Velocity = Energy


PDA






Lucky Jim
September 9, 2003, 09:29 AM
I have been shooting 165 grain cast bullets in my 30-30 at a local silhoutte match. Most of the guys use 180 grain bullets. I use the 165 grain bullets simply because I do not have to seat the bullet with the gas check past the case neck. Some of the old time shooters were telling me the 165 grain bullets would not knock down the rams at 200 meters. I have been hitting and knocking them down. That is to say when I hit one it will fall about 90% of the time. I am shooting at about 1800fps (I do not have a chronograph) using what reloading info I have. My question is - If the old time shooters are using 165 grain bullets say at 1200 to 1500fps and I am using the same bullet at 1800fps does the fact that I am firing the bullet at a higher fps translate into enough energy at 200 meters to knock down the ram where the slower bullet will not?

If you enjoyed reading about "Velocity = Energy" here in TheHighRoad.org archive, you'll LOVE our community. Come join TheHighRoad.org today for the full version!
Steve Smith
September 9, 2003, 10:27 AM
Kinetic Energy
Kinetic energy is a function of the bullet mass and motion. This motion involves the velocity and rotations. It is represented by the following equation.

KE = 1/2 M v2 + 1/2 I w2
NOTE: this assumes a single rotational axis and forward motion of the bullet. For multiple rotations and non-symmetric bodies, things get a little more complicated. For most bullets, the contribution of the rotational kinetic energy is on the order of a few foot pounds and is neglected, leaving the following equation for energy of the bullet:

KE = 1/2 M v2
where the mass is calculated as above.

So for the same 300 grain bullet, traveling at 3000 ft/sec, the energy (neglecting potential and rotational kinetic energy) is:

KE = 1/2 M v2 = [0.5] [0.00133 lb sec2/ft] [3000 ft/sec]2 = 5985 ft lbs
If we assume units of grains for the bullet weight and feet/sec for the speed, we can combine the constants to get a simplified equation:

KE = M v2/450800
where

450800 = 2 * 7000 * 32.2
Variables
PE potential energy KE kinetic energy
g vector gravitational acceleration M bullet mass
h bullet height v bullet speed
I moment of inertia about rotation axis w rotation speed

Jim Watson
September 9, 2003, 10:53 AM
And it is usually taken that momentum = MV
is a better measure of knockdown of a silhouette target than kinetic energy.
So yes, more velocity means your bullet hits the target harder.
There is considerable discussion in BPCR whether a 200 grain .30 caliber could be gotten up to 2000 fps with black powder, and if so, whether it would knock the targets down (Rams at 500m.)

Steve Smith
September 9, 2003, 10:57 AM
But MV alone doesn't do it. MV does not equal momentum. Mass increases momentum exponentially, while velocity only acts as an enabler.

You also have to keep bullet placement in mind. If you hit the top of the ram, it will be easier to knock over than if you hit the bottom.

MoNsTeR
September 9, 2003, 11:59 AM
MV does not equal momentum.
As far as I can tell, yes it does.
http://www.hazelwood.k12.mo.us/~grichert/sciweb/formulas.htm

Steve Smith
September 9, 2003, 01:01 PM
Velocity will increase momentum, yes. However, if you increase the mass of the object, you will increase the momentum more. Think if this in practical terms. What is more likely to penetrate through an elk, a 40 grain .22 magnum at 1800 fps, or a 330 grain .45 at 1200 fps?

Mike Irwin
September 9, 2003, 01:15 PM
If muzzle velocity equaled momentum, then a 124-gr. 9mm slug moving at 1,300 fps should be very nearly the equal of the 230-gr. .45 ACP slug at 850 fps when it comes to knocking down a bowling pin.

Unfortunately, it's not.

The 9mm simply stinks when it comes to clearing bowling pins.

One trick used by shooters to figure a load that will clear the intended target is to look at the specs for a round that will knock down the intended target 100% of the time with a solid hit, and then multiply the velocity times the bullet weight and divide by 100. Power factor, I belive it's called.

