised the weight is 1 Joule per second or 1 watt. In English units, the device that raised a 1-pound weight 1 foot in 1 second is 1 foot-pound per second.

The time it takes to propel our bullet is one second for 3000 ft in the above examples. It is not 1 second TIMES .0012 seconds. Just 1 second.

Can we agree up to this point? I know this is a toughy cause we have time in there now, but really, just read the damn steps one at a time.

Ok, we need to clarify that english units equation. They do mean, 1 pound, times 1 foot equals foot pounds of work, yes? Am I making a horrible mistake here by just multiplying those figures? Perhaps I should square some of them or all? Maybe the rest of the upright standing world missed something in the equations?

So, so far, work = lbs lifted x feet lifted x time in seconds to do it?

Still ok here? Should we now calculate the residual energy before calculating the horsepower? That would allow us to make astronomical claims. Somehow, I don't see in the examples where they calculate work, then take energy and add that in too. Did I miss something?

So, work is still just work? Now Vibe,

I have a fair handle on the differences between power, energy and work.
Oh yea, well I beg to differ with you. I think you should say, I SHOULD HAVE a fair handle on them, but don't.

Now, to convert that to power, we need to know how long it took to do THAT AMOUNT OF WORK.

We express that in seconds. Yes??? No squares or square roots here eh?

So, WORK = LBS WEIGHT x FEET DISTANCE LIFTED x SECONDS ELAPSED

Gee, where's the how fast it's going part? Did the engineering world leave that out? What the hell do they know about work anyhow.

SO, correct the incorrect line PLEASE. Or, just point out where I'm headed the wrong direction, OK?

1 Lb x 1 ft in 1 second = 1 ft lb / sec

2 lb x 1 ft in .5 seconds = 1 ft lb / sec Hmmm. faster,but still the same work?? Why?

1 lb x 2 ft in .5 seconds = 1 ft lb / sec Hmmm. Faster yet, hmmm, odd...

100 lb x .005 ft in .5 seconds = 1 ft lb / sec Slower? same work?

.005 lb x 100 ft in .5 seconds = 1 ft lb / sec

1 lb x 3000 ft in 1 second = 3000 ft lb / sec

WHOA!!!!! Wait a minute, a POUND at 3000 ft per second is only 3000 ft lbs / second?????? Impossible! Damn equations!

How can that be? Gee 4Mesh, You can't possibly mean that math continues to work even when you change the values can you? Hell, even a .03 lb object lifted the same distance in the same time causes more ft lbs than that in Vibes example?

Ok Vibe, am I wrong yet? Lynn? How am I doing?

I know what it is, I used 3000 ft. I should go back and change that.

3000 lb x 1 ft in 1 second = 3000 ft lb / sec

Ah, that's better. Now it's moving slower so when we square the distance, it is still 1. Vibe likes this equation as long as it's all 1's, but when you throw in numbers greater than 1 and some decimals, it's gets significantly tougher.

Now Here we go


Step 5
Convert foot-pounds per second into horsepower by dividing by 550. You can also convert watts to horsepower if you know that 746 watts equals 1 horsepower.

You mean, all we have to do is divide by 550? Man, I want to do more math than that. How bout I square the 550 just for the hell of it? Nah, that would make the HP sound too small. I think I'll take the square root of 550 and use that instead. That's better, now our HP is higher! Now quite high enough, but better'n ole 4Mesh's example. Using the accepted equations just doesn't give the result I want damn it! :D

Bill, man, we gotta wonder bout these folks...

Ok. Now, is there anyone out there with a REAL degree in mechanical engineering who can tell me why my 163.63 hp over .001 second example is technically incorrect given the stated input parameters? Hint. I already told you above that there's more to it. However, it will not change the result appreciably. It is however technically incorrect.

Vibe, did you do any work on that Mars Lander project a few years back? You know, the one that made a half mile deep hole in Mars? :D :D

Almost an hour and a half of my time wasted to produce this for someone who's parents probably paid significant money for an education that was never received...

dmoran65
09-23-2008, 10:54 AM
With modern day dynamometer's, horsepower is a calculated value from the measured TORQUE and RPM.
<< HP = TORQUE * RPM / 5252 >>

Happy Shooting
Donovan Moran

4Mesh
09-23-2008, 11:03 AM
Check out that last formula. By distance moved do you mean dragging a weight is as as much effort as lifting it?:)

Concho BillNo no Bill, :) I covered the lifting thing enough times before that I figured I didn't need to type that out again. If you see that I was a bit redundant elsewhere, now you know why :D But thanks for pointing that out and no, as you know, I did not assume them to be the same thing.

Actually, what's really funny is, I did a calculation with a spreadsheet in Excel and applied 44XX horsepower that was stated above to a 210 grain bullet. Guess what the muzzle velocity of that projectile was?

LMAO, EIGHTY EIGHT MILLION SEVEN HUNDRED THOUSAND FEET per Second! It's gonna 3/5ths circunavigate the earth in 1 second!

Yowwzzzaaaaa! I don't think I can get that much powder into a WinMag! A good dose of uranium might not do it!

We aim east at the range here in PA. With good aim, that load might propel me to the Ohio range at Thunder Valley!

Folks, a big part of understanding these types of things is to relate one number to another known quantity and result and say, huh, that sounds like it could be right, or that doesn't seem like it adds up. Then go from there. When a quantity is off by 5 orders of magnitude, that should ring a bell with someone.

Let's consider how we could determine something is wrong without doing calculations and just using common sense.

How many have ever been to a Top Fuel Drag Race? Ok, you're standing in the stands (cause everyone has stood up to see) and the tree goes green. They light the candles on the cars. You and 6000 other people are hundreds of yards away, and the concussion blows ALL of you back in the stands and drives a shockwave through you that makes your heart skip and your hair blow back. The pressure against your chest physically moves you back. Keep in mind, this is only the residual energy LOST by applying 5000 HP TO THE CAR.

Now, my gun doesn't hit me that hard. So I say, it doesn't make that much horsepower for any appreciable time.

Mister when you talk about thousands of horsepower, that's a lot of horsepower.

4Mesh
09-23-2008, 11:06 AM
With modern day dynamometer's, horsepower is a calculated value from the measured TORQUE and RPM.
<< HP = TORQUE * RPM / 5252 >>

Happy Shooting
Donovan MoranHead on over to Wikipedia and look at the explanations there for the various "Standards" of measuring HP and you will see that that torque calculation is nothing more than measuring the torque arm and then lifting the weight a measured distance over time.

Jetmugg
09-23-2008, 11:42 AM
4 Mesh:
Go, man, go. You are on a roll, and I believe that you are right on track. I also have a BS degree in engineering (Metallurgical). I cannot find a flaw in your logic or execution.

One of the issues that I have trouble with is the application of the HP measurement to a rifle in general. HP is not a first-principles type of measurement, it's a derived unit that was originated to describe the ability of a machine to do work once performed by animals. HP is well suited to that type of comparison - a machine to a mule for example. People can understand that a 5HP lawn mower engine has an amount of power that is not to be trifled with, but is nothing like a 200HP Evinrude outboard engine, or a 5000 HP top fueler.

Since a rifle does not perform the type of work that we are accustomed to being able to see and estimate, it makes the HP value much less tangible, and in my opinion, less appropriate for use. We tend not to think of rifles as tools for performing mechanical work in the same way that we do for lawn mowers, outboard engines, or 500 cubic inch Hemi's. It seems to be a distortion of the original intent of HP measurements to apply them to something that is as instantaneous as a rifle shot.

SteveM.

Bill Wynne
09-23-2008, 11:44 AM
I called a casual acquaintance of mine (It is a long story). He is Dr. Clamdunker who has some weird titles that mean the is the Dean of the School of Engineering at Texas A & M University or says he is.

If I may I would to paraphrase what he told me because, frankly, many of you would have trouble understanding his technical jargon. In simple english Dr. Clamdunker says that Mr. 4mesh and and me and those who agree with us are right and everyone else is wrong.

You can question me all you like, if you like, but you surely would not presume to question Dr. Clamdunker because he is real smart and knows a lot about horses and power and stuff.

Concho Bill

dmoran65
09-23-2008, 12:49 PM
Just some thought ---

If RPM is not to be factored into a HP equation of a bullet in flight, then I would think that the standard equation for "Momentum" would be similar to a HP rating. Momentum (lbs/sec.) = W * V (@ target) / 225200


Happy Shooting
Donovan Moran

4Mesh
09-23-2008, 01:41 PM
Donovan,

You can't really use the V number at the target because that number assumes drag loss from the atmosphere. Remember, to calculate the net HP of the gun, you'd have to disclude that and remaining velocity would be the same as starting V less the 9.8m/s^2 loss of gravity since we're shooting straight up.

If you wish to also include the rotational energy created and generate a work number for that, you need to define a swing arm distance which will be only a portion of the .308 bullet dia before engraving, and that would be a rather difficult number to arrive at indeed.

With or without that figure included, the HP is not very much.

Bill, I'm sure that Dr. Clamdunker got a laugh out of my semantic errors but probably still got the idea of what I was saying. I have no doubt he's forgotten more about horses than I'm ever gonna know!

I know, we had one once that Id'a preferred to feed benzine! My sister long ago had one that was a barrel racer. She was cool and all but when my sister went away in the military I got to deal with the horse. I can tell ya that when she'd run the cows through the electric fence and go on walkabout around the neighborhood, we'd come gallop'n back and through the orchard, I got my first experience with 1 horsepower! She'd see a nice apple tree and say, "wow, that looks like a great spot for a U turn!" You'd be surprised how hard you hit a big apple branch when a 1 horsepower horse runs under a tree at relatively low velocity! :D:D Especially when she had to duck to miss the branch herself!

Vibe
09-23-2008, 01:58 PM
Vibe, Lynn, those who are still disputing this... mother of gawd guys, you have to come away from this acceleration thing....4Mesh is WRONG HERE
No can do- it's an integral part of work, and therefore a required part of power.


Now, I want you to read this message real close like, and at each line, determine if my math is correct or incorrect. Now, please quote this entire message, and in the middle where I make my mistake, I want you to put a big bold red message that says, 4Mesh is WRONG HERE. Ok?....OK :D


Now.

Formula for HP is --------- WORK / TIME = HP4Mesh is WRONG HERE
Since you force me to be picky
WORK / TIME = ft-lbs/sec
1 HP=550 ft-lbs/sec

Are we still ok to this point?

Other than that you're fine so far. :D






Work is defined as
Step 1
Calculate how much work is done by the device. Work is the measure of force times distance. For instance, raising an object weighing 1 Newton 1 meter results in 1 Newton-meter (or 1 Joule) of work. In English units, raising an object weighing 1 pound 1 foot results in 1 foot-pound of work.

There are better ones
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/DefinitionWork.html
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/energy/u5l1a.html

http://en.wikipedia.org/wiki/Mechanical_work

And possibly the simplest one

If you put energy into an object, then you do work on that object.at

http://id.mind.net/~zona/mstm/physics/mechanics/energy/work/work.html



I do not see any velocity in that calculation, do you? ]
Yep 4Mesh is WRONG HERE
It's in the acceleration of Gravity term used to define a Newton.



Now, above Vibe, you were kind enough to mention that I am unable to understand this text I've provided. Would you like for me to quote your "degree in BS" statement about that? It looks to me like you multiply one value times the other. WEIGHT in POUNDS times DISTANCE LIFTED = WORK PERFORMED.
While technically correct - that is a gross over simplification of MgH (Mass*acceleration due to Gravity*Height)




Are we ok so far folks? Problems here?
Only those noted. :D




Ok, so weight is expressed in POUNDS and it already has gravity included into it. We don't need to include it again do we?
We do and we don't, but I'll cover that in a bit.


Are we OK to here? We know what a pound is?

1 Pound = 1 Slug * gravity acceleration.
4Mesh is WRONG HERE
OK here you are unconditionally wrong

Slug (definition)
7. The gravitational unit of mass in the foot-pound-second system to which a one pound force can impart an acceleration of one foot per second per second and which is equal to the mass of fan object weighing 32 pounds


? Still Ok? Should be, there's no calculations to do if we weigh the object we are propelling. Nature does that math for us! Whoopeee.
Need I comment?




? Only the distance YOU lift the box counts as work. .
Reassignment of emphasis = mine. :D


For our purpose, let's just say, ONLY THE DISTANCE YOU LIFT THE BULLET COUNTS AS WORK

Any troubles to this point Vibe?
As noted.



The time it takes to propel our bullet is one second for 3000 ft in the above examples. It is not 1 second TIMES .0012 seconds. Just 1 second.
How long is your barrel and how are you propelling anything once it leaves the (lets use 26") confines of the bore?
As soon as the bullet leaves the barrel 1/2MV^2 starts being converted to MgH...and heat. In a vacuum 1/2MV^2=MgH


Can we agree up to this point?
Oh Snapps no. :D


I know this is a toughy cause we have time in there now, but really, just read the damn steps one at a time.
:D


Ok, we need to clarify that english units equation. They do mean, 1 pound, times 1 foot equals foot pounds of work, yes? Am I making a horrible mistake here by just multiplying those figures? Perhaps I should square some of them or all? Maybe the rest of the upright standing world missed something in the equations?
:cool:


So, so far, work = lbs lifted x feet lifted x time in seconds to do it? 4Mesh is WRONG HERE
It's actually
work = lbs lifted x feet lifted / time
But hey...carry on.


Still ok here? Should we now calculate the residual energy before calculating the horsepower? That would allow us to make astronomical claims. Somehow, I don't see in the examples where they calculate work, then take energy and add that in too. Did I miss something? I'd say quite a lot...but carry on. :D


So, work is still just work? Now Vibe,

Oh yea, well I beg to differ with you. I think you should say, I SHOULD HAVE a fair handle on them, but don't. :D:D:D We'll see


Now, to convert that to power, we need to know how long it took to do THAT AMOUNT OF WORK.

We express that in seconds. Yes??? No squares or square roots here eh?

So, WORK = LBS WEIGHT x FEET DISTANCE LIFTED x SECONDS ELAPSED4Mesh is WRONG HERE
Is there a larger font....
4Mesh is WRONG HERE
DIVIDED by seconds elapsed. As in PER second.


Gee, where's the how fast it's going part? Did the engineering world leave that out? What the hell do they know about work anyhow.

SO, correct the incorrect line PLEASE. Or, just point out where I'm headed the wrong direction, OK?

1 Lb x 1 ft in 1 second = 1 ft lb / sec

2 lb x 1 ft in .5 seconds = 1 ft lb / sec Hmmm. faster,but still the same work?? Why?
Well because 4Mesh is WRONG HERE
2*1 ft-lbs of work
2*1/.5 = 4 ft-lb /sec of power


1 lb x 2 ft in .5 seconds = 1 ft lb / sec Hmmm. Faster yet, hmmm, odd...
Still 4 ft-lbs/sec


100 lb x .005 ft in .5 seconds = 1 ft lb / sec Slower? same work? 4Mesh is WRONG HERE
100*.005=0.5 ft-lb of work
0.5/0.5 = 1 ft-lb/sec of power


.005 lb x 100 ft in .5 seconds = 1 ft lb / sec
you're correct here, if you only knew why that was we'd be done.

1 lb x 3000 ft in 1 second = 3000 ft lb / sec


WHOA!!!!! Wait a minute, a POUND at 3000 ft per second is only 3000 ft lbs / second?????? Impossible! Damn equations!

How can that be? Gee 4Mesh, You can't possibly mean that math continues to work even when you change the values can you? Hell, even a .03 lb object lifted the same distance in the same time causes more ft lbs than that in Vibes example?
:D



Ok Vibe, am I wrong yet?
Well past wrong, and solidly into amusing. :D




Lynn? How am I doing?

I know what it is, I used 3000 ft. I should go back and change that.

3000 lb x 1 ft in 1 second = 3000 ft lb / sec

Ah, that's better. Now it's moving slower so when we square the distance, it is still 1. Vibe likes this equation as long as it's all 1's, but when you throw in numbers greater than 1 and some decimals, it's gets significantly tougher.
Sticks and stones Rosanna Rosanna Danna. :D


Now Here we go



You mean, all we have to do is divide by 550? Man, I want to do more math than that. How bout I square the 550 just for the hell of it? Nah, that would make the HP sound too small. I think I'll take the square root of 550 and use that instead. That's better, now our HP is higher! Now quite high enough, but better'n ole 4Mesh's example. Using the accepted equations just doesn't give the result I want damn it! :D

Bill, man, we gotta wonder bout these folks...

Ok. Now, is there anyone out there with a REAL degree in mechanical engineering who can tell me why my 163.63 hp over .001 second example is technically incorrect given the stated input parameters? Hint. I already told you above that there's more to it. However, it will not change the result appreciably. It is however technically incorrect.

Vibe, did you do any work on that Mars Lander project a few years back? You know, the one that made a half mile deep hole in Mars? :D :D

Almost an hour and a half of my time wasted to produce this for someone who's parents probably paid significant money for an education that was never received...
4Mesh? Do you feel like an ass yet? You worked so very hard at it I'd be disappointed if you didn't. :D

Vibe
09-23-2008, 02:35 PM
LOL. No worries Lynn. I probably needed the exercise.
In checking my notes on this subject, last night at work, I even surprised my self.
Since 1/2MV^2=MgH (in a vacuum)
that 200 grain bullet, launched at 3000ft/sec would travel 26.6 miles straight up....if it were not for air resistance. But as a long range shooter, you know that well before that first 3000 ft (1000 Yards) is traveled, your bullet will have lost almost 2/3 of it's initial velocity due to air drag alone, and that's not even straight up.

But I am really not going to tackle the task of explaining the HP involved in spinning the bullet up to speed. If I had this much trouble with 1/2Mass*Velocity^2
just think what 1/2 I(rotational moment of Inertia) * w (Omega)^2 would do to them. :D

4Mesh
09-23-2008, 02:59 PM
Do I feel like an Ass Yet? Not at all Vibe. This is fun to me!

I'll accept my due crow to eat as you have made a very few well founded corrections. However, Under no circumstance have you any idea what the correct answer is to the original problem if you still believe the gun makes 4000 hp for any >1 number of milliseconds.

Taken from your own references Vibe. Look at the first one you referred to, and down, I don't know, about 3 or 4 paragraphs you'll find this note.


Work is related to the distance a force moves an object and not the time it takes to move the object

Discluding your first reference in the message header, the first 4Mesh is wrong, is wrong. I won't mention most of them which are wrong due to wasting time and effort. (not to be confused with work, measured in recreation!)

Next.

4Mesh is WRONG HERE
Since you force me to be picky
WORK / TIME = ft-lbs/sec
1 HP=550 ft-lbs/sec
That one is correct! Score one for Vibe. Still a damn site closer than Energy / time / 550 = HP! Lol.


If you put energy into an object, then you do work on that object.Technically correct. However, I am not putting energy into an object thatis moving laterally. Our object is moving up against gravity. Forget then applying mass and velocity as it no longer applys Vibe.


Quote:Originally Posted by 4Mesh
I do not see any velocity in that calculation, do you?

Yep 4Mesh is WRONG HERE
It's in the acceleration of Gravity term used to define a Newton.I'm not even sure what you're talking about here. I asked if velocity is in that particular calculationand it is not. I ask because you apply it again later in THIS calculation and I do not. I already said it is included in the pound unit and semantically, you are grasping at straws to say it is in there when you know exactly where I am referring to. I do not include it because as I said, gravity is included in weight, we are lifting straight up, not moving a cart on wheels or an air hockey puck, we are lifting an object.

Referring to the slug definition.
OK here you are unconditionally wrongYou use a reference that words exactly what I said in different words and then say my references wording is incorrect because it differs from your reference. They say the same thing man! Weight includes gravity. Remember, this is why the example specifies lifting the object vertically, not accelerating it horizontally.


Originally Posted by 4Mesh
So, so far, work = lbs lifted x feet lifted x time in seconds to do it?

4Mesh is WRONG HERE
It's actually
work = lbs lifted x feet lifted / time
But hey...carry on.Score another for Vibe. Yes, it is divided by. 4Mesh eats another bite of crow gleefully!

Examples used stating calculated work did have errors and did not calculate work but did calculate power. 4Mesh used the incorrect term there and takes another bite of crow. Good job Vibe for catching the again, semantic error. Also incorrectly arrived at the results which concluded 4 ft lb sec of power. Vibe however in denial is not willing to accept that this is the correct calculation for power input figure to the HP equation. Vibe used some gawd awful abortion of a value that was substituted for the power number.


you're correct here, if you only knew why that was we'd be done.

1 lb x 3000 ft in 1 second = 3000 ft lb / sec
Yes, and 3000 Lb x1 ft in 1 second is also 3000 ft lb / sec. No matter how you add it up, you were wrong and I was right. The gun does not even come close to making 4000 hp does it?

Run the calculations for yourself and tell me what you come up with given the original input parameters. I want to know how many HP the gun makes in .1 seconds, and in 1 second if expressed that way. Will you do that for me please? If you are honest with us all, you will re-do your calculations and show that your arriving at 4000+ ft lbs / sec was so grossly wrong, it isn't funny. Not only did I do it right, but I told you 40 or 50 posts ago what it was that you did wrong and you didn't get it. Fact is, we're not sure you get it yet.

I have throuroughly enjoyed the exchange and my hat is off to you for catching the semantic and mathematical errors above, even if the math errors were only me typing the wrong operator but arriving at the right answers. I'll give you that the ones I did in my head quick were wrong and ate my crow accordingly.

Now, I want you to take your due bites of crow and tell the readers who calculated the HP correctly (or more correctly if you wish to stay in denial and say that the additional parameters that were not described at any time still affect the result and make me technically incorrect, as I already said is the case.

Ok Vibe. Tell us. Who came up with the right answer? You or Bill and I? We'll make this a multiple choice question and just say, who was within 20% and who was off by THOUSANDS of times over?

I thought you could change the font size too but ??? I don't know where that went to.

4Mesh
09-23-2008, 03:10 PM
Just so you don't stay in denial and use drag as a modifier to the calculations, just do gross HP and forget trying to now cloud things with the drag number associated with the air that affects the bullet for the 999 milliseconds it's coasting. I see that in your last post you are grasping at straws again trying to now compute net HP with drag factored in, in an effort to somehow show my example to be "technically incorrect".

Bottom line is Vibe, you are wrong. Admit it.

Vibe
09-23-2008, 03:29 PM
On the 300 Win Mag, I'll give another try.

