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Vibe
Guest
Whatever. Energy and power caculations are basic physics, not limited to (or even normally included in) benchrest.okay my bad.
I will leave you benchrest boys alone.
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moemag
Guest
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?
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?
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Vibe
Guest
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.
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.
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moemag
Guest
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)
Work=
(((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...
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.
Okay here we go.
Horsepower = (work)/(time)
Work=
(((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)
(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.
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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.
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.
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moemag
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Vibe
Guest
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).
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Vibe
Guest
1/2 M V^2= Muzzle energy of 2674 ft-lbsSo 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.
Time in barrel =(approx) 0.001323Sec
Delta Energy/time=2020620.391ft-lb/sec
HP=2020620.391/550=3673.85
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moemag
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moemag
Guest
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 • vgu² / 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.
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Bill Wynne
Active member
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
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Vibe
Guest
Ditto here.
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.
This is still a momentum equation. So everything that follows is false.
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/
http://www.remington.com/products/ammunition/bullet_energy.asp
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Vibe
Guest
Should I post my humble resume? I could use a good paying job these days, and there might be an opening soon.
O
Old Gunner
Guest
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.
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.
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moemag
Guest
http://en.wikipedia.org/wiki/Free_recoil said:
So you want the energy of the bullet...not the rifle?
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Vibe
Guest
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.
)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).
7) Wikipedia articles are often written by people who do not know what they are talking about.
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7) Wikipedia articles are often written by people who do not know what they are talking about.
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
Active member
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tobybradshaw
Guest
Maybe the muzzle brake thread should be resurrected, too, since the physics-differently-abled had trouble with that one, too.
Toby Bradshaw
baywingdb@comcast.net