BR Posted September 20, 2008 Posted September 20, 2008 Hello all -- specifically scientists and/or engineers in the group? In 9mm minor, shooting from the same gun at the exact same power factor -- whether 126, 136, or whatever -- do 147 grain bullets knock down steel, specifically properly calibrated poppers, "better" "faster" or "with more authority" than 124/25 grain bullets? I personally have no empirical or even subjective knowledge of this. I load and shoot 147s, because I have for a long time and they work for me and in my gun. Also, Dave Sevigny uses 147s, and he is my hero. But, I have read comments that 147s take down steel better or with more authority on steel than lighter bullets, and I wonder whether, from a scientific basis, this is true. Question #1: What is it -- whether momentum or energy or something else -- that takes down steel? Question #2: Is "power factor" (essentially mass x speed, right?) equivalent to some scientifically-defined term, such as momentum or energy? Question #3: Depending on the answers to 1 and 2, does and/or can equivalent power factor bullets -- really of any caliber -- have different affects when shooting steel? If so, what? (question 4, I guess). Again, I do not know the answer. As a reverse thought, I can feel the difference between shooting equal power factor 147s compared to 124s. Does this relate to affects on steel? Time for another drink. Cheers. -br
Merlin Orr Posted September 20, 2008 Posted September 20, 2008 Don't have to be a weatherman to know which way the wind blows..... Get your bud to hit you with a fastball then with a medicine ball and see which knocks you down. Use your own eyes (as I have) and you will plainly see - Heavy is better.
Flexmoney Posted September 20, 2008 Posted September 20, 2008 (I like heavier just fine, but..) Guys that have trouble with steel are: - shooting at steel that isn't set well - running too close to the 125pf floor - aren't shooting the steel in the middle of the circle. (edge hits and low hits...at minor...can lose out on steel) I shoot 124/5's out of my Production Glock...at about 135-140 power factor. They take down steel as well as my major loads in Limited used to (40's at 180g and 170+pf)...at any match that has the steel set properly. Load up to 140pf and forget about it. They won't slow you down. Here is that load in action:
gm iprod Posted September 20, 2008 Posted September 20, 2008 What Flex said. It is inertia (Power Factor) that knocks things down, not muzzle energy. Theoretically the 125gr has more foot pounds energy than the 147gr. But the 147gr loses less velocity than the 125gr between muzzle and target, therefore they lose less power factor. From the computer. 9mm 124gr MV 1050 PF130 Ft/Lbs 304 25Y/V 1012 PF 125.5 Ft/Lbs 282 9mm 147gr MV 890 PF131 Ft/Lbs 259 25Y/V 869 PF 127.7 Ft?Lbs 246 Advantage os to the 124gr for a flatter trajectory. Which, unless you are going to 50Y, is irrelevant for all IPSC purposes. How each loads shoots for you is way more important. But if you hit them in the middle the argument is settled, for plates or anything else that requires a 9mm bullet in it.
Aircooled6racer Posted September 20, 2008 Posted September 20, 2008 Hello: If a bus and a car are going 60mph which one will have a harder time stopping? Answer your question? Or how about a 300 pound lineman hitting you and a 100 pound girl hitting you. The girl would be more fun but the lineman will move you more. Use the 147's at 137pf and it will knock down all steel if set properly. Back to my cocktail now Thanks, Eric
lugnut Posted September 20, 2008 Posted September 20, 2008 M*V is momentum. If a 124gr and 147gr bullet with the same PF hit the same mass (popper) and since momentum in conserved it will have the same effect on the popper.
gm iprod Posted September 20, 2008 Posted September 20, 2008 I got hit by a 100lb girl. And it took weeks for the swelling in my groin to go down. It also hurt me way more than anything else. My friends still laugh at me behind my back.
G-ManBart Posted September 20, 2008 Posted September 20, 2008 Hello: If a bus and a car are going 60mph which one will have a harder time stopping? Answer your question? Or how about a 300 pound lineman hitting you and a 100 pound girl hitting you. The girl would be more fun but the lineman will move you more. Use the 147's at 137pf and it will knock down all steel if set properly. Back to my cocktail now Thanks, Eric But in this case the bus and the car are going different speeds so your analogy doesn't fit....