210 grain Bullet. = .03 Lbs.
Launched at 3000 FPS

Muzzle pointed vertically and plumb. Firing straight up.

.03x3000=90 Ft Lbs per second of energy created. (though the energy is created in far less time than 1 second.)

Time already factored by the muzzle velocity. Power is only created for the hypothetical 1 or 2 ms.

90/550 = 0.1636HP

Now, lets say the estimate of ~1ms bullet in barrel is accurate for our discussion. You could then say the gun produces 163.6HP for one thousanth of a second, or you could say it produces 81.8HP for 2 thousanths of a second (milliseconds, ms).



This one?
To tell you the truth I modeled my math on some easier numbers (just because I didn't remember exactly last night). But here it is. I used a 200grain bullet launched at 3000ft/sec
200grain / 7000 grains/pound = 0.0285 pounds
but pounds is not mass
so 0.0285 pounds/(32 pounds/slug)
So bullet Mass = 0.0008929 Slugs

V=3000ft/sec
so V^2 =9000000ft^2/sec^2
and pound force is in units of slug-ft/sec^2 according to F=MA
1/2MV^2 =4018 ft-lb of muzzle energy

I guestimated a 26" (2.166 ft)barrel and the average speed inside the barrel would be (V2(muzzle vel)-V1(zero))/2 or and average of 1500ft/sec - which allows a calulated time in barrel of 2.166/1500 = 0.001444444 sec

Work done was 4018 ft-lbs of energy added to the bullet, time taken to do the work was 0.001444444 sec

4018/0.001444444 = 2781726.923 ft-lbs/sec

(2781726.923 ft-lbs/sec)/550 ft-lbs/sec =5057.69 HP

I think someone else posted a time in barrel of 0.0017 sec which is probably more realistic since the acceleration inside the barrel is not constant.

Using that time in barrel would result in
4018/0.0017 =2363558.824 ft-lbs/sec

(2363558.824 ft-lbs/sec)/550 ft-lbs/sec =4297.38 HP

These would be net HP ratings and not the peak instantanious values..those would be higher.

:D

What happens after the bullet leaves the barrel is not applicable,as YOU are no longer doing work on the bullet.

alinwa
09-23-2008, 03:33 PM
The figgerer is the guy who considers what a thing is going to look like or what a thing is going to do or not do before he starts doing. In the Boy Scouts we had a term for this, "Be prepared".

Al, I figger I am a doer and a figgerer if that term means thinking and planning before doing and I kinda think you are too.

You wouldn't set up some forms that you intend to pour concrete into without figgering what the concrete thing is going to look like after you remove the forms, would you?

Would you call a figgerer to tell you how to brace the forms?

And I don't think that you would need to rely on a figgerer to tell you how much concrete to order to fill your forms.

Of course not! It would delay the project by several days.

I figger that this horsepower thing as applied to a gun is like a crossword puzzle to some of us. Nothing more and nothing less. Just a mind game, that's all.

By the way, me and old 4Mesh are still right.:)

Concho Bill




I know that Bill, ;) you've been right all along. I've argued with Phil before you see :D he's an incisive thinker too. It just threw me to see Henry Childs' post wherein he stated "horsepower of" blah blah blahh...... Henry don't THINK like that! But once I'd read it in context I realized what he was saying......and why. :) ...... sad thing is, I MISSED the whole horsepower reference the first perusal, hence the dribbling yoke....... I've never known Henry to be involved in improper application of terminology before. He's an Engineer Capitalized.......


I love the mind games, I've been a subscriber to Games magazine since 1977........31yrs of doing this stuff just for FUN!

al

4Mesh
09-23-2008, 03:39 PM
OK. If we're going to be civil again, let's work together on this and get to the spot where we disagree, then discuss that until we agree.


This one?
Nope.
I'm Not gonna quote everything and waste space. We both know how to read the last post! :)


but pounds is not mass
so 0.0285 pounds/(32 pounds/slug)
So bullet Mass = 0.0008929 Slugs
Right there we disagree. Not that pounds are not slugs, but why are you using that unit when we alread know the weight of the bullet in pounds and have an equation that uses that unit? You are confusing the issue without reason.


V=3000ft/sec
so V^2 =9000000ft^2/sec^2
1/2MV^2 =4018 ft-lb of muzzle energy
Technically speaking, you are correct and the muzzle energy is absolutely near this number. However, that value is 4018 ft lbs of energy, NOT 4018 ft lb sec of work.

Where is the weight OR the mass in your muzzle energy calculation. What are you using? Certainly not the slugs value above. It looks to me as if you went back to 200 grains of weight instead, despite calculating the slugs value.

And just for the record, I'm not going to nit-pick if you do not. I am not interested in that, I want the answer.

Now, I'm not going to quote more until you address this and we will go on. If we meet an impass here, this needs corrected before we can agree and go on.

4Mesh
09-23-2008, 03:46 PM
I've argued with Phil before you see :D he's an incisive thinker too.

al
Yes you did Al, and as I remember, I ate a rather healthy dose of crow then too! My hat is off to ya man. Luv-ya-brother!

Vibe
09-23-2008, 03:53 PM
Right there we disagree. Not that pounds are not slugs, but why are you using that unit when we alread know the weight of the bullet in pounds and have an equation that uses that unit? You are confusing the issue without reason.
There is a very good reason, but it is one that confuses a lot of people. Pounds are not Mass and the equations F=MA , E=MgH, and 1/2MV^2 ll HVE to have the Mass in specific units. If you will notice - I arived at very nearly the same answer for the "weight" of the bullet that you did..in pounds...but Pounds are not a unit of mass, and we have to rectify that to move on. Which is why we have to continue withthe conversion to Slugs (which is a PITA)


Technically speaking, you are correct and the muzzle energy is absolutely near this number. However, that value is 4018 ft lbs of energy, NOT 4018 ft lb sec of work..
Work has the same units as Energy because that's what it represents Ft-Lbs...Ft-Lbs/sec is a description of Power - Work per unit time.


Where is the weight OR the mass in your muzzle energy calculation. What are you using? Certainly not the slugs value above.. Yes it is.
1/2*0.0008929 Slug* 9000000ft^2/sec^2 =4018 ft-(Slug*ft/sec^2) = 4018 ft-lb



And just for the record, I'm not going to nit-pick if you do not. I am not interested in that, I want the answer.
That's all I've been trying to give you.


Now, I'm not going to quote more until you address this and we will go on. If we meet an impass here, this needs corrected before we can agree and go on.
Are we there yet?

4Mesh
09-23-2008, 04:01 PM
Are we there yet?
Sorry but No.

Listen, I disagree with a fundamental part of your position.

You are saying that ft lb energy is the same thing as ft lb sec of power. I disagree.

Now, which of these is correct.

.03 pounds LIFTED against gravity 3000 feet in one second = 90 ft lb sec of power.

Or.

.03 pounds LIFTED against gravity 3000 feet in one second = 4000+ ft lb sec of power.

You have to show me why you say that the very simple pounds and feet and seconds is grossly incorrect. It isn't. Sorry.

Think about this more deeply. I'm saying you are trying to use acceleration on a stationary mass moving it on a level plane and then factoring in gravity as if it existed when it does not. I'm going to try to demonstrate where your other formula is wrong but it may take a few minutes.

EDIT ----------
Also, in your above post, you show square feet and square seconds. Please review that.

EDIT again -------------

Ok. I have a better idea. I'm going to take values of units to do this calculation again and you tell me which is wrong, if either.

Both cases. 10 lbs. Lift it 10 ft. Do it in 1/2 a second.

10 x 10 = 100 foot pounds of work.

10 x 10 / .5 = 200 foot pounds EDIT seconds of power. Sorry there.

Ok to here?


Now next, I do an energy calculation on that same scenario.

70,000 grains x 10 feet in 1/2 of a second.

I'm going to use 20ft per second as an average velocity.

The answer is, 62.19 foot pounds of energy. Given that our velocity is small in ft/sec, the square does not become astronomically large and is only 400.

That is, 70,000 x 20^2 / 450240 = 62.19

These values are not the same as you propose they are. This is where I've said your error has been since the onset.

Vibe
09-23-2008, 04:35 PM
Sorry but No.

Listen, I disagree with a fundamental part of your position.

You are saying that ft lb energy is the same thing as ft lb sec of power. I disagree.
Please point out where I hade that statement, becasue that is exactly what I've been trying to disuade you of.

Now, which of these is correct.


.03 pounds LIFTED against gravity 3000 feet in one second = 90 ft lb sec of power.

Or.

.03 pounds LIFTED against gravity 3000 feet in one second = 4000+ ft lb sec of power.

You have to show me why you say that the very simple pounds and feet and seconds is grossly incorrect. It isn't. Sorry.
Well first we are not "lifting" 0.3 pounds to 3000 ft in one second - we are accelerating it to 3000ft/sec in 0.0014 sec (or 0.0017 depending upon the example) - after the first 26" we are no longer imparting energy into the equation and gravity begins "doing work" unto the bullet. And I would seriously doubtthat it would even reach 3000 ft in that first second.


Think about this more deeply. I'm saying you are trying to use acceleration on a stationary mass moving it on a level plane and then factoring in gravity as if it existed when it does not. I'm going to try to demonstrate where your other formula is wrong but it may take a few minutes.

EDIT ----------
Also, in your above post, you show square feet and square seconds. Please review that.
The ft^2/sec^ is entirely correct, it's the only terms that can result from a V^2 and is another of the main reasons you have to work with Mass units instead of pound (force) units.
Acceleration is a Key component, be it the acceleration due to gravity in the calcualtion of "static" work - IE changes in potential energy MgH, or "dynamic work" changes in kinetic energy 1/2MV^2. The change in energy is the work done, the TIME it took to do the work is what determines the Power involved.

I'll look at you edit added examples, later..we posted at the same time.

4Mesh
09-23-2008, 04:38 PM
Please point out where I hade that statement, becasue that is exactly what I've been trying to disuade you of.

You make this statement every time you take an energy number stated in foot pounds of energy, and plug it in to the HP equation calling it ft lb sec. You have done it in EVERY example you have done so far. Please see my edited example above. No need to go past this line of disagreement till we get this right.

Ramsh00ter
09-23-2008, 04:42 PM
Pete,



I read in the Speer Reloading Manual a situation involving the .223, I believe-- not sure -- where a 1/7 twist barrel would put the spin on a 3500 fps bullet of -- get this -- 330,000 rpm. In this instance, most varmint bullets would literally blow up out of the muzzle due to incredible angular rotation.

Just for fun, I thought I would post a Excell Spreed Sheet I put together awhile back to calculate RPM of a bullet out of different twist rates at different velocities.

Just a little part of the puzzle...

Randy

Vibe
09-23-2008, 04:43 PM
You make this statement every time you take an energy number stated in foot pounds of energy, and plug it in to the HP equation calling it ft lb sec. You have done it in EVERY example you have done so far. Please see my edited example above. No need to go past this line of disagreement till we get this right.
No. I do not think so
Ft-lbs is Work
Ft-lbs/sec (Foot pounds PER second) is Power

4Mesh
09-23-2008, 04:49 PM
Well first we are not "lifting" 0.3 pounds to 3000 ft in one second - we are accelerating it to 3000ft/sec in 0.0014 sec (or 0.0017 depending upon the example) - after the first 26" we are no longer imparting energy into the equation and gravity begins "doing work" unto the bullet. And I would seriously doubtthat it would even reach 3000 ft in that first second.

We're supposed to forget about the drag of air decelerating the bullet and just deal with the power involved. Lets just say we're looking for gross HP not net.

This comes under the heading of nitpicking I think. Since the barrel length has never been defined, let's just say that the barrel length is 1 foot to keep numbers round. Now, lift the bullet 1 foot in 1/3000th of a second. Yes, this increased the HP considerably, but for our purpose, it is the same thing and no where near approaches > 1 HP. In fact, I do believe it is exactly the same as the figures we've had before. If you decide to now use 26" as the lift distance instead of 3000', yes the numbers will change and we both agree on that. If you'd like, we can go back through this with 26" of lift instead and say we move the bullet 2.16 feet instead. Whatever.

The figures we used were used to simplify the example and they did. Until now, we worked within them. Now you're disputing barrel length which is being pulled out of the air. You'll be hard pressed to show me where you used 26" of barrel in your previous calculations.

I won't do another edit as I can see I've made that somewhat confusing on the long post before. Sorry.

4Mesh
09-23-2008, 05:02 PM
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

This was from what is today, post #30.

Above you take the 4195 value stated in foot pounds of energy and use it as if it is ft lb / sec. Even if we only use that without making it worse, you are saying there is 7.63 hp for one full second. But worse yet, now you divide by .0017 barrel time and raise the value into the stratosphere.

Your example says 24 and a half million foot pounds per second. That's a lot! ;)

Lynn, get ready to cough up the dollar.

4Mesh
09-23-2008, 05:07 PM
Before you say that dividing that by .0017 changes the units, that is not how it works.

What you have said is, that over one second 4195 ft lbs / second has been done and that you are now going to take what that would represent IF we say that much was done in .0017 seconds. When you divide by .0017 seconds, you are assuming that the total energy value was a total power produced for 1 full second continously.

You have taken an energy value and used it as power

Jetmugg
09-23-2008, 05:22 PM
Again, I just don't think that HP is a good unit of measure for a rifle shot. Firing a rifle involves releasing a single discrete package of energy, not the kind of continuously applied power that is normally associated with HP measurements.

We don't try to measure the horsepower of a single combustion cycle in one cylinder of a V8 engine, we measure the macro-scale power of the entire engine while it is performing meaningful work against a dynomometer (sp?).

The duration of time that the power is being applied during the firing of a rifle is so short that the original intent of the Horsepower unit of measure becomes almost unrecognizable.


Another 2 cents,

SteveM

4Mesh
09-23-2008, 05:37 PM
Al
Did I hear you say Henry was not correct?
Lynn
P.S. $1 says Vibe wants the 6 pack before the day is over
Lynn, I'd say leave henry out of this as I can assure you had someone questioned this and done it in a constructive fashion, he would have taken a serious look at his work and accepted any error had he made one.

He's in good company with his position. He has had an email quoted without his say, and has made no effort to defend nor deny anything.

This discussion as I see it is between Vibe and I, and I think both have agreed we have made mistakes, though I'll say Vibe hasn't yet owned up to his.

I can say that we have pretty much all made mistakes in this thread, so he who is without mistake throw the first stone.

FYI, you have not done any proofs of your work either.

alinwa
09-23-2008, 06:36 PM
Al
Did I hear you say Henry was not correct?
Lynn
P.S. $1 says Vibe wants the 6 pack before the day is over


Henry was not "incorrect" for the person to whom he was writing.

Henry was unclear, or more accurately, incomplete in his explanation. It was implied in Henry's post that the bullet achieved 4000+ "horsepower equivalent rate" but only for a very short time. Horsepower isn't used that way though, horsepower is DEFINED as work over time with "1 Horsepower" being defined as the force required to move 550lb a distance of one foot in the time of one second.

Put bluntly, the 300win Mag WILL NOT do that, therefore it simply does not generate even ONE horsepower.

This is just what horsepower IS.........

Henry was talking about acceleration, his use of "horsepower" was an attempt at laymanese, Henry's not very GOOD at conversing with mortals :D and this example illustrates that. If you were to flat-out ask Henry how many horsepower could be generated by the energy produced by the powder charge he would probably SWAG it to within +- 1% off the top of his head. (Although knowing Henry a little, he'd probably have to qualify his answer with "at xxxx barometric pressure" and "assuming stoichiometric coefficient was" and "provided the expansion of the pressure vessel was kept to within"......) What Henry's actually saying is that this ACCELERATION rate is tremendous, SO tremendous in fact that IF YOU KEPT IT UP for the required time you'd have to be generating (4000+) horsepower. Or, "for just the tiniest fraction of a second this thing is generating 4000+ horsepower equivalent" but unfortunately it can't sustain it for any length of time....... so in FACT it doesn't even generate ONE horsepower unit. It's like a hammer, a hammer "hits" very "hard" but getting it to do work is well nigh impossible. It would take many hammerblows to move the 550lb block the required foot in one second.


I think that Henry should have used the term "horsepower equivalent", expressing it as a rate.....but then it becomes "technical" ;)


It's a FUN question but until everyone achieves the same remove it's hard to come to a common viewpoint.


KUDO's on both vibe and 4mesh for perservering.......even though it aint winter yet :)


Hey, winter's coming. Might be time to reopen the "Mother of all Winddrift Threads" soon. I've got some more ammunition.


But FIRST I've got to get with Boatright and Rinker and find out why they both insist that a bullet assumes a point-down attitude in a left-to-right crosswind........ We might just MAY get The Great Ballistic One involved in this next one ;) and it could get HUGE.... :D:D:D:D:D


LOL


al

4Mesh
09-23-2008, 06:45 PM
My e-mails to Henry are many months old and why this one was put in storage alludes me at this moment.I believe Henry included a witty remark and tat is whar I thought was being saved.Henry never sent me this after reading this thread.It is likely 6 months old or older.
I didn't do any proofs because none of the numbers presented here are close.I've seen 5,000 hp and I've seen 0.669 hp given.
Hp is a rating just like the rating of a firearm.Has anyone here ever fired 700 rounds in one minute using an M1? The answer is of course "no" because it is a rate of fire even though nobody shoots one for 10 minutes straight and uses 7,000 rounds.
I would also not expect a 220 gr bullet with a rated
hp of 4500 move a 550 pound weight 3,002 feet.
If Jettmug has ever been on the track a simple push with your hand will move the car during the burnout.Very little of the 5,000 hp rating is ever applied to the track.
Lynn

Lynn, I saw long ago when you first talked about the email before it was even yet posted, that it was many months ago. I did read the thread, though I have missed some minor things. This is why I said it was quoting him without his say. That was LONG before this thread ever existed.

This units conversion mistake has been floating around this board for a VERY long time. Well before 6 months ago, and I mean well before. I have avoided discussions such as this in the past, not because I had nailed down this error, but because I could see with my eyes closed that something was wrong, but didn't know what. And I didn't want to get in some over the top discussion like this one. Unfortunately for me, I did get in this one and tried an example of my own. Suffice to say, I dont' think that will happen again.

I'll be honest, I do not like this sort of discussion. The time and effort it takes is insane, and then just as has happened here, someone changes the rules halfway and makes everyone wrong, in spite of there being a clear answer with the original parameters.

I have no doubt that Vibe went to work again today and won't reply for some time. I got a bit ahead of myself (for lack of better words) and removed a post that wasn't ready yet for discussion since we had not addressed the last issues.

Believe me when I say, when this is all over with, Vibe is going to look back and re-read this entire thread in detail, and very early on will say, OMG!!!! What was I thinking?

If I had to guess, Vibe is the unfortunate victim of a professor who used some coined phrase somewhere along the lines, and that phrase stuck with him. Something to do with energy and work or energy and power being the same thing. Some PART of the phrase is misleading, and he has said that several times in this thread about the energy = work or something to that effect. Let's not nitpick and go show that this isn't the word for word quote, it's the general meaning that he's expressed. The professor was wrong with some wording, or left out something and Vibe bit, line hook and sinker as he should, assuming the professor was right.

A similar such phrase that becomes a semantic hole of quicksand is quoted here on this board regularly. I choose not to point it out, as I think it is more fun to see how it affects the entire shooting community as a whole! :D And boy does it!

alinwa
09-23-2008, 08:55 PM
BTW vibe, I owe you a big THANK YOU!!!

I have thought for years that a slug was a pound-equivalent, kinda' like the poundal.

I WAS WRONG!

Thank you man, a slug is just over 32lb @MSL etc etc.

I love this stuff!

al

Jetmugg
09-23-2008, 09:26 PM
If Jettmug has ever been on the track a simple push with your hand will move the car during the burnout.Very little of the 5,000 hp rating is ever applied to the track.
Lynn

If Lynn has ever been on the track, he would know that top fuel cars do not do a "line lock" type of a stationary burnout. Let the fueler finish the burnout, back up, stage, and start the christmas tree. Now try to hold it back. Perhaps you could use a water ski rope, and just yell "HIT IT" when you feel like you have your feet planted.

SteveM.

4Mesh
09-23-2008, 11:11 PM
Vibe,

When you get back from work, or tomorrow, please re-read my posts 16 and 26. These address the fundamental mistake that's being made by an entire host of people who are "cheating" if you will by taking an energy number from their balistics program and then trying to use it as a parameter for this problem.

In your post #66, you clearly state (though you did it a number of times earlier) that the units for energy and work(power) are the same thing with seconds strapped on to the power one. This is the error you and others are making. Those units are not the same, despite the fact that they are expressed the same way. Ft Lbs can be both an energy, or a force, and the two are not even close to the same thing. As an energy it is intantaneous, but as a force, it relates to time. They are not interchangeable.

Man, this thread is too long to go backtracking through all the time. I nearly installed an old copy of MathCAD I have laying around here! hehe.

EDIT =====

Oh, and by the way, I do see now where we have been working with a defined barrel length since the onset, even if we didn't know it. If you use my simplified example and the 3000 ft / sec MV, and any one of the time in barrel examples for calculating HP, the barrel length can be calculated by using the velocity and the time we state the bullet is in the barrel. No matter though, regardless if we say the barrel is 1 foot or 3000 feet, it does not change the numbers because time adjusts with it inversely proportionate. (greatly simplified of course)

alinwa
09-24-2008, 12:16 AM
BTW, since't we're regressing, digressing, revisiting and gener'ly recessing....... you'se were comparing ballistic pendulum stories :) cool.

I never built the pendulum but I DID build and execute the "Monkey in a Tree" experiment in my high school physics class........using a dartboard, an electromagnet and a Huarache blowgun.

There, got MY brag in too :D


LOL


al

4Mesh
09-24-2008, 12:39 AM
Al,

When I think back to my younger days and remember all the stuff I built, I'd have to say it's a miracle I'm alive!

:D

Vibe
09-24-2008, 09:15 AM
In your post #66, you clearly state (though you did it a number of times earlier) that the units for energy and work(power) are the same thing with seconds strapped on to the power one. This is the error you and others are making. Those units are not the same, despite the fact that they are expressed the same way. Ft Lbs can be both an energy, or a force, and the two are not even close to the same thing. As an energy it is intantaneous, but as a force, it relates to time. They are not interchangeable.
You mean here?

