The science of recoil


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tackleberry45
April 18, 2008, 07:34 PM
I was asked why a .40 "snaps up" for recoil and why a .45 "pushes back" I really did not have a great answer other than the only thing I couod think of which was powder burn rate. I have a Smith M&P .45 and STI .45 that are tame on the recoil side for .45. I also shot a Beretta PX4 .45 which was not so tame, at least to me. I guessed becuase of a little higher bore axis. So THR folks what is the correct answer as to why a .40 tends to snap up and a .45 pushes back??

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easyg
April 18, 2008, 07:40 PM
So THR folks what is the correct answer as to why a .40 tends to snap up and a .45 pushes back??
The question itself is flawed.
I don't even agree with the notion that the .45 "pushes back", and the notion that the .40 "snaps up".
How they react is largely dependent upon the specifications of the pistols and the strength and ability of the shooter.

makarovnik
April 18, 2008, 07:50 PM
A lot of it has to do with velocity, bullet weight, barrel length and weight of the gun. I agree that the .40 is very snappy. You really feel like you need to hold on tight. Problem is that a lot of .40's are packaged in a small light frame. This is a mistake in my IMHO. A .40 in a large frame is quite controllable.

The .45acp doesn't seem as snappy when you are shooting it but if you video taped yourself shooting one you would see they have more muzzle rise than you expected.

ccmdfd
April 18, 2008, 07:51 PM
I'm with easyg on this one.

There's so much subjectivity when it comes to perceptions of recoil that it's almost laughable reading about them on the net.

So much depends upon the type of gun, it's weight, its bore axis. How well the grip fits the particular shooter (a very big thing IMHO) as well as any torque or rotational forces also play a role.

Gun Slinger
April 18, 2008, 10:12 PM
See for yourself:

Just "plug in" the numbers...

Where:

M = bullet weight in grains
V = muzzle velocity in feet per second
E = weight of powder charge in grains and
W = weight of the rifle or pistol in pounds

([MV + 4700(E)] 7000) (64.34809711) (W) = "free" recoil energy of the pistol or rifle (in foot-pounds)

Example:

A Glock 17 fires a 147 grain bullet with a muzzle velocity of 1000 feet per second using a cartridge that uses 4.0 grains of propellant and the Glock 17 weighs ~1.57 pounds unloaded. What is it's "free" recoil energy in foot-pounds?

([147 gr. x 1000 f.p.s. + 4700 x 4.0 gr.] 7000) (64.34809711) (1.57 lbs.) = 5.55312 ft.-lbs. of "recoil"

After that, how you "percieve" the recoil is pretty subjective, but the gun does the same thing everytime regardless of who is firing it.

Hope that this helps,

presspuller
April 18, 2008, 10:19 PM
Can't argue a bit with what has been said here.
The only thing I would add is that the different grip angles on different pistols makes a difference in the way recoil is felt.

Rustynuts
April 18, 2008, 10:49 PM
It's all physics. F=MA

The 40 bullet isn't all that lighter than a 45, yet has to be accelerated (A) to a much higher velocity in essentially the same distance (gun barrel). Slightly lower mass (weight, or "M"), but much higher A = higher F (force). That's the simple notion of it, then you have the different shapes of the brass/bullets, burn characteristics of the powder, etc., and it starts getting more complicated. Then you get into the grip and gun characteristics as well.

WinchesterAA
April 18, 2008, 10:52 PM
Wow.. After reading and doing much of the math in "Understanding Firearms Ballistics", I actually understood EASILY what Gun Slinger said..

I'll go ahead and recommend that book to everyone then.

I sucked at math, or so I thought, but when firearms are involved it's easy for some reason...

makarovnik
April 19, 2008, 12:41 AM
I would like to add that when someone asks a question about recoil they are usually asking about perceived recoil. I think that should be assumed. If it isn't "perceived" then it doesn't bother a person and shouldn't be an issue unless you're concerned about wear and tear on your gun.

Walkalong
April 19, 2008, 09:45 AM
I was asked why a .40 "snaps up" for recoil and why a .45 "pushes back" I really did not have a great answer other than the only thing I couod think of which was powder burn rate.

Velocity my friend, velocity. ;)

1911Tuner
April 19, 2008, 11:02 AM
Muzzle velocity has little to do with recoil energy. Most of the recoil momentum has hit your hand or shoulder long before the bullet reaches the muzzle.

You can have high velocity and low recoil impulse...as with the .22-250 rifle...and you can have low velocity and high recoil...like with the .45-70/405.

If we accept Newton's 3rd Law that force forward equals force backward, we can start to unravel the mystery.

The pressures generated by the burning, expanding powder gasses create a force vector between the bullet and the breechblock. Whatever force is imposed on the bullet is imposed on the breechblock in like measure...and at the same instant...so the gun is in recoil as soon as the bullet moves.

Force vector is defined as having magnitude and direction.


Here's a clue to consider:

If we ascribe an average of 30 fps per inch of barrel length...gained or lost...and the muzzle velocity from a 5-inch barrel is 850 fps...and roughly an inch of that barrel is chamber...we have 120 fps of the total coming from the acceleration in the barrel. That leaves 730 fps unaccounted for.

When does that 730 fps occur?

During the violent rise to peak pressure, and the initial punch of sudden acceleration...resisted by the bullet's entry into the rifling...and the resulting rise in force necessary to push it forward.

Likewise, the initial punch of equal force backward against the breechblock creates the greatest percentage of what we call recoil.

Accelerating the bullet from that point doesn't generate the level of equal and opposite force that occurs at the onset of the event.

Here's more:

http://www.thehighroad.org/showpost.php?p=4417450&postcount=8

ccmdfd
April 19, 2008, 11:11 AM
I'd love to see more science put into recoil.

I'm somewhat recoil sensitive due to some arthritis. But I've found there are certain .45's I can shoot with no trouble, while some 9mm's cause me to wince.

As I stated above, I think there's more to the story than just magnitude of recoil, which is easily calculated. There's directional issues-does the gun go straight up and down, or straight back into the hand, and torque. There's probably issues in terms of the rapidity of the impulse-a fast, sharp peak vs a slow round hump on the scale. There's certainly other issues I haven't thought of yet.

I'd love to see studies done which look into these different aspects. Thus if one is sensitive to muzzle climb, they can look and see which guns/calibers have the least amount of climb. If one is sensitive to torque, same story.