Matt Cheely Posted September 20, 2008 Posted September 20, 2008 The slower bullet has more time to impart its energy on the popper before blowing apart on the popper face. They might both have the same momentum but it's all about energy transfer to the target.
HSMITH Posted September 20, 2008 Posted September 20, 2008 My buddy shoots 147's at 135 power factor, I shoot 115's at 177 power factor, I can't tell the difference from watching a popper and neither can the popper. Both loads put down anything close to reasonably set steel. Plates are another story, I smack plates down HARD where his doesn't put them down as hard.
38supPat Posted September 20, 2008 Posted September 20, 2008 Split the difference, shoot 130/135gr bullets
lugnut Posted September 20, 2008 Posted September 20, 2008 Does it really matter? Most matches are calibrated so that the steel with drop with any of the appropriate loads. Call your shots (still working on this) and it's not an issue.
XRe Posted September 20, 2008 Posted September 20, 2008 (edited) The terms you want are momentum (Power Factor is a momentum number in weird units), inertia, inelastic collision, and conservation of momentum. Energy has nothing to do with it. (eta, assuming I'm recalling particle dynamics correctly - if energy were a factor, everyone one of us would be running the lightest bullet possible at a given power factor, if we were worried about taking steel down...) Edited September 20, 2008 by XRe
tk4 Posted September 20, 2008 Posted September 20, 2008 The slower bullet has more time to impart its energy on the popper before blowing apart on the popper face. They might both have the same momentum but it's all about energy transfer to the target. How much difference could this make. Is it enough to offset 1 or 2 pf difference?
G-ManBart Posted September 21, 2008 Posted September 21, 2008 The slower bullet has more time to impart its energy on the popper before blowing apart on the popper face. They might both have the same momentum but it's all about energy transfer to the target. If the bullet stops completely, it's transfered the exact same amount of energy to the popper regardless of whether it's over a shorter or longer time period (dwell time), but that really doesn't matter....it's not about energy, it's about momentum which would make it an impulse function.
steel1212 Posted September 21, 2008 Posted September 21, 2008 Ummmm.....how about call you shot, hit the popper, move on? If your making PF, If the steel is calibrated correctly, you should have no problem no matter what weight your pill is. I can't stand the feel of the 147s 9s, 200 grain 40s, or 230 .45s so I go down one step and push them faster. I shoot my 9mm minor 124s at 138 PF, when I hit them it goes down, you don't worry about chrono, and just shoot. Yeah, they don't go fly down, but I don't need them to, I just need them to drop. If its an activator there is usually something else to shoot at while your waiting. Pick which bullet feels best and shoot it.
AikiDale Posted September 21, 2008 Posted September 21, 2008 Just to add to the confustication let me add that the collision of the bullet with the steel target is an elastic collision and while some of the momentum is being transferred to the target some of the projectile's kinetic energy is being lost as heat generated by its deformation and resultant vectoring. In our little subset of the world any 9mm / .355 inch projectile from roughly an 85 grain .380 to a 158 grain .38 wadcutter can be driven fast enough to take down the steel. The lighter the projectile the faster it must go to move targets of the same mass. (and remember, I was an English major and a drop out)
chizzle Posted September 21, 2008 Posted September 21, 2008 Check out the following ballistics calculator: http://handloads.com/calc/quick.asp I'm a big fan of bowling pin shooting, and this calculator helps explain how different loads vary on Energy (ft/lbs) and momemtum. High momentum numbers help us knock bowling pins off tables, even when high energy levels fail. Enjoy! Chuck
BR Posted September 21, 2008 Author Posted September 21, 2008 (edited) I appreciate those who actually responded to the questions asked, especially those who answered that momentum, rather than energy, is what knocks over steel. For those others who responded about calling my shots, shooting well over power factor, and/or other sundry issues, well thanks for your advice. Anyways, if momentum is what knocks down an object, here a steel popper, then it follows that bullet weigh is irrelevant if the power factor -- the amount of momentum -- is equal. I think the ballistic calculator chizzle cited supports this as fact. Cheers. -br Edited September 21, 2008 by joker22
lugnut Posted September 21, 2008 Posted September 21, 2008 Anyways, if momentum is what knocks down an object, here a steel popper, then it follows that bullet weigh is irrelevant if the power factor -- the amount of momentum -- is equal. I think the ballistic calculator chizzle cited supports this as fact. Cheers.-br Someone did post a good point though. Momentum is conserved and IF all the momentum was directly transfered to the popper that it wouldn't matter whether you used a 124 or 147 gr bullet. However with EITHER a 124 or 147gr bullet ALL the momentum is not transfered since there is always splatter. That being said I don't know how to validate how much momentum of each is transfered from either of the bullets. I've seen large chunks of large bullets ricochet and I've seen smaller bullets deform/flatten and hardly move when you hit steel.... soo... I think calling the shot is the best way to solve any of these problems.
