[tsafa]

I have been doing a bit of research to try to understand if measuring Momentum or measuring Energy is more useful in ballistics. I understand the calculations of each... but I am not fully sure what they mean to the target being hit. I am interested in understanding the meaning of the calculated result.

For this discussion, I would like to consider two measurements I have taken.

.36 cal (80 grain) black powder revolver at 1250 ft/s
.44 cal (140 grain) black powder revolver at 945 ft/s

The calculated energy of both of these is 277 Ft/bs. (E=VxVxM/450,240)


Here is where it does not make sense. 277 foot/Lb is defined as the amount of force necessary to move 277 lbs one foot vertically in the air against gravity. This is something I can easily understand and relate to. If I lift a 20 lb weight vertically a distance of one foot I have exerted 20 ft/lbs of Energy. If I suspended 20 lbs in the air and shot it with either of these loads it would not move one foot vertically against gravity. It will probably not move vertically even 3 inches Hence the Energy calculation has no meaning other then in comparing one caliber to another.

The 450,240 divider seems to have no purpose other then converting a V²M to Ft/lbs. I would think that some other divider could be used to give a product that would indicate how much weight is actually being moved vertically against gravity. I like the idea of gong vertical because resistance is limited only to air resistance.


As a result of the Energy calculation giving a number that has no real meaing as far as I can tell, I have decided to take a look at Momentum and figure out if that will give me some meaningful product I can use to predict the effect on a target.

The Momentum calculation I have been able to find is:
Momentum= Mass x Speed /7000

Again the 7000 seems to be some time of converter to get to a lbs/s product.

The calculated momentum on my earlier example seems to be:

14 lb/s for the .36
18.9 lb/s for the .44

So now the million dollar questions.

What is a lb/s unit?? And what does it mean to my target?? What effect real effect can we predict??

[ebaines]

First you need to get your units straight. The unit for energy is lb-ft, not foot/pound. Regarding units for momentum: it's pound-seconds, not pounds per second. One pound-second is equivalent to applying a 1 pound force to an object for 1 second.

You are correct that not all of the kinetic energy of the bullet is transferred into kinetic energy of the block. A significant amount of the bullet's energy is lost to heat as the bullet hits its target and deforms. The way to calculate how the block moves is by using the principle of conservation of momentum. If we assume that the bullet buries into the body of the target and does not ricochet then the formula to use is this:



where = mass of bullet, = mass of block, = velocity of bullet, and = velocity of bullet & block together after impact. This lets you solve for :



The initial kinetic energy of the bullet is:



The final kinetic energy of the block and bullet together after impact is:



Note that if is large compared to , the kinetic energy of the bullet + block becomes a low value. This is why when you shoot a heavy target it doesn't move very much. And again, the difference between the initial kinetic energy of the bullet and the final kinetic energy is accounted for as heat and the energy of deformation of the bullet and block. This lost kinetic energy is:



Note that the larger the value of , the greater this loss of kinetic energy.

Whether you should be thinking about energy or momentum depends on what you're trying to figure out:

a. If you want to know how much the target moves, use momentum

b. if you want to know how badly the target is destroyed, use energy.

[tsafa]

Thank you Ebaines for your response. I has helped increase my understanding considerably.

If a very hard mettle was used in the bullet, like steel instead of lead (or something harder like depleted uranium bullets), could we get an energy transfer that approached the bullets energy if the bullet did not deform? Do you think it is possible to shoot a steel core bullet with a muzzle energy of 300 lbs into a 300 lb suspended steel weight and actually get the weight to move vertically one foot against gravity? Or is the loss to heat so much that it is not practical to even consider?

[jcaron2]

I think you have an error in your calculation of KE2. If you plug in your correct value for v2, the (m1+m2) term cancels, leaving KE2=1/2*m1v1^2=KE1. No energy is lost due to the assumption of perfect momentum transfer.

[ebaines]

Jcaron - I disagree. If you agree with me that:



then



Kinetic energy is only conserved in a perfectly elastic collision - and the case of a bullet burying itself into a target is not perfectly elastic.

[ebaines]

If a very hard mettle was used in the bullet, like steel instead of lead (or something harder like depleted uranium bullets), could we get an energy transfer that approached the bullets energy if the bullet did not deform? Do you think it is possible to shoot a steel core bullet with a muzzle energy of 300 lbs into a 300 lb suspended steel weight and actually get the weight to move vertically one foot against gravity? or is the loss to heat so much that it is not practical to even consider?

In you could get a perfectly elastic collision between bullet and target - such that the bullet bounces off the target without deformation, then it's possible to get complete energy transfer from the bullet to the target but only if the bullet and target have the same mass. This is what happens when you play billiards - the cue ball hits the target ball and if done properly the cue ball completely stops while the target ball moves on with the same velocity (hence energy) as the cue ball originally had. But if the target is heavier than the bullet then the bullet bounces backward (in a perfectly elastic collison) and some of the bullet's initial kinetic energy goes toward the ricochet velocity of the bullet backwards. Consequently there is less than 100% transfer of energy to the target.

[jcaron2]

Yup, you're right. Pay no attention to my jibber-jabber. Apparently late in the day before an extra long weekend I can't even do simple algebra. ;-)

[DrBob1]

Further thoughts. Remember that the basic purpose of the ballistics is to transfer as much energy as possible to the target and wreck havoc upon it.