measuring Momentum or measuring Energy is more useful in ballistics.
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??
Comment on ebaines's post
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.
Comment on ebaines's post
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. ;-)