Rifle: A machine rated in horsepower

Bill...

While were waiting on a reply from Lynn I decided to give you a little more to think about. I think I see why you aren't able to get a handle on this whole horsepower thing. Those guys that worked out a definition for a horsepower, could have used the force it took to drag the weight over the ground instead of lifting it.

Take a look at the lifting of the car. If instead of calculating what it take to lift the car, you calculate what horsepower it takes to drag or push the weight 220 ft horizontally in one second. For that matter you can determine what it takes to push it down a 220 ft hole in one second! When you lift the car weight adds to the horsepower, when you push the car horizontally the weight doesn't effect the power, when you push the car down a hole the weight adds to power. If you did the math you would find it takes 7874 hp to lift the car, 7086 hp to push it horizontally, and only 6299 hp to push it down the hole.

WHAT ALL THIS MEANS IS THAT THE INERTIA FORCE PLUS THE WEIGHT IS WORKING AGAINST YOU WHEN YOU'RE LIFTING, THE INERTIA FORCE IS AGAINST YOU BUT THE WEIGHT FORCE IS AT 90 DEG TO THE DIRECTION YOU'RE PUSHING WHEN GOING HORIZONTALLY, AND THE WEIGHT FORCE IS GOING SAME DIRECTION YOU ARE BUT INERTIA FORCE IS STILL AGAINST YOU WHEN PUSHING DOWN A HOLE.

I looked back at what I just wrote and maybe it'll just confuse you more, but I went to all trouble to write it so I'll let it fly and see what happens!
 
pacecil said:
Those guys that worked out a definition for a horsepower, could have used the force it took to drag the weight over the ground instead of lifting it.
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Pacecil,

I would think that they used something like a spring cotton scale that retracted to the 180 pound mark as the horse walked his circle. The scale would not know if it was measuring a lifted weight or a resistance being drug along.

I am enjoying this interplay. There is no way we can be too serious about this.:)

Concho Bill
 
When you guys get done figuring out what "Work" really is (Becasue that is all that you have discussed), THEN you can determine how much POWER it takes to do that work in 0.002 seconds. :D
 
After a life time in construction I know what work is. I come from a time before forklifts and nail guns and even before air condition pickups. I know what work is and I know something about manual labor and mechanical advantage. We worked with our backs and our legs and our arms. We also used our heads or else we wouldn't be here to keep you in line.:)

Concho Bill
 
Bill Wynne said:
After a life time in construction I know what work is. I come from a time before forklifts and nail guns and even before air condition pickups. I know what work is and I know something about manual labor and mechanical advantage. We worked with our backs and our legs and our arms. We also used our heads or else we wouldn't be here to keep you in line.:)

Concho Bill
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The term "Work" in physics is not exactly the same as what we think of as "Labor" in the real world. You can push all day on a wall to keep a pig from settling into a newly laid row of brick, and you will be tired and hot when it gets to where you do not have to steady it. But you will have done no "Work" as far as the kenematic physics are concerned.
My momma didn't have much luck keeping me in line. :D You, my friend, don't have a shot. :D
 
Vibe said:
The term "Work" in physics is not exactly the same as what we think of as "Labor" in the real world. You can push all day on a wall to keep a pig from settling into a newly laid row of brick, and you will be tired and hot when it gets to where you do not have to steady it. But you will have done no "Work" as far as the kenematic physics are concerned.
My momma didn't have much luck keeping me in line. :D You, my friend, don't have a shot. :D
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Well, at least I know what a horsepower is.:)​

First I would get the pig out out of the row of bricks so someone who hadn't done any work at all wouldn't come by and ask a dumb question like, "What is that pig doing in that wall?" Then I would learn how to spell kinematic physics before I tried to use the word.:)

My mother kept me in a straight line and I have no trouble calmly taking another shot after a miss.

Concho Bill
 
Bill Wynne said:
Well, at least I know what a horsepower is.:)​
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Well you certainly know how to SAY that you do anyways.:D

Bill Wynne said:
[First I would get the pig out out of the row of bricks so someone who hadn't done any work at all wouldn't come by and ask a dumb question like, "What is that pig doing in that wall?"
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LOL. See I do know what work is, and what a pig in a wall is.