Work has the same units as Energy because that's what it represents Ft-Lbs...Ft-Lbs/sec is a description of Power - Work per unit time.


No mistake there. The units are correct.
Ft-lbs can be Work, or Energy...but NEVER a Force as those two are definitely NOT the same thing. Force is simply Pounds, or the real units of
Slug-ft/sec^2 in the pound-foot-second system.
(The ft-lbs used to describe torque is another matter and we don't want to go there right now)
And the units of Power are the units of work (and or energy) divided by the time involved in doing the work or changing the energy.

4Mesh
09-24-2008, 09:49 AM
I'm gonna be perfectly honest with you. You have me so frustrated with trying to dream up new ways to say this that I am begining to make errors myself. Kinda like the post where I asked you to write in there 4Mesh is wrong, when 4Mesh isn't. Please be sure to gain entertainment value from this and don't take me too seriously as I fear may be happening. Those who know me will tell you that I'm really a hell of a good sport about stuff and it's all a friendly rib'n that I take probably better than I give out.

Now. I'm going to say this. You have avoided my math for so long it's starting to wear thin with me. My language skills are not broad enough to come up with many more ways to tell you the same things I've said. Without learning German or Dutch and saying it that way I'm "runnng out of bullets" Vibe.

I implore to you, PLEASE answer my questions. Now, I'm going to re-word and re-post a question I asked you some time ago and ask you to address it.

Also, Have you re-read this thread from start to finish? When the light clicks on, I know you're going to fall off your chair.

Here goes.
removed quotes because it's really not a quote, and it does make it hard to read.

Listen, I disagree with a fundamental part of your position.

You are saying that ft lb energy is the same thing as ft lb sec of power. I disagree.

Fixing that statement, I should have said (and I am sorry for this)
you say ft lb of ENERGY is the same thing as ft lb of WORK. But if you look at what you did, you DID turn it into power without even knowing it. Now, am I using the correct term on the second half there with "work" or should that be different? If that is a semantic error on my part, accept that and just answer the question that you know I'm asking. Please. Before typing anything about this, please read the next statements that I'm going to make more clear again hopefully.

Ok. I have a better idea. I'm going to take new easy values of units to do this calculation again and you tell me which is wrong, if either. we are going to take the same values for two items and calculate both the work in ft lbs and the energy in ft lbs.

Both cases. 10 lbs. Lift it 10 ft. Do it in 1/2 a second.

Now, don't nit pick the lift term. I'm trying to make this understandable for everyone.

10 x 10 = 100 FOOT POUNDS OF WORK. WORK WORK WORK WORK

Ok to here?

Now next, I do an ENERGY calculation on that same scenario. ENERGY ENERGY ENERGY

Remember, 70,000 grains is the same as 10 pounds.

70,000 grains x 10 feet in 1/2 of a second.

I'm going to use 20ft per second as an average velocity.

The answer is, 62.19 FOOT POUNDS OF ENERGY.
That is, 70,000 x 20^2 / 450240 = 62.19

62.19 <> 100 Correct?

Given that our velocity is small in ft/sec, the square does not become astronomically large and is only 400.

In this particular case, the velocity is low, and the calculation actually shows less energy in Ft Lbs than it does work in Ft Lbs. The number we use as a multiplier which is the SQUARE OF THE VELOCITY as stated MANY times before, does not show this example to be wrong by nearly as much as a larger velocity squared like 3000 ft/sec. With 3000 ft/s velocity, the error is so incredibly large that Stevie Wonder would be pointing this out in the dark.

BOTH THE EXAMPLES ABOVE USE Ft Lbs as an end result but the QUANTITIES DESCRIBE DIFFERENT THINGS. THIS IS THE ERROR I'VE POINTED OUT ALL ALONG. Please re-read post #16 at this time.

Ok. If you still do not see this, I am afraid that... I'm just going to keep my fingers tied and say that while someone may be able to help you, I am not the one. I'm trying to be as nice as I am capable of and show this but you're frustrating me with your pre-supposed answers and not working through my examples. You have not worked through MY examples even one time, and I have done yours COUNTLESS times and showed you so many different ways I'm exhausted.

All in fun. I do hope you've got it.

Please, promise me you will re-read this thread after you understand what it is I'm saying and after we have come to an agreement about the calculation of HP.

------------------------------------------------------
Sorry, but I have got a new way to add to this.

Think of it like this. Energy is directionless. Work is vertical. Does that help. Now do you understand why I kept referring to your examples working on a level plane and mine are in a vertical plane?

Also, think of the one ft lbs value as containing time, and the other does not. Again, on the energy value, you need to turn that back around into work, then come forward with the Power and HP calculations.

Vibe
09-24-2008, 09:53 AM
You are saying that ft lb energy is the same thing as ft lb sec of power. I disagree.
So do I ..And as far as I know I have NEVER said that. They are NOT "the same thing"

Vibe
09-24-2008, 09:59 AM
Ok. I have a better idea. I'm going to take new easy values of units to do this calculation again and you tell me which is wrong, if either. we are going to take the same values for two items and calculate both the work in ft lbs and the energy in ft lbs.

Both cases. 10 lbs. Lift it 10 ft. Do it in 1/2 a second.

Now, don't nit pick the lift term. I'm trying to make this understandable for everyone.

10 x 10 = 100 FOOT POUNDS OF WORK. WORK WORK WORK WORK

Ok to here?

Now next, I do an ENERGY calculation on that same scenario. ENERGY ENERGY ENERGY

Remember, 70,000 grains is the same as 10 pounds.

70,000 grains x 10 feet in 1/2 of a second.

I'm going to use 20ft per second as an average velocity.

The answer is, 62.19 FOOT POUNDS OF ENERGY.
That is, 70,000 x 20^2 / 450240 = 62.19

62.19 <> 100 Correct?

Given that our velocity is small in ft/sec, the square does not become astronomically large and is only 400.

In this particular case, the velocity is low, and the calculation actually shows less energy in Ft Lbs than it does work in Ft Lbs. The number we use as a multiplier which is the SQUARE OF THE VELOCITY as stated MANY times before, does not show this example to be wrong by nearly as much as a larger velocity squared like 3000 ft/sec. With 3000 ft/s velocity, the error is so incredibly large that Stevie Wonder would be pointing this out in the dark.

BOTH THE EXAMPLES ABOVE USE Ft Lbs as an end result but the QUANTITIES DESCRIBE DIFFERENT THINGS. THIS IS THE ERROR I'VE POINTED OUT ALL ALONG. Please re-read post #16 at this time.


OK. I see what you're asking I think...Give me a bit to work it out and make it clearer...but the main issue is in the difference in the "acceleration term"
The 10x10 example is working against 32 ft/sec^2 while the 70000 grain example is taking a difference in velocity.
One is an MgH problem, the other is an MV^2 calculation...you cannot simply equate the two.

10/32 Slug * 32ft/sec^2 * 10ft = 100 ft-lbs WORK
70,000grain * 1 pound/7000 grain * 1 slug/32 pounds * 32ft/sec^2 * 10ft = 100 ft-lb WORK

100 ft-lb WORK done in 0.5 sec = 100/0.5 = 200ft-lb/sec power

the 20ft/sec example is a bit different. IF you are saying that a 70,000 grain mass is traveling at an average of 20ft/sec and goes 10 ft...then no work was done, as the velocity did not change. IF on the other hand you are saying that it started at 0 velocity, and traveled 10 ft under constant acceleration in 0.5 sec resulting in an average speed of 20 ft/sec...then the final speed is 40 ft/sec at the end of that time period V(ave)=(V2-V1)/2 then if V1 =0, then V2 has to equal 2V(ave)

10/32 Slug * 40 * 40 ft^2/Sec^2 = 0.3125 * 1600 Slug ft/sec^2 - ft = 500 ft-lbs work

500ft-lbs work done in 0.5 sec is 500/0.5 = 1000 ft-lb/sec power.

If you are saying that this happened in a straight up direction then the previous 100 ft-lb of work due to change in potential has to be added to the 500 ft-lb kinetic energy increase due to velocity.
SO total work in this particular case would be 600ft-lbs
done in 0.5 sec would be 1200 ft-lb/sec of power.

Sorry this took so long. I had a virus crash during posting and have been diagnosing that as I posted.

4Mesh
09-24-2008, 11:16 AM
100 ft-lb WORK done in 0.5 sec = 100/0.5 = 200ft-lb/sec power

the 20ft/sec example is a bit different. IF you are saying that a 70,000 grain mass is traveling at an average of 20ft/sec and goes 10 ft...then no work was done, as the velocity did not change. IF on the other hand you are saying that it started at 0 velocity, and traveled 10 ft under constant acceleration in 0.5 sec resulting in an average speed of 20 ft/sec...then the final speed is 40 ft/sec at the end of that time period V(ave)=(V2-V1)/2 then if V1 =0, then V2 has to equal 2V(ave)

10/32 Slug * 40 * 40 ft^2/Sec^2 = 0.3125 * 1600 Slug ft/sec^2 - ft = 500 ft-lbs work

500ft-lbs work done in 0.5 sec is 500/0.5 = 1000 ft-lb/sec power.
Look there, you just took the same input units and came up with 200 <> 1000 units of power. It's off by a factor of 5 and this is a really small error as compared to what's been done here since page one of the thread.

========= 4Mesh leaves a huge sigh, shakes his head, closes his eyes and asks for God to intervene here. ==========

Will you PLEASE LET ACCLERATION OUT OF IT. FORGET IT. Nada, it is not there. The vessel is not accelerating in example two, it is floating in space. Stop trying to figure how it got up to speed.

Listen, when we lift the weight we lift it 10 feet. Might take a day, might take .01 seconds. The same amount of work is done. Time is imaterial.

When we calculate velocity, we do not need to know how fast the bullet accelerated. When we calculate downrange energy of a bullet. Let's say we have a MV of 2000 ft per second. Is the energy any different in that bullet than it is in a bullet that started at a MV of 10000 fps and slowed to 2000. When they both reach 2000 fps, they have the same energy if they weigh the same. In all truth, I hate that we have to qualify obvious statements like, yes the same bullet.



We used 20 ft per second as an AVERAGE velocity only because during that time it moved 10 ft in 1/2 second. We do not care how much time it took to get up to speed. Technically here YOU ARE CORRECT ON A POINT, the ending velocity will not be 20ft/second if there was an acceleration phase. Here's two better examples and you choose the one you want to use. They are the same end result. The 10 pound weight is sliding at 20ft per second across a perfect frictionless plane and has been so at the same speed for a decade. How much energy does it have? The acceleration rate is NO PART OF IT.

Or, the 10# weight magically accelerated to 20ft/sec exactly in NO TIME and traveled the 10 feet at the exact rate of 20ft/s. Use either. Sorry to keep changing my mind but you keep bringing in more things that do not apply.


If you are saying that this happened in a straight up direction then the previous 100 ft-lb of work due to change in potential has to be added to the 500 ft-lb kinetic energy increase due to velocity.
SO total work in this particular case would be 600ft-lbs
done in 0.5 sec would be 1200 ft-lb/sec of power.

Sorry this took so long. I had a virus crash during posting and have been diagnosing that as I posted.

That last paragraph I'm not going to address as in my mind it is not even applicable to our discussion. It is unfortunate that I need to keep adding more default parameters to the same problem eliminating parts that you keep bringing into it. All I can say is you are making this vastly more difficult than it is.

No problem on the computer issues. Deal with work first. (not to be confused with lifting boxes!).

4Mesh
09-24-2008, 11:24 AM
And Al, Bill, or someone here, please take over this discussion with a fresh point of view before ole 4Mesh is eating a can full of DynaCirc CR's for lunch :D:D:D I don't deal well with these sorts of things anymore. My teaching days were over a LONG time ago!

Vibe
09-24-2008, 11:29 AM
Look there, you just took the same input units and came up with 200 <> 1000 units of power. It's off by a factor of 5 and this is a really small error as compared to what's been done here since page one of the thread.
No errors, just trying to cover all of your scenarios.



Will you PLEASE LET ACCLERATION OUT OF IT. FORGET IT. Nada, it is not there. The vessel is not accelerating in example two, it is floating in space. Stop trying to figure how it got up to speed.

Listen, when we lift the weight we lift it 10 feet. Might take a day, might take .01 seconds. The same amount of work is done. Time is imaterial.

When we calculate velocity, we do not need to know how fast the bullet accelerated. When we calculate downrange energy of a bullet. Let's say we have a MV of 2000 ft per second. Is the energy any different in that bullet than it is in a bullet that started at a MV of 10000 fps and slowed to 2000. When they both reach 2000 fps, they have the same energy if they weigh the same. In all truth, I hate that we have to qualify obvious statements like, yes the same bullet.



We used 20 ft per second as an AVERAGE velocity only because during that time it moved 10 ft in 1/2 second. We do not care how much time it took to get up to speed. Technically here YOU ARE CORRECT ON A POINT, the ending velocity will not be 20ft/second if there was an acceleration phase. Here's two better examples and you choose the one you want to use. They are the same end result. The 10 pound weight is sliding at 20ft per second across a perfect frictionless plane and has been so at the same speed for a decade. How much energy does it have? The acceleration rate is NO PART OF IT.

Or, the 10# weight magically accelerated to 20ft/sec exactly in NO TIME and traveled the 10 feet at the exact rate of 20ft/s. Use either. Sorry to keep changing my mind but you keep bringing in more things that do not apply.

I covered that possibility

IF you are saying that a 70,000 grain mass is traveling at an average of 20ft/sec and goes 10 ft...then no work was done, as the velocity did not change
And if it was somehow traveling in an upwards direction at a constant 20ft/sec, again only the 100 ft-lbs of work is done against gravity requiring the output of 200 ft-lbs/sec of power.

Jetmugg
09-24-2008, 11:44 AM
I think that 4Mesh is pointing his rifle straight up when it is fired.

Vibe is firing his level with the ground.

Are these assumptions correct?

SteveM.

Vibe
09-24-2008, 11:48 AM
I think that 4Mesh is pointing his rifle straight up when it is fired.

Vibe is firing his level with the ground.

Are these assumptions correct?

SteveM.I've been wondering who would ask this first. :D
It really doesn't make much difference. After the bullet leaves the barrel, we are no longer "doing work" to it...It's on it's own. And the change in potential energy of a 200 grain bullet from chamber to 26" above the chamber is really really small compared to the energy in the velocity. I suppose I could do the math on it...if anyone was REALLY interested....but it ain't much. Much less than one ft-lb.

4Mesh
09-24-2008, 11:57 AM
Alright, let's do completely fresh examples.

I see again that you're grasping at straws or it sure looks that way to me. You continue to skirt the issue quite well actually and my hat is off to you for laying down a line ....... what was the General's line used in Good Morning Vietnam?

It looks like I have to make this more clear yet. While everyone else in the world knows what was being assumed, you are now playing the witty engineer who is going to try to use technicalities which are not relavant to the discussion, to say someone else is wrong (however slightly) and hope that they just go away with you not admiting you are the one who is off by roughly 50,000 times or so. That of course in the earlier examples. Now you're just off by 5 times because I made the velocity smaller. Here again, it's gonna get bigger.

I'll use one second as my elapsed time for the work and velocity is going to be expressed in ft/sec with no seconds specified.

Ok. Here goes. Once again, Please correct the part I do wrong.

1 pound. Lift 100 ft against gravity. Do the work in 1 second. 100 ft lbs of work. 100 ft lbs/sec of Power.

1*100 = 100 ft lbs.
1*100/1 = 100 ft lbs /sec

Now. 1 pound travelling at 100 feet per second. Calculate the energy we are dealing with. Remember, this is what YOU have done since the onset. Please don' t go back and delete all your posts to bury the dog dirt as others have done, it lacks character.

1 pound. 100 ft/s
1 pound = 7000 grains used for input to this equation.

7000 * (100^2 or 10,000) / 450240 = 155.47 FOOT POUNDS ENERGY.

Once again. 155.47 ft lbs energy does not equal 100 ft lbs of work.

Now, just to save a lot of time, which one of us used the WRONG one since this thread began? Which one Vibe.

Will you ever own up to this?

4Mesh
09-24-2008, 12:00 PM
Ummm, what is this BS about when is someone going to ask about horizontal vs vertical????? I've said that till I was sick of saying it. Matter of fact, I said about everything here enough times that it would have sunk in to ANYONE ... LONG ago.

I think it is time for Vibe to begin his long day re-reading this thread and then get back to us. I'm sure he will especially enjoy all the red "4Mesh is WRONG HERE" lines. Especially when they are WRONG.

Vibe
09-24-2008, 12:13 PM
1 pound. Lift 100 ft against gravity. Do the work in 1 second. 100 ft lbs of work. 100 ft lbs/sec of Power.

1*100 = 100 ft lbs.
1*100/1 = 100 ft lbs /sec

Now. 1 pound travelling at 100 feet per second. Calculate the energy we are dealing with. Remember, this is what YOU have done since the onset. Please don' t go back and delete all your posts to bury the dog dirt as others have done, it lacks character.

1 pound. 100 ft/s
1 pound = 7000 grains used for input to this equation.

7000 * (100^2 or 10,000) / 450240 = 155.47 FOOT POUNDS ENERGY.

Once again. 155.47 ft lbs energy does not equal 100 ft lbs of work.

Now, just to save a lot of time, which one of us used the WRONG one since this thread began? Which one Vibe.

Will you ever own up to this?
Well the first one works out right. :D

Grains is not a unit of MASS
1 pound * 1 Slug/32 pounds = 1/32 Slug
(100ft-sec)^2 = 10000 ft^2/Sec^2

1/2 MV^2= 1/2 * 1/32 * 10000 Slug-ft/sec^2 - ft = 156.25 ft-lbs of energy but represents no "work done" as there was no velocity change.

I think I left out the 1/2 factor in the virus crash post. Sorry. :D

And no the two values are not equal...why would they be?

I'm assuming that you want the 100 ft/sec to be a constant velocity in the upwards direction. In which case there was 100 ft-lbs of work done in 1 sec for 100ft-lbs/sec of power. No additional power was required to increase the velocity so 1/2M(delta)V=0 ft-lbs. 100 + 0 =100 Ft-lb work done.

4Mesh
09-24-2008, 12:16 PM
Grains is not a unit of MASS
1 pound * 1 Slug/32 pounds = 1/32 Slug
(100ft-sec)^2 = 10000 ft^2/Sec^2

1/2 MV^2= 1/2 * 1/32 * 10000 Slug-ft/sec^2 - ft = 156.25 ft-lbs of energy but represents no "work done" as there was no velocity change.

I think I left out the 1/2 factor in the virus crash post. Sorry. :D

And no the two values are not equal...why would they be?

I'm assuming that you want the 100 ft/sec to be a constant velocity in the upwards direction. In which case there was 100 ft-lbs of work done in 1 sec for 100ft-lbs/sec of power. No additional power was required to increase the velocity so 1/2M(delta)V=0 ft-lbs. 100 + 0 =100 Ft-lb work done.

Forget that crap. I KNOW GRAINS ISN'T A MASS. That's why we use a constant of 450240

4Mesh
09-24-2008, 12:19 PM
And no the two values are not equal...why would they be?
Because you have used these interchangeably SINCE THE DAWN OF THIS THREAD.

No, I'm not nitpicking about the 1/2 left off... We know where the thing is supposed to be headed. This entire excercise is suppose to be showing what you did wrong in the earliest example. PLEASE go BACK to the beginning and you will see answers off by 40 or 50 THOUSAND times the real quantity.

Vibe
09-24-2008, 12:24 PM
I'm afraid you will have to copy and paste it here. The only MAJOR error I've seen is that you still seem to think that you are "doing work" to the bullet 2997.8 ft after it has left the barrel...you are not. You "Did Work" to it ONLY for the 26" (or so) that it was IN THE BORE. Afterwards it is not being "worked upon" by you.

4Mesh
09-24-2008, 12:38 PM
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

Working backwards
.668 Hp = .668 * 550 ft-lb/sec = 367.4 ft-lbs/sec

367.4 ft-lbs/sec * 0.001711 sec =0.6286214 ft-lbs muzzle energy = pretty anemic.
From what is at this moment, post # 27 from you.

Now.

The only MAJOR error I've seen is that you still seem to think that you are "doing work" to the bullet 2997 ft after it has left the barrel...you are not. You "Did Work" to it ONLY for the 26" (or so) that it was IN THE BORE. Afterwards it is not being "worked upon" by you.

You say, this is not a MAJOR ERROR?????????????????????????????????

Please, let's start over here and don't make me copy and paste every instance I told you this is wrong because we don't have enough space on the internet.

Now that we know the muzzle energy has been used grossly wrong, I'll admit that the hypothetical barrel length does alter the numbers but for OUR DISCUSSION, we were using average values to make the math easier.

As far as that causing any appreciable error as you keep grasping at straws with, gravity will only affect the bullet by 9.8 Meters as we are using one second for the bullet to move 3000 feet. So, it will be 296x.x feet of travel instead and I SAID THIS LONG AGO.

So, if our barrel is one foot long, or 3000 feet long, that was not ever defined and does not change the answer does it.

Some time ago, you made a statement about my not being able to understand my own reference. Let's see, would you like to correct that now?

In the above example, you arrived at
2451782.583 ft-lb/secNOw, is that number correct? Is that considered a MAJOR error? Yes or no.

Let me ask a better question. If your bank inadvertantly took a check of yours for $3000 (the amount you wrote the check for) and charged your bank account $2,451,782.58, would you ask them to reconsider?

Jetmugg
09-24-2008, 12:47 PM
I agree that horizontal versus vertical will not make a difference in the final answer. However, I can see that the difference in direction is causing some difficulty.

The other thing that seems to be causing problems is related to the SAE units of measure. In particular, pounds of force as compared to pounds of weight.

Normally, I prefer SAE units of measure also. However, in this case, might it be more transparent to use metric units? We all know that a Kilogram is not a Newton, but it's not always apparent that a pound isn't always a pound.

The units have to come out right. That's the main thing that got me through Thermodynamics and Heat Transfer classes in college. I knew that if the units worked out right, at least there was a possibility that I had the correct answer. If the units didn't worrk out, there was no way that my answer was correct. (Metallurgical Engineering students have some overlapping curriculum with Mechanincal Engineering students).

Anyway, my suggestions are to agree upon a single direction (up or horizontal) so that the effect of gravity is understood clearly, then switch to metric units of measure to avoid issues with pounds of force and pounds of weight.

SteveM.

