Rifle: A machine rated in horsepower

[Montana Pete]

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]

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]

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]

Hmmmm. Do tell.

I would be afraid to say publicly!

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]

I will never eat crow

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]

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]

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]

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


LOL


al

 

[MistWolf]

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:

Where F = Force (lbs)
V = Velocity (ft/min)

I also found this:

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]

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:

Where F = Force (lbs)
V = Velocity (ft/min)

I also found this:

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]

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:

Where 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]

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]

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]

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]

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]

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!
E=MgH=1/2MV^2

Edited to ad the 1/2 factor...someone should have corrected me on it before now.

 

[MistWolf]

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]

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]

absolutely!
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]

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,