Do 147s really knock down poppers better than 124/5s?

[DrShooter]

You are probably looking for "impact" as the possible difference. If a bigger bullet takes more time to crush against the steel before sliding off, etc., it can deliver more impact or impulse. Checking out the famous X-ton crash of a duck in a head on collision with a 747 (on the internet, of course), you will see that the time it takes for the duck to crush against the plane makes a big difference. Mainly the length and stiffness of the duck are important. It's why they can take down a plane sometimes.

So there could be a difference even at the same powerfactor because they will be different length. This would be the mechanical engineering version of "dwell time."

There are square roots involved, so be warned before you go to wikipedia, searching for impact or collision.

[BR]

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.

OK. Now we are getting somewhere.

(a) Do we have agreement that "momentum" [mass x speed], as opposed to "energy" [mass x speed squared], is what knocks down steel?

( Do we agree that "power factor" is essentially a measurement of momentum?

© If one eliminates the elasticity/inelasticity and/or the sectional density and/or the hang-time/splatter variables, and assumes the exact same bullet construction in a different weight, i.e., zero 147 jhp v. 125 jhp, does the same power factor, of same caliber but regardless of specific bullet weight, have the same affect on a steel popper?

I believe that science says "Yes." Also, I do understand the concept of "calling the shot," and personally try every day to better effectuate that concept. But, this question is seeking an objective, scientific answer about a frequently espoused opinion that 147 bullets knock down steel better than 125s or other lighter bullet weights. Based on science, I think I disagree with that opinion.

So, assuming the exact same type of bullet, and taking out the elasticity/sd/hang-time variables, and assuming the shot is called and landed in the exact same spot on the metal target, also taking out that variable -- does science not dictate that "power factor," irrespective of bullet weight, is what affects knocking down steel?

Cheers.

-br

[AikiDale]

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.

OK. Now we are getting somewhere.

(a) Do we have agreement that "momentum" [mass x speed], as opposed to "energy" [mass x speed squared], is what knocks down steel?

( Do we agree that "power factor" is essentially a measurement of momentum?

© If one eliminates the elasticity/inelasticity and/or the sectional density and/or the hang-time/splatter variables, and assumes the exact same bullet construction in a different weight, i.e., zero 147 jhp v. 125 jhp, does the same power factor, of same caliber but regardless of specific bullet weight, have the same affect on a steel popper?

I believe that science says "Yes." Also, I do understand the concept of "calling the shot," and personally try every day to better effectuate that concept. But, this question is seeking an objective, scientific answer about a frequently espoused opinion that 147 bullets knock down steel better than 125s or other lighter bullet weights. Based on science, I think I disagree with that opinion.

So, assuming the exact same type of bullet, and taking out the elasticity/sd/hang-time variables, and assuming the shot is called and landed in the exact same spot on the metal target, also taking out that variable -- does science not dictate that "power factor," irrespective of bullet weight, is what affects knocking down steel?

Cheers.

-br

Yes, it is 'power factor' or momentum that knocks down the steel. And if you want to ignore all the other variables then mass of the bullet makes no difference. In the real world the heavier bullet will work better because in the real world you get all the extra variables....

[XRe]

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.

You didn't read the definitions, did you?

An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision).

In a perfectly inelastic collision the colliding particles stick together.

You'll also find a perfectly inelastic collision being referred to as a "plastic collision". Your animation is a perfectly inelastic collision, but that is not the only type of inelastic collision.

An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only (emph added by XRe) if there is no conversion of kinetic energy into other forms.

Quite clearly, we convert kinetic energy to heat, sound, etc, as the bullet smashes into the popper. In terms of the stuff that happens outside the atomic level, its hard to imagine any sort of collision that's actually elastic (though someone can probably think of one). If such a thing actually could happen, we could probably build a perpetual motion machine... but we can't...

[XRe]

So there could be a difference even at the same powerfactor because they will be different length. This would be the mechanical engineering version of "dwell time."

This is exactly what I was getting at regarding sectional density.

[XRe]

So, assuming the exact same type of bullet, and taking out the elasticity/sd/hang-time variables, and assuming the shot is called and landed in the exact same spot on the metal target, also taking out that variable -- does science not dictate that "power factor," irrespective of bullet weight, is what affects knocking down steel?