Vibe
09-24-2008, 12:49 PM
You say, this is not a MAJOR ERROR?????????????????????????????????
Not only is it not a "major" error....it is completely correct in that it represents the CHANGE in energy from chamber to muzzle upon firing.

I demonstrated this in post#62 as well and even provided all of the math to support it.



Now that we know the muzzle energy has been used grossly wrong, I'll admit that the hypothetical barrel length does alter the numbers but for OUR DISCUSSION, we were using average values to make the math easier.
Not only was it NOT used "Grossly wrong" it's a well documented "short cut" to the correct value of energy CHANGE between chamber and muzzle. What happens past the muzzle is of no consequence to the issue. At least not the issue of Horsepower.




Is that MY avatar I'm hearing?

Jetmugg
09-24-2008, 01:22 PM
Is 4000 horsepower applied for 0.001 seconds the same amount of work as 4 horsepower applied for 1 full second?

SteveM.

Vibe
09-24-2008, 01:26 PM
Is 4000 horsepower applied for 0.001 seconds the same amount of work as 4 horsepower applied for 1 full second?

SteveM.
Yeah..pretty much. (Pretty much exactly 2200ft-lbs if I calculated correctly.) :D

4Mesh
09-24-2008, 01:32 PM
Ok Vibe,

Using the assumptions in my post #12 that seemed to be good enough for you to take a first stab at this, how about you do a proof on this. You have supported this for a rather long time now. Please re-read the thread if you missed or need to see all the times I've asked you to support why you are using energy to calculate power, really just questioning the end result.

Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

In there, you used my hypothetical 210 grain bullet in a 300 win mag and a velocity of 3000fps included in the energy, but also used as an estimate of distance for work (that is within roughly 99% of the actual value if you use only the barrel time and disclude drag as that has nothing to do with what HP the gun made. Another point both of us have made multiple times. No other reference mentions 4195 ft of energy so it was my example you were "correcting" shall we say.

Now. Discluding other insignificant or irrelavant/ignored forces as you did in this example, do you still say the 2451782.583 ft-lb/sec power is an accurate estimate for this hypothetical situation and accurate enough for this discussion? Yes or no.

Now, Is it within 100 THOUSAND ft-lb/sec of the truth, yes or no?
Is it within 500 THOUSAND ft-lb/sec of the truth, yes or no?
Is it within A MILLION ft-lb/sec of the truth, yes or no?
Is it within 2 MILLION ft-lb/sec of the truth, yes or no?

Vibe
09-24-2008, 01:53 PM
210 grain * 1 pound/7000 grain * 1 slug/32 pounds = 0.0009375 Slug
(3000ft/sec)^2=9000000 ft^2/Sec^2

1/2MV^2= 1/2*0.0009375 *9000000 Slug-ft/sec^2 -ft

1/2MV^2= 4218.75 ft-lbs energy

Energy in chamber = 0
Energy at muzzle = 4218.75

(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

How much time did it take to do the work?
Again I'm going to assume a 26" barrel
Velocity at muzzle = 3000ft/sec
Velocity in chamber = 0
Assuming constant acceleration V(ave)=(Vmuzzle-Vchamber)/2=1500ft/sec
distance =Vt
so
T=d/V =(26" * 1ft/12")/(1500ft/sec)=2.166666667/1500=0.001444444 seconds

Work done/time to do the work= 4218.75/0.001444444=2920673.077 ft-lbs/sec

Work done in terms of HP
(2920673.077 ft-lbs/sec)/550ft-lbs/sec=5310.314685HP


Now. Discluding other insignificant or irrelavant/ignored forces as you did in this example, do you still say the 2451782.583 ft-lb/sec power is an accurate estimate for this hypothetical situation and accurate enough for this discussion? Yes or no.

It was a bit low. The 0.001711 sec. barrel time get's it a bit closer though, but I don't really know where that came from.

4Mesh
09-24-2008, 02:05 PM
(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

Wrong again. Here you go with energy = work. energy does not equal work.

Didn't I just get done spending a day teaching you that these two are not the same thing? I think your answer was, "Why would they be?"

Vibe
09-24-2008, 02:10 PM
Wrong again. Here you go with energy = work. energy does not equal work.

Didn't I just get done spending a day teaching you that these two are not the same thing? I think your answer was, "Why would they be?"

Energy does not equal work...but the CHANGE in energy DOES.
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/gifs/Work07.gif
* The Energy transferred into a system by the action of a Force is the Work done on the System.


According to the work-energy theorem if an external force acts upon an object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) is given by:
http://upload.wikimedia.org/math/7/e/9/7e9050971f758b530cab178701beff0d.png

4Mesh
09-24-2008, 02:29 PM
Vibe, check your PM's for a phone number so I can type less please.

Vibe
09-24-2008, 02:36 PM
Vibe, check your PM's for a phone number so I can type less please.
I hate long distance calls. :D

Jetmugg
09-24-2008, 02:41 PM
Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.

4Mesh
09-24-2008, 03:41 PM
Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.

Steve, give me a minute and I'll put a description of why this is incorrect. I have just had a conversation with Vibe, and he is absolutely correct (at least for this discussion and we're not nit picking the details.). The error is not in the calculation of this particular part of the equation, it is in the fact that we are not taking into account ALL of the things that are happening. These things assume that the projectile is fired in a vacuum (no atmospheric losses).

Hypothetically speaking, this bullet WILL ONLY DECELERATE 32fps due to gravity in the first second. Now, at that time, the remaining energy imparted in the bullet from being fired, (ENERGY TO ACCELERATE THE BULLET TO OUR SPEED) IS STILL IN THERE.

Gravity is our only source of hypothetical loss. So most of the energy is still in the bullet getting carried to the next 3000 feet of elevation. In fact, a VERY large portion of it.

Vibes calculations show the bullet will travel a hypothetical 26+ miles straight up (ignoring atmospheric losses) and this is correct.

The time to impart the energy at our hypothetical barrel length which he approximated at 26" but is another difficult number to nail down, represents a lot of energy transfer in a very short amount of time.

Short answer without going into TONS of math is,

VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:

Sorry Vibe, I tried size=x inside the UBB tags but it won't make it bigger:D:D

Horsepower on this hypothetical cartridge is definitely somewhere north of 4000 HP!

Very good Job VIBE! And THANKS for TEACHING ME that energy does not equal work BUT BUT BUT BUT ENERGY IMPARTED DOES = WORK!

4Mesh looks at HUGE plate of raw crow in rather large pieces and prepares for long happy meal.

Steve, the only analogy I can give that will explain where the rest of the HP is, is that we didn't LIFT the bullet (as Vibe said was an incorrect term I beleive), we accelerated it. I could not understand why that was getting included as work when we only move the bullet 3000'. Moving the bullet 3000 feet up is one part of the work. Making it go 2968fps PAST that point is the rest.

4Mesh
09-24-2008, 04:31 PM
After re-reading the thread again



OK. I see what you're asking I think...Give me a bit to work it out and make it clearer...but the main issue is in the difference in the "acceleration term"


the 20ft/sec example is a bit different. IF you are saying that a 70,000 grain mass is traveling at an average of 20ft/sec and goes 10 ft...then no work was done, as the velocity did not change. IF on the other hand you are saying that it started at 0 velocity, and traveled 10 ft under constant acceleration in 0.5 sec resulting in an average speed of 20 ft/sec...then the final speed is 40 ft/sec at the end of that time period V(ave)=(V2-V1)/2 then if V1 =0, then V2 has to equal 2V(ave)

10/32 Slug * 40 * 40 ft^2/Sec^2 = 0.3125 * 1600 Slug ft/sec^2 - ft = 500 ft-lbs work

500ft-lbs work done in 0.5 sec is 500/0.5 = 1000 ft-lb/sec power.

If you are saying that this happened in a straight up direction then the previous 100 ft-lb of work due to change in potential has to be added to the 500 ft-lb kinetic energy increase due to velocity.
SO total work in this particular case would be 600ft-lbs
done in 0.5 sec would be 1200 ft-lb/sec of power.

Sorry this took so long. I had a virus crash during posting and have been diagnosing that as I posted.

And right here is where I completely missed the boat on this. This was explained to me and I blew right past it.

Rad Mrdal
09-24-2008, 04:46 PM
Whatever the exact HP calculated figure is, it will be in several thousands of HP.

Rad

4Mesh
09-24-2008, 04:46 PM
Vibe, I'm still in denial about why this is coming out to be true when some of these figures are what they are. First, the .03 pound bullet, and I guess what I consider a reasonably low velocity...

In attempting to apply some common sense to this, I thought about how much fuel is consumed on the dragster vs bullet comparisons.

In a reference way above, a dragster uses 1.5 gallons of fuel per second.

Our Win Mag uses about 75 grains of 4831 to make 3k fps. If the .0017 Sec follows, and actually there were estimates somewhere of .0012, those two values could mean we are burning powder at a rate of between 6.3 and 8.9 pounds of powder per second.

Now this kinda shows a real world example that someone not doing the math could look at and determine the winner, if you will!

Nice going again. And thanks again.

Bill Wynne
09-24-2008, 05:29 PM
210 grain * 1 pound/7000 grain * 1 slug/32 pounds = 0.0009375 Slug
(3000ft/sec)^2=9000000 ft^2/Sec^2

1/2MV^2= 1/2*0.0009375 *9000000 Slug-ft/sec^2 -ft

1/2MV^2= 4218.75 ft-lbs energy

Energy in chamber = 0
Energy at muzzle = 4218.75

(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

How much time did it take to do the work?
Again I'm going to assume a 26" barrel
Velocity at muzzle = 3000ft/sec
Velocity in chamber = 0
Assuming constant acceleration V(ave)=(Vmuzzle-Vchamber)/2=1500ft/sec
distance =Vt
so
T=d/V =(26" * 1ft/12")/(1500ft/sec)=2.166666667/1500=0.001444444 seconds

Work done/time to do the work= 4218.75/0.001444444=2920673.077 ft-lbs/sec

Work done in terms of HP
(2920673.077 ft-lbs/sec)/550ft-lbs/sec=5310.314685HP


It was a bit low. The 0.001711 sec. barrel time get's it a bit closer though, but I don't really know where that came from.

Vibe,

Using your method, I have another problem for you to consider:

A 210 grain bullet strikes an immovable block of hardened steel at 3000 feet per second and makes a cavity 1/4" deep and drops to the ground. How much horsepower is required to slow the bullet to a full stop in 1/4"?

Concho Bill

4Mesh
09-24-2008, 06:18 PM
552273HP unless I'm wrong again.

Which is entirely possible.

Actually Bill, I discussed this very thing while on the phone with Vibe today. I was concerned about my earlier supposition that the barrel length had nothing to do with the calculation. That was not a correct supposition. Barrel length means a lot and that's why he was estimating it.

Unless I've learned nothing, or somehow there's more to the story, HP would be inversely proportional to the barrel length and time used to generate the energy, or the distance and time used to stop it. Yes?

alinwa
09-24-2008, 08:48 PM
OK OK, you'se guys have surpassed me with the math........I still need a couple of clarifications here :)


So in other words if one were to somehow gear down a 300WinMag it would really lift 550lbX4000 or 2,220,000lb one foot in one second? ONE 300WinMag round?

A 300WinMag round generates enough energy to lift over 2 million pounds over the course of one second?



Put it in simple terms for me.......am I WRONG too?????


Gearing up for my dish of crow here :D:D:D


al



BTW, Lynn, do you realize that your posts aren't showing up on the board?

Bill Wynne
09-24-2008, 09:03 PM
552273HP unless I'm wrong again.

Which is entirely possible.

Actually Bill, I discussed this very thing while on the phone with Vibe today. I was concerned about my earlier supposition that the barrel length had nothing to do with the calculation. That was not a correct supposition. Barrel length means a lot and that's why he was estimating it.

Unless I've learned nothing, or somehow there's more to the story, HP would be inversely proportional to the barrel length and time used to generate the energy, or the distance and time used to stop it. Yes?

552,273 Horsepower! Have you guys finally gone over the edge or around the bend? Get a grip and hold on real tight because you are losing it. This is not launching the Space Shuttle or powering and aircraft carrier. It it is but a pebble.

The term foot pounds is just that (feet x pounds) and 550 foot pounds per second or 3300 foot pound per minute = 1 horsepower.

I just can't deal with this anymore.

Concho Bill

4Mesh
09-24-2008, 09:09 PM
OK OK, you'se guys have surpassed me with the math........I still need a couple of clarifications here :)


So in other words if one were to somehow gear down a 300WinMag it would really lift 550lbX4000 or 2,220,000lb one foot in one second? ONE 300WinMag round?

A 300WinMag round generates enough energy to lift over 2 million pounds over the course of one second?



Put it in simple terms for me.......am I WRONG too?????
Unless I'm really off base still as well Al, Yes, we are wrong.


I am absolutely open to the guys who have other opinions on this at this point and believe me, they've got me just hoping I've learned my lesson.

At this point, I'm only making new posts in hope that my answers are now correct (or in the ball park) and that if they are not, I learn what it still is I'm missing.

A good ole plain english explanation seems to carry more weight. It's interesting eh.

Again Al, that was my guess putting the numbers into Vibes example above. If Vibe were here to correct you now, I think what he would say is you are imparting enough energy in the steel plate to do that much work IF IF IF IF IF you imparted that energy over one second. IN this example, the time is getting really slight because the bullet stops in 1/4" (or 1/2 or whatever it was above). 1/4 I think.

I did the time calculation (estimate) and you could basically say that if Vibes 1.4 millisecond time was correct, and it is pretty darn close with 3Kfps and 26" bbl. Then the time you are using to take the energy back out of the bullet is 1.4 / 26" * .25" = .013 milliseconds. =.000013 seconds.

So you can see, the HP number becomes inversely proportional to the time, despite the fact that the energy we are dealing with is a "constant".

Again, someone chime in if I'm off base. Bill, I see you've posted that you seem to still prefer my first examples. I gotta tell you guys, even if I was right at this point, it matters not because I've been disuaded from my original point of view and either Vibe's really good, or he's really good. If he's really good at fooling me, then it worked because I am seeing the simplicity of his example now, even if it is wrong, it sure seems ok to me.

ROFLMAO guys.

4Mesh
09-24-2008, 09:28 PM
No Al.

If you want to say it like that (how much weight will it lift in one second) then yes, it would lift 2.2 million lbs if the bullet pushed for 1 second.

However, if it pushes for only .000013 seconds, it will lift 28.86 pounds one foot.

Note....... Al, once you multiply by seconds, you have to remove it from the result. Now it's just how much weight and how far. No time is involved, though you could calculate it if you were really a sick pup.

Hows that Bill?

alinwa
09-25-2008, 02:17 AM
No Al.

If you want to say it like that (how much weight will it lift in one second) then yes, it would lift 2.2 million lbs if the bullet pushed for 1 second.

However, if it pushes for only .000013 seconds, it will lift 28.86 pounds one foot.

Note....... Al, once you multiply by seconds, you have to remove it from the result. Now it's just how much weight and how far. No time is involved, though you could calculate it if you were really a sick pup.

Hows that Bill?


If this is true (which has been MY contention all along) then the bullet really only generates a fraction of a horsepower........ I don't NEED the math, I can do it but I'd have to drag out my books and that's 'waaayyyy too much hassle.

What you just stated in this post 4mesh is what I believe to be true.....IF IF IF the 300WinMag case would SUSTAIN it's level of out put for a full second then and only then would it produce "4000 HP".


THAT SAID........ if vibe is saying that the 300WM actually is producing the horses then it follows that if one could "capture the light'ning in a bottle" and parse it out over a seconds time . . . . . THEN (ignoring inertial, heat, hysteresis, etc losses) . . . . . . the liddle barsti'd WOULD BE CAPABLE of lifting over 2 MILLION pounds....


IF I'M WRONG and that liddle wad of nitrocellulose really does contain that much energy then I've got to re-think some of my mostest basic tenets regarding energy transfer! And that little Chinaman who sucked the ocean through a drinking straw had NUTTIN' on the venerable 300winnie!

I've found that a nice puree is the easiest way to down old Corvus Commonensus..... cold, with lots of lemon and cod liver oil.

Otherwise the feathers get all stopped up at the entrance to my larynx....


BTDT :D

hacked for days.....

al

Bill Wynne
09-25-2008, 06:19 AM
4mesh my crow eating friend,

Unless you really want to eat a crow, you don't have to. You were right in your post 12 on page one. It really is that simple.:)

Consider this: In order to keep that kind of pressure on a bullet for one second you would need a much bigger cartridge and a much longer barrel.:)

As big as it is, it is only a 300 Magnum and it will only do so much.

Concho Bill

Vibe
09-25-2008, 09:23 AM
Vibe,

Using your method, I have another problem for you to consider:

A 210 grain bullet strikes an immovable block of hardened steel at 3000 feet per second and makes a cavity 1/4" deep and drops to the ground. How much horsepower is required to slow the bullet to a full stop in 1/4"?

Concho Bill
210 grain * 1 pound/7000 grain * 1 slug/32 pounds = 0.0009375 Slug
(3000ft/sec)^2=9000000 ft^2/Sec^2

1/2MV^2= 1/2*0.0009375 *9000000 Slug-ft/sec^2 -ft

1/2MV^2= 4218.75 ft-lbs energy

Energy in chamber = 0
Energy at muzzle = 4218.75

(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

How much time did it take to do the work?

Up to this point everything is the same...just opposite in direction, since we are losing energy to the block.

Time here is figured the same.
Ave Vel during decel is 1500 ft/sec -
distace traveled = 0.020833333 ft
0.020833333/1500=1.38889E-05 sec

4218.75/1.38889E-05 = 303750000 ft-lbs/sec

303750000/550=552,272.7 HP

But this is probably a good time to mention that what we see in this example is why Lasers can cut stuff. For a given amount of energy inputed into a laser tube, the shorter the pulse, the more Power it has. You can get the same amount of light from a WalMart flashlight, but focused and in very short pulses that same amount of light can cut steel.

As far as the bullet lifting anything...Nah..Not really. In completely inelastic collisions, Energy is not conserved...but momentum is.
So
if (m) is the mass of the bullet, and (V)is the velocity
and (M) in the mass of the target (that is free to move afterwards)
then (v) will be the final velocity after the impact.
The equation is this

mV=(M+m)v

Then if you take
1/2(M+m)v^2 = (M+m)gH

H=((v^2)/2)/g
v = Sqrt(2gH)

You have the beginings for calculating the results of the balistic pendulum.

:D :D :D :D :D :D

Bill Wynne
09-25-2008, 10:09 AM
210 grain Bullet. = .03 Lbs.
Launched at 3000 FPS

Muzzle pointed vertically and plumb. Firing straight up. Firing in a vacuum.

.03x3000=90 Ft Lbs per second of momentum created.

Time already factored by the muzzle velocity. (3,000 feet per second)

90/550 = 0.1636HP

This is the right answer. Everything else fuzzy thinking.:)

Life is good when your right.

Concho Bill

Vibe
09-25-2008, 10:18 AM
This is the right answer. Everything else fuzzy thinking.:)

Life is good when your right.

Concho Bill
Momentum has the terms MV = (Slug)(ft/sec).........No pounds here

(90 Slug-ft/sec)/(550 Slug ft^2/sec^2) = 0.1636 ft/sec
Looks like some sort of Velocity term to me instead of HP.

4Mesh
09-25-2008, 10:32 AM
Bill, Al,

Please send me your addresses. I make this really good poultry seasoning and it works on the raw bird as well as cooked! :D:D You need to let me know how many crows yer having for dinner though so I can send the right amount. It takes about 200 grams of seasoning for each bird (give or take). Just tell me how many birds and I'll try to do the math on this end as another excersize! I been having rather a lot of it lately though so you may want to act quick before I run out! :D:D

It's funny thinking about it, all this pounds, slugs, feet crap and here I am a 1K shooter who can say honestly, without converting, I could not tell you how many grains of powder I shot in any gun at any time last year! I can tell ya how many grams though!

I am laugh'n myself to tears on this cause I know where yeeer com'n from.

Al, send the PM (plain english explanation one) to vibe. Have him take a look at that and see if I'm even close on my analogies. I think they are now. I think it explains where the numbers come from a good bit better than this thread. Vibe, IF it's in the ball park, please post it. I didn't because I thought this was over and didn't want You to have to waste any more time on math problems to disprove any mistakes in it, should there be any.

Vibe, 552,272.7, I say 552,273! I'll accept that! I think I'm finally get'n it! Did you think it would ever happen? :D:D

Vibe
09-25-2008, 10:35 AM
Vibe, 552,272.7, I say 552,273! I'll accept that! I think I'm finally get'n it! Did you think it would ever happen? :D:D

Never had any doubts about it...(As to when it would happen...well I was startin' to worry about that part. :D)

4Mesh
09-25-2008, 10:51 AM
I'm just gonna remove some of the confusing stuff I said at the top which was addressing the different ways Vibe was trying to separate the calculations. Reading this over again, saying those things will confuse someone as much as the calculations did me.

So this is a slightly corrected version of that attempt. I sure hope it's understandable now.

Here's the PM I sent to Al to Attempt to say in plain english why Vibe is CORRECT.


Howdy Al.

I'm assuming that at this point, you and Bill are thinking I've hit my head pretty hard. I'll be honest and say I thought through all of this that Vibe was actually trying to trick me with some hidden misnomer and that he really knew all along I was right and he was wrong. Then he just skirted my questions to make me attempt to find a new way to ask the same thing.

Well, that wasn't the case. He genuinely was giving me all the info I needed, I just wasn't catching it. Kinda like trying to explain something that's beyond the understanding of a persons math, and they can't grasp what is being said. Yes, it's all there. No I could not see it.

Here's the mistakes I made that make this difficult to understand.

First. I tried to use simple numbers like 3k fps. Then used assumptions on that which were rough estimates good enough for our purpose. Things like, MV = 3000fps, and then bullet travels 3000'. These two assumptions were contradictory when considering real world physics.

YES, we HAVE to have acceleration (accurately) and time in barrel (accurately). Without these, the vast majority of the work number is lost or at least bogus.

Think like this. We cannot propel a bullet to 3000 fps, and then have it travel ONLY 3000' in one second. It would have to hit something at the 3000' point to stop it.

When we define work, it is timeless. It is only pounds over feet. The work is against gravity. Trouble is, if you do that work in such a time that gravity can't stop the bullet at the 3000' point, then the bullet still contains energy. That energy needs accounted for. I was ignoring that energy.

