Ballistic Pendulum

[colonelhogan44]

For one of my senior engineering classes, a group of us constructed a ballistic pendulum. We shot it today, and I have to say, it was the most fun I've ever had on a school project. Ever.

http://www.thehighroad.org/attachment.php?attachmentid=141102&d=1303712210

http://www.thehighroad.org/attachment.php?attachmentid=141103&d=1303712210

http://www.thehighroad.org/attachment.php?attachmentid=141104&d=1303712221

Results are attached.

http://www.thehighroad.org/attachment.php?attachmentid=141101&d=1303711497

Looked very reasonable to me, without having a chronograph to check against (unfortunately I don't even know anyone who has one...never even seen one used. After nearly 10 years of shooting and 2 handloading, that's pretty pathetic). The Winchester 3 inch slugs were very impressive performers, giving our 123 pound pendulum almost 5 inches of deflection (versus about 3 for a battle rifle class cartridge.

Has anyone ever ran a similar load out of a similar rifle/shotgun over a chrono?

As a side note, my cast 158 grain slugs over 13.5 grains of 2400 registered at a consistent 1600ish fps with zero leading; at 14 grains, they lead like no tomorrow.

[stubshaft]

IIRC - P. O. Ackley wrote about it's construction and subsequent use in one of his books. I always meant to try and construct one but never got around to it.

[303Guy]

That could only have been fun!:mrgreen:

Did you make any attempt to compensate for heat losses? Like a battle rifle bullet might 'read' lower than a 357 by virtue of greater penetration and hence greater heat losses. How would one know how much compensation to apply?

Heck, that would be great to use as a target at longer ranges too (as long as the suspension line or pen doesn't take a hit. [smilie=1:)

Great project!:drinks:

[colonelhogan44]

There is no compensation needed for losses (for heat, noise, deformation, etc.). Those are all losses in energy, which is never conserved in real collisions. Momentum, however, is conserved in an inelastic collision, which is the principle on which a ballistic pendulum operates. It would be impossible to get decent results if you had to compensate for those losses.

Conservation of Momentum:
(mass of bullet)*(velocity of bullet) = (mass of pendulum)*(velocity of pendulum)

[colonelhogan44]

I never even thought of putting it at long range...that would be a blast. After the report is written (BIG job, probably 15 hours...ugh), I'll have to play with it some more.

[Wayne Smith]

I get no pics, just red x's.

[mdi]

I get no pics, just red x's.

Me too! Is my 'puter lacking some kind of program to make it compatable with colonelhogan44's pics?

[theperfessor]

Hey ColonelHogan, shouldn't the second part of your equation include the mass of the bullet+the mass of the pendulum? I'm assuming your bullet stays in the pendulum.

[colonelhogan44]

The bullet mass is accounted for in the equations in the spreadsheet. Even if I neglected it, though, it would make less than 1 fps difference, due to the huge difference in weight (pendulum is ~124 pounds)

Let's try the pics again.

http://i586.photobucket.com/albums/ss301/colonelhogan44/100_3639.jpg

http://i586.photobucket.com/albums/ss301/colonelhogan44/100_3640.jpg

http://i586.photobucket.com/albums/ss301/colonelhogan44/100_3643.jpg

http://i586.photobucket.com/albums/ss301/colonelhogan44/Results-Full.jpg

http://i586.photobucket.com/albums/ss301/colonelhogan44/Results-Breif.jpg

I forgot that the pictures were hosted on another forum's server, and I tried to cheat and not have to upload them twice...so those of you who are not members and/or not logged in over there could not see them.

[Kraschenbirn]

Back in high school, a friend and I built solid-fuel rockets (the real thing...using machined deLaval nozzles and shaped solid-fuel charges we cast ourselves!) as an ongoing science project. In the process of developing/recording test data, we constructed a static test chamber that used the principle of the ballistic pendulum to measure thrust. Worked really well and the numbers derived from it matched quite closely the empirical results from actual free-flight test firings.

