Anyone Physics majors or professional on the Hide?

Trental

Trental

Sergeant
Full Member
Minuteman
Jan 30, 2011
261
8
Boise, ID
#1
I have a question on <span style="font-weight: bold">Momentum </span>in regards to bullets striking "armor plate" steel, and I don't have the background to prove/disprove the theory.

Here is the scenario... I shoot in Metallic Silhouette competitions where we shoot at metal shaped animals at varying distances. They are cut from a type of armor plate (no target damage at 550yds). There is always a constant debate in regards to the best caliber (.308, 7mm and 6.5mm). In regards to the 6.5mm projectiles at our 550 yard targets there is a theory that there is a velocity threshold that by driving the bullets too fast that they will not expend their momentum into the target, instead the bullets will disintegrate more rapidly and the direction of momentum will change to the sides instead of knocking the target over.

In layman's terms, we HAVE to knock the targets over for the hit to count towards our score. Sometimes we will "ring a Ram" which means that we hit the Ram (550yds) solid but it doesn't fall down. It just stands there and you hear the ring of the hit and see the bullet splash. Even though you hit it you don't get to count it.

So, like I said there is a theory that there is a fine line between "fast enough" and "too fast". Myself and a buddy want to build a Ballistics Pendulum to test the theory in real life but I want to think out loud before going through the exhaustive process of testing.

If I understand momentum correctly (p=m*v) (m=mass in Kg, v=velocity in meters/second), and I input the numbers for a 140gr projectile traveling at approximately 1874fps (6.5mm 140gr bullet at 550yds). Converting the bullet to kilograms I get .009072kg and converting velocity to meters/second I get 571.1952m/s. Out of the momentum formula <span style="font-weight: bold">I get (5.1819kg * m/s)</span>. So the BIG question is... is it possible that by going <span style="font-style: italic">faster</span> that the momentum could actually decrease due to bullet disintegration?

Now this doesn't take into account the angle of impact, nor does it take into account that there is nothing at the target to absorb(capture) all of the bullets momentum (the targets are undamaged and the bullets do disintegrate).

I guess, is there a point of diminishing returns?<span style="font-weight: bold"> At what velocity does momentum actually stop going up... and start going down?</span>

Hopefully there is someone on the hide that can help me theorize this before I get balls deep in fabricating a ballistics pendulum to test it.
 
#2
Re: Anyone Physics majors or professional on the Hide?

Cant answer you question but great post. Looking forward to the answer/explanation.
 
#3
Re: Anyone Physics majors or professional on the Hide?

Energy is lost when the bullet disintegrates. That is energy that is not delivered to the target.

Test this in your head by picturing someone throwing and egg at you at 90mph. Now picture the same big league pitcher throwing a steel ball the same weight as the egg at 90 mph. The effect on the target will be radically different.
 
#4
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Tim K</div><div class="ubbcode-body">Energy is lost when the bullet disintegrates. That is energy that is not delivered to the target.

Test this in your head by picturing someone throwing and egg at you at 90mph. Now picture the same big league pitcher throwing a steel ball the same weight as the egg at 90 mph. The effect on the target will be radically different. </div></div>

That is a painful analogy.
grin.gif


I want to change your statement around a little though to fit it to my scenario. Imagine a big league pitcher throwing an egg at you at 90mph and then also imagine the same big league pitcher throwing another egg at you except at 120mph. Will the egg at 90mph carry more momentum than the egg at 120mph due to a higher rate of disintegration of the 120mph egg?

You also said "Energy is lost when the bullet disintegrates. That is energy that is not delivered to the target.". This is true and ties into my question. What I am really looking for though is... is it possible that the <span style="font-weight: bold">momentum </span>lost increases as velocity also increases. Hope that makes sense.

Thanks for your input!
 
#5
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: maggot</div><div class="ubbcode-body">Cant answer you question but great post. Looking forward to the answer/explanation. </div></div>

Yeah... I have been trying to wrap my head around it for a while now. I think it is helping me by writing it out and reading through my thoughts a couple times Cheers!
 
#7
Re: Anyone Physics majors or professional on the Hide?

This would come down to "IF" the bullet disintegrates or not. A disintegrating bullet will displace momentum outward as the bullet displaces (as the front of the bullet stops the back wants to keep going and will deflect until all the materiel has shifted its energy to the side and back) as long as it does not penetrate and carry some of that energy past the target.

Bottom line is if you have a non deformable projectile then energy is energy no matter how fast or slow. Faster=more more mass=more
 
#8
Re: Anyone Physics majors or professional on the Hide?

