Power, Energy, and Momentum
The HORSEPOWER thread got locked, so I'll start a new thread to address some misconceptions. The original question was how much horsepower does a 30-06 cartridge develop in a bolt-action rifle? This is a difficult question to answer and none of the answers given in the original post were correct. Warning, this is going to be a long-winded explanation, be prepared for the MEGO effect. (My Eyes Glaze Over) First, some definitions.
Power is a measure of how much energy is produced per unit time. (Actually, energy only changes from one form to another. That's another topic.) In physics, we normally measure power in metric units, watts. Horsepower and BTU/hr. are traditional English units. The approximate conversion factors are 1,000 watts (1 kilowatt) = 1.34 horsepower and 1 watt = 3.41 BTU/hr.
If power is constant, meaning it doesn't change with time, then it is easy to calculate how much energy is produced: just multiply the power by the time lapsed. For example, a 100 watt bulb burning for 1,000 hours uses 100,000 watt-hours or 100 kilowatt-hours, a convenient unit if you are in the electrical power industry. The metric unit for energy is the joule, which is 1 watt for 1 second.
The more usual unit of energy for talking about firearms is the foot-pound, which as the unit says is the energy required to lift a one pound weight one foot. If you can generate 550 of these in one second, you are generating one horsepower.
Unfortunately, the power generated in a firearm is anything but constant. It starts at zero right at the instant of ignition, reaches a peak in a fraction of a millisecond, and then decreases. During this time, energy is changing from chemical energy to thermal energy in the combustion gasses to kinetic energy of the bullet, heat of the barrel and other parts of the gun, sound energy, and energy of the escaping gasses. (Since the original question asked about a bolt-action rifle, we won't worry about the kinetic energy in the action of an automatic or semiautomatic, that's really complicated.) What we mostly care about is the final kinetic energy of the bullet as it exits the barrel. I have read accounts that put this energy at 10% or less of the original chemical energy in the cartridge, the rest going to waste heat and sound.
While there is no single number for the power going into the bullet, because it varies with time (higher at the beginning, lower at the end of the barrel), we can estimate an average power applied to the bullet by taking the bullet's kinetic energy as it exits the barrel and dividing by the time it takes from the ignition of the cartridge to the bullet exiting the barrel. Since I have a Marlin Camp Carbine that shoots 9mm bullets, I'll do the calculation for that; you can do a similar one for your gun. Keep in mind that we are calculating the average power. The maximum power will be higher than that.
First, you need to find the kinetic energy in your bullet. You need the mass of the bullet (usually measured in grains) and the velocity (usually in feet per second). The easiest way to do the calculation to go to a table on this web page: http://www.naaminis.com/energy.html . Or, multiply the mass in grains by the square of velocity in ft/sec and divide by 450450 . For my example, I am shooting 115 grain JHP ammo and the muzzle velocity in my carbine is 1200 ft/sec. So my bullet energy is 115 x 1200 x 1200 / 450450 = 368 foot-pounds.
The calculation of how long it takes the bullet to exit is a little trickier and you have to estimate. If you assume that most of the acceleration of the bullet happens in the first four inches, then you can approximate the average velocity in the barrel as slightly less than the exit velocity. In my case, I'll estimate 1,000 ft/sec. The length of my barrel is about 16" or 1.33 ft. So it will take 1.33 milliseconds for the bullet to get out, 1.33 ft / (1,000 ft/sec) = 0.00133 seconds.
Finally, I get my estimate of average power, 368 foot-pounds / (0.00133 seconds) = 276692 foot-pounds/sec or 503 horsepower. As I said, the instantaneous value just after ignition will be higher than that and the instantaneous value as the bullet exits will be lower. We are only calculating the useful power going into the bullet. The waste power going into heating the barrel is about ten times that much.
Before you get excited about how much power that is, remember it's only for a very short time. Once the bullet is out of the barrel, the power is zero until you take the next shot.
Now, I want to say something about momentum. Newton's third law, which physicists have never found to be violated, says that the force applied to the bullet to shoot it out the barrel is equal and opposite to the force that the bullet kicks back against the gun and you holding it. Again, it is easier to look at the situation of a bolt-action rifle. The bullet starts out with zero momentum and ends up with momentum equal to its mass times velocity. The laws of physics state that the average force on the bullet multiplied by the time the force is applied (this is called "impulse") is equal to the change in the bullet's momentum. This in turn is equal to the average force pushing back on the gun and you during that time. Therefore, the best measure of how hard the shot kicks back is to multiply the mass of the bullet (you can use grains) by its muzzle velocity. By this estimate, a 173 grain 30-06 bullet going 2750 ft/sec will kick back about 3.45 times harder than my 9mm carbine. It will also hit the target that much harder, neglecting loss of speed in air.