In a vacuum, using a popper on frictionless bearings, and a point bullet (ie, no sectional density, etc), yes. Its all momentum.

Now, when you figure out how to shoot in that environment with that bullet, let me know

[SA Friday]

OK, don't know about all the scietific voodoo going on here, but I've seen WWB 115gr 9mm fail to drop steel too many times vs 147gr 9mm to not know that there is a difference. The pf of WWB is higher than my regular 147gr reloads, so I think you would be hard pressed to convince me that, "130pf is 130pf on steel regardless of bullet weight".

Sorry for the interruption, please continue with your Tom Cruise voodoo science.

[DrShooter]

http://www.efm.leeds.ac.uk/CIVE/CIVE1140/s...ll_notes02.html

Some review for those interested...

Remember it's not the momentum, it's the momentum transferred...

[AikiDale]

Okay, I may...may change my mind but only because I am so easily influenced.

My previous, perhaps errant, understanding of elastic and inelastic collisions in the real world as explained to me by the physics major to the English major was that if one were to throw a lump of clay at a wall and it stuck that was an inelastic collision and that if the cue ball hits the rack and the balls move that is an elastic collision. If that analogy is sorta correct (?) then would not the bullet knocking down the popper be an elastic collision? Or did I just get it all backwards? (Dyslexics of the World Untie!)

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.

You didn't read the definitions, did you?

An inelastic collision is a collision in which kinetic energy is not conserved (see elastic collision).

In a perfectly inelastic collision the colliding particles stick together.

You'll also find a perfectly inelastic collision being referred to as a "plastic collision". Your animation is a perfectly inelastic collision, but that is not the only type of inelastic collision.

An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only (emph added by XRe) if there is no conversion of kinetic energy into other forms.

Quite clearly, we convert kinetic energy to heat, sound, etc, as the bullet smashes into the popper. In terms of the stuff that happens outside the atomic level, its hard to imagine any sort of collision that's actually elastic (though someone can probably think of one). If such a thing actually could happen, we could probably build a perpetual motion machine... but we can't...

[XRe]

Okay, I may...may change my mind but only because I am so easily influenced.

My previous, perhaps errant, understanding of elastic and inelastic collisions in the real world as explained to me by the physics major to the English major was that if one were to throw a lump of clay at a wall and it stuck that was an inelastic collision and that if the cue ball hits the rack and the balls move that is an elastic collision. If that analogy is sorta correct (?) then would not the bullet knocking down the popper be an elastic collision? Or did I just get it all backwards? (Dyslexics of the World Untie!)

The cues are elastic (basically - in the real world, they aren't purely elastic, either, as energy is lost...), but a bullet striking a plate is not. Its also not perfectly inelastic. (that is, kinetic energy is lost to heat, sound, etc, as the bullet comes apart and all that, but the bullet doesn't generally stick to the plate, either). However, interestingly, I found several sites using the terms to mean different things Apparently, event the physicists can't agree on what to call a collision where kinetic energy is lost, but the objects in the collision do not stick together Apparently, this is one case where English is actually simpler than something else....

[Merlin Orr]

OK, don't know about all the scietific voodoo going on here, but I've seen WWB 115gr 9mm fail to drop steel too many times vs 147gr 9mm to not know that there is a difference. The pf of WWB is higher than my regular 147gr reloads, so I think you would be hard pressed to convince me that, "130pf is 130pf on steel regardless of bullet weight".

Sorry for the interruption, please continue with your Tom Cruise voodoo science.

+1

You can draw all the squiggly lines on a chalk board you want but I have seen and shot both loads a number of times side by side...The heavier bullet takes steel down better.

What is so difficult about believing your own eyes?

[chirpy]

Just a story from first hand experience...I was shooting CDP at my local IDPA club. They were using a non-standard popper which hadn't been calibrated - no problem. I was shooting 200 gr lead swc at 175+ pf. It took 5 rounds (at least) all good hits to drop it. Did a re-shoot with ESP .38 super 147 gr. jacketed loaded to minor and took it down with one shot much to the RO's and my surprise.

FWIW

Richard

[ranger]

This thread is bringing back ugly memories of Dynamics and Deformable Bodies classes at GA Tech.....best I can do is focus on hitting the center of the target because I never want to do that math again!