It would take some of the subjectivity out of the discussions.

1911Tuner
April 19, 2008, 11:58 AM
I'd love to see more science put into recoil.

Basically, the science is no more complex than Newton's 3rd Law...but there are other factors to plug into the equation when we bring up the recoil of a firearm.

Torque occurs because of another equal and opposite created by the direction and pitch of the rifling.

Muzzle flip is a function of the bore axis above the centerline of the wrist and the grip angle itself.

Then, there's that one point that so many people fail to consider...and one that I've argued for years.

The delaying/slowing effect of the bullet as it passes through the bore.

We understand that the bullet being forced through the bore under high frictional resistance. All that's required to understand how high those forces are is to try to push one through by hand.

This resistance causes the bullet to exert a forward drag on the barrel...and that drag is present as long as the bullet is there and moving.

This forward drag exists while the gun is recoiling backward...and since the barrel is moving backward at the same instant that the bullet is moving forward...the rearward acceleration is being resisted...buffered, if you will...by the bullet's influence.

Imagine a cork attached to a rope...fitted tightly into a plastic pipe.

Grasp the pipe in one hand and the rope in the other...and pull. The instant that the cork slips and starts to move in one direction...the pipe slips and starts to move in the other.

Keep pulling. The cork is being resisted by the pipe...and the pipe is being resisted by the cork. It requires continuing equal and opposite force to pull them apart...all the while, each one under resistance by the other.

The "Equal and Opposite" thing works in reverse, too. When you push on an object, you are immediately pushed BY that object through the vectored force provided by your arm.

Likewise, when you pull on an object...you are immediately pulled BY that object.

Whatever resistance is imposed on the bullet by the barrel is also imposed on the barrel by the bullet.

This delaying effect is only present in locked or fixed-breech weapons. Straight blowback designs aren't affected because the breechblock and barrel aren't connected. In these, what we feel as recoil is a function of the recoil/action spring, and the slide striking the impact abutment in the frame.

Gun Slinger
April 20, 2008, 01:47 PM
Originally posted by makarovnik:
I would like to add that when someone asks a question about recoil they are usually asking about perceived recoil. I think that should be assumed. If it isn't "perceived" then it doesn't bother a person and shouldn't be an issue unless you're concerned about wear and tear on your gun.


Since perception is neither objectively quantifiable (except in subjective terms and it has no standarized unitary value assigned to it) nor a constant from one person to another, it is not likely that someone will have the same perception to a given quantity of recoil that another person has so the best means to convey the recoil of a given gun/load combination remains that as expressed in terms of foot-pounds of KE. Unless one has an objective unitary measuement with which to refer to a level of recoil generated by a particular gun/load combination, you can only convey what you feel and the person to whom you are describing such a sensation to, will still need to "percieve" it for himself and will do so, most likely, with a different outcome than you experienced, rendering your "evaluation" of the effects 'moot' for him.
If nothing is percieved then there is nothing to quantify in the first place and such a comparison would be irrelevant since even airguns and the gentlest of rimfires generate a level of recoil (albeit a miniscule level) that can still be percieved by anyone.

Gun Slinger
April 20, 2008, 02:23 PM
Originally posted by 1911Tuner:
Muzzle velocity has little to do with recoil energy. Most of the recoil momentum has hit your hand or shoulder long before the bullet reaches the muzzle.

Actually, muzzle velocity has everthing to do with recoil energy.

Since momentum (ρ) is the product of mass (m) times velocity (v):

ρ = mv therefore mv = mv

then we can conclude that the mass of the bullet times the velocity of the bullet divided by the mass of the gun will equal the "recoil" velocity imparted of the gun:

m(bullet) x v(bullet) / m(gun) = v(gun)

Since there has been a resultant "recoil" velocity (v) imparted to the gun, the gun now also possesses momentum (ρ) as well as a certain amount of kinetic energy that is related to its momentum ρ = mv via the integration of the momentum equation (ρ = mv) with respect to v as below:

∫ mv δv = m ∫ v δv = m v = mv = (KE) Kinetic Energy

Kinetic energy and momentum are undeniably related and neither momentum (ρ) nor kinetic energy (KE) is time dependent (Δt) until such momentum (ρ) or kinetic energy (KE) is applied to a second body. At that point, momentum (ρ) will be conserved and only KE of the second body will differ since only momentum is (mathematically) conserved in such a "Newtonian" system.

Regards,

1911Tuner
April 20, 2008, 02:54 PM
Actually, muzzle velocity has everthing to do with recoil energy.

Nope. You didn't read anything except the one sentence...and ignored the rest.

Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit.

That initial punch is where 90% of the velocity occurs...for pistols, it's within the first half-inch of bullet travel...and probably more than 90% of the total recoil impulse.

Go back and re-read the whole thing.

Don't believe it? Try this. Warning:
It's illegal...but it will demonstrate the reality in neat fashion.

Go buy a cheap single-shot 12 gauge shotgun. Cheap is good, because you'll be destroying the barrel shortly after you do this.

Lop off the barrel flush with the end of a fired shell, so that no projectile acceleration will be gained after the initial punch. Add weight to the stock so that the gun's mass will be equal to the original.

Go fire the gun.

I don't recommend placing your face on the stock's comb unless you relish the thought of a black eye and/or a bloody cheek. Ask me how I know...

Cheers!

Gun Slinger
April 20, 2008, 04:39 PM
Originally posted by 1911Tuner:
Nope. You didn't read anything except the one sentence...and ignored the rest.

Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit.

That initial punch is where 90% of the velocity occurs...for pistols, it's within the first half-inch of bullet travel...and probably more than 90% of the total recoil impulse.

Go back and re-read the whole thing.

Don't believe it? Try this. Warning:
It's illegal...but it will demonstrate the reality in neat fashion.

Go buy a cheap single-shot 12 gauge shotgun. Cheap is good, because you'll be destroying the barrel shortly after you do this.

Lop off the barrel flush with the end of a fired shell, so that no projectile acceleration will be gained after the initial punch. Add weight to the stock so that the gun's mass will be equal to the original.

Go fire the gun.

I don't recommend placing your face on the stock's comb unless you relish the thought of a black eye and/or a bloody cheek. Ask me how I know...

Cheers!


I read your entire post prior to making my first post (#15) above.