Alas, though, the apparatus didn't survive our final series of test firings at the end of our senior year. A rocket exploded in the stand (possibly a cracked fuel charge) and blew the carrier assembly and a goodly portion of the pendulum arm into large hunks of shrapnel. Somewhere I've got a photo of the explosion, taken from a single frame of the 8mm movie shot by our project advisor.

[ItZaLLgooD]

This is a great idea. About how long did it take to build? Does break down enough to move or is it not worth the hassle?

[HammerMTB]

Back in the day ('80-'81) (that was the last century) we did not have access to a reliable, cheap chrono.
So for the purpose of detecting major/minor power factors in IPSC, we had a pendulum. Ours was very basic- if you moved the pendulum past the mark, you made major.
They certainly are interesting. But these days a chrono is so easy to come by and use, the pendulum seems to me to have gone the way of the slide rule. I DO have a slide rule in the bookcase next to me. I just don't use it any more.
Cool project tho! I bet it was fun to thump with 12 ga slugs!

[colonelhogan44]

Kraschenbirn:
I actually dabbled in rockets just like that as a sophomore, we were making good progress when the spring semester ended, and summer break swept our rocket program into the history books. It's fun stuff. We made fuel out of sugars and potassium nitrate cast into cylinders with a small hole through the center using machined nozzles as well.

ItZaLLgooD:
It took about 10 hours to build and get the testing done over two Saturdays. The pendulum part disconnects easily from the frame and then both can fit in the bed of a pickup without much trouble.

HammerMTB:
More like the 12 gauge slugs had fun thumping me...:holysheep those things kick like a mule from the prone position...50ish ft-lb of recoil of energy is no laughing matter.
I wish I had a case of those things. So awesome.

[nanuk]

is it true that a boolit that bounces back, makes the pendulum move more?
and to find a true comparative value, the boolit must stay with the pendulum?

[303Guy]

More like the 12 gauge slugs had fun thumping me... :holysheepthose things kick like a mule from the prone position...Holy Sheep - you didn't shoot a 12 gauge slug prone! Hee hee! Bet that hurt.:bigsmyl2:

Hey you guys had more fun than the legal limit, building rockets!:bigsmyl2:

[colonelhogan44]

is it true that a boolit that bounces back, makes the pendulum move more?
and to find a true comparative value, the boolit must stay with the pendulum?

No and Yes.

If the bullet bounces back in a perfectly elastic collision, both momentum and energy are conserved, but the bullet would come back at a very high speed, and the pendulum would move less.

In the real world, there are no perfectly elastic collisions, but there are perfectly inelastic collisions (bullet being captured) in which momentum is conserved. So, yes, the bullet must be captured to get accurate results.

[montana_charlie]

Looks a little like a Montana anemometer ...

[nanuk]

No and Yes.

If the bullet bounces back in a perfectly elastic collision, both momentum and energy are conserved, but the bullet would come back at a very high speed, and the pendulum would move less.


kind of went over my head....

but I do recall reading about that very thing, regarding which bullets knock heavy Rams down in the sillouette game. and the bullets that bounced back faster, actually imparted more energy into the target than ones that flattened and fell down.

something about equal force in the opposite direction.

thus the pendulum would swing MORE if there was a solid plate where the bullet bounced back. thus that was why the pendulum to be accurate had to capture the bullet.

I believe I read this in an NRA document.

[Beerd]

Looks a little like a Montana anemometer ...

or a North Dakota weather gauge:

"if it's wet, it is raining"
"if it's frozen, it is cold out"

you get the idea
..

[fourarmed]

Nanuk is correct. If the bullet bounces back, it imparts more momentum to the pendulum. You can get a lot of argument over this from silhouette shooters and others who slept through that part of the physics course, but it is true. The easiest way to think about it is by Newton's third law. The pendulum hits the bullet with the same force that the bullet hits the pendulum. The pendulum has to hit the bullet harder to reverse it than it does merely to stop it or slow it down. Therefore the bullet hits the pendulum harder when it gets reversed.