Bill Wynne said:
[Then I would learn how to spell kinematic physics before I tried to use the word.:)
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Hey. My spellchecker doesn't seem to work real well in IE, and I've only been posting in between batches of processing the honey I robbed from the bee hives today. ('bout 3 gallons from 2 hives). Besides, I know you read typonese alot better than you can cypher. :D

Bill Wynne said:
My mother kept me in a straight line and I have no trouble calmly taking another shot after a miss.

Concho Bill
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And yet you refuse to "calmly take another shot" at understanding this horsepower stuff. :D

Hard headed old coot. :p
 
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Lynn,

How would I know why these two smart people can't get their stories straight? Maybe they were educated beyond their IQ.

This just proves that even intelligent people can come up with different conclusions over what would seem to be a simple problem that surely has one simple answer.

Vibe,

It seems that a 1500 watt iron uses the equivalent of 2.0115331 horsepower or 85.0303543 BTUs (British thermal units) per minute or nearly 1/2 ton of air conditioning. I did not know that. That is a powerful hot iron. A lady can do a lot of work with an iron like that.

Concho Bill
 
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I think I see the problem.♠ Everyone has a feel for how much power a horse could put out - or an automobile. But no body can understand in their gut what a rifle will do. If I show an example everybody can agree with and have a feel for all will be made right. That thing we went through about pushing a 2500 lb car up to 440 fps (300 MPH!!!!!) in one second was kinda stupid! (Was that your idea, Bill?)

So lets figure what it takes to accelerate my 2500 lb car up to 60 mph in, let's say 5 sec. We'll do it in the same 220 ft. The method we'll use is this:
Divide the final speed 60 mph(88 fps) by the time (5 sec) to get the acceleration. This turns out to be 17.6 ft/sec/sec. Then we'll use F=Ma to find out how hard we have to push against the car (or let the engine "push") to get the car up to speed. The mass of the car is 2500/32, multiplying this by the acceleration will give us the force which turns out to be 1375 lbs. If we apply this force over a distance of 220 ft and do it for 5 sec, 1375 x220/5/550, we find we'll need 110 hp to take the car from zero to 60 mph. (If you add a little more power to accommodate wind resistance and friction most people would consider this a pretty reasonable figure.)

Now if we follow the same procedure for a bullet we want to accelerate from zero to 3000 fps in 24 inches of barrel, lets just see how much power is required. It's going to be tough for anyone to have a feel for the numbers we come up with for the rifle because they are so far away from what the car was but they are figured just like the car was. Our bullet will cover the two feet of barrel at an average velocity of 1500 fps and take .0013 sec to do it. This equates to an acceleration of 1153846 ft/sec/sec. This is one of those numbers kinda hard to get a feel for, but consider this: we are making the bullet go from zero to 2045 mph in .0013 sec! In any case we go through the F=Ma thing again and find we must apply 1009 lbs of force to the base of the bullet over the 2 ft of barrel length. If it's a 30 caliber we only need apply an average pressure to the base of 13635 psi. Cranking all the numbers around just like we did with the car we finally will arrive at 2282 hp required to get that little sucker from zero to 3000 fps! Seems like a big number, but remember we only applied that power for .0013 sec.

THERE YOU HAVE IT - THE CORRECT ANSWER AND HOW YOU ARRIVE AT IT SHOULD END THIS THREAD.
 
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pacecil said:
THERE YOU HAVE IT - THE CORRECT ANSWER AND HOW YOU ARRIVE AT IT SHOULD END THIS THREAD.
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One can always check the results by dividing the published muzzle energy by the time in barrel to get ft-lbs/sec and divide that by 550ft-lbs/sec/HP.
But that's a lot closer. :D
 
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You guys are right!

I just put that in there to see if anybody was watching! I used half the acceleration rate I should have. The correct horsepower is 4564
 
Montana Pete said:
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
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Look guys, This is a question that can be answered with either a simple yes or no.

In my case yes would be the correct answer. I will let it go there.
 
Bill Wynne said:
Look guys, This is a question that can be answered with either a simple yes or no.