The experiment that you propose (and no I won't try it 'cause I an awfully fond of my freedom :D ) simply changes the location of the muzzle so that it is at the same point that the shotgun shell ends. In other words, it is still the "muzzle" (a.k.a. the end of the barrel) and it is the point at which the highest velocity of the projectile will be achieved since the shot/wad is no longer being accelerated by the high pressure propellant gasses remianing trapped behind it. The only change effected will be that the bullet (or shot charge) will not exit the "muzzle" as fast as it would as if it had the entire length of the barrel in which to continue to accelerate from the continued force (F) of the expanding gasses produced by the burning propellant. While the shot and wad will most certainly leave the 'new muzzle' (created by cutting the barrel off so that it is now the same length as the shot shell) at high velocity that veocity will not be as high as it would have been had the shot and wad been allowed to accelerate the full length of the barrel. I could not have come up with a better example with which to illustrate the point that I have been trying to make.

Arguably, there will be some recoil with the arrangement that you suggest because you still have a mass (m) namely the shot and wad leaving the 'new muzzle' that you have created at a lower velocity (v) that will still result in momentum as well as K.E. being generated in both directions.

The fact is, that regardless of how short the barrel is made, there will always be velocity (v) since the mass (m) is moving out of the barrel regardless of its length which means there will always be momentum (ρ) and K.E.

I even find you (quoted again for the sake of clarity) to be in agreement with what I just explained above:


"Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit."


The "bullet exit" is the muzzle.

The acceleration of a bullet (or shot and wad) is a single impulse of exceedly short duration and momentum (not to mention K.E.) is a result of the final velocity (v) obtained of the mass (m) as it exits the muzzle even if the "muzzle" coincides with the end of the cartridge casing or shotshell mouth. It also a fine example of Newton's Second Law of Motion:

"The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."

Expressed mathematically, Newton's Second Law of Motion is:

F = ma

where "F" equals force (this is the "PUNCH" to which you refer above in your quoted statement), "m" equals mass and "a" equals acceleration.

Additionally, if you substitute the value 0 (fps) for the velocity variable (v) in any of the following equations you will see that since any number multiplied by zero equals zero that you cannot possibly have momentum or kinetic energy without having velocity.

∫ mv δv = m ∫ v δv = m v = mv = (KE)

ρ = mv

which proves the following statement false:

Muzzle velocity has little to do with recoil energy.

because it would violate the Newton's Third Law of Motion:

"For every action, there is an equal and opposite reaction."

Mathematically expressed Newton's Third Law of Motion is:

mv = mv

which proves again that muzzle velocity has everything to do with momentum (ρ) as well as the kinetic energy (KE) of both the bullet (or shot/wad) and the recoil of the gun.

Regards,

1911Tuner
April 20, 2008, 06:27 PM
I read your entire post prior to making my first post (#15) above.

Then you understand the physics of it without understanding internal ballistics.

Force forward equals force backward.
The maximum level of recoil occurs at peak force. Peak force occurs with peak pressure...and peak pressure doesn't occur when the bullet is about to clear the muzzle. It comes within a very small amount of bullet movement...even with slow rifle powders With some extremely quick pistol powders, it may peak before the bullet has even left the case.

See Newton 1A: Objects at rest tend to remain at rest. (Inertia)

Any object requires more force to get it started than it does to keep it moving afterward. It's during this rapid climb to peak pressure that the greatest percentage of the total muzzle velocity occurs...and likewise for the recoil impulse.

As the bullet accelerates ever faster, it requires less force to accelerate it...ever faster. Thus the rapid pressure drop...and hence less force forward/backward...results in only a small amount of recoil after the punch.

Consider again: (And these figures are representative rather than absolute...but experimentation will prove them to be pretty close.)

If the barrel is 5 inches long, and the chamber accounts for an inch of the total...and we can figure on the 30 fps of velocity per inch pf barrel gained or lost rule of thumb...and the muzzle velocity is 850 fps:

120 fps comes from the barrel length.
That leaves 730 fps unaccounted for.

If the initial punch takes up a half-inch...that leaves 3.5 inches of barrel to accelerate the bullet to the final velocity. Now we're down to
105 fps after the punch..and we have 745 fps unaccounted for. Since it can't come from the 3.5 inches of barrel...it has to occur early in the bullet's travel...during the violent acceleration as the pressure climbs toward peak.

And, now we've returned to force forward and force backward.

I understand that as long as the bullet is in the barrel and accelerating, that an equal and opposite force will be created. It's just not very much force compared to that generated at the outset.

Finally...This isn't theory without substance. I've proven it with real guns and real ammunition in blind tests. Revolvers with barrels and with barrels removed, leaving nothing but the chambers, and barrels cut off, leaving the forcing cones and a half-inch of rifling in order to keep the bullets from tumbling and destroying my chronograph. Then...with the forcing cones and the small rifled portion reamed to leave just barely enough rifling to spin the bullet. You'd be surprised at how small the difference in velocity was with the same ammunition lot.

And even more surprising was that nobody who participated in the tests could detect any real difference in recoil. One maintained that the recoil in the barrel-less gun was sharper...three times in a row...even with lower bullet velocity.

Gun Slinger
April 20, 2008, 10:12 PM
"Then you understand the physics of it without understanding internal ballistics."

Hey, you're the one arguing against Newton's Laws of Motion. I understand internal ballistics far better than you surmise above.

This statement violates Newton's Second Law of Motion: F = ma ...


"As the bullet accelerates ever faster, it requires less force to accelerate it"


...because in order to require less force (F) to continue to accelerate (a) the bullet would then require the body (m) being accelerated to actually become lighter, an absolute physical impossiblity at best.

I have, however, learned from this thread, that I cannot educate the ineducable. :banghead:

I give up. :)

1SOW
April 21, 2008, 12:20 AM
Is this the resulting conclusion?
View muzzle velocity as the 1/4 mile drag strip speed:
At a drag strip the slower top speed car may beat the higher top speed car through the 1/4 mile due to faster early acceleration. That faster accelerating car would be harder to keep the front end on the ground.

.45s are usually heavier steel guns with a bigger slower bullet that accelerates slower. Powder type has a big affect here.
That slower accelerating bullet would have less percieved recoil given all the other factors mentioned are equal.
The mass of the gun itself also follows the laws of inertia/how fast the muzzle will accelerate in an upward direction from a given force.
Along with bore axis etc., different strength recoil springs in the same gun will change percieved recoil/flip by changing the rate at which the recoil is felt by the shooter.