Power is a measure of how much energy is produced per unit time. (Actually, energy only changes from one form to another. That's another topic.) In physics, we normally measure power in metric units, watts. Horsepower and BTU/hr. are traditional English units. The approximate conversion factors are 1,000 watts (1 kilowatt) = 1.34 horsepower and 1 watt = 3.41 BTU/hr.
If power is constant, meaning it doesn't change with time, then it is easy to calculate how much energy is produced: just multiply the power by the time lapsed. For example, a 100 watt bulb burning for 1,000 hours uses 100,000 watt-hours or 100 kilowatt-hours, a convenient unit if you are in the electrical power industry. The metric unit for energy is the joule, which is 1 watt for 1 second.
The more usual unit of energy for talking about firearms is the foot-pound, which as the unit says is the energy required to lift a one pound weight one foot. If you can generate 550 of these in one second, you are generating one horsepower.
Unfortunately, the power generated in a firearm is anything but constant. It starts at zero right at the instant of ignition, reaches a peak in a fraction of a millisecond, and then decreases. During this time, energy is changing from chemical energy to thermal energy in the combustion gasses to kinetic energy of the bullet, heat of the barrel and other parts of the gun, sound energy, and energy of the escaping gasses. (Since the original question asked about a bolt-action rifle, we won't worry about the kinetic energy in the action of an automatic or semiautomatic, that's really complicated.) What we mostly care about is the final kinetic energy of the bullet as it exits the barrel. I have read accounts that put this energy at 10% or less of the original chemical energy in the cartridge, the rest going to waste heat and sound.
While there is no single number for the power going into the bullet, because it varies with time (higher at the beginning, lower at the end of the barrel), we can estimate an average power applied to the bullet by taking the bullet's kinetic energy as it exits the barrel and dividing by the time it takes from the ignition of the cartridge to the bullet exiting the barrel. Since I have a Marlin Camp Carbine that shoots 9mm bullets, I'll do the calculation for that; you can do a similar one for your gun. Keep in mind that we are calculating the average power. The maximum power will be higher than that.
First, you need to find the kinetic energy in your bullet. You need the mass of the bullet (usually measured in grains) and the velocity (usually in feet per second). The easiest way to do the calculation to go to a table on this web page: http://www.naaminis.com/energy.html . Or, multiply the mass in grains by the square of velocity in ft/sec and divide by 450450 . For my example, I am shooting 115 grain JHP ammo and the muzzle velocity in my carbine is 1200 ft/sec. So my bullet energy is 115 x 1200 x 1200 / 450450 = 368 foot-pounds.
The calculation of how long it takes the bullet to exit is a little trickier and you have to estimate. If you assume that most of the acceleration of the bullet happens in the first four inches, then you can approximate the average velocity in the barrel as slightly less than the exit velocity. In my case, I'll estimate 1,000 ft/sec. The length of my barrel is about 16" or 1.33 ft. So it will take 1.33 milliseconds for the bullet to get out, 1.33 ft / (1,000 ft/sec) = 0.00133 seconds.
Finally, I get my estimate of average power, 368 foot-pounds / (0.00133 seconds) = 276692 foot-pounds/sec or 503 horsepower. As I said, the instantaneous value just after ignition will be higher than that and the instantaneous value as the bullet exits will be lower. We are only calculating the useful power going into the bullet. The waste power going into heating the barrel is about ten times that much.
Before you get excited about how much power that is, remember it's only for a very short time. Once the bullet is out of the barrel, the power is zero until you take the next shot.
Now, I want to say something about momentum. Newton's third law, which physicists have never found to be violated, says that the force applied to the bullet to shoot it out the barrel is equal and opposite to the force that the bullet kicks back against the gun and you holding it. Again, it is easier to look at the situation of a bolt-action rifle. The bullet starts out with zero momentum and ends up with momentum equal to its mass times velocity. The laws of physics state that the average force on the bullet multiplied by the time the force is applied (this is called "impulse") is equal to the change in the bullet's momentum. This in turn is equal to the average force pushing back on the gun and you during that time. Therefore, the best measure of how hard the shot kicks back is to multiply the mass of the bullet (you can use grains) by its muzzle velocity. By this estimate, a 173 grain 30-06 bullet going 2750 ft/sec will kick back about 3.45 times harder than my 9mm carbine. It will also hit the target that much harder, neglecting loss of speed in air.
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In your recoil calculation you forgot to factor in the mass / velocity of the remaining ejecta (the powder gas itself) the powder gas is traveling at about 5200 fps. Add the energy of the powder gas to the bullet energy. Same rifle and bullet weight and muzzle velocity, but the 2 different powders, the load that uses more powder will kick harder
Some guys like a mag full of lead, I still prefer one round to the head.