If your bullet is traveling so slowly it retains 100% of its weight, then momentum is conserved perfectly. Increase the velocity to the point where fragmentation occurs and momentum is no longer conserved perfectly, some is lost to the disintegration.

You understand that already. Your question, I think, is this:

Assume the two velocities in question are both in the area where the bullet is fragmenting at impact. Does the amount of energy absorbed by fragmentation change as velocity changes?

Good question, and physics is not going to help you. Physics is perfect for idealized situations where impacts are perfectly elastic. The only way to get to your answer is to take some physical measurements, or find someone who has.

Here's my guess, and it's worth what you're paying for it.

I'll guess that the disintegration is more violent (smaller pieces flying farther)with the higher velocity. Duh. I'll also guess that the ratio of energy absorbed is fairly constant , regardless of velocity. For example, if 10% (a made-up number) of the energy is lost at the lower velocity, then 10% will also be lost at the higher velocity.

If all those guesses are true, then increasing the velocity should increase the energy delivered to the target as well as increasing the energy lost to fragmentation.

All that baloney depends on my assumption that the percentage of energy lost is fairly constant. That's a guess, but I'll bet it's safe. My suspicion is that the disintegration gets more efficient as velocity increases, meaning the percentage of energy used up actually decreases.

All guesses and speculation. If you really want to know the answer, mount an accelerometer to the back of a target and shoot it with the different loads. That'll clear it up once and for all. I'd personally be interested in the answer.
 
#9
Re: Anyone Physics majors or professional on the Hide?

You need to know what the energy required is to completely disintegrate the bullet, at what speed/angle, and if you want to be really picky, the time it takes to destroy the bullet. Without that information, you cant do a true energy transfer comparison between a non destroyed bullet (Mushroomed) or a destroyed bullet.

My personal opinion is that the argument is moot. If there is enough energy to destroy the bullet then the same amount of energy has already been transferred to the target as what a bullet would that mushrooms. The amount of time that the energy transfer is taking place is the real difference as the mushroomed bullet may impart the energy for a greater period of time. (However we are talking thousands of a second so again, I think its a bit moot)
 
#10
Re: Anyone Physics majors or professional on the Hide?

You are overthinking it with all the math. It comes down to the old Wiley Clapp vs Elmer Kieth argument of fast and light vs big and slow. FAst and light will not knock it down, but big and slow will. This is because a larger, heavier slower bullet will transfer all it;s energy into the target, where a small faster bullet will not.
This is most noticable in fleshy targets. Small and fast go through cleaner/clean. Bigger and slower go into the target making a larger hole and transferring more energy. ON a metal target, the speed and velocity of the smaller bullet has to go somewhere if not straight forward. This is to the outer ring, 360*, the bullet splatters, the energy follows the splatter. The bigger slower bullet does not shatter, but compresses and transfers more energy into the metal thus knocking the target down.
You can run a 175FMJBT down to 2400fps and watch it knock the target over each and every time. Speed it up, or use a faster lighter bullet, and all you get is DING!
See, no math at all, don't need it. The argument has been settled by men much older than us years before we were born, or at least while we were young enough to chase 19 year old tail and do it without insults
wink.gif
 
#11
Re: Anyone Physics majors or professional on the Hide?

Great posts guys!

Tim, your post is awesome and tend to fall in line with what I am thinking. I especially like the idea of thinking of this in a "percentage lost" type of way.

I think in the end there will only be one way to prove this, but the theory is interesting too. Just wrapping my head around it has been fun.
 
#12
Re: Anyone Physics majors or professional on the Hide?

Switchblade, yes your post makes sense. This is strictly based towards steel targets. Big and slow would be great if it weren't for recoil and accuracy. We shoot A LOT which is why a lot of us are moving to the 6.5mm versus the 30 cals. We aren't looking for hyper velocity light bullets, but we do need them to expend their momentum into the target.

Just for clarity, as far as I know... all bullets in the common calibers are disintegrating upon impacting the Ram at 550yds.
 
#14
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: flounderv2</div><div class="ubbcode-body">You need to know what the energy required is to completely disintegrate the bullet, at what speed/angle, and if you want to be really picky, the time it takes to destroy the bullet. Without that information, you cant do a true energy transfer comparison between a non destroyed bullet (Mushroomed) or a destroyed bullet.