Vibe did somewhere tell a rough estimate of how far the bullet would travel straight up, in a lossless atmosphere and all that. It was 26.6 miles I think. Don't quote me there but 26.x for sure. So, if we take 26.6*5280=140448 feet. Now, we have 4213 ft lbs of energy (potential, if it falls back to earth) 140448*.03=4213. THIS is the total energy if we want to "cheat" as I did and use how far it will go to calculate ft/lbs. See, we have to remove the "per sec" from the results and that can't be done while the bullet is still moving or contains energy.

Think of it like this.

Kinetic energy is inside the bullet because it's moving at some velocity and has mass. Mass x V^2 = kinetic energy.
Potential energy is how far above the ground the bullet is times how much it weighs. Weight x distance = potential energy.

As the bullet travels skyward, it looses kinetic energy, and gains a potential energy. All of that muzzle energy needs accounted for if we expect to come up with a horsepower number.

Weight includes gravity. Mass does not. That's why mass needs V squared... What confused me is that he was separating the two calculations or doing it all as MV^2 and that threw me for a loop. (using the two methods depending upon my questions)

If we expect to use feet traveled to estimate how much energy is in a bullet as I was doing, then we have to WAIT longer than the 1 second for nature (gravity) to pull all that Kinetic energy out of the bullet, and replace it with Potential energy (height above the ground). When the bullet reaches 26.6 miles, then we can say pounds times feet = foot pounds. Until it stops, we still have seconds in there!

That is where I was making my mistake. I wasn't letting the bullet come to rest. Vibe and I were not on the same wavelength with why he was so damn interested in the residual energy. It was I who was wrong on that and was blind to not see the other energy was in there. Funny part was, I can remember thinking really early on about how the physics did not work on the 3000 feet lift, and blindly just accepting that as gravity is already in there and that's it.

So, when the bullet gets launched, and we decide to stop pulling energy out at 3000 feet high, we need to also consider how much higher that bullet is gonna go.

Jetmugg
09-25-2008, 10:52 AM
Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.

My example above is just a "snapshot" of the time that the bullet is in the barrel. It has the correct terms (weight of the bullet in Newtons, distance traveled in Meters, time elapsed in seconds.) What's wrong with this analysis?

4Mesh
09-25-2008, 11:00 AM
What's wrong with this analysis?You are trying to quantify work against gravity.

But you do not allow gravity to bring the bullet to rest.

Vibe
09-25-2008, 11:11 AM
Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.
Same problem as in English units....Grains and Newtons are units of Force(against 9.8 Meters/sec^2)...not mass. Which is fine I guess if you are pushing the bullet through the barrel at 9.8 M/S^2 for 0.0017 seconds....But it won't make it that far with that force in that time.

3000 ft/s = 914.4 M/s You really should see that term in the calcs somewhere. (Preferably squared)

SBH
09-25-2008, 11:27 AM
Im glad we got that straight. :D

Jetmugg
09-25-2008, 11:34 AM
OK:
It's the excess unaccounted for energy which causes the problem with the horse, pulley, and weight example. The horse does not throw the weight up into the air with any acceleration, other than slightly overcoming gravity. I can see that now.

It still seems strange to express the power of a 300 Win Mag as 5000 HP, but only applied for 0.0017 seconds. It feels too much like a distortion of the original intent of the concept of HP. I'm not sure that James Watt would approve.

I have heard of folks attempting to build "gunpower" powered engines, but don't know of any operating engines in existence. There was an episode of Mythbusters where they tried to build such an engine. Does anyone know of such an engine?

SteveM.



.

jackie schmidt
09-25-2008, 02:09 PM
This is one of those times when I am absolutly certain that I do not have the educational background to follow all of this.

A similiar discussion comes up when the issue of Top Fuel Dragsters is discussed. Talk about big numbers being thrown around. anywhere from 6000 to 8000 HP.

The difference is, you can mathematically figure how much HP it takes to propel a 2300 pound object 1/4mile in 4.45 seconds from a standing start. You can also get a good idea from the amount of energy that is expelled as the large amount of fuel is consumed. It does take a certain amount of BTU's to make a given amount of power.

This is probably a pretty accurate figure. Then, you look at a typical Top Fuel Engine at 7000 HP as compared to a big EMD, (or similiar), Locomotive Diesel. Heck,the starter motor on the EMD is about the same size as the Top Fuel Engine. Could that Top Fuel Engine power that Locomotive with 10,000 tons of train behind.

The answer is yes. If you geared it down to transfer the torque, it would, in theory, run the train.

For about 4.45 seconds..........jackie

4Mesh
09-25-2008, 02:32 PM
More Math, feel free to correct.

As a note to those who were wondering, the time we had to wait for gravity to stop our 300 win mag 210 bullet with 3000fps V is 93.5 seconds.

Here's an example to quantify this all (and the reason I was lost)

Hopefully Steve this kinda shows a V8 300 Win Mag.

Ok, we have 8 revolving barrels, or 8 immaginary cylinders with 210 grain pistons and connecting rods!

Ok, 1 second divided by the time in the barrel. 1.4 milliseconds being used now cause it's a little closer to the 26" barrel estimate.

1/.0014=714.28 (cyl firings per second). This represents continous power output.

Divide by 8 cause we have 8 cyl (8 barrels) on our engine. = 89.28 times each barrel fires Each second. (call them revolutions) If this was a gattling gun, it would be firing 89 shots from each barrel each second, and would have 8 barrels doing that.

Times 60 = RPM's = 5357rpm's.

So, if we had 8 rotating 300 Win Mag Barrels firing in succession so that there is always a bullet being fired, the "machine" would have to run at 5357 RPM's to generate our hypothetical "Horsepower".

Don't forget, this is a great deal different than the dragster engine cause the way I figure this, the V8 300 Win Mag, is a 2 stroke at 5357rpms! It haint got no valves either!

I suppose it's safe to say I would not want to hold the trigger on that gun for one second. Least not in a sporter weight anyhow.

That's a cyclic rate of 42800 rounds per minute from a 300 Winnie. That's gonna kick.

Bill Wynne
09-25-2008, 05:48 PM
Montana Pete! This was your question. What is the answer?

Concho Bill

Vibe
09-25-2008, 11:00 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?Bear in mind that a rifle is a machine that produces useful work. Work is measured in foot/pounds. That's a simple calculation and can be found in most reloading manuals.

Now . . . horsepower is a measure of work plotted vs. time. In other words, the power of this machine that is the rifle. A machine rated at 10 horsepower can perform more useful work in a given unit of time than a machine rated at 5 horsepower.

Now, take any rifle and any load you want, and see if you can determine the horsepower produced by said rifle.

I would try this, but I'm an English major, unfortunately, and incapable of the analysis.

I've often wondered about this . . . hope someone takes me up on it.

Montana Pete
The answer to his queation is just "Yes". :D

Montana Pete
09-26-2008, 10:48 AM
Wynne, Vibe, and Others--

Yes, I realize it was my question. I wish I could say exactly who is right. Sometimes the analysis differs because not always the same thing is being measured.

A rifle begins to look a bit different when it is analyzed as a machine, versus the way Outdoor Life magazine tends to portray rifles, or the way a "western" movie portrays them.

Anyway, I am glad that some of us enjoyed this discussion and want to thank everyone who posted on the topic.

4Mesh
09-26-2008, 11:21 AM
I wish I could say exactly who is right.
VIBE WAS, AND STILL IS RIGHT!

4MESH WAS WRONG WRONG WRONG WRONG

Really. If espressed in instantaneous HP, the 4000+ number is right for our 300 Winnie. My instantaneous number of 163HP is incorrect.

I will say the discussion was enlightening, and that through this, I have learned something else that I feel will help my rifle to shoot better (if I learned what I think I learned).

Vibe, I bought a new copy of Math Cad 14 Today. Check just went out. Do you use it? Or, have you used it?

Vibe
09-26-2008, 11:39 AM
I will say the discussion was enlightening, and that through this, I have learned something else that I feel will help my rifle to shoot better (if I learned what I think I learned).
Hmmmm. Do tell.


Vibe, I bought a new copy of Math Cad 14 Today. Check just went out. Do you use it? Or, have you used it?
I used to use AutoCad quite extensively, and a few other Cad programs...but not that one.

4Mesh
09-26-2008, 12:01 PM
Hmmmm. Do tell.
I would be afraid to say publicly! :D



I used to use AutoCad quite extensively, and a few other Cad programs...but not that one.

I have a buddy who is a structural engineer who gave me a copy of it back in either the very late 80's, or very early 90's. Awesome program. Awesome. You'd love it. There is no program even close out there. Makes MatLab look like old technology.

I am betting I gave my Disks to someone and never got it back cause I looked last night and can't find it. I'm getting a new one anyhow.

In any case, it works like a spreadsheet, where every time you make a change it updates automatically. You build from top to bottom, and graphically build equations. Throughout, you can define symbols and variables. and the system re-calcs on the fly. Virtually all the examples of equations you see on the web, are screenshots of that program showing the symbols and proper representation.

I had it back in the MS-Dos days and it was very powerful then. It's a shame nobody makes a good low end copy of it. When it get's here, I'll see about putting your equations in and show an example of how it works. If you've ever used remote desktop, you could even play with it yourself if you like. (bout 1 min to set up)

When you go to work today, tell them they have to buy you a copy of that program.

Bill Wynne
09-26-2008, 05:12 PM
An old high school friend of mine that I have not seen for a very long time is coming to San Angelo for our Class of '58 reunion. Robert has spent his life as an engineer and has been very successful and he is very smart like me. He called me the other day and during the conversation this subject about horsepower of a rifle came up.

It is funny how such a bright guy who gets a engineering degree from the University of Notre Dame and go on to have a lifetime career in engineering can also be so wrong about this simple problem. Such a waste.

The .16 horse power is accurate if you consider the average horsepower for one second.

Anyway, this is the text of the E-mail that Robert sent me. He also included the horsepower of a 22 LR in the package.

Bill,

The bullet kinetic energy, E, is given by 0.5 m v^2 where v^2 means v to the power 2, m is the bullet mass and v is the muzzle exit velocity. This energy is also equal to the explosive force, F times the distance, d, which is the barrel length so that E = Fd. Thus F = 0.5 m v^2/d. Now power, P, equals the time rate of change in energy or P = time rate of change of Fd. Assuming force is constant, we have P = Fv = 0.5 m v^3/d.

I prefer to do engineering calculations in metric units and convert to English at the end. Let the gun barrel have a length of 0.75 m (~30") or d = 0.75 m. One grain equals 0.0648 grams or 0.0000648 kilograms (kg). Thus 40 and 210 grain bullets have masses of 0.00259 and 0.0136 kg respectively. Muzzle velocities of 1000'/second (sec) and 3000'/sec equate to 304.8 m/sec and 914.4 m/sec. Using the final formula for P, a 22 rifle bullet and a high power bullet have powers of 48.9 kilowatts (kW) and 6.94 megawatts (MW) respectively. Using the conversion factor of 746 watts (W) = 1 horsepower (HP), the 22 and high power bullets have horsepowers of 65.6 HP and 9298 HP respectively.

See you and Kay soon.

Robert

Vibe
09-26-2008, 05:29 PM
An old high school friend of mine that I have not seen for a very long time is coming to San Angelo for our Class of '58 reunion. Robert has spent his life as an engineer and has been very successful and he is very smart like me. He called me the other day and during the conversation this subject about horsepower of a rifle came up.

It is funny how such a bright guy who gets a engineering degree from the University of Notre Dame and go on to have a lifetime career in engineering can also be so wrong about this simple problem. Such a waste.

The .16 horse power is accurate if you consider the average horsepower for one second. for one second.this is probably true.


Anyway, this is the text of the E-mail that Robert sent me. He also included the horsepower of a 22 LR in the package.

Bill,

The bullet kinetic energy, E, is given by 0.5 m v^2 where v^2 means v to the power 2, m is the bullet mass and v is the muzzle exit velocity. This energy is also equal to the explosive force, F times the distance, d, which is the barrel length so that E = Fd. Thus F = 0.5 m v^2/d. Now power, P, equals the time rate of change in energy or P = time rate of change of Fd. Assuming force is constant, we have P = Fv = 0.5 m v^3/d.

I prefer to do engineering calculations in metric units and convert to English at the end. Let the gun barrel have a length of 0.75 m (~30") or d = 0.75 m. One grain equals 0.0648 grams or 0.0000648 kilograms (kg). Thus 40 and 210 grain bullets have masses of 0.00259 and 0.0136 kg respectively. Muzzle velocities of 1000'/second (sec) and 3000'/sec equate to 304.8 m/sec and 914.4 m/sec. Using the final formula for P, a 22 rifle bullet and a high power bullet have powers of 48.9 kilowatts (kW) and 6.94 megawatts (MW) respectively. Using the conversion factor of 746 watts (W) = 1 horsepower (HP), the 22 and high power bullets have horsepowers of 65.6 HP and 9298 HP respectively.

See you and Kay soon.

Robert
Looks like he included the 1/2 factor in his equation, but forgot to include it in the actual calculations-I got about 1/2 of that answer for the 210 grain @ 3000fps. - and the 30" barrel assumption vs mine of 26" made a bit of a difference as well.

Bill Wynne
09-26-2008, 06:05 PM
for one second.this is probably true.


Looks like he included the 1/2 factor in his equation, but forgot to include it in the actual calculations-I got about 1/2 of that answer for the 210 grain @ 3000fps. - and the 30" barrel assumption vs mine of 26" made a bit of a difference as well.

Robert has not been privy to our fun discussion of this matter, I am not sure if he even has a rifle. If he is wrong, at least he got an answer in the range of some of yours.

Concho Bill

alinwa
09-26-2008, 08:01 PM
Ok OK, You'se guys that want to figure "rate of horsepower produced during the explosion only".................... why not apply the same logic to the internal combustion engine which IS commonly rated in horsepower?


Spend all of this time to figure the "rate" of just the ignition cycle of the combustion stroke.......this will put the "horsepower" of your Pinto wagon somewhere in the neighborhood of "1500 HP" too :D:D:D:D:D


LOL


al

MistWolf
09-28-2008, 10:21 PM
I'm going to throw my hat into the ring, just for fun. To define horsepower, it's a measurement of work. Work is how far you move a weight over a distance in time. We have to know three things:

1) Bullet mass. This is the object we are interested in moving, not the weapon or the ejecta (burnt gunpowder).

2) The distance we will be moving the bullet. Just to the end of the muzzle? or to the target 100 yards away?

3) The time it will take the bullet to arrive from point A to point B. This can be called flight time

We need to know one more thing- the proper mathematical formula. A little searched turned up this: http://www.iprocessmart.com/techsmart/HP_2.gifWhere F = Force (lbs)
V = Velocity (ft/min)

I also found this: http://www.web-cars.com/images/math_img/horsepower.gif

Using this formula and using .0014 seconds of dwell time in the barrel, I find that the bullet itself has done .156 HP of work or 116 Watts. I used a 200 grain bullet, which is 1/35 of 1 pound, a velocity of 3000 fps (x 60 for velocity for one minute) and divided it by 33000. My calculator called the results as above. This is in the simplest terms and obviously after losses due to friction and heating the barrel. If this motion were sustained for one full minute you get 5143 lbs-feet compared to 33,000 lbs-feet for one horsepower.

From www.webcars.com- To help sell his steam engines, Watt needed a way of rating their capabilities. The engines were replacing horses, the usual source of industrial power of the day. The typical horse, attached to a mill that grinded corn or cut wood, walked a 24 foot diameter (about 75.4 feet circumference) circle. Watt calculated that the horse pulled with a force of 180 pounds, although how he came up with the figure is not known. Watt observed that a horse typically made 144 trips around the circle in an hour, or about 2.4 per minute. This meant that the horse traveled at a speed of 180.96 feet per minute. Watt rounded off the speed to 181 feet per minute and multiplied that by the 180 pounds of force the horse pulled (181 x 180) and came up with 32,580 ft.-lbs./minute. That was rounded off to 33,000 ft.-lbs./minute, the figure we use today.

Bill Wynne
09-29-2008, 09:35 AM
I'm going to throw my hat into the ring, just for fun. To define horsepower, it's a measurement of work. Work is how far you move a weight over a distance in time. We have to know three things:

1) Bullet mass.
2) The distance we will be moving the bullet.
3) The flight time

We need to know one more thing- the proper mathematical formula. A little searched turned up this: http://www.iprocessmart.com/techsmart/HP_2.gifWhere F = Force (lbs)
V = Velocity (ft/min)

I also found this: http://www.web-cars.com/images/math_img/horsepower.gif

Using this formula and using .0014 seconds of dwell time in the barrel, I find that the bullet itself has done .156 HP of work. I used a 200 grain bullet, which is 1/35 of 1 pound, a velocity of 3000 fps (x 60 for velocity for one minute) and divided it by 33000. My calculator called the results as above. This is in the simplest terms and obviously after losses due to friction and heating the barrel. If this motion were sustained for one full minute you get 5143 lbs-feet compared to 33,000 lbs-feet for one horsepower..[/i]

MistWolf,

I am so glad that you jumped in. Your figures agree with mine. The 33,000 foot pounds per minute=550 foot pounds per second. We were using a 210 grain bullet, so there is slight difference. You arrived at .156 HP and I arrived at .16 HP (rounded off).

If these super horsepower answers are correct than you could power an aircraft carrier by hanging a large block of steel off its' aft end and pull a dingy with a rope with a guy shooting bullets at it every so often.

I think that alinwa sums up the problem that some are having.

"Spend all of this time to figure the "rate" of just the ignition cycle of the combustion stroke.......this will put the "horsepower" of your Pinto wagon somewhere in the neighborhood of "1500 HP" too."

Concho Bill

Vibe
09-29-2008, 10:46 AM
I'm going to throw my hat into the ring, just for fun. To define horsepower, it's a measurement of work. Work is how far you move a weight over a distance in time. We have to know three things:

1) Bullet mass. This is the object we are interested in moving, not the weapon or the ejecta (burnt gunpowder).

2) The distance we will be moving the bullet. Just to the end of the muzzle? or to the target 100 yards away?

3) The time it will take the bullet to arrive from point A to point B. This can be called flight time

We need to know one more thing- the proper mathematical formula. A little searched turned up this: http://www.iprocessmart.com/techsmart/HP_2.gifWhere F = Force (lbs)
V = Velocity (ft/min)
In order to use that formula, you need to know the Force requred to GET the bullet going 3000fps in the 26" of the barrel. I really do not think that 0.03 Lbs is going to do that. Do you? Because that is the force number you are using.
And unless you've found a way to continue to apply the propellant force AFTER the bullet leaves the barrel, you are no longer "doing work to it" once the compressed gasses escape from behind the bullet. So "Flight time" is not an issue.

Bill Wynne
09-29-2008, 12:26 PM
In order to use that formula, you need to know the Force required to GET the bullet going 3000fps in the 26" of the barrel. I really do not think that 0.03 Lbs is going to do that. Do you? Because that is the force number you are using.
And unless you've found a way to continue to apply the propellant force AFTER the bullet leaves the barrel, you are no longer "doing work to it" once the compressed gasses escape from behind the bullet. So "Flight time" is not an issue.

Mr. Vibe,

I respect you and your determination so please don't take this wrong. Let me explain what MistWolf and I are doing.

The .03 pounds in this formula = 210 grains (The weight of the bullet). If you shoot straight up in the air at a muzzle velocity of 3,000 feet per second, the bullet will move 3,000 feet in the first second, less any external forces (gravity, air, birds, tree limbs, airplanes, and etc.). The rifle has applied 90 foot pounds of force for one second on the projectile. 550 foot pounds per second= one horsepower. 90/550 = .1636.

MistWolf only calculated the horsepower exerted during the time to get the bullet out of the barrel and I used the first full second and we came to the same answer because they are the same. There is no additional force from the rifle after the bullet has left the barrel. The difference is merely the time factor.

We are doing something with momentum and you are doing something with kinetic energy.

There now! I surely hope I have helped your condition.

Concho Bill

Vibe
09-29-2008, 12:50 PM
Mr. Vibe,

I respect you and your determination so please don't take this wrong. Let me explain what MistWolf and I are doing.

The .03 pounds in this formula = 210 grains (The weight of the bullet). If you shoot straight up in the air at a muzzle velocity of 3,000 feet per second, the bullet will move 3,000 feet in the first second, less any external forces (gravity, air, birds, tree limbs, airplanes, and etc.). The rifle has applied 90 foot pounds of force for one second on the projectile. 550 foot pounds per second= one horsepower. 90/550 = .1636.
90 foot-lbs of force??? When did FORCE get to where it could be measured in ft-lbs?
Force = MA= Mass*Acceleration
So HP=MAV/33000
Force is measured in pounds - specifically Slug-ft/sec^2


MistWolf only calculated the horsepower exerted during the time to get the bullet out of the barrel
At least he constrained himself to the time period when force was actually BEING applied by the rifle.



and I used the first full second and we came to the same answer because they are the same. There is no additional force from the rifle after the bullet has left the barrel. The difference is merely the time factor.

We are doing something with momentum and you are doing something with kinetic energy.

There now! I surely hope I have helped your condition.

Concho Bill
Yep. Momentum has not a thing to do with work....nor Horsepower. HP is defined as the change in ENERGY with respect to time.

http://www.benchrest.com/forums/showpost.php?p=449345&postcount=29

Bill Wynne
09-29-2008, 06:02 PM
Yep. Momentum has not a thing to do with work....nor Horsepower. HP is defined as the change in ENERGY with respect to time.


Back to basics

Foot-pound A unit of energy, the amount required to raise one pound a distant of one foot.

Horse power A unit for measuring power, equal to 33,000 foot pounds per minute, or 550 foot pounds per second.

Are we in agreement?

Concho Bill

alinwa
09-29-2008, 09:43 PM
I think that what vibe is pointing out is that the HP is only being PRODUCED (change in energy) during the "power stroke" from ignition to muzzle, that time during which there's a change in state......... after that there's no more horsepower being produced as the bullet's just coasting to entropy, it's bleeding energy but that's it.


al

Vibe
09-29-2008, 10:31 PM
Back to basics

Foot-pound A unit of energy, the amount required to raise (against the acceleration of gravity) one pound a distant of one foot.

Horse power A unit for measuring power, equal to 33,000 foot pounds per minute, or 550 foot pounds per second.