In my case yes would be the correct answer. I will let it go there.
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:D But in your case Bill (and only specifically in your case) the correct answer (according to your posts) is no. YOU can't. :p

We, on the other hand, have been able to do it. :D
 
I cant believe I just sat here and read this entire thread, but my interest got peaked. I was waiting to see if anyone tackled the fact that it takes substantially more effort to push a bullet down the bore. Wouldn't that artificially increase the weight of the bullet for purposes of determining horsepower, or would it just be akin to where the horse power is measured...ie rear wheels or flywheel. I am guessing the answers given would be considered rear wheel horsepower, other than the one poster who used muzzle pressure to derive his answer.

Very interesting thread, thanks for the information. I am very interest to know the answer to this question using these parameters: 3000fps, 210 gr bullet, 28 inch barrel, peak chamber pressure of 64,000 PSI. I guess I am wanting to know peak horsepower which would probably be calculated some where within the first three of four inches of travel. If I ran this through quickload and emailed someone the pressure curve would they give me an estimation?

I do not understand the math well enough to even take a shot, wish I could. While I am not very good at math, I was always good at taking standardized exams. I would have just marked the old standby "D. not enough information to answer" and moved on with this one.

Thanks again to all those who contributed to this interesting thread.
 
eddybo said:
I cant believe I just sat here and read this entire thread, but my interest got peaked. I was waiting to see if anyone tackled the fact that it takes substantially more effort to push a bullet down the bore. Wouldn't that artificially increase the weight of the bullet for purposes of determining horsepower, or would it just be akin to where the horse power is measured...ie rear wheels or flywheel. I am guessing the answers given would be considered rear wheel horsepower, other than the one poster who used muzzle pressure to derive his answer.
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Just like we usually do not consider the HP losses in your engine due to the drags from the alternator, power hydraulic pumps, or even how much it takes just to turn the engine as an air compressor - those are all internal, and (luckily) not part of this problem. :D

eddybo said:
Very interesting thread, thanks for the information. I am very interest to know the answer to this question using these parameters: 3000fps, 210 gr bullet, 28 inch barrel, peak chamber pressure of 64,000 PSI. I guess I am wanting to know peak horsepower which would probably be calculated some where within the first three of four inches of travel. If I ran this through quickload and emailed someone the pressure curve would they give me an estimation?
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Assuming that your pressure curve is approximately triangular - Approx linear rise to a peak, approx linear drop to muzzle exit, we can make some assumptions (it's not exactly that shape, but close enough for a few SWAGs)

The "average" of a triangle is 1/3 up. So the "peak" HP is going to be roughly 3x the "average" HP. Also the Average velocity will be closer to 2/3 your muzzle velocity rather than the 1/2 that I've been using for an estimate, so you time in barrel will really be shorter and you "average HP" quite a bit more than what's been stated here.

eddybo said:
I do not understand the math well enough to even take a shot, wish I could. While I am not very good at math, I was always good at taking standardized exams. I would have just marked the old standby "D. not enough information to answer" and moved on with this one.

Thanks again to all those who contributed to this interesting thread.
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I'd be quite a bit surprised that a rifle pushing a 210gr slug to 3000fps in a 28" barrel would generate any less than 20,000 PEAK HP. (For a few nanoseconds anyways) :D

But I didn't put a calculator to it.
 
eddybo said:
I am very interest to know the answer to this question using these parameters: 3000fps, 210 gr bullet, 28 inch barrel, peak chamber pressure of 64,000 PSI.
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Eddybo,

Try to keep your mind clear and stay away from extraneous details. All you need to know is the velocity and the weight of the projectile.:)

Remember there is more that one position on this problem but only one correct answer. Let me try one more time.

Horsepower = 550 foot pounds per second.

210 grain bullet / 7000 grains in a pound = .03 pounds

At the time of the bang, your bullet leaves the muzzle at the rate of 3000 feet per second.

.03 pounds X 3000 feet per second = 90 foot pounds per second (a healthy recoil)

90 foot pounds per second / 550 foot pounds per second = .1636 horsepower

Concho Bill
 
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