My personal opinion is that the argument is moot. If there is enough energy to destroy the bullet then the same amount of energy has already been transferred to the target as what a bullet would that mushrooms. <span style="font-weight: bold"><span style="font-size: 14pt">The amount of time that the energy transfer is taking place is the real difference as the mushroomed bullet may impart the energy for a greater period of time. </span></span>(However we are talking thousands of a second so again, I think its a bit moot) </div></div>

I think this is spot on. (Except the mushroom part. All the bullets are coming apart regardless of caliber. In this scenario anyway)
 
#16
Re: Anyone Physics majors or professional on the Hide?

An object put in motion tends to stay in motion until acted upon by an outside force. Other than environmental factors and gravity, the outside force here is the steel.

All the momentum is translated to the steel, it does not matter if the built disentigrates or not. The object has its momentum, derived by its weight times mass, until it hits something.

If a lighter projectile going fast enough to equal the momentum of a heavier projo going slower, it will deliver the same amount of energy.

Something else to consider is where the ram is hit. A straight hit will transfer the energy at the same vector as the tragectory. A side hit will do the same, but you now have a rotational component. This gets into moment of inercia. If most of the mass of the target is located in the center, a side hit will twist it more than if most of the mass is located at the edges. Same theory applies if it is hit high or low.

On a target that is hindged at the ground, it requires more energy to be transferred at the bottom than the top to overcome its static position.

All that said, a 6.5 mm projo that leaves the muzzle at 2600 fps can flat out mess some shit up...you should not have any issues there.
 
#17
Re: Anyone Physics majors or professional on the Hide?

gugubica,

Yes, "marginal hits" almost always take the animals down regardless of caliber due to leverage. I've taken down the Rams with my .22-250 with 55gr bullets by hitting them near the very edge. I should have clarified this in my original post but I didn't want to get too long winded.

Where we see most "ringers" is when the Ram is hit in the center mass which is just up in the body above the front leg (heart shot). I also want to note that this isn't terribly common, but in a sport where absolutely every point counts...

DollarRamMedium.jpg
 
#18
Re: Anyone Physics majors or professional on the Hide?

Looking at the photo, it appears that the target is not hindged, but sitting on a beam.

I this case, a side hit would be more effective. The reason is you are rotating the target enough for it to fall over. If it were hindged across the entire bottom, it would probably be a different story.

Curious if the target is ever sitting on a large flat surface? Would be interesting to see how much a side hit rotated the target.
 
#19
Re: Anyone Physics majors or professional on the Hide?

No, they are always on a rail. For the hit to count at least one foot has to be touching the ground. They almost always fall completely off the rail except in the cases of "ringing".

We shoot offhand with absolutely no support and no slings in silhouette, so bullet placement is a pretty tough thing to control as you can imagine. I think on a good day I probable can manage a 3 to 5 MOA hold. Trigger control is key.

I too would be curious how much they would rotate. The Rams weigh 53 pounds with an obvious bias towards the front.
 
#20
Re: Anyone Physics majors or professional on the Hide?

When Chicken Lady and I play Wii bowling I can get more strikes if I slow down and don't throw the ball too hard..........if that helps.
 
#21
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Switchblade</div><div class="ubbcode-body">You are overthinking it with all the math. It comes down to the old Wiley Clapp vs Elmer Kieth argument of fast and light vs big and slow. FAst and light will not knock it down, but big and slow will. This is because a larger, heavier slower bullet will transfer all it;s energy into the target, where a small faster bullet will not.
This is most noticable in fleshy targets. Small and fast go through cleaner/clean. Bigger and slower go into the target making a larger hole and transferring more energy. ON a metal target, the speed and velocity of the smaller bullet has to go somewhere if not straight forward. This is to the outer ring, 360*, the bullet splatters, the energy follows the splatter. The bigger slower bullet does not shatter, but compresses and transfers more energy into the metal thus knocking the target down.
You can run a 175FMJBT down to 2400fps and watch it knock the target over each and every time. Speed it up, or use a faster lighter bullet, and all you get is DING!
See, no math at all, don't need it. The argument has been settled by men much older than us years before we were born, or at least while we were young enough to chase 19 year old tail and do it without insults
wink.gif
</div></div>

+1

Physics terminology: Momentum Transfer. Or as we look for in hitting live targets.. Energy Transfer.

Also, Even in real bowling I see a slower ball get more pin action than a faster ball in most instances as the deflection angle off the pins with a slower ball will have more contact with other pins during the distance of the set. (It also does the same thing on my PS3)
 
#23
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: VA Gentleman</div><div class="ubbcode-body">Then again, thinking of Big and Slow and Mass and Energy... I am sure Shankster could probably offer an insight with visuals of which words would not be needed for explanation. </div></div>

The Quiver factor.