TMann
April 21, 2008, 12:34 AM
Another factor that is difficult to measure is the relationship between the size of ones hand and the gun's grip. It's pretty obvious that a nice, wide, rubberized grip will feel better than a small, narrow, hard grip. However, I believe that the shape and size of the shooter's hand will play a part, too. I am a short guy with small hands, and I think that some small guns just fit my hand better than they would someone who has a big hand. For example, there are many people who hate shooting the KT P3AT, and will tell stories of having a bloody thumb by the end of their range session. And yet, I went out yesterday and shot 75 rounds through mine yesterday with no problems.

If the grip doesn't fit your hand, you won't enjoy shooting it.

TMann

1911Tuner
April 21, 2008, 06:09 AM
Hey, you're the one arguing against Newton's Laws of Motion.

No, I'm not. Actually you are...or at least you seem to be.

You're still not grasping my point. I understand that recoil energy/momentum is a function of Mass x Velocity. It's that one phrase..."Muzzle Velocity"...that doesn't ring. Remove that from your statement, and we're in agreement. The speed that the bullet is moving when it exits the muzzle has very little to do with recoil impulse, because the gun is in hard recoil well before that point. By saying that muzzle velocity determines recoil, it sounds like you feel that recoil doesn't start until the bullet reaches the muzzle. Kunhausen made that same mistake in his early theory of function...which he amended in his later editions. If that is, in fact what you're saying...then you don't understand the physics of it, either.

Plainly stated: Action and reaction begin at the same instant. No?

Pressures peak, early...quickly...and with some powders...violently. Pressures don't continue to climb until the bullet exits the muzzle. The gun is recoiling well before the bullet exits, and whatever added recoil occurs after the pressure peak amounts to a nudge compared to the initial impulse. The only exception would be if the mass of the powder charge is equal to or greater than the mass of the bullet. Then, you would have a readily measureable recoil impulse after bullet exit, though it would still be relatively small because with such a powder charge weight/bullet mass combo...recoil impulse comes from the acceleration of the bullet AND the powder gasses, plus any unburned powder that remains after the onset. Essentially, double the bullet's mass is factored into the mass/velocity equation.

Let's try a hypothetical:

You've got a rifle with a 30-foot barrel, made of a new space-age metal that weighs the same as a standard 22-inch barrel. Let's assign .30-06 as the caliber, and use a slow powder...like IMR 4350.

Even as slow as this powder is, it would burn completely up in the barrel before the bullet could hit the air. When you fire this long rifle...will it kick?

Of course it will. Even though the bullet never reaches the muzzle...the rifle will kick, because whatever force is imposed on the bullet is imposed on the breechblock in equal measure.

saltydog452
April 21, 2008, 11:16 AM
I wish that I could find the photos, but I remember seeing high speed photos of the newest Federal HST ammunition being fired. The slide was clearly in recoil before the projectile had ever left the barrel.

Bullseye shooters knew this decades ago when they stressed 'followthrough' and 'not giving up on the shot'. 'Course this was with softball loads moving along at a leisurely pace.

salty

1911Tuner
April 21, 2008, 11:22 AM
Salty...I'll go find the link to Tripp's stop-action videos that'll show it. Might take a while, so sit tight.

You'd be surprised at the number of people who really believe that recoil starts after the bullet exits...even after ol' Isaac Newton explains in detail that it can't happen like that in this universe.

Gun Slinger
April 21, 2008, 11:37 AM
"I understand that recoil energy/momentum is a function of Mass x Velocity."

Tuner,

That's the point that I was trying to convey when I said that your statement,


"Muzzle velocity has little to do with recoil energy."


was contrary to Newton's Third Law. I don't dispute at all that the bullet makes an impressive initial gain in velocity during its first few inches of travel as internal pressure "spikes" or that as the maximum expansion volume of the burning proprellant is reached that pressure falls off rapidly towards equlibrium at atmospheric pressure or that while such a pressure decrease is occuring that the bullet continues to make velocity gains (accelerate) albeit not as great as that made during the first few inches of barrel dwell time and that recoil begins at the moment of bullet movement and ends when it leaves the muzzle (ΣW). If you'll take a look at the posts that I have made above, never have I made such assertions contrary to what I have just stated.

Indeed, the only disagreement that I had with what you posted was the statement quoted above because when we calculate free recoil energy the only velocity that matters is muzzle velocity since we cannot choose to stop the being effected by the recoil impulse when the bullet is only a third of the way down the barrel and ignore the smaller, but not necessarily insignificant, incremental gains made in velocity after that point. Calculating free recoil energy is a function of the cummulative (total) acceleration of the mass down the entire length of the barrel or essentially:

∫ ma δa = m ∫ a δa = ma = ΣW = the sum of all the work done to the bullet during its travel down the entire length of the barrel

So after all of this, we find ourselves in agreement...:eek:

Who'd a thunk? :scrutiny:

:)

Regards,

1911Tuner
April 21, 2008, 11:57 AM
So after all of this, we find ourselves in agreement

Yep. I started to realize that we were on the same page...but we just weren't finishing all our sentences sometimes.

Many people focus so hard on Newton 3 that they fail to consider the full impact of Newton 1 or factor it in.

Here's the video. Pay particular attention to the last 1911 in stop-action...just before the revolver. Place a pencil point at the end of the recoil spring tunnel in the frame, and watch. First you'll see a light puff of blowby gas...followed by the nose of the bullet peeking out of the muzzle, but still not out...followed by the exit gas and fireball.

You'll notice that the slide is moving even before the blowby gasses appear.

Equally interesting is the cutaway showing how much the ammunition bounces around in the magazine during recoil.

The really fascinating clip is the bolt bouncing on the AR15...three times...before it settles down.

Watch it several times. You'll always see somethin' that ya missed.

http://www.trippresearch.com/media/movement/hispeedgateway.html

Gun Slinger
April 21, 2008, 05:52 PM
Neat site, the video is like nothing that I've seen before.

Very cool.

1911Tuner
April 21, 2008, 06:24 PM
Very cool.

Yep. Watch the 4th clip closely...the one that doesn't have the pistol anchored. Notice how little the frame moved on cartridge ignition...and that the frame doesn't really get moving in recoil until the slide has moved an inch or so, and started putting a load on the recoil spring...and how the muzzle didn't really flip all that much until the slide hit the impact abutment?