[colonelhogan44]

I stand corrected. I did the calcs, and the pendulum would move almost twice as fast after hit in the case of a perfectly elastic collision.

[nanuk]

Nanuk is correct.


first time for everything I guess..... :bigsmyl2:

[303Guy]

Hang on - Total momentum before the collision equals total momentum after the collision. So the projectile bounces back at the same speed it struck and therefore posesses the same momentum as it did before but now the pendulum has double the momentum .... ? That can't be right. The law of conservation of momentum states: momentum before = momentum after. I'm seeing three times the momentum in this scenario!:confused:


This is what a perfect elastic collision should look like.
http://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Newtons_cradle_animation_book_2.gif/200px-Newtons_cradle_animation_book_2.gif

[colonelhogan44]

This is what a perfect elastic collition should look like.
http://upload.wikimedia.org/wikipedia/commons/thumb/d/d3/Newtons_cradle_animation_book_2.gif/200px-Newtons_cradle_animation_book_2.gif

For equal masses, yes. for unequal masses, you'd get the smaller one bouncing back at a slightly lower speed than it initially had, and the larger mass with a small velocity.


So the projectile bounces back at the same speed it struck and therefore posesses the same momentum as it did before but now the pendulum has double the momentum .... ? That can't be right. The law of conservation of momentum states: momentum before = momentum after. I'm seeing three times the momentum in this scenario!:confused:
It would appear that way, but momentum is a vector quantity, meaning it is direction sensitive! if the bullet comes back in the opposite direction, with nearly the same speed (nearly opposite velocity -- vector quantity as well..) momentum is conserved.

[fourarmed]

Unlike energy, momentum is a vector quantity. The momentum of the bouncing bullet is negative.

Oops! Didn't see Colonel Hogan's last statement, which is correct. Back in the late 70s Speer did experiments with a ballistic pendulum that had a steel face instead of wood, so that the bullets striking it shattered, and part of the bullet material was thrown backward, just as in silhouette shooting. They tested a large number of their bullets in different weights, velocities, and calibers, and found that there was remarkably little difference in the momentum transferred to the pendulum. It was about 110% of the incoming bullet momentum regardless of other parameters.

[CATS]

Is that one pine board in front of a steel box?!!! If that is so you run the risk of getting hit with back spatter. Not safe. Not safe. Not safe. No way that should be shot from that distance. Ok, so what did I miss that makes this cool rig safe? I am not trying to flame you, I just must have missed something. Neat idea by the way....
CATS

[mpmarty]

Momentum is energy. Thus if bullet has X velocity, strikes pendulum and bounces back at say .5 X then the pendulum only receives half as much energy. The returning bullet has what's left of the X momentum.

[colonelhogan44]

Is that one pine board in front of a steel box?!!! If that is so you run the risk of getting hit with back spatter. Not safe. Not safe. Not safe. No way that should be shot from that distance. Ok, so what did I miss that makes this cool rig safe? I am not trying to flame you, I just must have missed something. Neat idea by the way....
CATS

It's got 6 phone books and a midway USA master catalog inside of it. Perfectly safe.

[colonelhogan44]

Momentum is energy. Thus if bullet has X velocity, strikes pendulum and bounces back at say .5 X then the pendulum only receives half as much energy. The returning bullet has what's left of the X momentum.

Momentum is completely different than energy. Knowing one and masses allows you to calculate the other, but by no means are they the same. ;-)

[nanuk]

what I took from my reading is this

if a bullet is trapped in the pendulum, the pendulum receives the x momentum from the bullet.

if a bullet bounces back, not only does the pendulum receive the momentum from the bullet, BUT also receives extra (energy/momentum?) from the bullet, as the equal and opposite thing rears it's head.... something caused that bullet to change 180degree, and accelerate, and that (energy/momentum) is therefore also transfered to the pendulum in the opposite direction.