Q = M/V

Q equals rate of quiveer
M equals mass
V equals velocity

Of course this does not allow for D (density of the mass being impacted), as I try to keep my distance from that much lard and had no opportunity to test it..
 
#24
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: maggot</div><div class="ubbcode-body"><div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: VA Gentleman</div><div class="ubbcode-body">Then again, thinking of Big and Slow and Mass and Energy... I am sure Shankster could probably offer an insight with visuals of which words would not be needed for explanation. </div></div>

The Quiver factor.

Q = M/V

Q equals rate of quiveer
M equals mass
V equals velocity

Of course this does not allow for D (density of the mass being impacted), as I try to keep my distance from that much lard and had no opportunity to test it..</div></div>

LOL... The only true physics equation that can rattle the intangible ID side of the brain and cause nightmares.

Sorry to hijack a great topic.. My bad.

Back to regular discussion please...
 
#25
Re: Anyone Physics majors or professional on the Hide?

Alrighty, give this a go:

Imagine you have your steel plate, and enclose this plate in an imaginary boundary so you can keep track of anything that enters or leaves the boundary. Any amount of energy entering the boundary has to leave the boundary also, officially called the Conservation of Energy.

Now, imagine your bullet traveling ~1800 FPS just outside this boundary. Once the bullet crosses the boundary, it brings with it all it's energy, which can be determined with the equation 1/2*m*V^2, or half the mass times the velocity squared. Now that that amount of energy has crossed the boundary, that amount of energy has to leave the boundary in any way.

Ideally, all that energy would go into pushing that plate through the boundary (using the same equation as before). For this to happen, no part of the bullet can pass through the boundary again because those fragments carry energy that could have gone into moving the plate (minus fragments that just fall due to gravity). So any bullet fragments that fly off after impact carry an amount of energy with them and that is energy that cannot go into pushing the plate, or in this case, wasted energy.

This means that the ideal situation would be a bullet that impacts the plate at the highest velocity possible <span style="font-weight: bold">without</span> splashing or having any fragments come off at any appreciable velocity. So, if all the bullets are splashing, the best would be the most massive bullet, traveling the slowest.

This is all ideal though, so it sounds nice, but is mostly useless. In a real situation, energy is lost to heating, sound generation, and deformation of the plate and the bullet. The best way to check is to do a field test of whether the above holds true or not.

As far as where to hit the target would be, it would depend on what the target is sitting on. Ideally the best would be at the center of mass (where it would balance on a point). An impact at the center of mass would push the target, whereas an impact anywhere else would push the target as well as try to rotate the target. If that rotation helps to tip the target off a narrow ledge, it might be advantageous to try that instead of a center of mass shot.

Sorry for the long-winded answer, hope it helps!
 
#27
Re: Anyone Physics majors or professional on the Hide?

Jackimho,

Nice post. Well thought out. I have been thinking about fabricating two ballistics pendulums because of the exact scenario in your post. One pendulum would be the typical "capture" type pendulum that the projectile will catch the bullet and all of it's momentum can be recorded by the pendulum. The other one, the one that will serve more for my purpose will be an armor plate pendulum that will basically allow me to record "momentum lost" due to bullet fragmentation.

There is no bullet retention at these distances. All bullets blow up when they hit the rams. Well... if you lobbed a .22LR out to 550yds and managed to hit the ram I imagine it would retain most of its weight.

In my sport (Metallic Silhouette) there is a rough estimate that floats around via word of mouth that to reliably knock down the Rams you need 1,000 foot pounds of "energy" on target.

My plan with the armor plate ballistics pendulum is to pick the most popular 6.5mm bullets and start at the low end of velocity and work my way up to max loads and see if at any point my ballistics pendulum records a downward trend in momentum as velocity increases. This will do two things; 1) It will tell me which bullets hold together the best and deliver the energy to the Ram instead of outwards. 2) It will tell me if in fact there is a velocity at which due to bullet disintegration the momentum decreases.

I am hoping to build the pendulum and test this before winter sets in here in Colorado.
 
#28
Re: Anyone Physics majors or professional on the Hide?

Sounds like a great idea! I'd be interested to hear the results.

As far as the 1,000 ft-lb of energy goes, that sounds about right. I don't know the weight of the Ram, but to deliver that much energy to it, with a 120 grain bullet, it would have to be traveling 1,935 feet/sec and a 165 grain bullet would have to be traveling 1,650 feet/sec (as long as you trust my late at night math), both of which seem to be in the range you were talking about earlier.