Testing has shown that in a 5-inch gun, the standard 16 pound spring's in-battery preload is about 9 pounds, on average. I'd be willing to wager that if the shooter really took a crushing, solid grip, that the frame would barely move until the slide was almost at the impact abutment.

DaveBeal
April 21, 2008, 10:16 PM
One of the points made in this thread is that the first inches of travel down the barrel impart more speed to the bullet than the last few. If you find a graph of chamber pressure versus time on the web (like in the Wikipedia article on "internal ballistics") you can predict how a bullet's speed changes as it moves down the barrel.

Change in momentum is equal to force integrated over time. If you ignore friction, the force on a bullet is equal to the pressure behind it times its cross-sectional area. Since the bullet starts at rest, if you take a pressure graph, the speed of the bullet at any time T is equal to the area under the pressure curve from time 0 to time T, times the bullet's cross-section, divided by its mass. Since the pressure curve peaks as (or a bit before) the powder is completely burned, this is the point of maximum acceleration for the bullet. It continues to accelerate along the rest of the barrel, but more slowly, because with no additional gas being generated the pressure drops as the volume behind the bullet increases.

Gun Slinger
April 21, 2008, 10:38 PM
Dave,

Good explanation of ΣF dt !!

Ain't Calculus wonderful? :D

1911Tuner
April 21, 2008, 10:51 PM
If you ignore friction, the force on a bullet is equal to the pressure behind it times its cross-sectional area.

But you can't ignore the friction...and that's the wild card in the game. The inertial and frictional resistance of kicking the bullet in the butt to get it moving, and swage down into the lands and grooves causes a violent pressure and force rise...and because force forward equals force backward...there is where the maximum recoil impulse is generated.

Even in a rifle caliber with a slow burn-rate propellant...slow compared to pistol powder...the pressure peaks early in the bullet's travel. Probably within a couple inches in the .308 or .30-06 class. The area under the curve represents maximum force...and how long it's sustained. Pistol powder pressure graphs show a sharp, fast peak...like the apex of a triangle... and rapid drop as the volume of the cylinder increases. As pressure drops, so goes force...so that at the point that the bullet is about to exit the muzzle...the recoil impetus is maybe 5% of the maximum, depending on...again...burn rate and barrel length. It's entirely possible to continue to increase barrel length until the bullet actually exits at a lower velocity that the maximum speed that it attained while still in the barrel if the powder is so fast that the gas plug runs out of steam before the bullet gets to the muzzle.

The other, often forgotten factor is the mass of the gasses and unburned particulate...which is roughly equal to the mass of the powder before ignition. So...in the typical .45 ACP cartridge...a 230-grain bullet actually is about 235 grains. So, that must be factored in as well.

The science of internal ballistics is a fascinating subject for discussion...and debate...as we've seen on this thread. It's at the same time more simple and more complex than one might suspect.

Good thread.

DaveBeal
April 22, 2008, 11:00 PM
But you can't ignore the friction...

Tuner, I was going to argue with you about this. Compared to 30000 PSI chamber pressure of a 9mm round, I figured the friction of the bore was insignificant, but thought I should do an experiment first. So I bought a box of 9mm JHP bullets and tried to push one down the barrel of my Beretta 92 with a cleaning rod. I couldn't budge it, even pounding my hand on the handle of the rod. So I'm not going to suggest that the friction is negligable.

But I'm a shooting newbie and probably the only person here who didn't know this. :o

Does anybody know what the bore friction force is of a 9mm round? I'm curious how much of the powder's energy gets used to overcome it.

1911Tuner
April 23, 2008, 06:08 AM
I couldn't budge it, even pounding my hand on the handle of the rod.

Howdy Dave. That'll give ya an idea of the magnitude of the forward drag that the bullet places onto the barrel during its break to freedom. :)

It varies. If you knew the exact construction of the bullet...jacked alloy and thickness and the makeup of the lead core...hardness...it could be closely calculated. It also tends to vary from one barrel to another because of dimensional differences and surface texture of the bore...smooth vs rough...land/groove depths...even rifling twist rates will affect the resistance of the bullet's passage. That's why we see different handloading data listed in loading manuals for different bullet manufacturers in identical bullet weights with outwardly identical bullets. The pressure figures came from a given charge of Powder X with their bullet. That can easily change with someone else's bullet.

JohnBT
April 23, 2008, 09:20 AM
I'm enjoying this immensely, thanks to all.

I haven't had any physics classes since I finished undergrad school in '72. Which reminds me of this one old prof who refused to call it anything but The Calculus. Five days a week of The Calculus this and The Calculus that was enough to drive me up a wall. But I digress.

John

1911Tuner
April 23, 2008, 10:46 AM
John...I had a sharp guy try to explain Calculus to me once. I blinked a couple times and said: "No, thanks. I have enough trouble with pie are square."

:D

brickeyee
April 23, 2008, 11:26 AM
The parameter that Tuner is describing is called 'jerk.'

It is the rate of change of acceleration.

It is not used in many applications, but crops up in all sorts of funny places when you start slicing time into smaller and smaller periods while doing analysis.

FieroCDSP
April 23, 2008, 11:50 AM
Ugh..related rate calculations. Brings back horrid memories from high school.

The physics of recoil are interesting, but the fact that there is every design under the sun out there means we don't have to understand the numbers to find a gun that works for us. :D

The 40 is "snappy" compared to the 45, but with a low bore-axis gun, good two handed grip, and properly balanced stance, it is easily controllable. Try the full size M&P 40 and compare it to a Glock 40cal (or Fo-Tay as some know it). Same round, different feel.

Thanks to everyone who posted math in this thread. At least I had a cup of coffee first, so I wasn't entirely in the dark.

tackleberry45
April 23, 2008, 05:27 PM
GREAT replies. This is exactly what I was looking for when I started this post.

Gun Slinger
April 23, 2008, 08:43 PM
Dave,

Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses, otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.

Looking at the frictional co-efficients for copper on steel and considering the total area in contact with the bore and making an educated guess, I would believe the magnitude of -F to be in the 150N - 200N (~40-45 pounds) range afer the bullet is engraved and in motion depending on the alloy composition of the bullet jacket and the barrel alloy.

Considering that the "average" (I know it is not a constant force and that it peaks then diminishes rapidly) force acting on a 115 gr. 9mm bullet during its trip down a 4.5" barrel that achieves a muzzle velocity of 1150 fps, is on the order of 900-1100 pounds for a little more than 6/10,000's of a second and the fluctuations that are possible due to gas blowing by the bullet that may reduce that frictional component, it is really not that significant and not something that anyone would notice or feel.

