Momentum of a .308 bullet

[Laserslug]

Momentum, force, and pressure of a .308 bullet, 146 grain

momentum has units of pound seconds ( b = mv )

in English
1/3 oz x 2400 ft/s = 1/48 lb x 2400 ft/s = 2400/48 = 50 lb sec

time for bullet to smash = 1 foot/12 inch x 1/2 inch / (2400 ft/s) =
.5/(12 x 2400) = 17 microseconds

force = b / time = 50 / 17 us = 2.9 million pounds

pressue = force/ area
area = .308 x .308 = .1 sq inch
2.9 MILLION LBS / .1 SQ INCH = 29,000,000 PSI
The pressure is 29 million pounds per square inch while the bullet is smashing.
___________________________________
notes on metric units to double check :
b = mv = 9g x 775 m/s in metric = 7kg m/s = 48 lb sec

 

[Freedom_fighter_in_IL]

LMAO, dude, you need to go back to math class.

 

[ChronoCube]

Nice calculation. I think the momentum value is a better representative of the bullet's power than the pressure is. I'd like to see similar figures for other calibers too.

 

[Norrick]

nothing wrong with your math, but its not an incredibly useful calculation.

That would be valid if the bullet had a blunt front end. The cross section you've used is for a spitzer shaped bullet. It would probably have to be a rigid bullet as well, alot of the energy is lost to bullet deformation which changes the cross section area, and since the bullet loses of mass, and velocity during the impact, the pressure would not be constant either (remember momentum is m*v).

 

[Freedom_fighter_in_IL]

Ok I will change what I said to "Dude you need to learn to apply real world physics to this subject". To put it as simply as it can possibly be put, if a bullet hit an object with 29 million ppsi it would completely obliterate ANY fluid matter that it came into contact with. This is what is so problematical with trying to apply paper formulas in REAL WORLD application. Your average .308 cartridge load IMPACTS at 100 yards with an AVERAGE of 1500-2000fp or less. You have to, As Norrick said, take into account the cross section of the bullet as well as the given mass and velocity.

 

[Laserslug]

Pounds x Seconds is Momentum

Yes, you are all right. The numbers are for a simplified model of the bullet 1 foot from the muzzle for a wadcutter made of tungsten hitting a diamond as large as a boar. But still, the units of momentum are pound seconds.

50 pound seconds of momentum for a .308 are lost in less than 1 second. That momentum is lost in about 17 microseconds.

 

[unwise11]

I'm sorry to be abrupt.. Why did this thread get made? Are you a math major?
I'm just confused as to what this is to explain..
So what does all this complicated stuff mean?

 

[Norrick]

momentum is velocity times mass

velocity is feet per second in your example... we could alternatively use meters per second, miles per hour, anything you wish...

mass is pounds mass (as opposed to pounds force, two very different things)


if you multiply feet per second by pounds mass, you get pounds mass and feet on the numerator, but the per second still exists in the denominator. The "feet" units do not simply disappear, and if they did you would still incorrectly have pounds per second, not pound seconds.

Dont get too worked up, just look it over. Multiply the units out as if they were fractions. Cancel the terms that are shared on the numerator and denominator (there are none in this case).

 

[unwise11]

Well, this is beyond an 8th graders comprehension and I'm in advanced math... so..
Yeah.

But cool, I have a .308 that rocks.

 

[Ignition Override]

How about energy at 200 yards, or even 400, if not further?
A guy on Youtube depicts energy for several rifles on his numerous videos. Just search for "420 yards".

Does any website display a simple comparison chart, whether in ft. lbs, or velocity, between an Enfield bullet: 150 grain at about 2,400 fps, after about 600 yards, and a Garand (or other rifles') 150 grain bullet at about 2,800 fps?

Any other examples with objects of similar weights (no matter what) are always interesting to check. The bullets will probably be different weights or shapes etc, not just the powder loads.

 

[MCMXI]

First off, there's a reason why ballistic calculators output an energy term which is kinetic energy. Because INSTANTANEOUS kinetic energy can be calculated easily.

Kinetic Energy = 1/2*mass (lbm)*velocity^2

In the OPs example, the INSTANTANEOUS kinetic energy is as follows:

K.E. = 0.5*(146gr/7000gr)/(32.2ft/sec^2)*((2,400 ft/sec)^2) = 1865.5 ft-lb

momentum has units of pound seconds ( b = mv )

No it doesn't. There's a mass term and a velocity term so the units are (lbm*ft)/s where lbm is pounds mass and not pounds force (lbf).

in English
1/3 oz x 2400 ft/s = 1/48 lb x 2400 ft/s = 2400/48 = 50 lb sec

If you account for g (32.2 ft/s^2) which you need to do since pound force (lbf) is not the same as pound mass (lbm) for a bullet in flight, you can calculate the INSTANTANEOUS momentum. Velocity is a non-linear function regardless of whether the bullet is in flight or impacting a target. For a bullet in flight, you can approximate the velocity function to be linear over a short period of time such as 17 microseconds but for a bullet hitting a hard target that model just won't work.