I know... it makes no sense to my finite mind, but it is calculatable using math

[303Guy]

"For any physical system, the mass, momentum, and energy of the system must be conserved. Mass and energy are scalar quantities, while momentum is a vector quantity."

The thing is, a vector quantity cannot be negative. You cannot have a negative velocity or a negative momentum. It's either stationary or it's moving - relative to something else. It's a vector quantity which means it has direction. It can't have a minus direction on top of that. It's like powder - you either have some or you don't. You can have a minus value for velocity or momentum relative to another to describe the direction is is moving in. So, the momentum of the bouncing boolit system must remain unchanged before and after the collision.


... if the bullet comes back in the opposite direction, with nearly the same speed (nearly opposite velocity -- vector quantity as well..) momentum is conserved - colonelhogan44.Here the pendulum takes on a very small velocity and momentum from the boolit.

[fourarmed]

Bullet mass m, velocity +v, Pendulum mass M, velocity 0. Initial momentum is +mv. If bullet bounces back with velocity -u (forward is positive, back is negative), then final momentum is

-mu + MU = +mv and U=(mv+mu)/M

If the bullet sticks in the pendulum then final momentum is

(m+M)U = +mv and U=mv/(m+M)

Clearly U (velocity of pendulum after collision) is greater in the first case.

This is basic physics, guys. It has been around since the time on Newton.

[garandsrus]

I haven't done any physics in quite a while, but I think that the formulas fourarmed posted are slightly incorrect.

If you solve the first formula, assuming a zero velocity for MU (really large weight that doesn't move):
-mu + MU = +mv
-mu + 0 = mv
0 = mv + mu

This can't happen. Twice the momentum of the bullet cannot equal to 0.

Here's what I think the formulas should be:

Bullet bounces back:
mu + MU = mv so U = (mv-mu)/M
This would mean that the momentum imparted from the bullet minus the momentum the bullet retains is what's transferred to the pendulum. In the above example where the plate doesn't move, 0 = mv - mu which works mathematically.

Bullet is absorbed:
(m + M)U = mv so U = mv/(m+M) (no change from above).

U would be larger in the second equation so the momentum of the pendulum would be greater if the bullet is absorbed.

I can understand where the sillywet folks prefer a bullet that bounces off the target. If the bullet splatters, much of the energy is dissipated in the "splattering" instead of the target. A bullet that bounces off would transfer that energy to the target instead, while the bullet is reversing course.

There is an experiment where you drop a basketball and tennis ball, with the tennis ball on top. The basketball doesn't bounce nearly as high as it does without the tennis ball, but the tennis ball shoots up much faster than it originally did. Here's a link I found: http://www.science-projects.com/Drop/DropBalls.htm. The basketball transferred some of it's momentum to the tennis ball. This should be similar to a bullet bouncing off a steel plate.

Here's another discussion that is interesting:

http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html Here's the summary:

It turns out experimentally that in any collision between two objects (where no interaction with third objects, such as surfaces, interferes), the total momentum before the collision is the same as the total momentum after the collision. It doesn’t matter if the two objects stick together on colliding or bounce off, or what kind of forces they exert on each other, so conservation of momentum is a very general rule, quite independent of details of the collision.

John

[fourarmed]

The equations are correct. The velocity of the pendulum after the collision is not zero. If that were true, what would be the point of a ballistic pendulum? Are you saying they don't work? People have been using them since 1740.

[303Guy]

Thanks garandsrus. You've saved me from having to actually think!:mrgreen: That's way beyond my capabilities!:roll:

On negative momentum or energy we would have to have either a negative velocity or a negative mass ...

See what I mean? An object (mass) is either at rest, i.e. has zero momentum or kinetic energy or it is moving, giving it kinetic energy and momentum. Just like South is not a negative North although it is in the opposite direction.