1911Tuner
April 23, 2008, 09:01 PM
Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses,

Well...Once engraved, it would have to drop...but I wouldn't call it insignifigant. Ever had a squib load stick a jacketed bullet halfway through the barrel? Still gotta have a hammer or an arbor press to back it out unless you soak it in penetrating oil for a couple hours.


otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.

See Newton 1B for the clue to that.
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed. The coefficient of friction doesn't change...only the magnitude of force required.

Gun Slinger
April 24, 2008, 12:04 AM
Once the bullet is underway and engraved by the rifling the frictional force (-F) is actually quite negligible as opposed to the force (F) generated by the expanding propellant gasses,


Originally posted by 1911Tuner:
Well...Once engraved, it would have to drop...but I wouldn't call it insignifigant. Ever had a squib load stick a jacketed bullet halfway through the barrel? Still gotta have a hammer or an arbor press to back it out unless you soak it in penetrating oil for a couple hours.

The "resistance" (-F) is of very little significance when compared against the propellant generated force (F) in that it is insufficient to overcome it and the bullet still develops enough velocity to leave the barrel at the desired velocity. That the bullet leaves the barrel is what is important here, since the internal frictional forces induced by contact with the barrel are insufficient to stop it.
A squib load is a vastly different situation than a bullet that is already in motion in the bore because the moving bullet is not static whereas the squib load is. There is considerable difference between the coefficient of friction of the two different materials (bullet jacket and barrel steel) involved in each situation (static vs. dynamic) and is why the squib load bullet requires the use of an arbor press or hammer and oil to remove it from the bore.




otherwise acceleration would really suffer and the bullet wouldn't achieve the speed that it does.


Originally posted by 1911Tuner:
See Newton 1B for the clue to that.
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed. The coefficient of friction doesn't change...only the magnitude of force required.


Never said that the coefficient of friction could change and it doesn't in either the 'static' or the 'dynamic' regime. I said that propellant gas blowby will alter the friction component (a different concept) by reducing contact via its interposition between the two materials, thereby 'floating' the bullet, or acting as a 'gas bearing', so to speak, and would require less force as a result to overcome the resistance (-F) induced by that contact. Two different elements, both of which dictate in different ways how the friction encountered will be modified.




Originally posted by 1911Tuner:
Once the object is in motion, it requires less force to keep it in motion...and the faster it goes, the less force it takes to keep it going and/or accelerate it to a higher speed.

Incorrect.

Once in motion, the object actually requires no force to keep it in motion at a certain velocity since it tends to remain in motion unless acted upon by another impinging force (friction in the barrel or air resistance being a fine example) so there is no additional force required to maintain the initial velocity except for the exceptions noted.

However, while velocity and acceleration are related, they are two different properties and in order to accelerate (ΔV) a body increased force is required because you are now adding K.E. by changing the velocity (accelerating) of the body in motion. It does not require anymore force to accelerate a body that is already in motion because it's mass (m) remains the same regardless of its velocity (v) and is still effected in the very same way by an accelerating (or decelerating) force, because acceleration is independant of any existing velocity (even 0) the mass may have as dictated by the equation:

F = ma*

*In examining the equation above, you will note that there is no variable in the equation for velocity (v) and that there is no modification of the variable for Force (F) or for mass (m) for any 'pre-existing' velocity that the body may (or may not) possess. Once in motion, a body remains the same mass (m) and experiences the same acceleration (a) by a given force (F) regardless of whether it is in motion (v) or not.

There is no such equation for what you propose below:


Originally posted by 1911Tuner:
("the less force it takes to keep it going and/or accelerate it to a higher speed.")


If you can provide either a mathematical expression or a proven documented textual description of such a phenomena, I would love to see it since it would set the world of Newtonian mechanics on its ear.

As it exists, the only increase of mass (m) that is quantifiably related to velocity (v) is that as described by one A. Einstein in his theories of General and Special Relativity and is encountered only at the higher "relativistic velocities" as they begin to approach the speed of light (c) and I do not think that you are referring to velocities in that range as they are presently unattainable in firearms and likely to be so for the foreseeable future and are irrelevant to this discussion anyway.
:)

1911Tuner
April 24, 2008, 05:56 AM
Once in motion, the object actually requires no force to keep it in motion at a certain velocity since it tends to remain in motion unless acted upon by another impinging force (friction in the barrel or air resistance being a fine example) so there is no additional force required to maintain the initial velocity except for the exceptions noted.


Agreed...but the frictional resistance is constituted as an outside force that is working to stop the bullet and restore equilibrium.
A bullet fired in outer space would continue to travel at the exit velocity unless and until it struck something or entered the gravitational field of something large enough to alter its constant state of motion.

"Objects tend to remain in motion *unless* acted upon by an outside force."

The "resistance" (-F) is of very little significance when compared against the propellant generated force (F) in that it is insufficient to overcome it

Agreed again...but the point wasn't compulsive force vs resistive force. It was simply that because the coefficient of friction doesn't change throughout the trip through the bore...that the forward drag on the barrel also remains constant as long as the bullet is still there and moving, and that it requires less compulsive force to accelerate it to a higher velocity than it took to initiate its acceleration. If it didn't...the bullet would slow down after the initial punch. (Which can happen in some barrels with certain powder charges and burn rates.)

Walkalong
April 24, 2008, 09:24 AM
Just got back to this post. Hmmm....

The original poster asked about being snappy, which I took as torquey because that is what many .40 shooters complain about, vs a push in the .45, and I said "velocity my friend"

That was totally dismissed, but I stand by it.

During the ride through the bore is when the sideways(rotational) torque happens as the bullet gets up to its rotational speed there.

To get up to its velocity with the same rate of twist in that length creates more sideways torque in the .40 than the .45. It gets up to speed much faster.

Am I wrong?

1911Tuner
April 24, 2008, 10:03 AM
velocity my friend

It's not all about the velocity. It's determined by mass times velocity.
As noted...you can have extremely high velocity and light recoil. The .22-250 with a 50-grain bullet at 3800 fps is an example.

Conversely, you can have fairly low velocity and heavy recoil. My 7.5 pound Sharps carbine in .45-70 will demonstrate that to you clearly with a 405-grain lead bullet at 1400 fps. :eek:

Walkalong
April 24, 2008, 10:30 AM
But we are comparing something much closer together than the .22-250 and .45-70.

