Math equations

Dmaxboy08

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Hey guys. I have a question regarding a math equation and I was wondering if someone was smarter than my 23 year old brain who could shed some light for me on figureing out how to figure out how much force a bullet has on impact. For example, if you are shooting a .45acp 230g round nose bullet that travels at 800fps, how could you figure out what the force of impact is? Thanks.
 
Here's what I do to keep it simple without having to convert any of the variables used.

Grain x velocity squared / 450436

I got 450436 by multiplying two times the acceleration of gravity (I used 32.174 fps) by 7,000 which is how many grains are in a pound.

Example: 255 grain at 1000 fps

255 x 1000 x 1000 / 450436 = 566 ft-lbs


Another example: 158 grain at 1450 fps

158 x 1450 x 1450 / 450436 = 737.5 ft-lbs

EDIT: It's been a while since I used this.. I double checked with calculators and they come out right
 
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Hey guys. I have a question regarding a math equation and I was wondering if someone was smarter than my 23 year old brain who could shed some light for me on figureing out how to figure out how much force a bullet has on impact. For example, if you are shooting a .45acp 230g round nose bullet that travels at 800fps, how could you figure out what the force of impact is? Thanks.

You're asking for the mathematician's approach to answering this question, which is way more troublesome than an engineer's solution, which is to simply look up the answer in a published table, such as are easily accessible online --- Federal Cartridge has a good one... (bullet shape, i.e., ballistic coefficient, is next to irrelevant at handgun ranges...).
 
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Hmmmm...I was always under the impression velocity decreased as distance increased.

Density altitude, or density of air mass....Humidity....A bullet will travel faster at Flagstaff Arizona, than it will at Miami Florida...

Force of impact??? Would not the object being hit have consideration of the impact...A 4 inch piece of solid steel will not give to the impact...Whereas a bullet hitting 40 inches of say jello, will give to the impact. Therefore, would not the softer material have a less of a impact?

Reverse my thinking...Would I want to get hit with a steel bar, or a bar made of foam...Which will hurt me more?


WuzzFuzz
 
The equations given above are for the kinetic energy of the bullet. Energy is the ability to do work (which is force x distance).
These equations do not give the "force of impact", they give the maximum energy that can be provided by the bullet if all the energy is dissipated in the target..
 
Now, how do you prove this?

It is just a variation of the kinetic energy equation adjusted to give the answer in ft lbs. However, the OP asked what the FORCE of the impact would be, which is actually a different question.
 
The equations given above are for the kinetic energy of the bullet. Energy is the ability to do work (which is force x distance).
These equations do not give the "force of impact", they give the maximum energy that can be provided by the bullet if all the energy is dissipated in the target..

Good observation. If we were calculating force, the answer would be in lbs. (only) and not if ft-lbs.
That said, I'm not sure the OP is clear on what he wanted. Generally speaking, energy is the accepted measure of on target capabilities and ballistics.
 
Yeah, my equation is for ft lbs at muzzle.... Or any point in the bullets trajectory if you knew the velocity. It's the same info you'll find on the back of some boxes where they give you the ft lbs.
 
Force = mass * acceleration. Acceleration is the first derivative of velocity wrt time but im thinking it can also be roughly calculated as the velocity divided by the time it took to stop the bullet. It's not zero but say in ballistic gelatin would be the time it took from when it entered to when it stopped. The best way would be to measure the velocity at impact using a chronometer or even from the video taken of the bullet's entrance. Maybe I'm over thinking.
 
F=MxV, where force = mass times velocity. Thats all i got for ya.

Not quite-mass times velocity is momentum, which is NOT a force.

The force is equal mass times acceleration. Alternatively, if you want to think of the force in terms of momentum, it's equal to the time rate change of momentum over the change in time(F=dp/dt), or treating mass as constant(probably close enough for our purposes) F=mdv/dt.

Depending on how you want to model the impact of a bullet, the math can get fairly complicated fairly fast. Knowing the velocity as a function of time or displacement in the target would be a good first step to calculating the force at any given time after impact. If you knew the force as a function of time, you could integrate it from v=initial to v=0 to get kinetic energy.
 
The equation you are looking for is the impulse-momentum equation. Impulse is defined as force x time, momentum is mass * velocity. The change in momentum is equal to the impulse:

force * time = mass * change in velocity

Solving for force,

force = (m * dv)/t

The trick is to determine how quickly the bullet stops, high-speed photography would be helpful. If the bullet were to hit a solid steel plate, it would stop very quickly, i.e. very small time. The force could potentially be very large. If you are shooting into ballistic gel, the bullet stops more slowly (larger t) and the force is therefore smaller.
 
The equation you are looking for is the impulse-momentum equation. Impulse is defined as force x time, momentum is mass * velocity. The change in momentum is equal to the impulse:

force * time = mass * change in velocity

Solving for force,

force = (m * dv)/t

The trick is to determine how quickly the bullet stops, high-speed photography would be helpful. If the bullet were to hit a solid steel plate, it would stop very quickly, i.e. very small time. The force could potentially be very large. If you are shooting into ballistic gel, the bullet stops more slowly (larger t) and the force is therefore smaller.

Yes! However integrating that over time would show the entire force of impact. In gel it's just spread out but more easily measured... on a steel plate you might need a rotating mirror camera to do that.

Nice description :)
 
m v = f t
Mass x velocity = force x time
momentum = impulse

The force of impact depends upon the time of impact. If the bullet hits a steel plate (small time of impact) then the force is great. If the bullet hits a bale of cotton and takes a relatively long time to come to a stop, then the force is small.

That's why the force of impact catching a baseball can vary. If hold your arms stiffly and stop the ball quickly the large force stings your hands. If you let your hands move backward while catching the ball the impact time is longer and thus the force is smaller.
 
i wish i knew about this forum when i was in trade school.
my take on the math question is: shoot the biggest slug you can get airbourne, just to make sure.or shoot it twice.
 
+P

Thank you very much.

Pay back for the headache :p

My math skills have gone the way of my sentence diagrams. :(
 
(cα ⋅ p̂ + Βmc2) ψ = iℏ δψ/δt

I think that's the one. :)

Can't be. All those tricky equations have factors of Pi in them.:D

Books that go back to first principles can be handy when needed, like you are short on sleep.;) The rest of the time I need quick fixes. As I once told somebody, "If the explanation descends into the realm of multiple nested integrals on a page, I KNOW somebody is trying to snow me."
 
(cα ⋅ p̂ + Βmc2) ψ = iℏ δψ/δt

I think that's the one. :)

Any equation that includes two Chief Bosun Mates (squared chief Bosun Mate? ) has to be the right one !

This from a former Sonar Tech - we alternated between Deck and Weapons depending on the whims of somebody in charge. BMCs mostly seemed to have their heads on straight.

Now, back to the math lesson of the day.
 
WAy cool thread!!

In gelatin, one could know the fps of the bullet at initial impact, and using the distance the bullet took to stop, figure out the rate of (de)acceleration.

Would that be useful?
 
WAy cool thread!!

In gelatin, one could know the fps of the bullet at initial impact, and using the distance the bullet took to stop, figure out the rate of (de)acceleration.

Would that be useful?

That would be a starting point, but it's making the assumption that the rate of deceleration is linear with respect to penetration. If there's any expansion or other deformation of the bullet, this would not be a valid assumption.
 
Oh man, this could be some kid's masters thesis eh? At least it seems to me it could, being 1/2 caveman and all.
 

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