[dodgyrog]

Back in the day ('80-'81) (that was the last century) we did not have access to a reliable, cheap chrono.
So for the purpose of detecting major/minor power factors in IPSC, we had a pendulum. Ours was very basic- if you moved the pendulum past the mark, you made major.
They certainly are interesting. But these days a chrono is so easy to come by and use, the pendulum seems to me to have gone the way of the slide rule. I DO have a slide rule in the bookcase next to me. I just don't use it any more.
Cool project tho! I bet it was fun to thump with 12 ga slugs!

I remember an old friend trying to make Minor with a H&K P7 9mm on a pendulum. We'd all gone through and made major easily enough.
He shot several mags full of ammo at it with absolutely no success. Posted a DNF for the comp.
Prior preparation prevents piss poor performance?

[fourarmed]

There are a lot of things in science that are counter-intuitive. Doesn't mean they aren't true. Physics teachers love those things because they make for dramatic demonstrations.

As I tell my students in sophomore physics lab when they disprove one of Newton's laws, "Congratulations, you have just disproven a fundamental principle that has stood for 300 years. Before you apply for the Nobel prize, however, I recommend you get a second qualified opinion."

[Jim_Fleming]

One thing that Parker Ackley did specify in his book about the ballistic pendulum is that to get more accurate readings the weight of the pendulum had to remain constant.

I.E. if you're going to be shooting 30 shots at it, with 200 grain bullets, you should have a container mounted on the pendulum with 30 200 grain bullets in it... When you shoot a round, you remove a bullet. This way the pendulum's weight never varies...

I had both of his books at one time, not sure if I still do or not... Might've lost them to an ex-marital problem. Now y'all got me thinking about trying to find them.

Thanks, gentlemen.





IIRC - P. O. Ackley wrote about it's construction and subsequent use in one of his books. I always meant to try and construct one but never got around to it.

[garandsrus]

The equations are correct. The velocity of the pendulum after the collision is not zero. If that were true, what would be the point of a ballistic pendulum? Are you saying they don't work? People have been using them since 1740.

There are a lot of things in science that are counter-intuitive. Doesn't mean they aren't true. Physics teachers love those things because they make for dramatic demonstrations.


Fourarmed,

No, I am not saying that ballistic pendulums don't work. They certainly do. We used one in High School Physics class. We actually shot a .22 rifle in the school building! Not something that could be done today. Yes, I took physics in college also, but as I said, it was a long time ago.

If you replace the ballistic pendulum with an air hockey table my note may make more sense. The air hockey table has a very low coefficient of friction so the puck doesn't slow down much as it travels short distances. When a puck hits the side of the table, the table doesn't move. The puck bounces off the table side at relatively the same speed as it struck the table with, based on my observations. It definitely does not speed up, again, based on observation. This is an elastic collision which should be roughly equivalent to a bullet bouncing off a LARGE pendulum, although the collision would not be as elastic since the bullet would deform to some degree.

If the puck bounced off the rail with twice the momentum it hit with, which is what you are saying based on the -mu + MU = +mv formula) , pretty soon it would be going at the speed of light. The velocity would need to increase since the momentum increased and the mass didn't change.

Using these assumptions:
air hockey puck - 1 oz (m)
Velocity 1 f/s (u)
Velocity when bouncing back 1 f/s (v) (assumes same velocity on bounce back)
Air Hockey table - 5000 oz (M)
Air Hockey table velocity - 0 f/s (U)

Your first formula results in:
-mu + MU = +mv
-1x1 + 5000x0 = 1x1
-1 = 1 which cannot be true

and U=(mv+mu)/M
0 = (1x1 + 1x1)/5000
0 = 2/5000 which is of course not true.