For the two rounds I noted above:

.45 - 850x230=195,500/100=195.5 (200 is considered optimal by many pin shooters)

9mm - 1300x124=161,200/100=161.2, or well below the optimal performance level that experience has shown to be good for bowling pins. That likely explains why the 9mm stinks at this sport.


Running the numers that you're showing, though, gives an interesting result... Your load at 1800 fps should be better at knocking down the rams at 200 yards than the 180 gr. load, at least at the muzzle.

What is happening with the bullet at the 200 yard mark, though, is open for speculation. Is it dropping enough velocity that it becomes marginal, where as the heavier bullet maintains just enough velocity to make it more effective?

Interesting situation.

Norm Lee
September 9, 2003, 02:19 PM
This seems to be drifting for some reason. Momentum (P) is the linear product of mass and velocity. P=MxV I think we started at one point here to think MV = muzzle velocity.

If velocity goes up 10%, so does momentum. If mass increases 10% momentum increases 10%. Increase mass and velocity each 10% and.....


Power factor is just momentum without the rigor of working the units.

When a projectile strikes a bowling pin the mv of the projectile becomes or is transformed into MV and mv=MV The new M is the mass of pin and projectile and we can calculate now the new V. Much more difficult trying to work things from an energy viewpoint since the energy can go into heat, light, etc in ways we won't be able to predict.

Cheers,

Norm

Steve Smith
September 9, 2003, 02:33 PM
Norm, I don't know why I'd lost that simple forumla, but you're right. P=M*V

renaissance
September 9, 2003, 02:59 PM
I am NOT doubting that "Rotational Energy" contributes to Knock down power.....................

{ If Steve Smith is the author of Steve's reloading Pages } I do not hesitate even for an instant to believe that it is true}

but

I AM having a hard time understanding HOW this comes about,
even (as it was said) "Minimaly" or "Neglectibly".

Looking at it mechanicaly:

The axis of rotation of the bullet is perpendicular to the axis upon which the falling target (hopefully) will rotate.


How does energy exerted around a given axis exert force upon objects limited to rotating around second axis which is perpendicular to the first?

Alternately stated

How does energy in one plane, contibute to the movement of an object which lies in a perpendicular plane?

renaissance would like to know!

Steve Smith
September 9, 2003, 03:11 PM
I am not the author of Steve's Reloading Pages.

I think that rotational energy would have an affect on the perpendicular center of gravity of the target.

Its affect would be almost negligible on objects with significant mass. The energy could be felt in the gun, however, when it torques.

Mikul
September 9, 2003, 03:53 PM
Bowling pins are quite different from steel plates. .45 bullets usually penetrate 1/2 way into the pin and stop. Nines are known for going straight through which will not transfer all of the energy to the pin. A pin that has been shot up make this even more likely. This is due to the frontal surface area of the bullets and the shape of the bullet. A semi-wadcutter is a good pin bullet, but they don't exist in 9mm, so you deal with tuncated cones.

Oh, and Steve, your example is flawed: Think if this in practical terms. What is more likely to penetrate through an elk, a 40 grain .22 magnum at 1800 fps, or a 330 grain .45 at 1200 fps?

You increased the mass of the bullet in your example by 825%, but only reduced your velocity by 33%. A proportional decrease in velocity would be a 330grain bullet traveling at 218fps. Either that or get the .22 to move nearly 10,000fps.

Back to the original question. Two bullets at different weights leaving the muzzle with the same momentum will not have the same momentum at 200 meters.

At 200 yards, your bullet will be traveling 1267fps. giving you a momentum of 209, while the old-timers will traveling at 1119 and giving them a momentum of 201. So, it looks like you have slightly more energy at that range than your shooting partners at THAT range. I'd hazard to guess at 300 meters the tables would be turned.