I still think the velocity difference in those two calibers has a lot to do with the torque felt by most shooters in the .40 vs the push in the .45.

1911Tuner
April 24, 2008, 10:58 AM
I still think the velocity difference in those two calibers has a lot to do with the torque felt by most shooters in the .40 vs the push in the .45.

Sure. But it's still mass times velocity, no matter how close the two are.

Be aware too, that only a tiny fraction of the felt recoil in an autopistol comes from the explosion of the cartridge itself the way that it does in a fixed-breech weapon...like a revolver or a single-shot pistol.

The recoil spring's resistance and the slide's speed in compressing the spring...and slide to frame impact determines what is transferred to your hand.

All else being equal...the stiffer the spring, the sharper the recoil. The faster the slide moves...the sharper the recoil with a given spring. The slide's speed and its mass when it hits the frame's impact abutment "flips" the muzzle upward the greatest amount.

But the actual internal event? Nada.

Hypothetical pistol with 30-foot rails with the slide mounted on the front. Weight of the gun is identical to a normal pistol. No recoil spring. Fire the gun, and you probably won't feel a thing...and since the slide has to move 30 feet to reach the impact abutment...you probably never will, aside from a slight disturbance from the rail to rail friction.

Matt304
April 24, 2008, 12:26 PM
If you want to talk about the actual physical profile of the recoil energy, the impulse curve, then just think of it in terms of delivered impulse to the projectile. That is the easiest way to look at it, because we can take a slice of the curve during any given time frame to determine how much impulse and thus total energy has been delivered in recoil. The recoil curve is essentially directly relative to the impulse curve delivered by the propellant energy release into the projectile.

The true recoil curve at hand is entirely dependent upon the burn time and energy release of the propellant. The impulse curve shapes and peak pressure times can vary widely between calibers and powders. The problem trying to quantify felt recoil this way is that it truly means diddily squat to a human perception of it. The reason is quite simple. We feel the total impulse curve of energy delivered to us in only a couple of milliseconds, making it impossible for us to distinguish the actual impulse profile itself when referring to two like total energy amounts with different pressure curves. However, what we can account for ourselves is the total impulse energy delivered by that event, as it hangs around much longer as it dissipates and decelerates in our hand and arms. Because of this human limitation, we must speak in terms of total impulse delivered as the basic means of our perception and comparison.

Think of two rocket motors.

One burns over the course of 1 second, and generates a steady 10lbs of thrust for that time duration.

The second rocket motor burns for 2 seconds, and generates a steady 5lbs of thrust for that duration.

At this point, we know that both rocket motors deliver the same total impulse, they just burn for different durations.

Now, if you had to hold each one in your hand, the motor burning for only 1 second is going to seem like it gives you more of a kick. If gunshots took 1 second for the pressure event to complete, this motor would represent a high pressure load with a very fast pressure rise time.

The second rocket motor is going to feel like a much more modest push, as its duration is longer. It would similarly represent a low-pressure load, with longer pressure event.

At these amounts of time for the events to occur, we ourselves can easily distinguish a difference in feel between the pressure events of the rocket motors.

OK, lets try something interesting. Speed the rocket engines' burnrates up, but keep them at the same total impulse. The 1 second rocket motor now burns for 1/20 of a second, and the 2 second motor is increased to 1/10 of a second, still twice as slow as the first motor. What do you think you will feel at this point? The rocket engines are getting much closer together in feel, both lasting for just a very quick burst of energy, instead of the large difference they had before.

Now, push those times even faster, down to the level of firearms. Down to the point where both of these events each occur over only a few thousandths of a second. Our "high-pressure" rocket motor #1 now lasts for a duration of 0.002 seconds, and our slower rocket #2 now lasts for a duration of 0.004 seconds.

Now imagine holding the two in your hand at these speeds, and trying to tell the 2/1000s of a second difference in pressure profile between them. Do you think you could distinguish the difference?

This is why the true pressure profiles themselves are far less relevant than the total impulse delivered by the energy involved.

I get what 1911Tuner was saying about "muzzle velocity", and I agree. The simple fact is that recoil calculators assume that the bullet is still accelerating, and has not lost any acceleration before muzzle exit. Otherwise, it's not really about muzzle velocity, it is rather about total impulse delivered during any given time frame. But, from muzzle velocity we can determine energy involved which had to occur at some point before the muzzle exit to achieve that projectile speed. Then, from that total impulse value we can then approximate feel, hence my example of why true profile curves are irrelevant.

Here is a little graph I drew to attempt and illustrate what the recoil on the bolt face or breech of the gun would look like over time and, as you would imagine, bullet travel:

http://img381.imageshack.us/img381/3235/recoilcn6.png

Also do notice the shape of the impulse curve in the graph. Since this profile is directly proportional to the shape of the acceleration curve on the projectile, each 1" of barrel removed will never do the same as any other inch of barrel. Think of printing that graph out on a piece of paper, and then taking scissors and cutting vertical slices off from the right side, making it shorter. Each slice, you would have a different amount of blue area, or pressure. This is the way you would be removing acceleration impulse as you cut the barrel shorter, closer to its maximum pressure point.

1911Tuner
April 24, 2008, 02:17 PM
Excellent illustration...and it identifies clearly the point of peak pressure and force delivered...to bullet and breechblock alike.

Again...The majority of the bullet's total velocity and the total recoil energy is delivered early in the event.

Walkalong
April 24, 2008, 02:47 PM
Maybe if the 1911 .45 was plastic on the bottom, it would torque more too. :D

GLOOB
April 24, 2008, 03:35 PM
I learned a good bit from reading your posts. But I can't resist the urge to correct you on one point.

"Peak force occurs with peak pressure...With some extremely quick pistol powders, it may peak before the bullet has even left the case."

Until the bullet is moving, there is obviously no recoil going on.

Force of recoil is directly related with acceleration. That first quick jolt of greatest acceleration (during which time the peak pressure may theoretically still be building, but in all probability is actually being reduced), would be coincident with the highest peak force of recoil, assuming the rest of your argument to be sound.

Take a bottle of compresssed air, for example. You take a hammer and break off the valve, and the bottle shoots across the room. The peak pressure is there before you break the valve. The acceleration and force occur afterward, during the time the pressure is reduced.

1911Tuner
April 24, 2008, 03:58 PM
Until the bullet is moving, there is obviously no recoil going on.