It seems like the heavier bullet would be a better choice there , since it requires less velocity to deliver the energy and would retain it's velocity better since it is heavier.
 
#29
Re: Anyone Physics majors or professional on the Hide?

I'm reaching way back. KE doesn't cut it. It like saying if you could get a bumble bee going fast enough you could stop a fraight train.

We did this in Accident Re-Construction Class. Figuring the speed a bumble bee weighing X would have to go to equal the KE of a fraight train weighing Y. Using KE only, in theory it could be done.

But, we know that's not going to happen. A good example is hitting said bumble bee while driving down the road. The Bee will splatter and do no damage to the windshield. Now take a piece of gravel, weight the same as the bumble bee. You are going to damage your windshield.

So, now we have to determine the damage done to the two objects. In the case of the Hardened Ram, there will be little damage. The damaged is all in the projectile.

Regardless of which object (or even both) you can determine the speed of the two objects and momentum needed to do the damage but you need the recovered objects (bullet and target). With the two, you should be able to determine the momentum needed to "knock over" without damaging the ram, with the bullet. And just short of that "knocking over" the same object (bullet) would have to be going.

I don't remember the fomular, I've been out of the game for quite a while.

It's all laid out in Baker's "Accident Investigation" put out by the Northwest Traffic Institute.

Mr Baker explained how to determine the angles, and modify those angle to work out a direct hit. Granted we were doing this on vehicles which we had to take into account drag factors, but you also have a drag factor for air, without the road surface.

What we did is find determine how fast two objects had to go for one object to move the second object backwards, or in this case, how much one object would move a stationary object.

Yes you need KE, and momentum, but you also need to know what it takes to damage the object, finding both direct and induced damage.

Like I said, its been a long time, but if one was to get Mr Baker's book, its laid out. Get ready to smoke your slide rule or calculator.

The 7-08 was designed for Metallic Silhouette in mind. It pushes a 150 grn 7mm bullet fast enough to tople the sheep. It seemed to be the perfect match for bullet weight, velocity, accuracy, and ease of recoil on the shooter for knocking over rams.

I see no reason to re-invent the wheel, but if we just want to know, to play with the math, Mr Wilson's "Traffic Investigation" would be the sourse I would go to.
 
#30
Re: Anyone Physics majors or professional on the Hide?

Fantastic post Kraig. I love the bumblebee analogy and it does actually put this into perspective. The Ram is the windshield and the bullet is the bee. Nearly identical results too! Splat! Your post brings a lot of light to the subject.

I currently shoot a 7mm-08 for my Hunter class gun and it performs great. I am building a second gun for my "Standard" class gun in 6.5 Creedmoor. Variety is the spice of life!

Definitely not trying to reinvent the wheel. The .308 was king in MS Silhouette back in the day, then it was replaced by the 7mm-08. Now it seems that the 6.5mm cartridges (.260rem, 6.5x47) are gaining more and more ground over the 7mm everyday for the same reasons it phased out the .308win. Less recoil and flatter trajectory.
 
#31
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Tim K</div><div class="ubbcode-body">Energy is lost when the bullet disintegrates. That is energy that is not delivered to the target.

Test this in your head by picturing someone throwing and egg at you at 90mph. Now picture the same big league pitcher throwing a steel ball the same weight as the egg at 90 mph. The effect on the target will be radically different. </div></div>


very well put Tim. This is more of a case of bullet construction than diminishing returns.

Try some 110gr Barnes Banded Solids. I bet these would knock these targets down better than the 140s.

http://www.midwayusa.com/viewProduct/default.aspx?productNumber=811212
 
#32
Re: Anyone Physics majors or professional on the Hide?

You should find that is comes down to energy deflection. You may be able to tell a little bit buy looking at the "splash" of the bullet marks on a white steel target. At higher velocity the splash may be measurably larger than at slower velocity. By using this theory you would think that F=A*P would show the pressure/energy being trasmitted to the area. The problem is that when the bullet expands faster, the fragments loose energy to heat and drag during the sideways expansion of the bullet. In theory the P-pressure would drop more rapidy than at slower velocity and a smaller "splash" All things being equal that is.

Just some more food for though. I may have to try to prove this out. Might be able to use a strain gauge to measure it.
 
#33
Re: Anyone Physics majors or professional on the Hide?

Kraig, I think you are talking about impulse of a force and collision types, which I agree should play an important role in this.