Using the formula without the negative,
mu + MU = +mv
1x1 + 5000x0 = 1x1
1 = 1, which is true

If you disagree with the above, please provide an example with the equations and values so I can understand what I am doing wrong. The only thing that would make sense is that you would substitute a velocity of -1 for the bounce back to imply direction. The multiplication of the two negatives would of course be a positive, resulting in the formula I proposed.

Here's the text from Wikipedia showing the difference between speed and velocity: In physics, velocity is the measurement of the rate and direction of change in the position of an object. It is a vector physical quantity; both magnitude and direction are required to define it. The scalar absolute value (magnitude) of velocity is speed, a quantity that is measured in meters per second (m/s or ms−1) when using the SI (metric) system.

I think that most people use speed and velocity interchangeably when they really mean the scalar "speed".

Thanks,
John

[303Guy]

It's funny that. We always speak of the velocity of the boolit (which we actually do not know) and the speed of a car (even when we do know the velocity). Mind you, we also speak of boolit weight ..... :mrgreen:

[colonelhogan44]

Hopefully this will clarify and end the debate over momentum, energy, velocity.

First, here are the equations which govern a direct central impact scenario, condensed for your reading and edification pleasure.

http://i586.photobucket.com/albums/ss301/colonelhogan44/GEBP.jpg

Next, I have worked through an example, running the calculations for a "bullet" of 1kg mass, and a "pendulum" of 100kg mass, to illustrate the codependent relationship of the equations.

http://i586.photobucket.com/albums/ss301/colonelhogan44/GEBPex.jpg

Lastly, a review (or lesson for those not familiar) of:

Vectors

http://en.wikipedia.org/wiki/Euclidean_vector

Momentum

http://en.wikipedia.org/wiki/Momentum

Kinetic Energy

http://en.wikipedia.org/wiki/Kinetic_energy

[fourarmed]

garandsrus, your air table analogy fails because the air table is sitting on a surface that can exert more force on it than the puck can, so it doesn't move. If the air table were suspended from strings, like the ballistic pendulum is, then it would behave just like the pendulum. That is, if the puck hit and stuck to the rail, the table would move. If the puck hit and bounced off the rail, the table would also move, but with greater speed than in the first case.

Likewise, the fallacy in your equations is your assumption that U is zero. Unless the mass of the pendulum is infinitely large, that is an impossibility, and results in the absurdity that 1=-1. The pendulum is large, but not infinitely large. Therefore it moves, and its positive momentum, MU, more than makes up for the negative momentum, –mu, of the rebounding object. Note also that the speed of rebound must always be somewhat less than the speed of approach.

[fourarmed]

If you look at the results of the derivation provided by Colonelhogan, you will see what I mean. In the elastic case, the pendulum moves with a low speed (but a bigger one than in the inelastic case), and the bullet rebounds, but with slightly less than its original speed.

[garandsrus]

Fourarmed - Do you agree with the formulas and results provided by Colonelhogan? The formula he provided for conservation of momentum doesn't match what you suggested. The equation Colonelhogan supplied doesn't have a negative sign in front of the bullet velocity bouncing back. The bouncing back velocity he calculates is also not negative.

It would appear that the conclusion Colonelhogan came up with doesn't conserve momentum for an elastic collission. He has the pendulum moving twice as fast, so it has twice the momentum as before. The bullet also retains nearly all it's momentum, so the result of the collision is almost three times the momentum as before the collision (2 pendulum + 1 bullet). My guess is that if the pendulum does have twice the momentum, then the bullet must have a negative momentum value, so that 2 - 1 = 1, conserving momentum. This would support the formula you provided.

Here's what you included in post 32: -mu + MU = +mv. This is really -mu + MU = +mv + MV where V = 0, causing the last part of the equation to drop out.

I suggested this is post 33: mu + MU = mv, which has the same part of the equation (+ MV) dropping out. This matches Colonelhogan's formula.

