Rifle: A machine rated in horsepower

[Vibe]

[COLOR="Red"]4Mesh is WRONG HERE[/COLOR]

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

Now.

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

4Mesh 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.

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.

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.

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.

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?

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.

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.

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 ELAPSED

4Mesh 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?

Ok Vibe, am I wrong yet?

Well past wrong, and solidly into amusing.

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.

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!

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?

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.

 

[Vibe]

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.

 

[4Mesh]

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 wrong

You 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]

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]

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.



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

 

[alinwa]

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 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]

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]

I've argued with Phil before you see 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]

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]

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]

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]

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]

Bullet RPM

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]

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]

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]

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]

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]

Duration of time.

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]

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]

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 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....


LOL


al