Are we in agreement?

Concho Bill
absolutely! :D
E=MgH=1/2MV^2

Edited to ad the 1/2 factor...someone should have corrected me on it before now.:D

MistWolf
09-30-2008, 03:32 AM
A few quick clarifications-

I only calculated the horsepower generated by the motion of the bullet, which is the net. I did not calculate how much horsepower was lost due to friction or heating the material of the barrel.

After the bullet leaves the muzzle, no horsepower is being applied to the bullet to increase or mantain velocity. However, the bullet is still producing horsepower albeit at a diminishing rate. Horsepower is a measurement of work being done. The definition of work is a weight moving over distance during time. The bullet after it exits the barrel is a weight moving over distance during time. Work is being done and it can be measured in horsepower.

Horsepower is measured in pounds-feet not foot-pounds. I do not understand the difference enough to explain it, but the website I got my information from defines 1 horsepower as 33,000 pounds-feet per minute. I may also be mistaken that there is a difference

Vibe
09-30-2008, 09:13 AM
A few quick clarifications-

I only calculated the horsepower generated by the motion of the bullet, which is the net. I did not calculate how much horsepower was lost due to friction or heating the material of the barrel.
My calculations also did not include the frictional losses.


After the bullet leaves the muzzle, no horsepower is being applied to the bullet to increase or mantain velocity. At this point the problem ends - all the HP produced by the rifle has been produced and it is doing no more work and the elapsed time has stopped.


However, the bullet is still producing horsepower albeit at a diminishing rate. No. It's not. The atmosphere and gravity are doing work against the bullet to dissipate the energy imparted to it by the HP produced by the rifle.


Horsepower is a measurement of work being done.With respect to time.


The definition of work is a weight moving over distance during time. The bullet after it exits the barrel is a weight moving over distance during time. Work is being done and it can be measured in horsepower.These statements are not true with respect to the original problem of how much HP is produced by the rifle. Work would be a weight (mass) BEING moved over a distance during time...The act of just moving is not enough, it's energy must be changing and Power is the rate of that CHANGE.


Horsepower is measured in pounds-feet not foot-pounds. I do not understand the difference enough to explain it, but the website I got my information from defines 1 horsepower as 33,000 pounds-feet per minute. I may also be mistaken that there is a difference
It's a convention only, in an attempt to not confuse it with torque. Both units involve the multiplication of Ft units and Lb units but describe much different conditions.

Bill Wynne
09-30-2008, 12:19 PM
absolutely! :D
E=MgH=1/2MV^2



Vibe, That the formula for Kinetic energy.

A foot-pound = the weight in pounds x the distance in feet it is raised (on earth). You know, foot-pound. Foot-pound = feet x pounds. Webster's New World Dictionary.

Current definitions (Wikipedia Encyclopedia)

The following definitions have been widely used:
Mechanical horsepower ≡ 33,000 ft·lbf/min
= 550 ft·lbf/s

I just don't see any thing about Kinetic energy is not used in the equation.

I know this may seem silly but I see this problem as quite simple.

Concho Bill

Vibe
09-30-2008, 12:41 PM
Vibe, That the formula for Kinetic energy.
Yes it is..Expressed in units of Ft-Lbs.


A foot-pound = the weight in pounds x the distance in feet it is raised (on earth). You know, foot-pound. Foot-pound = feet x pounds. Webster's New World Dictionary.
Which describes Potential energy


Current definitions (Wikipedia Encyclopedia)

The following definitions have been widely used:
Mechanical horsepower ≡ 33,000 ft·lbf/min
= 550 ft·lbf/s

Work is change in ENERGY/time...any kind of energy, not limited to Potential


I just don't see any thing about Kinetic energy is not used in the equation.
Because you have limited yourself to looking at the change in potential energy...over a much longer time than is applicable.


, I know this may seem silly but I see this problem as quite simple.

Concho Bill
True it really isn't that complicated. However this problem of finding the HP of a rifle cannot be solved by finding the change in potential energy as you keep trying to quote, it must be solved by finding the change in kinetic energy. Work done is the change in total energy of a system on an item, that includes potential, kinetic, rotational, even thermal..et cetra. But we have limited ourselves to finding the HP of a rifle in the terms of the most meaningful values. Even shooting straight up, the change in potential energy from chamber to muzzle (MgH where Mg=0.03 lbs and H=26"=2.17ft) is minuscule(0.065 ft-lb really is insignificant here), the change in rotational energy is larger (and a bit harder to calculate, but should be included) however the change in velocity represents the largest change in energy of a fired rifle. So that is the HP that is most meaningful to solve for. And since the HP is the CHANGE in energy per unit time, we look at the energy at the muzzle vs the energy at the chamber. in the chamber 1/2MV^2 when V=0 is just zero, so the CHANGE is equal to the muzzle energy of 1/2MV^2 (where V=3000fps in this case).

But we have to divide the muzzle energy by the time it took to get from chamber to muzzle, and in this case that is between 0.0014 and 0.0017 (depending upon the estimate you use).

THEN, and only then, we can divide by the HP conversion of 1HP=550 ft-lb/sec,

MistWolf
09-30-2008, 01:26 PM
If something is moving, work is being done and it's generating horsepower. If you simply drop an object, as it's falling it's doing work. So yes, there is horsepower being generated by the bullet as it travels, although it is diminishing. Just like winding up a flywheel. You wind it up, let it spin and can use the motion to power things although no more horsepower is applied to the flywheel.

If foot/pound is potential energy of an object (which does NOT require motion or work) is pounds/feet what we get when it's moving?

Vibe
09-30-2008, 01:53 PM
If something is moving, work is being done and it's generating horsepower.
As a statement, by itself, this is not true. If it is moving at a constant velocity, under no acceleration, it's energy is not changing, thus no work is being done.


If you simply drop an object, as it's falling it's doing work. So yes, there is horsepower being generated by the bullet as it travels, although it is diminishing. Just like winding up a flywheel. You wind it up, let it spin and can use the motion to power things although no more horsepower is applied to the flywheel.
Stored energy is not power. nor horsepower. In the case of the dropped bullet, you would be very hard pressed to measure the work done BY the bullet. It would be easier to measure the work done TO the bullet by the atmosphere by it's performance in air vs it's calculated performance in vacuum. However that is not the problem at hand. The problem under discussion is the work done BY the rifle. That work ends at the muzzle.

If foot/pound is potential energy of an object (which does NOT require motion or work) is pounds/feet what (do) we get when it's moving?
Pounds-feet of kinetic energy. Energy is measured in the same terms (or can be) no matter what the type. Likewise the CHANGE in energy is also measured in the same units.

In fact - in the case of a frictionless situation (M)ass*(g)ravity*(H)eight=(1/2)(M)ass*(V)elocity*(V)elocity
Both sides will be equal AND measured in the same units. A Mass (M) dropped from Height (H) will achieve Velocity (V) when dropped in a vacuum under the influence of gravity (g). But this only hold true under truly frictionless conditions, or when frictional losses can be assumed to be very small. A rapidly traveling bullet in normal atmosphere will drop from 3000fps to somewhere just above 1000fps in the first 1000 yards...and that's not even pointed up. There are just too many losses to consider the use of that as even a secondary gauge of the actual HP results or numbers.

MistWolf
09-30-2008, 03:17 PM
You may want to take a look at this page
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/DefinitionWork.html

After a bit of research, I must say I have a clearer understanding of what Vibe is saying. As long as the bullet in flight or the spinning flywheel are not transferring their energy to anything, work is not being done. However, both have energy that can be transferred, thus measured as horsepower when energy is transferred.
http://id.mind.net/~zona/mstm/physics/mechanics/energy/work/work.html

Vibe
09-30-2008, 03:30 PM
You may want to take a look at this page
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/DefinitionWork.html

It does talk about gravity doing work
I've seen it. In fact I've posted a link to it in this thread. :D


I still think a body in motion, such as a bullet in flight is a force, though it be small.
Not until you try to stop it. :D


Same for a flywheel. They are both a force and can cause a change if/when harnessed. Matter of fact, you can hook up a spinning flywheel to a dyno, measure the torque and calculate the horsepower. Well I suppose you can measure the HP required to stop it in a certain time period, and derive the amount of stored energy based upon the HP required. But it only slightly has any bearing upon our rifle problem.



The power of a falling object can be harnessed to power something. A good example of this is the trebuchet
Which almost directly converts MgH into 1/2 MV^2


PS- I don't understand all the squiggly math symbols they're using on that page,
http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/DefinitionWork.html
Indicates the integration of the Force Vector (F=Ma) with respect to the change in position (Differential Displacement Vector).


but I betcha someone could use them to give us a clearer picture on how much HP our 300 Mag generates

I wouldn't bet much on it...I've used it here and there are those who still argue about it. The 210grain slug, in the barrel for 0.00144 seconds, exiting at 3000 fps produces just over 5000 HP using that math. I've shown all of the math here, and explained the logic behind why it has to be that way...yet here I am 11 days and 11 or 12 pages later, still trying to convince those who will not believe the physics of it.

MistWolf
09-30-2008, 03:34 PM
:) I was typing, posting and editing my last reply as I was thinking it out and Vibe quoted me while it was still formulating. My final edited post is as above. As you can see, I came to the conclusion Vibe was leading me to. If this causes any confusion to any following this thread, I apologize

Vibe
09-30-2008, 03:43 PM
LOL. No worries MistWolf. I'm of the opinion that if that confuses anyone that has read the rest of the thread...well they were bound and determined to be confused anyway. :D

MistWolf
09-30-2008, 03:44 PM
Vibe, it may not be so bad as you think. The 5000 HP may be the GROSS HP. The .156 or whatever the figure was may what's needed after overcoming all other HP eating forces, such as barrel friction

Vibe
09-30-2008, 03:48 PM
Vibe, it may not be so bad as you think. The 5000 HP may be the GROSS HP. The .156 or whatever the figure was may what's needed after overcoming all other HP eating forces, such as barrel friction



How was it Winnie the Pooh put it?????


Oh yeah.




Oh Bother

I give up. You may continue your blissful life now. :D

MistWolf
09-30-2008, 11:36 PM
*Shrugs* I just don't know how much HP is lost to friction in this case. Just something to think about, not to prove a point.

...and yes, I'm having a blissful life. My sons love me and are doing well and my new wife loves me too. (I think she may even like me!:eek:)

Bill Wynne
10-01-2008, 05:59 AM
*Shrugs* I just don't know how much HP is lost to friction in this case. Just something to think about, not to prove a point.

...and yes, I'm having a blissful life. My sons love me and are doing well and my new wife loves me too. (I think she may even like me!:eek:)

Mr. MistWolf,

You also have me to agree with your .156 hp answer. With all you have going for you in your life with your family and now me, you would be greedy to ask for more.

As for me, I don't really care if the facts run all over the place, I have my own opinion and it is not changing. The force of 5,000 hp would make much more noise and kick a lot more but I won't tell Vibe or 4mesh because they are happy thinking they are right too.

Concho Bill

speedpro
10-01-2008, 06:32 AM
I remember when simplicity and happiness was a "cork gun"... :)

Jetmugg
10-01-2008, 07:35 AM
I haven't tried myself - but how's this for an exercise.

Calculate the HP of a single cylinder 4-stroke engine (i.e. Briggs and Stratton) during the "power" stroke only. Compare this value with the stated HP (from the manufacturer) of the same engine.

SteveM.

longshooter
10-01-2008, 10:03 AM
If the 300 Mag. were fired, such that the bullet
was exiting the cartridge case and entering the barrel bore as the
fuel dragster was about 80 yards from the finish line, the bullet would reach
the finish line first.

This thing keeps running thru my mind. If the hp of the bullet were not substantial, how would it be able to close the distance on the dragster? I keep thinking hp to weight ratio. I can't shake this thought.

Bill Wynne
10-01-2008, 10:51 AM
This thing keeps running thru my mind. If the hp of the bullet were not substantial, how would it be able to close the distance on the dragster? I keep thinking hp to weight ratio. I can't shake this thought.

It might make you feel better if it was just a .222 Remington with a 50 grain bullet.

Look at it this way, bullets are faster than dragsters. I find it amazing that the dragster does so well in the contest.

Concho Bill

MistWolf
10-01-2008, 01:02 PM
Mr. MistWolf,
You also have me to agree with your .156 hp answer. With all you have going for you in your life with your family and now me, you would be greedy to ask for more.

Concho BillThank you Mr. Bill, for helping me see things in their proper perspective. You are right, to ask for more would constitute greed and ingratitude on my part. I shall take my riches and my Briggs & Stratton rifle, and retire to a quiet life by the sea

Rad Mrdal
10-01-2008, 04:05 PM
Mr. MistWolf,

You also have me to agree with your .156 hp answer. With all you have going for you in your life with your family and now me, you would be greedy to ask for more.

As for me, I don't really care if the facts run all over the place, I have my own opinion and it is not changing. The force of 5,000 hp would make much more noise and kick a lot more but I won't tell Vibe or 4mesh because they are happy thinking they are right too.

Concho Bill

Pal whatever the correct figure is, it's in the thousands of HP. You will do much better job if you try to limit yourself to the bullet barrel BR issues.Rad

Rad Mrdal
10-01-2008, 04:40 PM
This is a math question with only 1 right answer.The correct answer was given along with the formula and was well explained.Anything else is a waste of time and more importantly it is wrong.
To put this into perspective run your dragsters engine for 0.001711 seconds and let us know how loud and powerful it sounded and how far the car went down the track.
Lynn

Champs, lets do it in simple figures. 300wm, 84gr Retumbo, 0.0017 sec.

1sec : 0.0017 = 588 firings per second

588 x 84gr of Retumbo = 49 392 gr

49 392 : 7000 gr = 7.056 pounds of Retumbo burned within one second.

Pal do you recon that would help to move your gragster? Rad

Bill Wynne
10-01-2008, 06:11 PM
Champs, lets do it in simple figures. 300wm, 84gr Retumbo, 0.0017 sec.

1sec : 0.0017 = 588 firings per second

588 x 84gr of Retumbo = 49 392 gr

49 392 : 7000 gr = 7.056 pounds of Retumbo burned within one second.

Pal do you recon that would help to move your gragster? Rad

Well Pal, I don't know if that would move a dragster but I will bet you could stop one with just some of that firepower if you wanted to.

Concho Bill

Rad Mrdal
10-21-2008, 09:42 PM
Pal, you need to understand what HP actually means. An engine will produce maximum HP at certain RPM. Regardless will it be running at these RPM for for 1 sec or 10hours. Theoretically if something like the .300wm is able to burn 3.5 pounds of fuel if it would run for a full 1 second it is obvious that the HP output will be in thousands of HP. Hypothetically in the theory if the dragster engine would used the gun powder as a fuel it would burn 4.33 sec x 3.5 pounds = 15.155 pounds of fuel for its 4.33 sec. record run. Rad

Al Nyhus
10-22-2008, 06:30 AM
I'll say this..it's been fun reading this thread. :)

Now, as to this quote:


That HP rating is at one point in the engines operating range and is extremely brief.The car during its burnout and getting to the lights is not producing 5,000 horsepower.After 440 yards it is not producing very much power either even though it is still moving forward.It is only producing peak power for a few brief moments in its run right before it shifts and near the end of the run.It is still rated at 5,000 horsepower even if it runs at 5,000 hp for only a brief part of a second during its 4.33 second run.

Current nitromethane Top Fuel and Funny Car rigs make well in excess of 7,000 horsepower..closer to 8,000. They don't "shift" in the conventional sense as they don't have a transmission. They use a multi-stage clutch assembly (5-8 stages) that is electronically timed and aplies the clutch in stages as the car goes down the track. Because of this, the engines actually run at or near peak r.p.m. for most of the run...not just "for a few brief moments". NHRA requires these cars to be equipped with a rev limiter (made by MSD) so the crew chiefs adjust clutch engagement rates, fuel flow and dozens of other things to keep the engines operating just below where they will hit the rev limiter.

And since Englishtown of this season, NHRA has shortened the track length to 1000 ft. for the TF and FC classes. They don't race to 1320 ft (440 yds) at this time.

In the interest of not having the answer dummied down. ;)

Rad Mrdal
10-22-2008, 04:13 PM
Hey Champ, you've missed the crucial words in the post such as ( hypothetically in the theory). Rad

Al Nyhus
10-23-2008, 07:30 AM
Lynn: I knew the info was dated....just didn't want anyone reading this thread to think that professional Drag Racing was still stuck with '80s technology. ;) When people realize that a nitromethane Top Fuel/Funny Car powerplant makes as much horsepower in one cylinder as an entire Formula One engine....it brings it into perspective.

And NASCAR rigs are positively anemic by comparison. :D 'Course...NASCAR as a whole sucks huge anyway. Might be a pun there. ;)

Now back to our regularly scheduled programming.......

Dew
11-01-2008, 11:13 AM
This has been a most interesting thread... All I want to know is if I get in stuck in 4 wheel drive, how many shots with my rifle do I have to take to get me out of the mud? I guess I would have to have a buddy sit in the bed of the truck and shoot to the rear so I could use the extra horsepower to push me out onto dry dirt.:D

Bill Wynne
11-01-2008, 08:22 PM
No matter how you cut it, the definition of a horsepower is the amount of work required to lift 550 pounds one foot in one second (550 foot pounds per second). To move your tiny bullet 3000 feet per second (that is 3000 feet in a second), you need about 90 foot pounds produced in a in a second. There you have it, a known weight, a known distance, and a known time.

Just concentrate on those numbers. That is all there is to it. Way less than one horsepower.

Don't worry about the dragster in your driveway or the engine in an aircraft carrier. They have nothing to do with the problem.

There now, now that I have explained it again, don't you feel better.:)

Concho Bill

Vibe
11-01-2008, 11:17 PM
No matter how you cut it, the definition of a horsepower is the amount of work required to lift 550 pounds one foot in one second (550 foot pounds per second). To move your tiny bullet 3000 feet per second (that is 3000 feet in a second), you need about 90 foot pounds produced in a in a second. There you have it, a known weight, a known distance, and a known time.

Just concentrate on those numbers. That is all there is to it. Way less than one horsepower.

There now, now that I have explained it (incorrectly) again, don't you feel better.:)

Concho Bill

They are not the ones that missed it Bill.

Bill Wynne
11-02-2008, 06:10 AM
They are not the ones that missed it Bill.

Now Vibe,:)

If you won't mention that kinetic energy thing, I won't mention anything about Arkansas physics. (I know that is not fair. I am just joking)

Concho Bill

Vibe
11-02-2008, 09:18 AM
Now Vibe,:)

If you won't mention that kinetic energy thing, I won't mention anything about Arkansas physics. (I know that is not fair. I am just joking)

Concho Bill
LOL. That "Kinetic energy thing" as you so eloquently phrased it - it a major part of the definition of Power, and thus horsePower. I'm sorry to inform you that you simply cannot have the one without the other.

Bill Wynne
11-02-2008, 11:37 AM
LOL. That "Kinetic energy thing" as you so eloquently phrased it - it a major part of the definition of Power, and thus horsePower. I'm sorry to inform you that you simply cannot have the one without the other.

I think you are wrong, my friend.:)

Just for grins, using your formulas consider your bullet to be twice it's weight and the velocity being 1/2 of what is was. You will have the same number of foot pounds per second, therefore the same horsepower by definition. However, you will have a different amount of kinetic energy.

Trust me, just try it. If you can prove to me that I am wrong, I will say so.

Concho Bill

longshooter
11-02-2008, 12:04 PM
This has been a most interesting thread... All I want to know is if I get in stuck in 4 wheel drive, how many shots with my rifle do I have to take to get me out of the mud? I guess I would have to have a buddy sit in the bed of the truck and shoot to the rear so I could use the extra horsepower to push me out onto dry dirt.:D



Champs, lets do it in simple figures. 300wm, 84gr Retumbo, 0.0017 sec.

1sec : 0.0017 = 588 firings per second

588 x 84gr of Retumbo = 49 392 gr

49 392 : 7000 gr = 7.056 pounds of Retumbo burned within one second.

Pal do you recon that would help to move your gragster? Rad


Why don't you have him fire 588 times, in the space of one second, and we'll see if this gets you out. Maybe he should wear an extra shirt.

Vibe
11-02-2008, 12:46 PM
I think you are wrong, my friend.:)
The opinion of 10,000 men is of no value if none of them know anything about the subject.
- Marcus Aurelius


Just for grins, using your formulas consider your bullet to be twice it's weight and the velocity being 1/2 of what is was. You will have the same number of foot pounds per second, therefore the same horsepower by definition. However, you will have a different amount of kinetic energy. Actually you will have 1/2 the Ft-lbs of energy in that scenario. And power is defined as the change in energy (Energy, not momentum) divided by the time it took to enact the change.


Trust me, just try it. If you can prove to me that I am wrong, I will say so.

Concho BillI have. Several times. And you still haven't.

Bill Wynne
11-02-2008, 01:02 PM
Actually you will have 1/2 the Ft-lbs of energy in that scenario. And power is defined as the change in energy (Energy, not momentum) divided by the time it took to enact the change.

I have. Several times. And you still haven't.

Vibe, You have been wrong several times, but I have not given up on you.:)

This is the definition of a horse power.

1 hp ≡ 33,000 ft·lbf/min by definition
= 550 ft·lbf/s since 1 min = 60 s

Do you see anything about kinetic energy in that definition.

Concho Bill

Bill Wynne
11-02-2008, 06:03 PM
Guys,

The definition of what a horsepower is was based on how much work a mill horse could do. I don't think it is a very good way to state the power of a rifle. It can be, however, a good comparison measure of an engine.

I will say that Montana Pete offered a good mind twisting question. My 50 year old college physics book (which I saved) offers less than a paragraph and a drawing of a horse to the subject of horsepower lifting a 550 pound weight to the subject. The author must have thought that covered it.:)

Concho Bill

Vibe
11-03-2008, 10:08 AM
Guys,

The definition of what a horsepower is was based on how much work a mill horse could do. I don't think it is a very good way to state the power of a rifle. It can be, however, a good comparison measure of an engine.

I will say that Montana Pete offered a good mind twisting question. My 50 year old college physics book (which I saved) offers less than a paragraph and a drawing of a horse to the subject of horsepower lifting a 550 pound weight to the subject. The author must have thought that covered it.:)

Concho Bill

Please refer back to post 107 of this thread
http://www.benchrest.com/forums/showpost.php?p=450092&postcount=107

Work equals a change in ENERGY - Potential, Kinetic, Thermal....doesn't matter, could be all of one or the other or a mixture of any or all (which is most often the case). If the ENERGY was changed, work was done. Either TO an object if the energy increased or BY the object if it decreased.