Colonelhogan suggested: MaVa + MbVb = MaVa' + MbVb'

Colonelhogan - Since you are in school, can you run your calculations and process by your professor and get his opinion as to whether or not they are correct? Since there are a mix of vectors and scalars, it is possible that a negative sign was missed somewhere to signify direction, like the velocity of the bullet bouncing back. Nowhere in the calculations did direction or vectors come into play. If we assume that the bullet hits and bounces back in a straight line, the only vector "influence" may be to change a + to a -. There would not be any partial velocities due to non 180 degree direction changes to contend with.

Thanks,
John

[nanuk]

the other day I was looking for the TV remote control, and found X

[colonelhogan44]

garandsrus,

It should be -980 for the bullet. I left this off to simplify it for those who may not be familiar with vector mechanics. The formulas and calculations are correct, they are straight out of a highly regarded Engineering textbook. The negative signs are arbitrarily defined, and as long as the same convention is adhered to, it won't matter in the end. I could call the incoming velocity -1000, and you'd get the same end result.

I can't vouch for anyone else's equations, and frankly don't have the time to check them, with finals and graduation next week. (I really should'nt even be on here...but you know how that goes :guntootsmiley:)

[303Guy]

So now, how do I construct a vertical ballistic pendulum that will fit inside my 'test tube'? Already I have a sand trap inside it. This sits on top of the original catcher which caught pass throughs from the wet wool test medium I was using. A suspension spring maybe?

[DrMik]

There is no compensation needed for losses (for heat, noise, deformation, etc.). Those are all losses in energy, which is never conserved in real collisions. Momentum, however, is conserved in an inelastic collision, which is the principle on which a ballistic pendulum operates. It would be impossible to get decent results if you had to compensate for those losses.

Conservation of Momentum:
(mass of bullet)*(velocity of bullet) = (mass of pendulum)*(velocity of pendulum)

Yes, up until this year when we moved labs we had a ballistic pendulum set up in a large box on a trolley It is great for demonstrating conservation of momentum and muzzel velocity. The gun carridge was smashed up by technicians who didn't like it. It was too big to have it in the new labs as we have less room. Our rifle was an air rifle using CO2 cartridges .22 calibre and we used different pellet masses. The fastest pellet recorded 430 metres per second (around 1300 ftps). I am making a new carridge at home this weekend and have come up with a much smaller version using the same rifle. The new one is mounted vertically. The block will be suspended using four studs and eyelets attached. (it is a wooden block) with velcro on one end so that a small cover block can be mounted to shoot at.
We used to take the ballistic pendulum out to education and skills fairs and we were very popular because of it.

[35remington]

The velocities look pretty likely for most of the readings, but I rather doubt the 1900+ fps of the 12 gauge slugs, even in three inch. Two hundred fps faster than claimed is not terribly probable.

Didn't read for the whole thread, but depending upon the slug's weight and the effect of the wads (and whether they stay attached, fall away, or are accounted for in the total weight of the slug or not) could possibly have influenced the reading. At the close range fired, the wads could still be nearly touching or "drafting" behind the slug, boosting the movement of the pendulum more so than if they were not present.

The wads can weigh up to 1/8 ounce or more.

[fourarmed]

I have been off the board for a while, and hadn't seen this thread was active again. The derivation by colonelhogan44 is correct except for the algebra mistake at the very end. If you check it, it does give V12 = -980, as he later added.

Part of the confusion is due to the fact that I chose to use the negative sign in the equation itself (hoping to make it clearer) while he wrote the formula with plus signs, causing the numerical value of V12, which I called u, to be negative.

[fourarmed]

Since we're on the topic, here (I hope) is a picture of the ballistic pendulum I designed for our demonstration apparatus collection. The chairman insisted after a notorious school shooting, that it had to be virtually impossible for anyone to get shot with it.

[lwknight]

Unless your pendulum actually traps the bullet , energy would be lost to radial deflection.
Bullets hitting steel generally splatter in all directions. There is a lot of energy left in the spray of liquefied lead.

