Recoil and point of aim

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Red Cent

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The facts
One revolver. Say a M19 357Magnum. 6" barrel. Factory stock.
Bullet. 125gr hollow point @ 1350 fps
Bench rested. Sandbags. Front of frame touching only. Nothing impededing the grip. Barrel not touching anything.
Target/x ring is 30 yards.

Second bullet. 125 gr hollow point @ 900 fps.


The pistol is sighted in dead on the X ring with the first load. Where will the second round hit and why? This is not a trick. It is not a trick question. Simply internal and external ballistics.
 
Hard to say without some very advanced physics calculations. On one hand, the slower bullet will drop slightly more by the time it reaches the target. On the other hand, the slower bullet takes longer to exit the barrel, while the barrel is tilting up ever so slightly from recoil.

Best way to tell is experimentation. Sure beats an hour of writing numbers on a chalkboard. :D
 
Since gravity is a function of acceleration over time, and the 900fps bullet has more time for gravity to act on its trajectory, then wouldn't the point of impact be lower than the 1350fps bullet?
 
Depends where the sights are set for and the velocity they are set for. The hit could be either high or low depending.
 
Where?

My guess is that the slower bullet will strike higher.

Why? That part of recoil which occurs up to the point the bullet exits the muzzle is called "jump". I'm a double rifle shooter. We double rifle shooters live and die by jump. Suppose I have a load such that both barrels are spot on. If I reduce the powder charge slightly the rifle will shoot wall eyed. If I increase the powder powder charge slightly the rifle will shoot cross eyed. This is due to longer barrel dwell time with the slower load, allowing more time for jump, and vice versa. So I expect the revolver muzzle to rise a bit higher before the bullet exits with the slower load.

This effect will be offset somewhat by the fact that the slower bullet will have more drop.

I ain't makin' no cash money bets.
 
Slower bullet will hit higher, unless it is so slow there is no appreciable recoil. I shoot 125's at 1300ish and 125's at 900ish in .357, and the slower one hits higher, but not by much.

Totally different weight gun though. Heavier.
 
Slower bullet will hit higher, unless it is so slow there is no appreciable recoil. I shoot 125's at 1300ish and 125's at 900ish in .357, and the slower one hits higher, but not by much.

Why? I'm not seeing the reasoning why as slower bullet would impact higher than a faster bullet.
 
I guess I'm stuck with the idea that the bullet would be exiting the muzzle faster than you'd begin to see the effects of recoil, akin to the way that the bullet is gone out of an auto before the mechanism begins to unlock. Interesting.
 
Simple physics. When the bullet starts to move forward,{action}, the gun starts to move rearward.{reaction}
 
As others have said, the slower bullet would hit higher at 30 yards because it is in the barrel longer while the barrel is rising due to the recoil. The barrel will be pointing higher when the bullet leaves the barrel.
 
It will still hit the X ring slightly higher but not much. The trajectory of the bullet at 30 yds has a slight arc traveling at 900 fps. The trajectory of the bullet travelling at 1350 fps is more of a straight line at 30 yds. Both will hit the X ring. Distance is too close to effect impact of poa.
 
Red, when the primer pops. It takes a rather amazing amount of force to jam the bullet into the bore. If you'll note, virtually no bullet is smaller than bore diameter. Takes some effort to get the bullet started.
 
Disagree with recent posts and believe John Parker had it right in post #4....slower bullet takes longer to reach the target...therefore gravity has more time to work and pulls it lower than the faster bullet.

Recoil and time in the barrel shouldn't matter in this instance; muzzle flip happens after the bullet leaves the barrel. Else, we'd never be able to aim at or hit anything without anticipating how much the muzzle would flip before the bullet leaves the barrel. Think about it. Muzzle flip only matters when we're "double-tapping" because we have to bring the barrel back down to the same spot to fire an identical round.
 
So many factors to consider here. The slower round will have dropped from it's trajectory more by time it reaches the target, but how much in only 30 yards? 1/4" maybe?