INSTANTANEOUS Momentum = mass * velocity (instantaneous)
= (146gr/7000gr)/(32.2 s^2/ft)*(2,400 ft/s)
= 1.55 lb-ft/s

time for bullet to smash = 1 foot/12 inch x 1/2 inch / (2400 ft/s) =
.5/(12 x 2400) = 17 microseconds

So you're calculating how long it takes for the bullet to travel 0.5" and calling this "time for bullet to smash". OK, so you calculated the time for a bullet with a velocity of 2,400 fps to travel 0.5" but how did you come up with 0.5" and what does it mean?

force = b / time = 50 / 17 us = 2.9 million pounds

force = mass * change in velocity with respect to time
force = mass * dV/dt

So basically, you're trying to sum up the total force exerted by the bullet in 17 microseconds. You can't do this without calculus and knowing what the non-linear velocity function is. You'd also have to model the bullet as having a meplat equal to the O.D. of the bullet.

pressue = force/ area
area = .308 x .308 = .1 sq inch
2.9 MILLION LBS / .1 SQ INCH = 29,000,000 PSI
The pressure is 29 million pounds per square inch while the bullet is smashing.

Again, pressure is a function of force which is a function of dV/dt. This simply doesn't work.

 

[barnetmill]

I see that you use 1/3 oz that is the bullets weight. Did you convert your weight or force units to mass for your calculations?
I see some one else has beat me to it.

 

[Freedom_fighter_in_IL]

Well, this is beyond an 8th graders comprehension and I'm in advanced math... so..
Yeah.

But cool, I have a .308 that rocks.

Don't worry son, I went through Physics 4 in college MANY moons ago and I STILL have to think hard to make sense of half this stuff.

I'll just be content with letting 1858 explain the hard stuff. I'll stick to keeping it simple!

 

[Laserslug]

You are correct that a pound is a force, not a mass. That is why in the Original Post I showed the metric numbers at the end :

notes on metric units to double check :
b = mv = 9g x 775 m/s in metric = 7kg m/s = 48 lb sec ~= 50 lb sec sanity checked

I will research the rigorous conversion of pounds force and pounds mass. (The English "slug" is the mass unit, as I recall.)

 

[Laserslug]

I checked some references and my statemnt is correct that momentum has units of pound second in the Imperial system and newton second in metric



newton second is metric momentum , pound second is imperial
impulse adds to momentum
http://www.unc.edu/~rowlett/units/dictN.html

pound mass is defined here
http://en.wikipedia.org/wiki/Pound_(mass)
The pound mass is correct to use in b = mv momentum where m is 1 pound of mass
The pound of force is correct in the units of momentum = pound second
That is why the whole world uses Metric and not imperial units, except the USA.
I intentionally used the archaic Imperial units here in THR because that is traditional here.
I also showed the metric version in the OP as a sanity check. I am right and you are wrong.

 

[MCMXI]

I checked some references and my statemnt is correct that momentum has units of pound second in the Imperial system and newton second in metric

The problem is that you need the mass of the bullet to calculate momentum or kinetic energy for a bullet in flight. Many years ago, it was understood that gravity accelerates everything downward at a rate of 32.2 ft/sec^2. Since all calculations on this planet face the same issue, it was decided to "bury" the acceleration into the definition of pound force (lbf).

1 lbf = (1 lbm * 32.2 ft)/sec^2

Similarly, the definition of a Newton in the S.I. system of units is

1 N = (1 kg*1m)/sec^2

So basically, the only thing you got right in your first post was the time it takes that particular bullet to travel 0.5" and even then, you brushed over the fact that the velocity over 0.5" isn't constant. For a bullet in flight it can be treated as a constant, but you use the same reasoning when it hits a target which is incorrect. Everything else is wrong and frankly, meaningless.

If anyone is interested in calculating bullet energy, instantaneous kinetic energy is easy to calculate. This it what every ballistic program I've ever seen puts out. If you calculate kinetic energy, you will soon discover that you need to account for g with 32.2 ft/sec^2. Don't be fooled into thinking that this is entry level math. Calculating the force of a bullet on a target as a function of time is definitely NOT simple or trivial.

As the saying goes, a little knowledge is a dangerous thing.

 

[Zak Smith]

Cool story bro.

One other problem not mentioned yet

pressue = force/ area
area = .308 x .308 = .1 sq inch

Might want to give your geometry teacher a call.

Unless you're shooting square bullets 0.308" on a side, the area is pi*r^2, or 3.14*(0.3/2)^2 = 0.07 in^2

 

[Freedom_fighter_in_IL]

Zak, step on my fingers and keep me nice for a change please

This guy is cracking me up. I mean seriously, 29 MILLION pounds per square inch? That would basically be like the space shuttle falling 4 stories and landing on a mosquito? The temporary cavity created by such an impact would end up being not so temporary! You hit a deer with THAT much energy and, to put it mildly, there would be no deer left!

 

[MCMXI]

If anyone has taken a chemistry class or any kind of engineering class, they will have encountered dimensional analysis. One of my college professors once told me " units are your friends". Many students HATE units but they keep you honest.

When you consider that the momentum equation is simply a special case of Newton's Second Law, the units DO NOT change regardless of whether you're using the S.I. system or the Imperial system.

F=m*(dV/dt) = m*a

Zak, I noticed the error in calculating the area of the bullet but given the fundamental errors in his thought process, it didn't seem that important. On a side note, MEs use (PI/4)*d^2 because shafts, bolts, rod, pins, cylinders etc are given with diameter measurements. Math teachers use PI*r^2 ... and maybe EEs.