Steve Smith
September 9, 2003, 04:18 PM
Weeeell, now we're getting into something that I have practical experience with. A heavier bullet, given equal BC, will retain its speed and energy at distance, when the lighter bullet will not. This is one of the biggest reasons Highpower shooters use heavy bullets at longer distances.

Zak Smith
September 9, 2003, 04:23 PM
We're mixing a few different things here:

1. Momentum of the bullet is simply its mass times its velocity (with appropriate units)

2. The momentum of a flying bullet changes as it slows down due to air resistance. Plug into your external ballistics program for particular values. A bullet with a higher BC will slow down less than one with a lower BC. A bullet that starts out slower will lose a smaller fraction of its initial velocity due to air resistance (because drag is proportional to v^3, if I remember correctly).

3. The ability to knock down steel or know a bowling pin off a table is not the same as momentum or kinetic energy. If the collision were "perfect" (in the physics sense), then it would be closest to momentum. (For an example of a "perfect" collision, think of billiard balls.) But as Mikul mentioned, the bullet actually has to impart momentum to the pin, not just zip right through it or blast it into pieces.

-z

Zak Smith
September 9, 2003, 04:26 PM
A heavier bullet, given equal BC, will retain its speed and energy at distance, when the lighter bullet will not.


Steve, I don't agree, or I'm misunderstanding you.

The external ballistics can be modelled accurately using only the initial velocity and the B.C. of the bullet. The bullet's mass is a component of calculating its BC -- the BC subsumes the information about mass. Once the BC is established, the mass is irrelevant to the external ballistics.

However, if you have two bullets with the same BC value, and one is launched slower and one is launched faster, I agree that the one that started slower will lose a smaller fraction of its initial velocity at any range than the one that was launched faster.

-z

edit: Oh, yeah, if the BC of the heavier bullet is greater - as you'd usually expect - then I agree that momentum and energy are better retained at a distance.

Steve Smith
September 9, 2003, 04:28 PM
But you can have different weight bullets with the same BC and that is what I am talking about. The heavier bullet's mass does not slow as quickly because of the higher momentum.

Zak Smith
September 9, 2003, 04:37 PM
Steve,

If the two bullets with the same BC but different masses start at the same velocity, their velocities at any range will be equal - because a bullet's mass is a component of the BC - but the heavier one will have more energy and momentum: both will be greater than the lighter one by the ratio of their masses.

When you "fire" a bullet through any modern, accurate, external ballistics calculator, besides environmental conditions, it merely asks for BC and muzzle velocity. These alone determine the rate at which it bleeds velocity. And downrange energy and momentum only depend on the instantaneous velocity and the bullet's mass (which does not change).

-z

Mike Irwin
September 9, 2003, 04:38 PM
Mikul,

I've been shooting bowling pins for years and I've never had a 9mm penetrate completely through a pin unless it hits the thing part of the neck or the top cone.

Even a .45 can go the whole way through those.

Through the body, though? Never seen that, and I've shot a lot of 9mm at a lot of bowling pins.

Mike Irwin
September 9, 2003, 04:46 PM
Also, another observation...

one of the things that I both learned and observed while both covering and shooting black powder cartridge rifle silhouette is that the smallest caliber that will reliably take down the plates all the way out to the 500 meter rams is the .40-60 Winchester.

The .38-55/.38-56, which fire a somewhat lighter bullet at a higher velocity, is occasionally seen to make a perfect hit on the plate, but fail to drop it.

Steve Smith
September 9, 2003, 04:47 PM
Hmm, perhaps you are right. Its been over a year since I modelled this in Quickload, which I no longer have. If so, i apologize.

Hmm...would be hard to have two bullets with same bc AND same diameter, AND same material at the same time.

Zak Smith
September 9, 2003, 04:58 PM
Steve,

Another source for this is that in the "ballistics" section of the Speer reloading book (the back third, if I remember correctly), they list trajectories based on BC and muzzle velocity, not caliber & bullet.