On one hand, the muzzle could have risen more due to recoil by time the bullet leaves the barrel. On the other hand, the slower bullet is producing less recoil, so maybe the barrel rises less than the faster projectile? or maybe these factors cancel out, and initial trajectory is the same for each projectile. I am not motivated to do the math, but I'm going to figure muzzle rise at the point the bullet leaves the barrel to be the same for each projectile. For the slower, the barrel has had more time to rise under recoil, but it is also recoiling slower.

The big difference would come if the bullet weights were different.

Muzzle flip does not wait for the bullet to leave the barrel. The instant the bullet is subject to an accelerating force, the gun is subject to an accelerating force equal in magnitude in the opposite vector.
 
What would happen to the bullet if:

Position the pistol as before except place a sandbag under the grip. Tie the barrel down solid just in front of the receiver. With the sights dead on the X ring, fire the 1350fps at the target. Will the bullet hit the same place as the first 1350fps round hit? If not, why?
 
Recoil and time in the barrel shouldn't matter in this instance; muzzle flip happens after the bullet leaves the barrel.

It may seem that way, but it doesn't work that way. :) At some specific distance, the slower round will print lower on the target due to velocity. But, if the sights are regulated for a faster cartridge, the slower round will strike the target above the POI of the faster at close range.

You can try this with a revolver (I.E., .38/.357) and really see the difference. :cool:
 
Recoil and time in the barrel shouldn't matter in this instance; muzzle flip happens after the bullet leaves the barrel.

It may seem that way, but it doesn't work that way

Both statements are correct. It depends on what type of gun is used.

Action and reaction begin at the same instant. If you push on an object, you immediately get pushed back by that object. If you pull on an object...you immediately get pulled.

With revolvers, muzzle flip starts as recoil starts...which starts with bullet acceleration.Wirh some cartridge/powder burn rates. With pistol calibers, about 95% of the recoil is in full effect at around a half-inch of bullet travel.

With autopistols, 90% or more of the muzzle flip occurs when the slide impacts the frame.
Locked breech/recoil operated pistol slides move nominally .010 inch at bullet exit. With a .45 ACP/230/830 fps load...by the time the slide hits the frame, the bullet is about 20 yards downrange.

This is a very old fluroscope photo of a 1911 pistol firing. Here, the bullet is within 3/4 inch of reaching the muzzle. Note the position of the link. The slide has moved approximately .070 inch, showing that recoil has begun. At this point, the only action against the frame and the shooter's hand is via the recoil/action spring, and very little if any muzzle rise is seen.

This is a live photo. The pistol is being fired by a human being. If you look closely, you can see the bones in the shooter's hand.

Gun20Fired.jpg
 
Higher or lower vs lighter or heavier.......

Well it depends upon the direction you are shooting: if true north or south there will be no difference,.

But if you are shooting east, into the earths rotation than the bullet will strike lower as the target is moving upwards in a complicated formula taking into account the co-sign of your angle of aim to your latitude, bullet velocity (time in flight) and dunken donuts*.

Shooting west into the setting sun will strike higher as the target is setting relative to the flight time, again a complicated formula including of course—dunken donuts*


* cause America runs on Dunken Donuts
 
RC, if the gun is immobilized so that it can't move in recoil...all shots would hit a pretty much the same spot until the distance became a factor The slower bullet then would hit lower due to being in the air longer, giving gravity more time to pull it down.

Assuming equal outside forces...dead consistent grip on the gun...and equal recoil generated by a slow/heavy and light/fast bullet...the slow, heavy bullet would strike higher on the target at a given distance.

Again...time is the deciding factor. The slow bullet is in the barrel longer. As the barrel rotates upward, the slow bullet will exit at a higher elevation than the light/fast bullet.

All this ties in to the reason that a consistent grip is important when shooting revolvers...and why offhand groups can print higher or lower than sandbagged groups...especially if the sandbag is placed under the butt of the gun instead of under the shooter's wrists. Change the grip and the POI changes. Ever notice how many single-action shooters hook their little finger under the grip? That provides a spot weld that gives the shooter a consistent hand placement on the gun. With that taken care of, all he has to concentrate on is the grip strength.
 
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