Impulse is essentially dissipating collision energy over a set period of time, the longer a collision takes, the less overall force is applied with the same amount of energy being dissipated. Imagine you fall off a roof, if you land on your concrete driveway, you will probably be hurt pretty badly, because the collision between you and the driveway takes very little time because there isn't much "give" in the concrete. Now imagine you fall from the same height, but this time you land in a snowbank that is a few feet deep. In all likely hood, you'll be fine because the snow allowed you to slow down from the same speed as before, just over a longer time.

As far as collision types go, there are plastic collisions, elastic collisions, and collisions that are a combination plastic/elastic. Plastic collisions are where two objects hit and there is no "bounce", or the objects stick together and stop all motion because their kinetic energy is used up through deformation (i.e. two clay balls hitting each other). Elastic collisions happen when there is a perfect bounce and after the collision, the objects continue on in a new direction with the same amount of energy as before. Elastic collisions are theoretical only, they cannot occur because some energy is always lost in the collision, plastic collisions very rarely occur. Most collisions are combinations, with the two objects deforming somewhat as well as continuing on in a new direction with a portion of their initial energy (like a car accident).

Going back to the bee and the windshield example, that is an example of a high impulse, mostly plastic collision. All the impact energy is used in a very short time and almost all is going into deforming the bee, while small amounts go into heat generation and generating the sound you hear as well as vibrating the windshield.

Now if a piece of gravel hits the windshield, that is again a high impulse collision, but this time it's mostly elastic, since the gravel takes a lot more energy to deform, it bounces off instead of splattering. Because the gravel is not deformed like the bee was, that extra energy is transmitted to the windshield and this is what breaks the windshield.

Going back to the bullet and the Ram, that is similar to the bee and the windshield. It's a high impulse, mostly plastic collision, with some of the energy going into deforming the bullet, and some going into moving the Ram, as well as some that is wasted in heat and sound generation.

Ideally, the most energy is transmitted to the Ram when the bullet strikes it and sticks to it, meaning that no energy is wasted in deforming it. Since this can't actually happen in the real world, the best solution is when the bullet strikes it and the least amount of energy goes into deforming it as possible. Since we know that all the bullets are splashing on impact, the best would be one that splashes the least, or the slowest round. Since you also want to maximize momentum, the highest mass bullet with the slowest speed would be best.

I initially ignored impulse because any bullet impact is a high impulse impact, whether it's going 1,600 FPS or 2,600 FPS, all the collisions will happen extremely fast.
 
#34
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: Jackimho</div><div class="ubbcode-body">Kraig, I think you are talking about impulse of a force and collision types, which I agree should play an important role in this.

Impulse is essentially dissipating collision energy over a set period of time, the longer a collision takes, the less overall force is applied with the same amount of energy being dissipated. Imagine you fall off a roof, if you land on your concrete driveway, you will probably be hurt pretty badly, because the collision between you and the driveway takes very little time because there isn't much "give" in the concrete. Now imagine you fall from the same height, but this time you land in a snowbank that is a few feet deep. In all likely hood, you'll be fine because the snow allowed you to slow down from the same speed as before, just over a longer time.

As far as collision types go, there are plastic collisions, elastic collisions, and collisions that are a combination plastic/elastic. Plastic collisions are where two objects hit and there is no "bounce", or the objects stick together and stop all motion because their kinetic energy is used up through deformation (i.e. two clay balls hitting each other). Elastic collisions happen when there is a perfect bounce and after the collision, the objects continue on in a new direction with the same amount of energy as before. Elastic collisions are theoretical only, they cannot occur because some energy is always lost in the collision, elastic collisions very rarely occur. Most collisions are combinations, with the two objects deforming somewhat as well as continuing on in a new direction with a portion of their initial energy (like a car accident).

Going back to the bee and the windshield example, that is an example of a high impulse, mostly plastic collision. All the impact energy is used in a very short time and almost all is going into deforming the bee, while small amounts go into heat generation and generating the sound you hear as well as vibrating the windshield.

Now if a piece of gravel hits the windshield, that is again a high impulse collision, but this time it's mostly elastic, since the gravel takes a lot more energy to deform, it bounces off instead of splattering. Because the gravel is not deformed like the bee was, that extra energy is transmitted to the windshield and this is what breaks the windshield.

Going back to the bullet and the Ram, that is similar to the bee and the windshield. It's a high impulse, mostly plastic collision, with some of the energy going into deforming the bullet, and some going into moving the Ram, as well as some that is wasted in heat and sound generation.

