OK, somethingis wrong with the math here!

[MeekandMild]

From the Pykrete website we find this quote:

Both were shot with a .243 rifle at 100 meters. A .243 bullet fired from 300 meters can push a 1 lb object for a distance of 1000 feet. Remington Catalog lists stats for a .243 bullet: 243 - 100 gr bullet 1,945 ft - lbs energy .30-06 - 150 gr bullet 2,820 ft - lbs energy .300 Win mag 150 gr bullet 3,605 ft - lbs energy

What is the math error here? If this was true then the converse is true that a .243 bullet could knock a 1000 pound object a foot. Or a hundred pound object ten feet. We obviously don't have .243 rifles blasting holes in backstops and knocking down their shooters. So what is the REAL force and how is it calculated?

I sem to remember something in the dim recesses of my memory of foot pounds per second squared.

 

[Greg L]

Perhaps if you tied a coconut to the bullet you could see how far it could fly?

Wait, no red .

 

[Kruzr]

Perhaps if you tied a coconut to the bullet you could see how far it could fly?

With the coconut, it can fly far enough to knock a migrating sparrow out of the air.

 

[Reno]

A projectile with 1000ft-lb of energy could theoretically lift a 1000lb object 1' in to the air (or a 1lb object 1000', or anywhere in the range where the weight * the distance liften = 1000ft-lb) if it were in a vaccum, AND it could transfer 100% of its energy in kinetic form to the projectile perfectly upwards.

 

[CrudeGT]

With the coconut, it can fly far enough to knock a migrating sparrow out of the air.

Are you assuming this is an African Swallow, or a European Swallow?

 

[GunnySkox]

The problem isn't the math. In the pure mathematical sense, in an elastic (no energy lost) collision between the bullet and our favorite 1 lb object in which the bullet transfers ALL of its energy to the object, said object would go hurtling off however many feet had the bullet ft-lbs of energy.

However, in practical situaitons, this never happens. If our bullet goes whanging into a hard object with 1000 ft-lbs of energy, it's extremely likely that when it hits, only a fraction of its energy will be transferred to the object, which will then shove back with equal force, sending our little .243 off in some random direction (newton's 3rd law). Also, some of the energy is turned into heat and sound, but that's negligible.

Basically, when the bullet hits the object, it just doesn't get the chance to shed all of its energy. During the initial collision, the exact amount of force the bullet hit the object with, hit the bullet right back. The object probably moved a bit in the bullet's direction, but when it hit back, the bullet, still retaining most of its energy, got whanged off to who knows where.

Alright, I think I got it in one of those...

~Slam_Fire
Feel free to tell me I'm wrong, I'm not too sure about all this, I'm tired.

 

[Reno]

It's really not an issue of the target "pushing back."

If the bullet does get captured by the target (let's say it's a big wood block), a LOT of its energy will be lost by the deformation of the target (the wood breaking and tearing and getting heated by friction). Unfortunately, it's impossible to retain 100% of the energy, or even a sizable fraction. One thing that is conserved, though, is the momentum.

 

[odysseus]

Perhaps if you tied a coconut to the bullet you could see how far it could fly?

Are you assuming this is an African Swallow, or a European Swallow?

I fart in your general direction!

 

[ietrash]

1,000 ft/lbs of energy is the amount of energy of 1,000 lbs. being dropped 1 ft.

 

[Jim K]

"...some of the energy is turned into heat and sound, but that's negligible."

Maybe not so negligible. It is the bullet's dynamic energy transformed into heat that causes cratering in steel plate and also allows armor piercing bullets to work. In the case of AP bullets, the dynamic energy of the bullet is converted to heat when the bullet stops, the steel is melted and the AP core goes through the hole or the momentarily soft steel.

Lead bullets striking a steel plate or a rock at high speed simply melt and lead splatter can be seen with obvious signs of molten metal.

Jim

 

[sumpnz]

Conservation of energy only exists in theory. Conservation of momentum exists in reality. To continue with the example of the .243 bullet, if it weighs 0.015lbs (105 grains) and impacts a target at 2500 fps, it has a momentum at the time of impact of .015lbs*2500fps=37.5ft-lbs/sec.

If that bullet strikes a 200lb stationary mule deer, the mule deer would react at (37.5ft-lbs/sec)/200lbs=0.1875 fps.

If that same bullet hit and was captured by a 1lb metal plate, that plate would fly off at 37.5/1=37.5 fps.

What!? A swallow carry a coconut?

It could grip it by the husk!

It's a question of where he grips it. It's a simple question of weight ratios! A one pound bird cannot carry a 5 pound coconut.

However, a swallow need to beat its wings 43 time every second to maintain air speed velocity, am I right?

I'm not interested!

well, if it's an african swallow ...

Edit: I thought about this a little more, and my math above isn't totaly correct. I neglected to take into account the mass of the bullet after impact. In the case of the mule deer, it doesn't make a difference until 5 decimal places, but in the case of the 1 lb metal plate, including the bullet meand that the velocity of the combined bullet/plate would be 36.95fps.

 

[G1FAL]

Are you assuming this is an African Swallow, or a European Swallow?

I dont know!
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AAAAAAARRRGGHHHHHHHHHHHH!

 

[Hardtarget]

this velocity/energy thing may be why Hollywierd thinks a 9mm will blow a guy over the hood of an F-150...or a 12ga will blow one across an alley.
Its just a mathmatical way to compare cartridges.Bigger/faster=more.
Mark.