Obviously.. No argument there.

What you haven't thought about is that the bullet has to travel a small distance before it leaves the case. Essentially, the case itself is a "gun barrel" of sorts...smoothbore, of course...and the case head makes up a "breechblock" for that smooth barrel. So, a metallic cartridge essentially makes it a "gun within a gun."


And...If there's a little headspace, the case starts to move rearward in "recoil" before the recoil force acts on the true breechblock...but a few thousandths of an inch can pretty much be disregarded. I doubt if the time could be measured without some very expensive and sensitive equipment. This is why rifle cases manage to stretch when the pressure nails the forward section to the chamber walls.

Matt304
April 24, 2008, 04:45 PM
If you look at my graph, that shows you directly how long the energy takes to be applied. There is no delay anywhere in reaction; it is instantaneous. The force profile looks theoretically exactly like that. The recoil accumulation represents a direct relation to surface area of that graph at any given time.

Imagine that you drew a line straight down from the peak of that graph. On the left of the line, compare the graph area to the area on the right side. Already about 40% of total energy has been released by the time the highest pressure peak is reached.

Assume that the total recoil velocity of your firearm reaches 10fps from each shot you fire. Based on my prediction, this would tell you that if we stopped time on that line I mentioned earlier, the gun would already be moving 4fps in your hand as peak pressure is occurring.

Right along the lines of what 1911Tuner was saying, soon we have a gun which is nearly at full recoil as the bullet is just beginning its journey down the barrel. The whole remainder of the time the bullet spends traveling down the barrel, it is riding inside of a gun which is already traveling at near its full recoil speed.

The moral here is that a proper and consistent hold is everything.

But there's more. The good side to this is that as the gun is at its full recoil velocity, our hands have much less of an influence on it. And at around 1 millisecond elapsed time, the barrel doesn't get to pivot very far off course. The hotter the load energy, the more of an influence it will have on the shot, as there will be more off-center velocity and gun movement in the same given timeframe. Hold and technique is still very much important at all times, just remember that. The final deceleration over a much longer time period is what allows for the gun to continue its headed course and finally kick up.

1911Tuner
April 24, 2008, 05:13 PM
There is no delay anywhere in reaction; it is instantaneous. .

Absolutely...and you'd be surprised at the number of people that I've fought with over the fact that action and reaction begin at the same instant...that the gun is in recoil before the bullet leaves the barrel. I even had one guy insist that the force vector "splits" with half of it pushing the bullet...then after the bullet exits...the other half starts pushing on the breechblock. In the same breath, he claimed to have had two years of college physics. I was laughin' so hard that I couldn't see to type.

Ah well. Onward...

Matt304
April 24, 2008, 05:42 PM
I even had one guy insist that the force vector "splits" with half of it pushing the bullet...then after the bullet exits...the other half starts pushing on the breechblock.

Oh boy, it looks like you've run into a real prodigy. :D

Yes, I myself have noticed many arguments about recoil physics in which people claim that there is no way the gun moves by the time the bullet leaves the barrel. I guess if you aren't inclined with physics whatsoever, it can be an easy incorrect assumption to make. So long as they accept the right answers and move on, no harm done. But this is definitely the way it works.

On a more interesting notion, this does make you think about the effect of barrels which are too long, and if they may have some negative effect on POI due to an unbalanced gun design. Revolvers come to mind. I suspect the amount of time may be too short to have much effect for a few extra inches, since they would be adding mere 1/10000s of a second, but it does make you wonder.

DaveBeal
April 24, 2008, 06:40 PM
Absolutely...and you'd be surprised at the number of people that I've fought with over the fact that action and reaction begin at the same instant...

You bet. Momentum is conserved at all times. Before you pull the trigger, nothing is moving. But as soon as the bullet starts to move forward, something must move backward. At first it may just be the case. Then it's the slide, then it's the frame and your arm. Eventually, the whole earth recoils. But not by much.

1911Tuner
April 24, 2008, 07:46 PM
Eventually, the whole earth recoils. But not by much.

That's a stretch... :D

Actually, "it" doesn't have to move. In order for both objects in an action/reaction pair to accelerate into motion...the force applied must be sufficient to overcome the resistance offered.

If you push on a house, you can't move it...but that doesn't mean you didn't apply a compelling force. It just means that the force wasn't large enough. The house "felt" acceleration, but its mass was too great. If you push hard enough to make yourself move...technically, there is an action/reaction event.

It's a little like the tree falling with no one around to hear it question. If you place the butt of a rifle against a tree and fire it...does the fact that the rifle doesn't move backward mean that it didn't kick?

DaveBeal
April 24, 2008, 09:47 PM
That's a stretch...

Actually, "it" doesn't have to move.

I admit that it's entirely theoretical, but if you could attach a gun with absolute rigidity to the earth and fire it, the earth would move a bit. It has to, because there was no net momentum in the earth/gun system before you pulled the trigger, and there has to be none after. Since the gun is fixed and can't recoil by itself, the earth has to.

In your house example, if I lean against a wall of my house and suddenly push with enough force to throw myself backward, something has to move forward. Probably just a portion of the wall will move by flexing, but theoretically it could be the whole house if it is rigid enough.

1911Tuner
April 24, 2008, 09:56 PM
Well...I guess hitting a gnat would alter the speed of a F118 at full throttle if ya get down to brass tacks.

;)

Walkalong
April 24, 2008, 10:03 PM
The earth would absorb the energy, just like the brick wall of the house, but the Theory is OK.

Matt304
April 24, 2008, 11:31 PM
Getting a little astray, but what the heck. It's fun to think about.

In theory, if the world was outside of an orbit and just floating out in space at a complete rest, and there was complete equilibrium of all other forces on Earth, a gunshot could put the planet into motion. It would start moving at a rate so slow it could nearly never be measured, but it would have to begin a motion at some extremely low speed.

Since we are being pulled upon by the sun's enormous gravity, nothing on earth could stand a chance at changing our path.

1911Tuner
April 25, 2008, 05:39 AM
Getting a little astray, but what the heck.

Okaaaayyyy. Since we've literally gone into orbit, it might be worth mentioning at this point that while weight and mass are related...they're not the same.

In the vast frontier of space, outside the gravitational pull of another body, a baseball and a mile-wide asteroid weigh exactly the same. Nothing.

BUT! You can easily accelerate a baseball with one hand in space...but not the asteroid. Mass is still mass.

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