Ideally, the most energy is transmitted to the Ram when the bullet strikes it and sticks to it, meaning that no energy is wasted in deforming it. Since this can't actually happen in the real world, the best solution is when the bullet strikes it and the least amount of energy goes into deforming it as possible. Since we know that all the bullets are splashing on impact, the best would be one that splashes the least, or the slowest round. Since you also want to maximize momentum, the highest mass bullet with the slowest speed would be best.

I initially ignored impulse because any bullet impact is a high impulse impact, whether it's going 1,600 FPS or 2,600 FPS, all the collisions will happen extremely fast. </div></div>

Excellent explanation, Jack....that even a layperson can clearly follw. Your ok for a Newbey. Is this area your field of work?
 
#36
Re: Anyone Physics majors or professional on the Hide?

<div class="ubbcode-block"><div class="ubbcode-header">Originally Posted By: maggot</div><div class="ubbcode-body">Excellent explanation, Jack....that even a layperson can clearly follw. Your ok for a Newbey. Is this area your field of work? </div></div>
Thanks, not my field of work, but I did just finish up my degree in Mechanical Engineering a couple years ago. Right now I don't really do much with it, so when I get a problem, I like to sink my teeth into it and see if I still remember anything! I will admit, I busted out more than one textbook to double-check myself on this one.
 
#37
Re: Anyone Physics majors or professional on the Hide?

Well, I'm 3 years out from getting my major, but here's a go. In simple physics (complicated adds impulse), bullet disintegration will not affect the momentum transfer as it regards to the target, which is what you need be concerned with; energy transfer to the target only matters in collateral damage to the target, and we are dealing with a collision which involves moving the target, not affecting it's integrity we are talking about momentum. The elasticity of the collision is irrelevant as no matter what linear momentum is conserved; energy is discharged as heat and sound, but linear momentum is conserved. In a simple model where the bullet is traveling on the x axis and the target plate is in the yz plane, when the bullet impacts the target and franges but does not deform the target in any manner, the fragments fly off parallel to plate and the sum of momentum of all the pieces cancel out, as for every piece of mass moving at some speed in some direction there is some moving in the opposite direction at the same speed in the big picture. This means that all the momentum the bullet had before it impacted is transferred to the plate. The main reason you're probably seeing some targets drop and some not is because you are talking about torque and angular momentum since the forces we're talking about are not necessarily in line with the center of mass of the target. If you hit low with a bullet of relatively low momentum you won't get the torque that will cause the target to tip over but instead push it off the platform, and may even increase the friction between the target and whatever it's sitting on if the bullet was traveling "downhill" when it impacted the target. I can make that more understandable if need be, I have pretty poor communication skills.
 
#38
Re: Anyone Physics majors or professional on the Hide?

Left out the part that if the bullet frags and some parts were to fly backwards towards the shooter then the momentum transferred to the target is even greater as the momentum delivered is equal the difference in the momentum of the projectile before and after the collision, and if fragments head back to the target then you have negative momentum. For example bullet hits target at 1500fps then returns to sender at 1000fps then (mass of bullet)*(-1000fps)-(mass of bullet)*(1500fps)= -(mass of plate)(xfps) (that is if the bullet hits in line with the center of mass of the plate). Same type of calculation if the angle of impact isn't 90 deg to the target, just in multiple coordinates. But that is all simplified. In the real world we need to talk about something called impulse, which is a measure of force imposed over a time period. As the fragmentation is very fast, the time period over which the bullet can put a force on the target is diminished. This is similar to mushrooming where the momentum isn't consistent, as not all the mass of the bullet impacts the target with the same speed. From my understanding these impacts are not very significant, but because we are dealing with small objects and large speeds, relativity says they are more significant, but it shouldn't be very much. The torque you gain or lose is much more significant, as you can easily multiply the effect you have on the target
 
#39
Re: Anyone Physics majors or professional on the Hide?

I like shooting more than maths. I think that the best way to sort this out is with some field testing.
The pendulum might be a bit hard to make and get accurate results so why don't you do this....

Put a solid steel target onto a flat smooth surface like an old table top. Mark the table at the front of the target with a pen or place the target at the edge of the table. (frag will not damage the table this way and the wife will be happier).
Then shoot the target at distance with load #1 and see how far back it is pushed.
Now reset the target back to its original position and try load #2 and see how far back it is pushed.
You can keep doing this as much as you like with different loads and it should answer all your questions.
 