 

[Iron Mike]

This whole African VS. European swallow is pointless . Remember that somehow those coconuts got to England and helped to alleviate the horse shortage . Peronally I like the giant African swallow theory(the ones with the 175 foot wingspan)

 

[entropy]

What if two swallows...

 

[Baba Louie]

But, would it move a 1 lb object 1000 ft.?
Back to the basics, class...
http://www.physicsclassroom.com/Class/newtlaws/newtltoc.html
Check out "The Big Misconception" in Newton's Second law page. Inertia, friction and gravity are a reality. It's all too much for me to figure out.

The real questions I want to know are "Does a sucking chest wound have the equivilency of moving said 1000# block one foot and does it really matter when I ponder 9mm v. .45 acp?"

 

[MrAcheson]

The problem isn't the math. In the pure mathematical sense, in an elastic (no energy lost) collision between the bullet and our favorite 1 lb object in which the bullet transfers ALL of its energy to the object, said object would go hurtling off however many feet had the bullet ft-lbs of energy.

I'm pretty sure any kind of collision involving two different masses must be inelastic. That means it is physically impossible for all the mechanical energy to be transferred. You can't have mass1/mass2=velocity2/velocity1 (conservation of momentum) and mass1/mass2=(velocity2/velocity1)^2 (conservation of kinetic energy) at the same time unless m1 and m2 are the same. I might have made an oversimplifying assumption somewhere but I'm pretty sure this is right.

So you need a 1lb .243 bullet which ain't gonna happen.

 

[WT]

Ballistic pendulum. See link.


http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

 

[B_Scott]

MrAcheson is correct - Plus you can't have friction, hence the pool ball on ice example of an almost elastic collision in physics 101.
Also, the popular example uses pool balls because the entire pool ball acts as one unit (for the most part). Meaning that there wil be no measurable deformation - whether elastic or inelastic.

So - the one pound bullet cannot deform and neither can the one pound block of steel. Oh and the bullet must impact the steel at a 0 degree angle which would imply no gravity since gravity will be forcing the bullet in a downward direction.

- Now to read up on those links posted...I love physics

 

[MeekandMild]

So what is the REAL force and how is it calculated?

Ahem. So what is the REAL force? Last time I shot a .243 it didn't kick me half the calculated distance (about 4 feet).

 

[FNFiveSeven]

Meek & Mild:

The violence of the recoil you feel from your rifle is directly related to the amount of recoil Energy you have to endure upon firing. The recoil energy you absorb, E = 1/2 (Mass of gun)* (Velocity of gun)^2.

The momentum of the gun smashing into your shoulder is equal (more or less) to the momentum of the bullet leaving the barrel. Momentum is = Mass * Velocity. So,

(Mass of Gun)*(Velocity of Gun) = (Mass of bullet) * (Velocity of bullet)

or (Velocity of Gun) = (Mass of bullet) * (Velocity of bullet) / (Mass of Gun)

Substitute back into equation one given above

Recoil energy, E = 1/2 * (Mass of gun) * (Mass of bullet)^2 * (Velocity of bullet)^2 / (Mass of Gun)^2

or simplifying we get

E = 1/2 * (Mass of bullet)^2 * (Velocity of bullet)^2 / (Mass of Gun)

The reason your rifle doesn't blow your arm to bits like it does the deer is due to the mass of the gun. Every time you double the mass of the gun you halve the felt recoil energy.

Furthermore, talking about how far your rifle kicks you back has very little to do with any energy, recoil, etc. If you were in space, even a .22 LR would throw you a thousand light years. OTOH, if you were braced against a concrete wall, even a 50 BMG wouldn't move you one inch.

 

[B_Scott]

OK the force on you:
F=mass*acceleration
Mass of bullet = .015 lbs (stated earlier) or 0.0068039 kilogram
Acceleration = 902.208 meters/second squared

6.138 Newtons of force total.
This is small.

 

[FNFiveSeven]

B_Scott,

A quick investigation of your equation reveals that it does not work, specifically because it does not take into account the weight of the rifle. In cases like this, the point is often best demonstrated by taking the extreme, so let's pretend that we were shooting a 1000 pound rifle. It should be obvious that in this case the force on your body would be practically 0, not 6.138 Newtons. Weight of the gun is everything in these calculations, it cannot be ignored.

-Robert

 

[sumpnz]

blackrazor - Don't mean to be overly picky, but you forgot to include the mass of the propellant in your recoil considerations. The mass of the burning powder is quite significant in the overall calculations, and is a big part of the reason why magnums kick so much harder. Think about it this way. A .300 Win Mag with an 20" barrell will give a muzzel velocity about the same as a .308 from a 26" barrel assuming the same bullet weight. But the recoil of the magnum is a lot higher (for these purposes assume that the magnum rifle is the same weight as the non-magnum). Without taking the powder into consideration, the recoil would be equal. But since the magnum holds about half again the powder, the recoil is noticibly higher.

 

[B_Scott]

Blackrazor - the force caused by the tiny bullet is still 6 Newtons.
It is however acting on the rifle AND the shooter.
Energy is a function of force - so you took it one step beyond my calculation.
I just wanted to answer Meek&Mild's question of how much FORCE.

AND

sumpnz, The mass of the powder is not the reason magnums kick more, the powder only adds to the velocity and velocity is what is used in the calculation.
If the mass was the reason all one would have to do is add 1/2 of a pound to there magnum rifle stock and it would kick like a .223