Edited by - Tailgunner1954 on 04/13/2002 14:13:480 -
Good point, tailgunner. I did mention the waste energy in the earlier discussion, but I assumed we are interested only in the energy of the bullet as "useful" energy. How much momentum do you figure the exiting gas carries? It's going faster than the bullet, but the mass is low. 0 -
Gordian,I am some kind of impressed with that answer.In good old South eastTexas.We would say it will blow the he&l out of em.Thanks for taking the time with a educated answer.
A unarmed man is a subject.A armed man is a citizen.0 -
A big A+ to Master Blade.
(Now, can you tell me how to get rid of this flu?)0 -
((8.5gr/7000)*4000^2)/450450 = .04 ft/lb (I lowered the gas velocity due to the bore / case ratio of your camp carbine.)
for a -06 however it works out a little higher
((60gr/7000) * 5200^2)/450450 = .51 ft/lb. 0r 12.75 times more gas energy in a 06
Have my math skills gone the way of my memory and I just calculated energy ? (yep)
8.5/7000*4000=4.88 ft/lb momentum
60/7000*5200=44.57 ft/lb momentum Or 9.13 times more gas momentum.
Help me out here Gordian, It's been to many years for me.
Some guys like a mag full of lead, I still prefer one round to the head.0 -
Your absolutley correct Gordian
My eyes did glaze over0 -
competentone, I have a sure cure for the flu. Drink a lot of cool fluids (not plain water, you need the electrolytes) and take acetaminophen (Tylenol). Suck a zinc lozenge a couple of times a day to help prevent further infection of the nasal passages and throat. Take cough syrup as required. Eat light meals, lots of fruit. Get plenty of rest. Stay warm. If you follow this plan, you should be better in 7 days. If you don't, it will take about a week to get better.
tailgunner1954, let's see if I understand your calculations. We are trying to account for the high-speed jet of gas escaping from the barrel, as this carries off waste energy and momentum that is doing us no good. To do this, we need to estimate the mass of the gas. It's easiest if we use the same units that we used for the bullet, that way we can understand the loss of energy and momentum as a fraction of the useful energy and momentum in the bullet.
My formula for energy assumes that we are measuring mass in grains, so we don't need to convert it to pounds by dividing by 7000.
For my 9mm carbine, as I understand what you wrote, you are estimating the mass of the gas as 8.5 grains exiting the muzzle at 4,000 ft/sec. This gives an energy of 8.5 x 4000 x 4000 / 450450 = 302 ft-lbs. This is almost as much energy as goes into the bullet, which was 368 ft-lbs. The gas has a kinetic energy 82% of the bullet's energy. That isn't surprising. I think even more energy than that goes into heating the barrel and the gas itself.
The momentum can be measured in many different units, the easiest in this case is grain-ft/sec (mass x velocity). The momentum of my 9mm bullet is 115 * 1200 = 138,000 grain-ft/sec. The momentum of the hot gas jet is 8.5 x 4000 = 34,000 grain-ft/sec. So this increases the kick of my carbine by about 25%.
Now for the 30-06. We have a 173 grain bullet going 2750 ft/sec. Energy = 173 x 2750 x 2750 / 450450 = 2904 ft-lbs. This is a lot more than the 9mm, not surprising. The momentum of the bullet is 173 x 2750 = 475,750 grain-ft/sec. For the gas, you give a mass estimate of 60 grains and a velocity of 5200 ft/sec. Gas kinetic energy is 60 x 5200 x 5200 / 450450 = 3602 ft-lbs. This is more than the bullet, or 124% of the bullet energy. The momentum is 60 x 5200 = 312,000 grain-ft/sec. This is about 66% of the momentum of the bullet, which increases the kick accordingly. So for a 30-06 compared to 9mm, the hot gas carries a lot higher fraction of the energy and momentum.
Whew, this sure gets complicated! Just for fun, let's calculate the kinetic energy in the action of my Marlin Camp Carbine versus a Browning BAR High Grade Standard in 30-06. Just kidding, let's not.0 -
How much of the energy was lost during the expansion of the bullet upon target impact?
AlleninAlaska aglore@gci.net0 -
Sorrey Capt Kirk. At least I didnt use a BB gun as an example.
Hay how many times will I need to pump my pellet gun to get 1/4 Horsepower?
I would like to thank the members for the well thought out and researched information. I really liked the twin Holley instalation. By the way where can I get a tunnel ram for an M1?
Seriously thanks TOOLS0 -
Thats a easy one Al, just give us the bullet weight and velocity at both the point of impact and exit from the target.
Gorden: thats where I messed up, converting grains to pounds
Test for the class: calculuate the numbers for a 400gr bullet @ 2400fps over 100gr of powder.
*Just kidding folks*
Some guys like a mag full of lead, I still prefer one round to the head.0
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