I've been plaing with my own hacked-up differential trajectory calculator for a while, based on the G1 model. It matches all the published trajectories I've found.

With regard to the same BC & diameter & material, this is generally true except when you consider some of the more expensive/exotic bullets. For example, a 165gr .30-caliber bullet from Lost River Ballistics has a BC of .636, while the Barnes X is .505, and the Sierra is .404

http://www.lostriverballistic.com/LRB_Start.htm

-z

Norm Lee
September 11, 2003, 04:15 PM
All this talk of shooting bowling pins! What fun! We’ve (my bride and I) shot a pin or two and still try to get after them at least once a month. There’s just one pin match that we know of within reach. Pity.

I started with a nine and when she finally decided to try her hand she used a .44 Spl. which she shared with me until she got her own. When she decided she wanted to shoot in the semi auto match as well as the revolver one she bought a 1911 in 45. We had started out looking at a .40 cal companion to my 9mm BHP but the seller would not work with her at all and she walked. Later, at Christmas, she got one for me as she knew it worked better than the 9.

Now I use a 45 in bottom feeder land and a 686 .357 wheel gun. She likes the .357 too (the caliber was her idea in the first place) but has been using a .40 S&W lately in the auto match.

I had been loading the nine to about a 165 PF and it would do the job with a well placed hit and no external disturbance. The pins were lookin’ decidedly lazy if struck a little off center and sometimes would simply lie down. If two pins collided on their journey off a table they would often both decide to stop and rest awhile until they could be motivated to continue.

Here’s what we think we’ve learned so far. Momentum (power factor) is the thing-a-ma-bob that does the job. No question. That 9 mm load has 409 ft lbs of energy and a 115 gr load at 1320 fps has 444. Meanwhile a 230 gr .45 ACP at 850 has but 368 ft-lbs energy but a pf of 195. Performance on pins for these loads is in inverse order of available energy. Coincidentally the 195 pf load is generally considered the minimum acceptable for decent performance on pins. Lots of margin over my 9mm load. We like to take out some extra insurance and run ours up to 202-207 pf (puff).

At our match, we have learned that velocities should be kept above 800 fps (preferably 850 ) else the probability of a bounce back is unacceptably high. Motivates pins more poorly and annoys the spectators. (Like putting lipstick on a pig)

At the other end of the speed range, it seems that projectiles moving at 1450 fps and above will too often zip on through a pin without shedding or transferring the requisite momentum. I am sure there may be some bullet styles or construction techniques that would help mitigate this but 1450 is a good enough spot to put your upper limit.

Reviewing the bidding we see that a 230 gr. bullet does the job at 850 fps and a 135 is good to go at the upper end. And I’m still maintaining that equal PF gives equal performance on the pins. There is no argument for heavy and slow over light and fast or vice versa or is there?

Well there might be. Felt recoil, too, is a momentum transfer effect and, as far as I can tell it all starts out by assuming that M1*V1=m2*v2 where the ones belong to the gun and the twos belong to the ejecta (bullets and other stuff expelled from the muzzle). When you look up the formulas that result from this you should notice that the velocity assumed for the gasses is pretty high, up around 4000 fps. Now these gasses come from burning powder and,, if perfectly converted their momentum contribution would be powder mass times gas velocity. That 4000 factor makes even relatively small differences in powder charge weight significant in the recoil department. Since we’re talking about equal pf there is no difference contributed by bullet mass. 200 gr. bullet at 1000 fps or a 250 at 800 fps, all same. But your performance with the heavier bullet may be better since the recoil will be lower and you’ll spend less time yanking it back down out of the sky. Did I forget to say that you will need more powder to get the 200 to 1000 fps than the 250 to 800 (same powder)?

Cheers,

Norm

If you enjoyed reading about "Velocity = Energy" here in TheHighRoad.org archive, you'll LOVE our community. Come join TheHighRoad.org today for the full version!