#40
Re: Anyone Physics majors or professional on the Hide?

wadcutter has a good idea.
I think what you are seeing with some hits knocking the ram over and some not has more to do with shot placement than velocity,providing your loads are over the energy threshold.
I also think that higher velocity will always increase your chances of knocking the ram over.The energy departed to the ram will increase with velocity increase.The ram has to cause the bullet material to deflect laterally,and this deflection is what imparts energy to the ram.The more "violent"the deflection, the more energy imparted to the ram ,( keeping the bullet weight the same).
 
#41
Re: Anyone Physics majors or professional on the Hide?

Oh, for sure bullet placement is critical. But not necessarily controllable. 550yds offhand is a difficult feat. Below is a star that marks where "ringers" usually happen. Hitting that spot isn't a guarantee of a ringer but it is where it generally happens. There are other outside forces that come into play as well, like wind from the back pushing against the backside of the Ram, or a stubborn Ram that might lean slightly toward the shooter because the feet are welded on slightly crooked. Even mud between the feet and the rail can be a stiction issue.

ea9a6bd9.jpg
 
#42
Re: Anyone Physics majors or professional on the Hide?

A few more things I thought of:

When we shoot IPSC and USPSA handgun matches, the guys shooting the 9mm have a more difficult time getting the steel targets to tip (yeah, they should probably be calibrated better, but target set-up isn't an exact science). Just today we had guys shooting 9mm that had to shoot the steel 3-4 times to get it to fall, whereas the guys shooting the heavier and slower .40 and .45 had no trouble getting the targets to fall.

RideHPD: I think you're on the right track here, but I would argue that bullet disintegration does affect momentum transfer. The only momentum in the system that can be conserved is the momentum of the bullet that is then transferred to the target. Ideally, all the momentum is transferred to the target, i.e. when the bullet hits and falls with no splash. When the bullet does splash, those parts of the bullet that fly off into the wilderness do so with momentum that can not be transferred to the target anymore, even the pieces that fly backwards (-x direction), any part that requires a force/energy to move, gets that energy from the initial momentum of the bullet, and that force/energy can no longer be applied to the target. Your example as written is the collision equation, just a little jumbled:

(mass of bullet)*(-1000fps)-(mass of bullet)*(1500fps)= -(mass of plate)(xfps)

should actually be:

(mass of bullet)*(1500fps)= (mass of plate)(xfps) + (mass of bullet)*(1000fps)

or:

m1*v1 + m2*v2 = m1'*v1' + m2'*v2' (notice the "primes" on the right side)

where m1 and v1 are the mass and velocity of the bullet and m2 and v2 are the mass and velocity of the target. Assuming the masses of the bullet/fragments and target are fixed, the m2*v2 term on the left cancels out because v2 equals 0 prior to impact. After impact (the right side), ideally the m1'*v1' term cancels out because v1' would be zero (bullet falls after impact), meaning v2' would be maximized. If the fragments still have velocity, the v1' would be non-zero and the v2' term would be reduced.

Yell at me if you still have a problem with it, I said I was an engineer, I never said I was a <span style="font-weight: bold">good </span>engineer.
grin.gif


Another thing I wanted to add, a lighter bullet traveling fast enough could potentially overcome the advantage of the heavier, slower bullet, which is math I don't want to do, so I'm all for field testing this!
 
#43
Re: Anyone Physics majors or professional on the Hide?

Exactly, that's what I was getting at in the second post regarding impulse and how long the fragments have to impose force on the target, how much this is I do not know off the top of my head, and having less experience I'll default to you. I really should have made that whole post clearer as I was dead tired and just kind of typing whatever ramblings were going through my brain. I'll try to write some code to model it on my computer and see how things go; this is really an interesting problem. Also, you're right that I did not write the momentum equation in the standard format but I my intention was to move things around for those on here who have a better understanding of the idea of "every force has an equal and opposite force," where the difference in bullet momentum before and after leads to an equal momentum change but in the opposite direction. Just didn't know if you caught that or not.
 
#44
Re: Anyone Physics majors or professional on the Hide?

Not trying to answer the Physics question, but rather a possible solution, have you tried a tipping die on your bullets? There are those that would have you believe a heavier bullet (of same general shape) will always have the higher BC-they live in their own little world, and are wrong. A tipping die can increase the BC quite a bit, maybe just enough to knock that ram over, due to the higher impact vel. When you look at the Sierra 155SMK Palma, you can see what happens when you run a bullet through a good tipping die(compare 155 SMK, with 155SMK Palma). The Smaller the Caliber, the greater increase you should be able to get with a tipping die.
 
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