Rifle: A machine rated in horsepower

[Bill Wynne]

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]

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]

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]

Lynn,

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

send it over

al

 

[alinwa]

bada-BING bada-BOOM

Ask and ye SHALL receive

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]

BTW Lynn,

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


al

 

[Bill Wynne]

I think your formula is flawed

"(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]

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]

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]

And the answer is...

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]

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 -

 

[4Mesh]

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]

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]

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]

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.

 

[4Mesh]

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]

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.
(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]

LOL. Actually in the complete derivation of the term....it does.
(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]

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]

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.