You might have 100 ft/lb energy differential to the pendulum but on impact 20 ft/lbs of the bullets energy goes sideways into oblivion and only 80 pounds goes into the pendulum.

As to whether the bullet bounces off or not is of infinitesimal consequence because the bounce velocity is so low that the energy could not be easily measured.

[XWrench3]

is it true that a boolit that bounces back, makes the pendulum move more?
and to find a true comparative value, the boolit must stay with the pendulum?

that type of thing happens to me a lot. i shoot into dead stumps, and find roughly 10% of my boolits laying in front of the log after shooting. and yes, they actually do hit the log, not just dribble out of the end of the barrel.

[fourarmed]

Lwknight, energy is lost in all sorts of ways unless the collision is perfectly elastic, which is an idealized situation which virtually never is achieved in the real world. The thing which is always conserved in collisions is momentum, not energy. Trying to understand the ballistic pendulum using energy considerations is probably why so many people misunderstand it.

[lwknight]

Momentun is energy. Or more accurately " ballistic energy" . or geomotive force in this case.
The bullet in itself has no energy unless its compared to the object that will be affected by it. The earth rotates at a very high rate of speed and the earth orbits the sun even faster. Speed as percieved on earth is realitive only to the observer and the objects that the observer referrs to.

[Jim_Fleming]

The average distance from the Earth to the Sun is 149,597,890 km. Therefore in one year the Earth travels a distance of 2*Pi*(149,597,890)km. This means that the velocity is about:

velocity=2*Pi*(149,597,890)km/1 year

and if we convert that to more meaningful units (knowing there is 365 days in a year, and 24 hours per day) we get:

velocity=107,300 km/h (or if you prefer 67,062 miles per hour)

So the Earth moves at about 100,000 km/h around the Sun (which is 1000 times faster than the speeds we go at on a highway!)

I swiped this from:

http://webcache.googleusercontent.com/search?q=cache:ve261is-nscJ:curious.astro.cornell.edu/question.php%3Fnumber%3D356+earth's+revolution+aro und+the+sun+speed&cd=2&hl=en&ct=clnk&gl=us&source=www.google.com

Ain't no way I could've figured this out.

The way this relates to this thread is where it gives a relative speed to highways. Compared to our puny feet per second velocities...

lwknight, I figured you'd appreciate this.

[lwknight]

I like it. Actually I knew the numbers but did not see any point that being technical would mean anything more than words like " really fast " and " a lot faster ". But for those who would be curious , a tidbit of interesting info.

[Ozarklongshot]

WOW This is an incredible gathering of the smartest group of critical thinkers one could imagine. That or it's a big ol pile of of very impressive B.S.ers. Either way I'm not smart enough to tell the difference. I feel a little smarter after reading that though.
But I think this is how it boils down. If I hit a big heavy thing(<scientific term) really hard like with a bullet. and the big heavy thing has method of marking it's movement when hit really hard. Then I can compare how different hard hitting things hit harder. O ya and the most accurate is if the bullet stays in the big heavy thing and I remove exact amount of weight after hitting big heavy thing for subsequent bullet impacts.
Did I get this right??? I got a reservation at a Holiday inn!

[Jim_Fleming]

Ozark, essentially I'd say you got it exactatackaly correct, sir.

Please everyone pardon my diverting this thread into a geeky kind of nerdish direction with trivial information.

Sent from my Droid

[303Guy]

So the Earth moves at about 100,000 km/h around the SunYou omitted the really meaningful bit. That equates to 29.805km per second! 29,805m/s or 97,787fps! That's faster than a speeding bullet!:2gunsfiring_v1::holysheep

[Jim_Fleming]

LMAO! 303 you've got a good point!

Take Care, Friend!

J.



You omitted the really meaningful bit. That equates to 29.805km per second! 29,805m/s or 97,787fps! That's faster than a speeding bullet!:2gunsfiring_v1::holysheep



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