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Old May 20, 2008, 08:02 PM   #1
pistol-whipped
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How do you figure a bullet's force on impact?

You know on the box of hunting ammo you get at your local walmart,it has that chart that shows you the bullet's ft/lbs. How can I do that for my hand loads?
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Old May 20, 2008, 08:53 PM   #2
joneb
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divide the velocity by 1000 and square it, then multiply that by the bullet weight, and then multiply that number by 2.219 this number is the energy in FT/LBS
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Old May 20, 2008, 09:02 PM   #3
roy reali
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I Dont Think So

I did not think that kinetic energy is the same as force. Otherwise when you hit a 100 pound deer at one hundred yards with a 7mm magnum round the animal should be put into low earth orbit.
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Old May 20, 2008, 09:46 PM   #4
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mass times velocity squared. remember your bullet weight is published in grains not pounds you must convert.
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Old May 20, 2008, 09:51 PM   #5
SL1
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pistol-whipped,

Jibjab's formula gives the correct transformation of a bullet's mass and velocity values to the kinectic energy value. But it is a little unclear what that formula comes from. It is an algebraic simplification of the fundamental formula for kinectic energy, which is 1/2 x mass x velocity-squared. Jibjab's formula contains constants of 1/7000 to convert the bullet's mass from grains to pounds and 1/32.2 to convert those pounds to "slugs" (which are the English system units of mass, although most people don't realize that). The answer (for bullet weight in grains and velocity in feet per second) comes out in foot-pounds. Of course, if you were doing this for a shotgun load with the bullet mass in onces, you would need to divide by 16 instead of 7000 to get pounds.

One fact to remember is that you need to reduce the velocity from the value at the muzzle to the value at the target to get the energy that strikes the target. That depends on the target's range and the bullet's ballistic coefficient. There are tables and computer programs that can give you the velocity reductions.

But, you need to know your starting velocity for your handloads. The reloading manuals give you values for teh muzzle velocity OF THE TEST GUN, but yours may differ by as much as a couple hundred feet per second. You really need to shoot your loads over a chronograph to get the answer to your question for your own handloads.

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Old May 20, 2008, 11:05 PM   #6
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Quote:
Jibjab's formula gives the correct transformation of a bullet's mass and velocity values to the kinectic energy value. But it is a little unclear what that formula comes from.
SL1, Are you making fun of my garage mathematics ? That 2.219 # can be a little more refined
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Old May 21, 2008, 08:25 AM   #7
Jim Watson
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You can also figure momentum, mass times velocity, which some think a better indication of performance.

If you are using the word "force" in a strictly scientific sense; that is calculated as mass times acceleration; or in this case, mass times deceleration as the bullet slows or stops as it penetrates the target. Figuring the deceleration would be difficult if, like me, your physics was limited to idealized class exercises and sterile lab experiments.

Every gunwriter on the planet has his own figure of merit for bullet effect on impact. Hatcher Factor (two different approaches), Hatcher as modified by people trying to make hollowpoints look good, Taylor Knock-Out, something called PIR, and I don't know what all else. Few come out in scientific units, but if one of them agrees with your observations, it might be useful
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Old May 21, 2008, 08:32 AM   #8
SL1
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Quote:
Quote:
Jibjab's formula gives the correct transformation of a bullet's mass and velocity values to the kinectic energy value. But it is a little unclear what that formula comes from.

SL1, Are you making fun of my garage mathematics ? That 2.219 # can be a little more refined
Not fun intended

Your formula is the quickest way to make the calculation, so that is the best way to actually do it. But, it looks like some sort of deep magic to people who do not understand the basic formula and where the numbers came from. On another thread, I found a very experienced person who did not understand the 1/32.2 factor, which is the acceleration due to gravity or "g" value. He had written that he was not clear on why gravity came into the kinectic energy formula, since it was acting in a direction perpendicular to the bullet's flight path. So, for this thread for a new guy, I thought it might be helpful to get the fundamental formual written out and explain the constants that end up giving you the 2.219 multiplier and the 1000 divisor.

Lecture over, NOW let's have fun

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Old May 24, 2008, 12:43 AM   #9
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??? I thought we WERE having fun! Gotta love the internet for it's ability to rake in a broad range of experience!
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Old May 24, 2008, 06:15 AM   #10
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I’m sorry Guys,, That’s just to diffi-colt for me at my advanced age.
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Old May 24, 2008, 11:25 AM   #11
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I found a very experienced person who did not understand the 1/32.2 factor
A lot of folks do not realize that weight is a force and depends on gravity, while mass is a property of an object.

A spring scale weighs things, and would not read correctly on the moon.
A balance compares masses and would work just fine.

It is possible to estimate the force a bullet could produce when striking an object, but you need to know how the bullet slows down as it penetrates.
Acceleration is the change in velocity.
If you make assumptions about the distance the bullet travels in the object and assume it is linear you can extract the average force exerted over the distance.

A more exact calculation is very similar to computing bullet drop, and requires detailed estimates of the slowing forces acting on the bullet (the thing we are trying to determine in this case).
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Old May 24, 2008, 11:56 AM   #12
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At impact?