Power is determined by how much time it took to do said amount of Work. Horsepower is simply a unit used to measure Power.

By your calculations how much higher off the ground should a 3000hp dragster be after a 4 second run? And yet it remains on the ground (if the driver is lucky and all goes as planned) and only gains Velocity really rapidly.

I would wager that even your 50 year old Physics book has a good chapter on Power.

moemag
06-06-2009, 01:33 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?

Bear in mind that a rifle is a machine that produces useful work. Work is measured in foot/pounds. That's a simple calculation and can be found in most reloading manuals.

Now . . . horsepower is a measure of work plotted vs. time. In other words, the power of this machine that is the rifle. A machine rated at 10 horsepower can perform more useful work in a given unit of time than a machine rated at 5 horsepower.

Now, take any rifle and any load you want, and see if you can determine the horsepower produced by said rifle.

I would try this, but I'm an English major, unfortunately, and incapable of the analysis.

I've often wondered about this . . . hope someone takes me up on it.

Montana Pete

Holy Thread Resurrection!

It just so happens I am an Aerospace engineer who happens to work at a certain company that makes very impressive electric Gatling guns, who happened to come across this thread doing some research. So just in case your still interested…

Yes, you can take any load you want with any rifle you want and figure out how much horsepower the RIFLE makes.

The rifles FREE (felt) RECOIL is measured in foot-pounds. I don’t know why you all are so concerned with the external ballistics of the round LOL! It looks like one or two may have figured this out… but I don’t think it caught on. I honestly didn’t read all the posts. If you want to find out how much horsepower the round creates... thats another question.

Once you figure out what that is, which is where your F=1/2MV^2 comes in

Look at this page to figure that out...http://en.wikipedia.org/wiki/Free_recoil

But being that a single gun shot is so short... we kind of assume it’s instantaneous and it’s just a force applied and there is no horsepower. But if you really want to divide the (recoil force)/ (time it takes for the round to leave the barrel) then you could find some* amount of horsepower.

Where it comes in handy for me is when my guns are shooting at over 40 rounds per second. in which case the rate of fire will allow for a useful reason to know horsepower.

In which case the M134d makes about 16.4 (and upwards to 22) horsepower when shooting.

Enjoy.


-
*Edit, on second look, it may not be very smal

Bill Wynne
06-06-2009, 01:57 PM
Holy Thread Resurrection!

It just so happens I am an Aerospace engineer who happens to work at a certain company that makes very impressive electric Gatling guns, who happened to come across this thread doing some research. So just in case your still interested…But being that a single gun shot is so short... we kind of assume it’s instantaneous and it’s just a force applied and there is no horsepower. But if you really want to divide the (recoil force)/ (time it takes for the round to leave the barrel) then you could find a very small amount of horsepower.

Where it comes in handy for me is when my guns are shooting at over 40 rounds per second. in which case the rate of fire will allow for a useful reason to know horsepower.

In which case the M134d makes about 16.4 horsepower when shooting.

Enjoy.

There now, see I was right!

And I don't even know Dr. Moemag. Thanks anyway.

Concho Bill

Vibe
06-06-2009, 03:33 PM
There now, see I was right!

And I don't even know Dr. Moemag. Thanks anyway.

Concho Bill
Not even close. First off his 16.4 HP is the net result of 40 very short high HP impulses (and 40 relitively long "rest" periods) during each second. This was not the question. The other thing is that he is working off of the "free felt recoil" of the rifle, and while the HP is "felt" on that end as well, the velocity measurement is a best guess based upon calculations from the measured velocity of the bullet. The question was

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?

That would have to be per single shot. The fact that he can scavange 16.4 HP from the gun to do other work is actually immaterial to the problem.

alinwa
06-06-2009, 03:46 PM
I think that the confusion here lies in the terms "generated," "rated" and "produces."

The challenge question uses "generated" which to me means actual work done.

........"to me means"...... herein lies the problem, we're each interpreting the (poorly structured?) original question differently.

To illustrate, I agree totally with vibe's last post :) yes "The fact that he can scavange 16.4 HP from the gun to do other work is actually immaterial to the problem. "

al

moemag
06-06-2009, 04:19 PM
Calculating the horsepower of my minigun is no different form the way a car sums up a whole bunch of relatively short cylinder bursts that occur in a given time frame to give horsepower of an engine. Yeah its not perfect. but thats what we do.

I offered a method. With an example.

Horsepower is force/time.
Horsepower of a rifle would be (force of the rifle)/ (time that the force is acting upon the rifle.)

Force of rifle = recoil however you define it.

I used free recoil because it is easy enough to determine. Sure if you want to take into account muzzle breaks, soft shoulders and limp wrists… it’s not going to be perfect, but it will be close.

Time that the force is acting upon the rifle= just what it is… but there are a lot of nasty little calculus things that come into play to figure that out.

Basically… finding the horsepower created of a single shot rifle is pointless. You have a foot-pound force, that’s sufficient enough is it not?

Here ya go. If you want to know how much horsepower a rifle can generate...

Next time your at the range, shoot from a shopping cart on a ramp of a given angle, weigh the cart with you and your rifle in it, and see how far it rolls up the ramp, all the while timing from fire till you stop rolling. If ya really want to get complex, throw in a wheel to ramp friction coefficient.

Vibe
06-06-2009, 04:31 PM
Calculating the horsepower of my minigun is no different form the way a car sums up a whole bunch of relatively short cylinder bursts that occur in a given time frame to give horsepower of an engine. Yeah its not perfect. but thats what we do.

I offered a method. With an example.

Horsepower is force/time.
Horsepower of a rifle would be (force of the rifle)/ (time that the force is acting upon the rifle.).

I expected better.
Horsepower is NOT force/time. Horse power is ENERGY/time



I used free recoil because it is easy enough to determine. Sure if you want to take into account muzzle breaks, soft shoulders and limp wrists… it’s not going to be perfect, but it will be close.

Time that the force is acting upon the rifle= just what it is… but there are a lot of nasty little calculus things that come into play to figure that out.
Lets just use published muzzle energy divided by time in barrel.
Barrel length/((Muzzle vel-initial vel)/2)
Barrel length divided by average velocity will be "close enough".



Basically… finding the horsepower of a single shot rifle is pointless. You have a foot-pound force, that’s sufficient enough is it not?
Nope. Force is in pounds - torque and energy are in Foot-pounds and pound- feet


Here ya go. Next time your at the range, shoot from a shopping cart on a ramp of a given angle, weigh the cart with you and your rifle in it, and see how far it rolls up the ramp, all the while timing from fire till you stop rolling. If ya really want to get complex, throw in a wheel to ramp friction coefficient.
Balistic pendulum = method of measuring momentum...Again..Not energy.
Horsepower is the CHANGE in ENERGY over a period of TIME.

moemag
06-06-2009, 04:35 PM
I expected better.
Horsepower is NOT force/time. Horse power is ENERGY/time


okay my bad. ;)

I will leave you benchrest boys alone.

Vibe
06-06-2009, 04:36 PM
okay my bad. ;)

I will leave you benchrest boys alone.
Whatever. Energy and power caculations are basic physics, not limited to (or even normally included in) benchrest.

moemag
06-06-2009, 05:08 PM
Wait.

Okay I did mess up part of it... I will get back to it tho. because I do need to figure this out. Like I said I am doing research to figure this out for something else.

but...

Power=((Mass of rifle+shopping cart+ shooter)(force of gravity)(vertical displacement traveled on ramp))/(time it takes)

How does that not equal horsepower?

Vibe
06-06-2009, 06:36 PM
Wait.

Okay I did mess up part of it... I will get back to it tho. because I do need to figure this out. Like I said I am doing research to figure this out for something else.

The powder charge accelerates the bullet from zero to muzzle velocity in the time it takes to clear the barrel. All actions afterward are the results of the momentum imparted from the dissipation of that energy. Momentum is conserved, energy is not. The Horsepower generated by the cartridge/bullet/rifle combination is only generated during the time the bullet is in barrel. Lots of things can be calculated from the rifle/shopping cart/ ramp...but horsepower is not one of them. The energy generated might possibly be approximated, but not the time it took.
If you generate enough energy in one second to generate one HP, generating the same energy in 0.1 seconds takes 10 HP, doing in in 0.01 seconds takes 100 HP. But all three are just as likely to roll your cart the same distance, since the same amount of energy is involved.

Your test approximates what you can learn from a ballistic pendulum, which measures the momentum of the bullet and target and then uses that resultant velocity to plug into 1/2(M+m)V^2=MgH. From this you can backtrack to find the original velocity of the bullet required to impart that momentum, and using that, the known mass of the bullet, and the approximated barrel time calculate the HP of the system during the shot...But it just adds a lot of unrequired sources of error and loss.



Power=((Mass of rifle+shopping cart+ shooter)(force of gravity)(vertical displacement traveled on ramp))/(time it takes)

How does that not equal horsepower?
MgH/t would equal a Power equation....Provided you used the correct Time value. The only "Time" involved in doing actual work is while the bullet is traveling from one end of the barrel to the other (with some leeway given to the powder charge gasses leaving). Everything after that is coasting and is doing work on the environment as opposed to having work done to it. Work energy is only being generated/done during the gas expansion.

moemag
06-06-2009, 06:56 PM
Okay, so its been a while since I have sat in physics class and I let some dumb stuff get by. Not no more! I have been working on this most of the afternoon...

Okay here we go.

Horsepower = (work)/(time)
http://upload.wikimedia.org/math/0/8/0/080770c45f43bee417ca10cf589328e9.png

Work=
http://upload.wikimedia.org/math/c/b/4/cb4e314c528c7d5c8dfff1781d5576a5.png

(((1/2)(Mass gun lbs.)(Final velocity of the gun f/s)^2)-((1/2)(Mass of gun lbs.)(Initial velocity of the gun ‘0 f/s’)^2))/Time S.

Time= (length of barrel in feet)/((1/2)(Final velocity) <-as we agreed upon.

Thus…
((Mass of gun)(final velocity of the gun)^3)/(4 • length of the barrel)

Then…
final velocity of the gun = {(mass projectile in grains • velocity projectile in feet/sec) + ( mass of powder charge in grains• velocity of powder charge in feet/sec )} / (mass gun lbs. • 7000)

So...


The Single shot statioinary rifle horsepower equation is…

(3·((Mc·Vc)+(Mp·Vp))^3)/((1.8865·10^14) ·L·Mg^2)

Where:
-Mg=Mass gun in lbs.

-Mp=mass of projectile in grains

-Vp=muzzle velocity in ft/s

-Mc=Mass of powder charge in grains

-Vc=Vp=Final Velocity of powder charge in ft/s. This is a fudged number and only hole in the whole thing. I have to assume that it is equal to final velocity of the projectile, I could have made it the velocity of the exiting ejecta as one variable but I left it incase you can figure it out!

-188,650,000,000,000 is the huge dimensional constant I came up with!

-L=length of barrel in Inch’s


So for a M1a with a 22 inch barrel
Weighing in at 9.9 lbs.
156 gr. Bullet
48 grains of powder
With a muzzle velocity of 2771 ft/s

It creates 1.332206213 Horsepower Upon firing.

EDIT: hmmm... still working on it.

pacecil
06-06-2009, 10:27 PM
This is really interesting. I read back over all the previous posts and am amazed at how difficult everyone has made this. Even this last post of moemag is kinda surprising. The original question was: how much horsepower does a gun generate when it fires? I don't think the original poster who posed the question got a straight answer from anyone! So here goes:

Just to use some of the characteristics set up previously: we've got a muzzle velocity of 3000 fps, an average chamber pressure of 20,000 psi, and a barrel length of 26". WORK done pushing the bullet from chamber to muzzle is equal to force on bullet base times the barrel length: 20,000 X .046 X 26 = 23920 in-lb = 1993 ft-lb. A typical time in barrel would be about .0015 sec, so this amount of work is done in .0015 seconds. The work done can then be considered as: 1993 / .0015 = 1328666 ft-lb/sec. Converting this to HORSEPOWER: 1328666/550 = 2415 hp. If you wanted to compare this with engine outputs which are often expressed in horsepower hours this gun would be putting out (.0015/3600) x 2415 = .001 hp-hours each time you fired it. GUNS ARE HIGH POWERED ENGINES THAT YOU RUN FOR VERY SHORT PERIODS!

Bullet weight enters in only as it affects pressure and time in barrel. The use of average pressure is an approximation - the force acting on bullet would vary as pressure in bore changed . To get the true time in barrel you have to look at pressure curve and develop an equation expressing how pressure varies as bullet moves up barrel and then integrate to determine true total pressure effect on barrel time. Using average pressure is much simpler and probably accurate enough for most purposes.

moemag
06-06-2009, 10:46 PM
Indicated horsepower
Indicated horsepower (ihp) is the theoretical power of a reciprocating engine if it is completely frictionless in converting the expanding gas energy (piston pressure x displacement)in the cylinders. It is calculated from the pressures developed in the cylinders, measured by a device called an engine indicator - hence indicated horsepower. As the piston advances throughout its stroke, the pressure against the piston generally decreases, and the indicator device usually generates a graph of pressure vs stroke within the working cylinder. From this graph the amount of work performed during the piston stroke may be calculated. It was the figure normally used for steam engines in the 19th century but is misleading because the mechanical efficiency of an engine means that the actual power output may only be 70% to 90% of the indicated horsepower.

.

Vibe
06-06-2009, 10:53 PM
This is really interesting. I read back over all the previous posts and am amazed at how difficult everyone has made this. Even this last post of moemag is kinda surprising. The original question was: how much horsepower does a gun generate when it fires? I don't think the original poster who posed the question got a straight answer from anyone! So here goes:

Just to use some of the characteristics set up previously: we've got a muzzle velocity of 3000 fps, an average chamber pressure of 20,000 psi, and a barrel length of 26". WORK done pushing the bullet from chamber to muzzle is equal to force on bullet base times the barrel length: 20,000 X .046 X 26 = 23920 in-lb = 1993 ft-lb. A typical time in barrel would be about .0015 sec, so this amount of work is done in .0015 seconds. The work done can then be considered as: 1993 / .0015 = 1328666 ft-lb/sec. Converting this to HORSEPOWER: 1328666/550 = 2415 hp. If you wanted to compare this with engine outputs which are often expressed in horsepower hours this gun would be putting out (.0015/3600) x 2415 = .001 hp-hours each time you fired it. GUNS ARE HIGH POWERED ENGINES THAT YOU RUN FOR VERY SHORT PERIODS!

Bullet weight enters in only as it affects pressure and time in barrel. The use of average pressure is an approximation - the force acting on bullet would vary as pressure in bore changed . To get the true time in barrel you have to look at pressure curve and develop an equation expressing how pressure varies as bullet moves up barrel and then integrate to determine true total pressure effect on barrel time. Using average pressure is much simpler and probably accurate enough for most purposes.
For using an approximation of area under the pressure curve - that result is not far off from what I was getting. But 1/2MV^2 with a known bullet mass and velocity is a lot easier to come up with. But the rest of your logic is pretty sound, particularly that which I highlighted in red.


But then again I would suppose that any gun firing a 200grain bullet at 3000ft/sec from a 26 in barrel would have a higher average pressure than 20,000psi. (Those were the numbers being used 09-23-2008).

Vibe
06-06-2009, 11:27 PM
So for a M1a with a 22 inch barrel
Weighing in at 9.9 lbs.
156 gr. Bullet
48 grains of powder
With a muzzle velocity of 2771 ft/s

It creates 1.332206213 Horsepower Upon firing.

:cool:
1/2 M V^2= Muzzle energy of 2674 ft-lbs
Time in barrel =(approx) 0.001323Sec
Delta Energy/time=2020620.391ft-lb/sec
HP=2020620.391/550=3673.85

moemag
06-07-2009, 12:52 AM
1/2 M V^2= Muzzle energy of 2674 ft-lbs
Time in barrel =(approx) 0.001323Sec
Delta Energy/time=2020620.391ft-lb/sec
HP=2020620.391/550=3673.85

...

3600hp=http://www.gnrhs.org/gn400cc.jpg
I seriously hope not.

moemag
06-07-2009, 01:36 AM
Is your Boss viewing this thread?
Lynn

Im sure it would be a good laugh. I goof stuff up... but I wont stop till I figure it out. Im kinda known for it.

That being said...

It is my opinion that “determine the horsepower generated by the rifle” means the power the rifle stock will hit someone with, and I will stand up for that.

I am working with the agreed upon assumption of the time interval being that of barrel length/(1/2)muzzle velocity. At .0013235 seconds.

I find that the energy/work of a recoil is determined as follows…
vgu = {(mp • vp) + ( mc • vc)} / mgu • 7000 → Etgu = mgu • vgu˛ / 2 • gc
Where as:

-Etgu is the translational kinetic energy of the small arm as expressed by the foot-pound force (ft•lbf).

-mgu is the weight of the small arm expressed in pounds (lb).

-mp is the weight of the projectile expressed in grains (gr).

-mc is the weight of the powder charge expressed in grains (gr).

-vgu is the velocity of the small arm expressed in feet per second (ft/s).

-vp is the velocity of the projectile expressed in feet per second (ft/s).

-vc is the velocity of the powder charge expressed in feet per second (ft/s). I have found further information stating... The NRA Fact Book (1988) gives some estimates for p. For small arms, the gas velocity is about 4000 fps for smokeless and about 2000 for blackpowder. Thus I will use 4000 ft/s.

-gc is the dimensional constant and is the numeral coefficient of 32.1739

-7000 is the conversion factor to set the equation equal to pounds.

-L is length of barrel (inches)


So using that equation for my before spec’d M1A , 22 inch barrel, 9.9 lbs, 156 gr projectile, 48 gr of powder charge, and a projectile velocity of 2771 ft/s

This gives me 12.48499291 ft•lbf. That over our time interval of .0013235 s is 9433.315383 ft•lbf/S . thus 1 horsepower (mechanical) = 550 ft•lbf/s giving me 17.15148252 HP.

So all simplified...

(3•{(mp • vp) + ( mc • vc)}^2 • vp)/(2.695•10^10•gc•L•mgu)

I’m going to call it a night on that one folks… and I would take that to my boss.

Bill Wynne
06-07-2009, 07:25 AM
No matter how you cut it, the definition of a horsepower is the amount of work required to lift 550 pounds one foot in one second (550 foot pounds per second). To lift your tiny bullet 3000 feet in a second (that is 3000 feet in a second), you need about 90 foot pounds produced in a in a second. There you have it, a known weight, a known distance, and a known time.

Just concentrate on those numbers. That is all there is to it. Way less than one horsepower.

Don't worry about the dragster in your driveway or the engine in an aircraft carrier. They have nothing to do with the problem.

There now, now that I have explained it again, don't you feel better.:)

Concho Bill

I haven't changed my mind since I wrote this. Are we sure that we haven't been educated beyond our IQ?

Concho Bill

Vibe
06-07-2009, 10:50 AM
Im sure it would be a good laugh. I goof stuff up... but I wont stop till I figure it out. Im kinda known for it.
Ditto here.


That being said...

It is my opinion that “determine the horsepower generated by the rifle” means the power the rifle stock will hit someone with, and I will stand up for that.
Who cares about the stock?? I mean really. The rifle is a device designed to launch a bullet, not a stock. Find a chronograph that will accuartely measure stock velocity and we could continue along that vein. Lacking that, find the data on the bullet and you can assume equal and opposite reactions are indeed equal.


I am working with the agreed upon assumption of the time interval being that of barrel length/(1/2)muzzle velocity. At .0013235 seconds.

I find that the energy/work of a recoil is determined as follows…
vgu = {(mp • vp) + ( mc • vc)} / mgu • 7000 → Etgu = mgu • vgu˛ / 2 • gc
Where as:..This is still a momentum equation. So everything that follows is false.




So using that equation for my before spec’d M1A , 22 inch barrel, 9.9 lbs, 156 gr projectile, 48 gr of powder charge, and a projectile velocity of 2771 ft/s

This gives me 12.48499291 ft•lbf. That over our time interval of .0013235 s is 9433.315383 ft•lbf/S . thus 1 horsepower (mechanical) = 550 ft•lbf/s giving me 17.15148252 HP.

So all simplified...


Let me ask one question....What is the published muzzle energy of the 156 gr round at 2771fps? From the manufacturer? I have 22LR rifles that generate more than 13ft-lbs of muzzle energy (in fact I would have to go to the Aguila Colibri which is 20 grains and has no powder at all to get that close).Dang, there are air rifles and pelletguns that generate more!

Handgun examples
http://www.chuckhawks.com/handgun_power_chart.htm

and from
http://www.remington.com/products/ammunition/ballistics/

Energy Calculator
This tool will help you estimate the energy of your favorite cartridge. If you are unsure of which bullet weight and/or velocity to submit, use the information from our on-line ballistics tables above.
Example: The muzzle energy of a 300 Remington Ultra Mag 180gr Core-Lokt Ultra bullet propelled at 3250 feet per second is determined using the follow formula:
M x V2 ÷ 450400 = foot pounds energy.

Step 1: Multiply M (M = bullet weight in grains) times V2 (V2 = the square of bullet velocity in feet per second): 180 x 3250 x 3250 = 1,901,250,000

Step 2: Divide the product of step 1 by 450400: 1,901,250,000 ÷ 450400 = 4221 foot pounds of energy.


http://www.remington.com/products/ammunition/bullet_energy.asp

Vibe
06-07-2009, 11:54 AM
Is your Boss viewing this thread?
Lynn
Should I post my humble resume? I could use a good paying job these days, and there might be an opening soon.:D

Old Gunner
06-07-2009, 01:00 PM
I started reading this thread and while its intriguing I couldn't handle all the math just now.
I'll go over the entire thread later, but for now I figured I'd mention something, which may have already been mentioned.

At some time after WW2 the Soviets experimented with a real "bullet hose" type weapon meant to guard tanks from being overwhelmed by infantry carrying anti tank charges of various sorts, magnetic mines or sticky bombs I guess.