[AustinMike]

+1

You can draw all the squiggly lines on a chalk board you want but I have seen and shot both loads a number of times side by side...The heavier bullet takes steel down better.

What is so difficult about believing your own eyes?

Took a break from my Calculus homework and read this thread. I'm with Merlin. Da bigger bullet make the popper fall down very nice. Just shoot a .45 "flying ashtray" and the air it displaces is enough to knock the popper down as it sails past if you miss.

I hate math, but unfortunately my homework isn't gonna solve itself. Now back to squiggly lines...

[mpeltier]

Answers,

question 1: Momentum of a properly placed shot, on a properly calibrated steel target is what knocks them down.

question 2: No not exactly; Power factor was originated to provide a means of calculating the power of the ammo used in competition to ensure all contestants were on an even playing field. Steel was then calibrated to the minimum of this power factor.

question 3: A heavier bullet at an equivelant power factor may give you a larger margin of error if the steel is set a tad on the heavy side or your shot is not as good as it should be (edge hit).

Personal observations: I have done unscientific comparisons with 124gr/ 147gr and 115gr reloads/wwb115. shooting them in practice and in competition. The 115 wwb were the worst of the lot, do in part I believe to the round nose fmj bullet. The 115gr HP reloads were more consistant in knocking down steel. When I loaded and compared the 124 vs 147, I found no differance I could attribute to a 23 gr differance in bullet weight when hitting steel. Doing side by side drills I was consistantly faster shooting the 124's and they are easier to reload. I have often followed respected competitors shooting .40 and .45 ammo that had trouble with a steel target and then I would proceed to knock it down easily with my 124's (see answer 1). Even the dreaded texas star at ft benning fell easily to my 124 9mm's.

[XRe]

question 2: No not exactly; Power factor was originated to provide a means of calculating the power of the ammo used in competition to ensure all contestants were on an even playing field. Steel was then calibrated to the minimum of this power factor.

Are you sure this wasn't because Colonel Cooper was a .45 bigot? It obviously gives advantage to .45 over 9mm.... (not that he was wrong or anything....)

[mpeltier]

I am sure that had something to do with the number that was originaly concieved. In later years Cooper was also quite fond of the 10mm. But none of it had anything to do with the OP's original question #2, or did it? Todays power factor is lower, I believe to accomodate the .40 as a major power factor.

[gm iprod]

As previously stated Power Factor was "invented" as we know it to level the playing field. It gave some advantages to the fast & light, and some advantages to the heavy & slow. But more importantly it gave us something to make sure no one was pulling a fast one.

Again all the numbers in the world are just that, numbers.

You have to find out what works for you. I know guys who swear by one theory or the other and get their arses handed to them one week and do the handing out the next.

[Harmon]

the bullets loaded to the same powerfactor should perform the same on the steel, but the lighter bullet moving faster converts more of its energy into heat and sound instead of transmitting the energy to the target.

the slower bullets seem to have more "hang time" on the steel.

this is why the silhouette shooters chose the 180 grain .357 magnum loads.

[Sean Gaines]

I noticed no one has said anything about hollowpoints, and the bullet profile. Hollowpoints stay on the steel just a split second longer dispersing energy on the plate, while its flattening for a split second longer, where as a roung nose hits and deforms a little and skips off dipersing some of its energy into the air.

[bufit323]

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.

This is where the truth lies.

What would be interesting is to see what "type" of bullet transfers momentum most efficiently. Would a SWC impart more or would a JHP, or a LRN?

[RePete]

I've never had a problem with the 147's - unless it was induced by me.

I've used 124/5's and haven't had many problems. My problems have been poor hits on the steel - my fault, too low a PF - my fault.

I think, mostly, that it's a urban myth.

My PF for 9mm is 138, both for 147's and 124/5's. I just like the feel of the 147's on recoil.

[TonyT]

My limited empirical evidence suggests that the 147's knock down steel with more authority than 124's. I also belive that the plated bullets knock steel down better than the FMJ's. I also prefer the feel of a 147 gr. FMJ at ca 900fps with a fast burning powder such as Bullseye. That being said I have never had any problems knocking down steel with 124's except for shooter malfunctions.

Edited November 8, 2008 by TonyT