"ballistic pendulum"

IMO all the math calculations for figuring "energy" aren't worth the you-guessed-it energy.

The Taylor TKO "formula" often mirrors real world results, but who cares (except math geeks and certain state Fish & Game rules-setters).

To answer the OP query, use the formula given for calculating energy.
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Old May 24, 2008, 12:31 PM   #13
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Quote:
The Taylor TKO "formula" often mirrors real world results,
Yes and no. The TKO formula favors heavy, large caliber bullets, the kinetic energy formula favors faster bullets, the momentum formulas show better results with heavy bullets. Pretty much everyone who comes up with one of those formulas designs it so that it applies to their favorite idea (notice I didn't say "theory", because a theory has to be supported by evidence).

If you use a TKO calculator ( http://www.handloads.com/calc/quick.asp ), you will notice it favors large caliber, heavy bullets and downrates lighter, smaller, high velocity rounds. So you get ratings like

44 Mag 240 gr @ 1,300 fps
TKO= 19

7X57 140 gr @ 2750 fps
TKO= 15

which would suggest that a 44 Mag out of a handgun is a "better" hunting round than the 7X57 out of a rifle. Check the results for any large caliber round, and it will typically outperform smaller caliber rounds on the TKO scale.

Whether you agree with that or not, I would challenge you to take the firearms out and shoot a few rounds at 300 yds and see which one you would rather hunt with. Real world experience says the faster rifle round will outperform the slower handgun round every time. Otherwise we would still all be shooting 60-caliber Tower muskets (that's a "Brown Bess" to some of you).
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Old May 24, 2008, 12:48 PM   #14
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exactly

Why I don't care about "formulas"; they meet the need of the poser (or advocate), often disregarding real world results.
Just math.

Please note I said I consider these 'formulas' worth less than the energy required to figure them.
But I am 'aware' of them LOL.




Funy; the only rifle I ever owned was chambered in 7x57.
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Old May 24, 2008, 05:17 PM   #15
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Physics Allert

Warning Warning This post contains Physics and Math. Some people are offended by math. If you are one of those go on to the next thread now or, read only at bedtime, as drowsiness is a known side effect of Physics

The force of impact of a bullet is not a simple problem. It is governed by the concept of momentum, not energy. The impact force, called ‘impulse’ in physics is simply the change in the momentum of the projectile divided by the length of time that the collision took.

As an example I will use a 150 gr. Bullet traveling at 2000 ft/s, which impacts a deer and penetrates 12 inches. These will be converted to SI [you know as metric] since it is easier than converting to the English equivalent of ‘slugs’. The bullet mass is 0.00972 Kg. and the velocity is 609.6 m/s. The bullet has a Kinetic energy of 1806.172 joules and a momentum of 5.925 Kg m/s. On impact the bullet will be brought to a complete stop in 0.001 seconds [12 inches or 0.3048 meters]. The force of the collision [impulse] will be 5125.212 Newtons. [One Newton is approximately the force that a McDonald Quarter Pounder exerts on your hands as you lift it to your mouth]

While this is a pretty good wallop, it is a far cry from the 1332 ft lbs of energy which, if applied correctly, would lift a 300 lb deer over four feet into the air.

If the bullet passes through the deer, the problem becomes more difficult.

Aside: The energy of a bullet is important in determining the amount of work necessary to get the projectile going, but on impact most of it will be converted into heat. The same thing happens if it does not encounter an impact. The air will be heated by its passage.

OK, time to wake up. I hope you enjoyed the nap. Yes, I do expect that there will be some who object to this analysis. Thirty-seven years of presenting this and other Physics concepts taught me to expect disbelief on first exposure.
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Old May 25, 2008, 12:21 AM   #16
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soporifically speaking

Does the composition or density of the impacted media enter the formula?
Does the (potential) deformation of the projectile enter the formula?
Construction of projectile?

Oh, forget it; I'm goin' back to sleep.


Besides, my favorite projectile is my Subaru.
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Old May 25, 2008, 08:01 AM   #17
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I recall that it is momentum (M * V) that is conserved in a collision, not Kinetic Energy (M * V * V).

It is far easier to increase KE a lot than it is to increase ME a little. Shill gunwriters have to write articles selling products, and it is far easier to shill an increase in KE. Just an increase in 10 fps will increase KE by 100. An increase in 100 fps will increase KE by 1000. Shills just make that increase seem galactic by proportions.

This is the basis for the sales campaign for the 327 Federal. Drop bullet weight, increase velocity, and claim you have something as effective as a 357.

You would think after reading all the garbage put out by shill gunwriters that anything under 3000 fps is totally ineffective, but 16 inch Battleship shells flew around 2000 fps. I heard the New Jersey’s shells made 60 foot divots in the soil of Lebanon.
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Old May 25, 2008, 10:22 AM   #18
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Energy= Velocity x velocity x bullet weight/ divided by 450240
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Old May 25, 2008, 11:04 AM   #19
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Impulse is the integral of force and time, and as such is NOT purely the force of an impact (in say pounds).
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