It had a curved barrel similar to that the Germans had tried with the STG and some SMGs, but was a purpose built weapon of unique design. It ran off a small engine that used part of the pressure of combustion to propell half inch ball bearings at high velocity down the curved tube barrel, and give a continuous spray of bullets over the surface of the tank. The rate of fire was awesome and limited only by how fast the ball bearings could be fed by the hopper.
Wish I knew more about how it worked. I don't know if they ever fielded it. I only heard of it once back in the 60's.

An automatic weapon could be used as a propelling engine for a space walk.
It would basically be a rocket with a solid element to its exhaust.

Gun Powder engines were designed a few centuries back, I can probably find info on those in my books.
I think they mainly used a charge to drive a weight upwards and let gravity supply the energy to the machinery the weight was connected to.

In Europe muzzle energy of a bullet is often expressed in joules, which could make conversion formulas a bit easier.

moemag
06-07-2009, 01:37 PM
This is still a momentum equation. So everything that follows is false.


Free recoil is a vernacular term or jargon for recoil energy. Free recoil denotes the translational kinetic energy (Et) imparted to the shooter of a small arm when discharged and is expressed in joule (J) and foot-pound force (ft·lbf) for non-SI units of measure.

Free recoil should not be confused with recoil. Free recoil is the given name for the translational kinetic energy transmitted from a small arm to a shooter. Recoil is a name given for conservation of momentum as it generally applies to an everyday event.

Free recoil, sometimes called recoil energy, is a byproduct of the propulsive force from the powder charge held within a firearm chamber (metallic cartridge firearm) or breech (black powder firearm). The physical event of free recoil occurs when a powder charge is detonated within a firearm, resulting in the conversion of chemical energy held within the powder charge into thermodynamic energy. This energy is then transferred to the base of the bullet and to the rear of the cartridge or breech, propelling the firearm rearward into the shooter while the projectile is propelled forward down the barrel, with increasing velocity, to the muzzle. The rearward energy of the firearm can be calculated and is called free recoil, the energy of the bullet can be calculated and is called muzzle energy.


So you want the energy of the bullet...not the rifle?

PEI Rob
06-07-2009, 01:38 PM
http://www.unitconversion.org/unit_converter/energy.html

Vibe
06-07-2009, 03:05 PM
So you want the energy of the bullet...not the rifle?
At this point, yes. For several reasons.

1) that's pretty much what the original question asked.

2) Given that "For every action there is an equal and opposite reaction" - the muzzle energy of the bullet should (in the simplest of cases) be in the same range as the rifles reaction (Give or take a bit depending upon other factors and the fact that this is a momentum function and not an energy function. :D)

3) Since that would not include the "recoil energy" effect of the escaping gasses, we should be able to agree that the muzzle energy of just the bullet should be quite a bit lower than the energy imparted to the rifle by propelling both the bullet mass and propellant mass forward.

4) But, by the use of different devices such as very efficient muzzle brakes, the recoil effects upon the rifle can be drastically altered, without affecting the muzzle energy of the bullet at all.

5) We have chronograph devices in wide use to measure the forward velocity of the bullet directly, what we have to measure the reaction of the stock is not nearly as accurate.

6) There are online calculators for this all over the web (So that those who don't actually know how to do the math can check the work). :D

7) Wikipedia articles are often written by people who do not know what they are talking about. :D

alinwa
06-07-2009, 06:20 PM
7) Wikipedia articles are often written by people who do not know what they are talking about. :D



Now HERE we have serious disagreement......... if you see a wiki that's inaccurate your job is to FIX IT! This is the ultimate expression of the Encyclopedia Britannica concept.

al

Bill Wynne
06-07-2009, 07:13 PM
Are there any engineers, science/math teachers, or others out there who love math?

Challenge: Can you take a specific cartridge in a specific rifle and determine the horsepower generated by the rifle?


Montana Pete

Thus began one of the best threads in a long time.

Concho Bill

tobybradshaw
06-07-2009, 09:09 PM
3) Since that would not include the "recoil energy" effect of the escaping gasses, we should be able to agree that the muzzle energy of just the bullet should be quite a bit lower than the energy imparted to the rifle by propelling both the bullet mass and propellant mass forward.


Maybe the muzzle brake thread (http://benchrest.com/forums/showthread.php?t=58233) should be resurrected, too, since the physics-differently-abled had trouble with that one, too. :)

Toby Bradshaw
baywingdb@comcast.net

Vibe
06-07-2009, 10:50 PM
I always wondered what my avatar sounded like....Now I know.

Vibe
06-08-2009, 10:27 AM
Maybe the muzzle brake thread (http://benchrest.com/forums/showthread.php?t=58233) should be resurrected, too, since the physics-differently-abled had trouble with that one, too. :)

Toby Bradshaw
baywingdb@comcast.net
You did ask for it. :D

tobybradshaw
06-09-2009, 12:43 AM
You did ask for it. :D

And now you're asking for it. :)

You seem to have more patience than I do. Please be charitable to those who think of Newton's Third Law as the Third Suggestion. ;)

Toby Bradshaw
baywinddb@comcast.net

Swamp Fox
06-09-2009, 05:15 AM
I dirrected a couple of guys from another forum to drop in and see what they thought.

Digital Dan came up with the formula you need.

"Not being schooled in engineering I have none of their structured methodology. Therefore, if a .22 and a .338 Elkinstuffer can both kill a pig I find in favor of efficiency, since the lead and powder consumed by the .338 will, if properly rearranged, kill many pigs. At least 10 times as many. Therefore, the .22 RF is 10 times more efficient. That means the gerbils don't have to work so hard generating those horsepowers. Or pig powers. Chops. That's it, we need a pork chop index!

Ci=D^2(V/BHN)/W-a

Where:
Ci = Chop index
D= bullet diameter
V= velocity in warp fraction
BHN=thickness of cranial plate
W=mass
a=acceleration due to gravity or in a Shelby Cobra

How much pork could a .22 chop if a .22 could chop pork?"


Who said you can't get good info on the net?

Bill Wynne
06-09-2009, 05:25 AM
Newton's Forth Observation of Motion

(doesn't quite rise to the status of a law)


"It is a wonder of the highest order that some of us know which direction to point a rifle, much less, find the trigger."

Concho Bill

John Kielly
06-09-2009, 06:24 AM
IHow much pork could a .22 chop if a .22 could chop pork?"
We export a certain number of feral pigs to Europe. They're rank enough, but the European laws require that they be shot (& carry the bullet hole to prove it) for them to be accepted & imported as game meat.

So, they are baited into large weldmesh traps, maybe emough room for 50 odd of them, with carrion baits, then put out of their misery & onto a German plate with a .22 rimfire shot to the head.

Vibe
06-09-2009, 08:46 AM
Please be charitable to those who think of Newton's Third Law as the Third Suggestion. ;)

Toby Bradshaw


Now there is a fairly apt analogy. :D



Newton's Forth Observation of Motion

(doesn't quite rise to the status of a law)


"It is a wonder of the highest order that some of us know which direction to point a rifle, much less, find the trigger."

Concho BillDang Bill. I have to give you that one.
That has got to be, hands down, one of, if not THE most accurate post you've made in this thread. :D

longshooter
06-09-2009, 10:42 AM
Now there is a fairly apt analogy. :D

Dang Bill. I have to give you that one.
That has got to be, hands down, one of, if not THE most accurate post you've made in this thread. :D


Hopefully, he feels better now!


Longshooter

Old Gunner
06-09-2009, 11:14 AM
Remembered another instance of a cartridge used to turn an engine, the starter cartridges used for some radial engines. They look more like a large shot shell sized blank cartridge.

According to hollywood a rifle shot can lift a Homeboy off his feet and throw his body through a plate glass window. So studying the film we might estimate the weight and velocity of said Homeboy and then calculate just how much energy in terms of Homeboys per foot is contained in an AK 47 magazine. Start off with X number of HPF per AK/M = ?.
Eliminating Gang Bangers being the useful work of the equation.

Vibe
06-09-2009, 12:00 PM
Remembered another instance of a cartridge used to turn an engine, the starter cartridges used for some radial engines. They look more like a large shot shell sized blank cartridge.


Reference the movie "Flight of the Phoenix".

moemag
06-09-2009, 02:31 PM
Vibe,
I still stand by my stock horsepower calcs...;)
as far as bullet horsepower, I will apolgize and say your ~3600hp is right.

I can't find any way around it hahaha! I really wasnt expecting it was going to be that high.

Cheers. It was fun.

Vibe
06-09-2009, 02:40 PM
Vibe,
I still stand by my stock horsepower calcs...;)
as far as bullet horsepower, I will apolgize and say your ~3600hp is right.

I can't find any way around it hahaha! I really wasnt expecting it was going to be that high.

Cheers. It was fun.
Your "Stock HP" may also be correct...for the few thousandths of inch that it's powered. While the Momentum of the stock is equal to the momentum of the bullet and gasses, the energy is not, so the change in energy is not the same - therefore the HP cannot be the same. Quite frankly I'd never thought to look at the HP from the perspective of stock movement. But, like you, I was a bit surprised at the disparity.

Yes. It has been fun. :D

Now..If you'd care to jump into the Muzzle Brake thread (http://benchrest.com/forums/showthread.php?t=58233), that one looks to have equally interesting "debate". :D :rolleyes:

Bill Wynne
06-09-2009, 09:18 PM
I respectfully submit one more time that the work done by a 220 grain bullet with a muzzle velocity of 300 feet per second is way less than one horsepower.

Horsepower is defined as work done over time. The exact definition of one horsepower is 33,000 lb.ft./minute (550 foot pounds / second). Put another way, if you were to lift 33,000 pounds one foot over a period of one minute, you would have been working at the rate of one horsepower. In this case, you'd have expended one horsepower-minute of energy.

Newton's third law of motion states for every action there is an equal and opposite reaction. Therefore: The weight of the bullet X it's velocity is equal to the weight of the rifle X it's velocity. They are exactly the same.

220 grain bullet = .0314 pounds (a good sized bullet)

.0314 pounds X 3000 feet per second = 94.2 foot pounds per second

94.2 foot pounds per second / 550 foot pounds per second =.1712727 and that is how much horsepower is produced. You can make this more complicated but you cannot make this more correct.

I know this to be true from my long ago study with a ballistic pendulum.

Consider this. No mortal man could withstand the impact caused by thousands of horsepower. Superman? Maybe.

Concho Bill

Vibe
06-09-2009, 10:31 PM
I respectfully submit one more time that the work done by a 220 grain bullet with a muzzle velocity of 300 feet per second is way less than one horsepower.
You have now had two people with engineering degrees tell you different. If you STILL chose to remain ignorant, it's now all on you.



Horsepower is defined as work done over time. The exact definition of one horsepower is 33,000 lb.ft./minute (550 foot pounds / second). Put another way, if you were to lift 33,000 pounds one foot over a period of one minute, you would have been working at the rate of one horsepower. In this case, you'd have expended one horsepower-minute of energy.


Newton's third law of motion states for every action there is an equal and opposite reaction. Therefore: The weight of the bullet X it's velocity is equal to the weight of the rifle X it's velocity. They are exactly the same.
This is true, however that is a statement of conservation of Momentum. The Energy is NOT similarly conserved.


220 grain bullet = .0314 pounds (a good sized bullet).
We'd been using 200, but what the heck.


.0314 pounds X 3000 feet per second = 94.2 foot pounds per second.
0.0314 under the influence of the acceleration of GRAVITY (32/ft/sec/sec)
Under the influence of the Acceleration due to Rifle this is 72.22 pounds.



I'v got to go to work now, so I'll get to the rest tomorrow.

pacecil
06-09-2009, 11:51 PM
you calculated:

220 grain bullet = .0314 pounds (a good sized bullet)

.0314 pounds X 3000 feet per second = 94.2 foot pounds per second

The pounds in the units of ft-lbs/sec is not the weight of the bullet. It's the pounds force you applied in accelerating the bullet up to 3000 fps
Also the feet in the units is not the feet in 3000 fps. It's the ft you moved the bullet thru to get it up to 3000 fps.

Multiplying the weight of bullet by it's velocity at the muzzle will not give you the work done on the bullet when you accelerated it out the barrel. You get this by multiplying the force applied over the distance the bullet traveled and then dividing by the time the force was applied. The force is constantly changing as the bullet travels up the barrel, which is why you must integrate or use an average.

Bill Wynne
06-10-2009, 06:21 AM
You have now had two people with engineering degrees tell you different. If you STILL chose to remain ignorant, it's now all on you.

Respectfully, we can disagree and not be disagreeable.

Not to make this personal, Vive, but on any test at the school of engineering that you attended did you miss any question?

What I am saying is that even people with engineering degrees are sometimes in disagreement. Could it be that you and others have complicated a simple definition of a horsepower into something James Watt would not recognize?

Concho Bill

pacecil
06-10-2009, 09:16 AM
You asked: Could it be that you and others have complicated a simple definition of a horsepower
This doesn't appear to be the problem. What is happening is you are simply not doing it right. What you have calculated as work, from which you will later calculate horsepower, is not that at all - it's momentum. This is an understandable error because momentum and work would appear to have the same units, ft-lbs/sec (or lb-ft/sec) You do work when you apply a force to a body and move it a distance in a period of time. In the units ft is the distance, lbs is the force and sec is the time.

A 200 grain bullet traveling at 3000 fps can only do work or have work done on it during the time it travels some distance. If it travels in a vacuum it will do no work nor will it have any work done on it. If it travels in air, the air will do some work on the bullet as it slows the bullet down. At the same time the bullet will be doing work on the air as it moves the air out of the way.

Vibe
06-10-2009, 09:19 AM
Respectfully, we can disagree and not be disagreeable.

Not to make this personal, Vibe, but on any test at the school of engineering that you attended did you miss any question?
LOL. Yep...But not this one. :D


What I am saying is that even people with engineering degrees are sometimes in disagreement. Could it be that you and others have complicated a simple definition of a horsepower into something James Watt would not recognize?

Concho Bill
I will freely admit that my numbers are not completely accurate, but that is due to the fact that I have simplified the math to the point that the Integral Calculus required to accurately determine the true and actual acceleration has been reduced to a simple averaging of velocity -other wise the HP number would really be quite a bit higher. The actual time in barrel is a bit less than my "estimate". Your new 220 grain bullet at 3000 fps from a 26" barrel would more likely be applying closer to 6000HP to the bullet.

Which brings me to another point. You keep refering to work done by the bullet.

I respectfully submit one more time that the work done by a 220 grain bullet with a muzzle velocity of 300 feet per second is way less than one horsepower.

We are not concerned with this because it was not part of the quetstion. This "work done by the bullet" is just he bullet bleeding off all of the energy that we applied to it during the firing process.

moemag
06-10-2009, 10:19 AM
engineering is the art of modelling materials we do not wholly understand, into shapes we cannot precisely analyse so as to withstand forces we cannot properly assess, in such a way that the public has no reason to suspect the extent of our ignorance.
- dr ar dykes -
:d

Vibe
06-10-2009, 10:34 AM
Now that is accurate enough. :D
And would describe the practices of most of the engineers I've come in contact with my modest career.

Bill Wynne
06-10-2009, 12:26 PM
"engineering is the art of modeling materials we do not wholly understand, into shapes we cannot precisely analyze so as to withstand forces we cannot properly assess, in such a way that the public has no reason to suspect the extent of our ignorance."

Good statement Moemag. I like your humor.

Let me add from my observation of engineers and their projects:

There two schools of engineering in the world.
Those who designed the pyramids and those who designed the Golden Gate Bridge. One uses the most materials that can fit into the space and the other uses the least materials that will adequately do the job.

Vibe,

Sorry about what I said about that bullet and work. You know the by the bullet. How about to the bullet. There now, I feel better.

Concho Bill

Vibe
06-10-2009, 01:01 PM
Vibe,

Sorry about what I said about that bullet and work. You know the by the bullet. How about to the bullet. There now, I feel better.

Concho Bill

No apology required, I'm just trying to discover the source of your confusion and why this is so difficult for you to grasp. You're as hard headed and argumentative as I am. :rolleyes:
But the bullet does exhibit a bare minimum of horsepower in flight (at least until it hits a solid target), and if that had been why you thought as you claim to, we could have addressed it.

Bill Wynne
06-10-2009, 04:08 PM
No apology required, I'm just trying to discover the source of your confusion and why this is so difficult for you to grasp. You're as hard headed and argumentative as I am. :rolleyes:
But the bullet does exhibit a bare minimum of horsepower in flight (at least until it hits a solid target), and if that had been why you thought as you claim to, we could have addressed it.

Don't worry about it Vibe. I have my story and I am sticking to it.

I have fired a 300 H&H Magnum Ackly Improved* into a 14 pound ballistic pendulum** and it swung out and lifted up a few inches at the most. Very loud report, healthy recoil, but very little horsepower.

* 150 grain bullet, MV = 3500 feet per second more or less.
** This was a steel pipe full of sand with a piece of rubber inter tube over the end of the pipe with cotton behind it. The pipe was hanging by two light weight wires.

Note: I was maybe 19 years old. I have lived 50 additional years. I must have been a little crazy.

Concho Bill

moemag
06-10-2009, 09:03 PM
http://i33.photobucket.com/albums/d53/MoeRoark/06-10-09_1818.jpg
Picked this up about an hour ago

I have some reading to do
:D

I will get back to this, and muzzle brakes in a few days.

Vibe
06-10-2009, 10:37 PM
I have fired a 300 H&H Magnum Ackly Improved* into a 14 pound ballistic pendulum** and it swung out and lifted up a few inches at the most. Very loud report, healthy recoil, but very little horsepower.

* 150 grain bullet, MV = 3500 feet per second more or less.
** This was a steel pipe full of sand with a piece of rubber inter tube over the end of the pipe with cotton behind it. The pipe was hanging by two light weight wires.

Note: I was maybe 19 years old. I have lived 50 additional years. I must have been a little crazy.

Concho Bill
Ballistic pendulums are neat and interesting, but have nothing to do with HP - they only measure momentum.
The one I built was only 3 lbs, but I was using 30 to 35 gr bullets at less than 2000fps. At 85" the pendulum was "calibrated" to swing almost exactly 1" for every 100 fps. (Made the math easier) :D

Bill Wynne
06-11-2009, 06:27 AM
Ballistic pendulums are neat and interesting, but have nothing to do with HP - they only measure momentum.
The one I built was only 3 lbs, but I was using 30 to 35 gr bullets at less than 2000fps. At 85" the pendulum was "calibrated" to swing almost exactly 1" for every 100 fps. (Made the math easier) :D

The size of yours was good thinking. You could demostrate the same results as I did with a pellet gun in your garage without near the danger. I tested four different centerfire cartridges with mine and the results were very close to their know velocities. I tested a .222 Remington, a 25/06 Ackley Improved, a 30/06 and that 300 H & H Ackley Improved.


How does this relate to our problem at hand?

To obtain the velocity of the projectile striking the pendulum, the movement of the ballistic pendulum against the force of gravity is measured in units of mass divided by distance such as foot/pounds. 550 foot/pounds per second = 1 HP.

You have a known weight of projectile and a known velocity in feet per second. Convert what you know to foot pounds per second and divide that number by the number of foot pounds per second that are one horsepower (550) and that is the horsepower developed by your rifle and transfered to the projectile or bullet.

It is no more complicated than that.:)

Concho Bill

Vibe
06-11-2009, 10:04 AM
How does this relate to our problem at hand?

To obtain the velocity of the projectile striking the pendulum, the movement of the ballistic pendulum against the force of gravity is measured in units of mass divided by distance such as foot/pounds. 550 foot/pounds per second = 1 HP.

You have a known weight of projectile and a known velocity in feet per second. Convert what you know to foot pounds per second and divide that number by the number of foot pounds per second that are one horsepower (550) and that is the horsepower developed by your rifle and transfered to the projectile or bullet.

It is no more complicated than that.:)

Concho Bill
You are right, it really is no more complicated than that. However - for the HP being applied to the bullet by the rifle what you have is the Ft-Lbs of energy of the bullet at the muzzle divided by the time it took to get it up to that speed. And for out 200gr bullet at 3000ft/sec that is a muzzle energy of around 4100ft-lbs( use the published data from the manufacturer, or an energy caculator in a load manual), and it took 0.001444 seconds to clear the 26" barrel (Distance traveled, under the applied force, divided by the average velocity).

So...What do you get when you divide 4100 ft/lbs by 0.00144 seconds?

4100/0.001444 = How many ft-lbs/sec?

And dividing that number by 550 gives you how much HP?

Bill Wynne
06-11-2009, 10:40 AM
Vibe, It doesn't matter that some work is done quickly. That amount is all the ever done. Wait a second before you calculate how much horsepower was applied. There would be a lot of horsepower if the initial powder charge could keep up for a full second but it can't.

Why? Because it just doesn't have enough power (horsepower that is).

I think I know why we just look at the bullet weight and bullet velocity and don't ever mention the horsepower.

Concho Bill

pacecil
06-11-2009, 10:47 AM
You said You have a known weight of projectile and a known velocity in feet per second. Convert what you know to foot pounds per second and divide that number by the number of foot pounds per second that are one horsepower (550) and that is the horsepower developed by your rifle and transfered to the projectile or bullet.

This is WRONG! Read my posts 233 and 235! You think that if your answer is to be in ft-lb/sec you can use any thing you have in front of you with these units and this will make your answer right. You cannot use bullet weight for the lb in the units. The lbs has to be the gas pressure force in lbs. You cannot use the ft from the fps velocity. This ft unit has to be the barrel length in ft, that is the distance over which you were adding work to the bullet.

You can't get horsepower from a bullet at a single point in space. You have to decide over what distance of travel you want to determine the horsepower was expended.

Vibe
06-11-2009, 10:53 AM
Vibe, It doesn't matter that some work is done quickly.
Other than that is exactly what the definition of "Power" is, and that we like barrels shorter than 1300" to get up to a useful velocity. :D


That amount is all the ever done. Wait a second before you calculate how much horsepower was applied. There would be a lot of horsepower if the initial powder charge could keep up for a full second but it can't.
Nothing in any physics book ever indicated that time HAS to be measured only in full second intervals. (Good thing too, or lasers would be useless)



Why? Because it just doesn't have enough power (horsepower that is).
If it COULD maintain that level of power for a full second, it would seriously maim the shooter. Even 16" naval guns don't take a full second to go off.


I think I know why we just look at the bullet weight and bullet velocity and don't ever mention the horsepower.

Concho Bill

The obvious answer is that the HP value is so esoteric as to not be of any real relevance to everyday life. But the question was asked, I answered factually...You argued. :D
And here we are. :p