AikiDale Posted September 21, 2008 Posted September 21, 2008 I appreciate those who actually responded to the questions asked, especially those who answered that momentum, rather than energy, is what knocks over steel. For those others who responded about calling my shots, shooting well over power factor, and/or other sundry issues, well thanks for your advice. Anyways, if momentum is what knocks down an object, here a steel popper, then it follows that bullet weigh is irrelevant if the power factor -- the amount of momentum -- is equal. I think the ballistic calculator chizzle cited supports this as fact. Cheers. -br Ummmm......depends. The more massive (and softer lead) projectile traveling at slower speed will more efficiently take down the (harder and considerably more massive) steel target than the faster lighter projectile because the higher velocities will lend themselves to the loss of momentum as it is converted to the deformation, splattering, process. Of course if you have a fast enough bullet it will work, but at the cost of more recoil, longer transition times. As noted above, bowling pin shooters seem to prefer heavier bullets to take the pins off the table. But do the experiment and let us know how it comes out! Load some 115s, 124s and 147s to the same PF (momentum) and knock down a popper with them. I'm looking forward to the report!
XRe Posted September 21, 2008 Posted September 21, 2008 Just to add to the confustication let me add that the collision of the bullet with the steel target is an elastic collision and while some of the momentum is being transferred to the target some of the projectile's kinetic energy is being lost as heat generated by its deformation and resultant vectoring. That would make it in inelastic collision, as I said... not an elastic collision (where the kinetic energy of the system is the same before and after collision... see: http://en.wikipedia.org/wiki/Elastic_collision The most efficient case here would be to make a bullet that would "stick" entirely to the popper face, making it a plastic collision... but we don't get that either. Instead, we have: http://en.wikipedia.org/wiki/Inelastic_collision - and its a momentum game + the ability of the projective in question to not distribute its momentum through fragmentation, etc. (those may not be correct physics terms, but....) Somehow, I imagine sectional density plays into all that somehow, but... I couldn't begin to give you how.. (that would imply that heavier bullets in the same diameter are going to be better at knocking down poppers, as well - and that bigger bullets of the same weight - or heavier - are also better)...
AikiDale Posted September 22, 2008 Posted September 22, 2008 Just to add to the confustication let me add that the collision of the bullet with the steel target is an elastic collision and while some of the momentum is being transferred to the target some of the projectile's kinetic energy is being lost as heat generated by its deformation and resultant vectoring. That would make it in inelastic collision, as I said... not an elastic collision (where the kinetic energy of the system is the same before and after collision... see: http://en.wikipedia.org/wiki/Elastic_collision The most efficient case here would be to make a bullet that would "stick" entirely to the popper face, making it a plastic collision... but we don't get that either. Instead, we have: http://en.wikipedia.org/wiki/Inelastic_collision - and its a momentum game + the ability of the projective in question to not distribute its momentum through fragmentation, etc. (those may not be correct physics terms, but....) Somehow, I imagine sectional density plays into all that somehow, but... I couldn't begin to give you how.. (that would imply that heavier bullets in the same diameter are going to be better at knocking down poppers, as well - and that bigger bullets of the same weight - or heavier - are also better)... I'm going to politely disagree and use the animation from the citation above to explain: http://en.wikipedia.org/wiki/Image:Inelast...r_sto%C3%9F.gif If the bullet stayed with the target and none of its mass bounced back and hit the spectators, shooters or RO's it would be an inelastic collision. Of course, I could be wrong about that and if you can explain it without using math I won't understand I'd be ever so gratefull.
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