Bullet drop question

[tombogan03884]

The slower bullet has a higher arc to it's trajectory.
Take a 45/70 against a .223 the 45/70 will hit higher because it has a trajectory like a rainbow, where as the .223 at higher speed has a much flatter trajectory.
If you want to get into the actual mathematics of ballistic science they can explain to you why the 45/70 will give accurate kills at up to a mile, while the .223 won't.
But you'd need to be a physicist to under stand it, and the answer would have to come from some place like NASA.

[crusader rabbit]

And just to add my two cents... 

If you have an essentially flat trajectory out to about 100 yards (i.e. .223, or .22mag, or .357 mag out of a carbine) bullet weight or bullet speed should not make any real difference--as long as the combination maintains the "flat to 100" trajectory.  After those "flat" distances are exceeded, either the heavier or slower bullet would of physical necessity drop lower... unless we happened to be firing from the Space Ship Enterprise and gravity ceased being an issue.  The rule here is that the longer gravity has to work on something, the greater the effect.  However, if it's flat to 100, then it's flat to 100, irrespective of the speed or weight.

Offered because I've had a couple of beers and I'm just feeling smart.

Crusader

[Timothy]

I don't know one way or another!

I'm just impressed by the proper use of the word irrespective!

 

Offered because I've had some cheap Canuk whiskey cuz that's all I can afford at the moment...

[Solus]

I knew I hadn't phrased my question properly.

Leave sighted in range out of the question. I understand about a bullet crossing the line of sight in two places, "on the rise" and "on the drop".  Let's just say that we are firing at a 4x8 sheet of plywood at say 50, 100, 150, and 200 yds. and there is a sighting point painted on the target so that we will always shoot at the same point The gun is not locked into any device which holds it from moving. It matters not who is firing the gun but it is being held in a typical firing position by a shooter. Now I understand about the effects of gravity pulling a bullet down and about the parabolic curve etc. Explain why the faster will hit higher on the target than a slower bullet. This is real world stuff not theory, guys. I don't mean to be argumentative here I just can't reconcile theory with what I've seen. Have patience I can be obtuse at times. (Fancy enough statement for you TB?)  


Thanks for all the help anyway.

Pecos

Thought about how to explain this over night....here is the best I can do.

Say the rifle is bore sighted with a laser, so the sight line is an extension of the bore.

The laser spot is marked on each of the plywood targets at the different ranges.

The rifle is positioned in the exact same location before each shot.

Both the fas and slow bullets will begin to drop the instant they leave the bore and neither will ever rise above that laser sight line.

Since it will take the slower bullet longer to get to the target, at any range, gravity will have more time to work on it and because of this it will drop farther than the fast one over the distance to the target.

That help?

P.S. I have my doubts about time of dwell of the bullet in the bore allowing a change in impact do to recoil.

I do not have the math to calculate how fast a bullet will accelerate in a barrel, but a workable value would be half of it's muzzle velocity...the faulty assumption here would be that it's acceleration from zero to full speed is linear.

So say a bullet has a muzzle velocity of 2000 fps and our assumption is that it's average speed in the bore is 1000fps.

Use a 24" barrel for easy of calculations and the bullet will leave the bore 1/1000th of a second after ignition.  A 1000fps muzzle velocity would give 500fps average, by our assumption, and it would leave the bore 1/500th of a second after ignition.

Again, I don't have the math to calculate it, but my hunch is that even the 1/500th of a second is not enough for the entire rifle to overcome it's inertia and move in recoil.  Some one might know this answer.

There might be barrel harmonics that could be able to "vibrate" the barrel in this time frame, but I don't know if that would be enough to show a significant change in impact point.

If the weapon is being held by a person, there are movements that might affect the impact point.

For instance, anticipation of recoil might be the reason a faster bullet of the same weight might hit lower than the slower one.  The barrel is dropped a bit in anticipation of the heavier recoil.

P.P.S.   The P.S. are just thoughts of what might be influencing all this...and does not mean I have any proof or or opinion other than they might be having an effect on all of this.

P.P.P.S  You know, I've not watched them with this in mind, but perhaps the slow motion shots on Shooting Gallery might show if the rifle/pistol has any muzzle movement due to recoil before the bullet leaves the muzzle.  Will check the reruns 

[Solus]

This is a link to a non-YouTube video.    http://www.rifle-accuracy.com/Rifle_Recoil.wmv     You will get a chance to save it if you wish.

It does show the rifle bullet leaving the muzzle and , to me, it looks to be well before muzzle rise begins.

Notice there is a secondary "muzzle blast" when the recoil seems to begin.  My guess is that this is a super-sonic round and the second blast is the sound wave??

[Sponsor]

[Timothy]

It stands to reason, based on my knowledge of physics, that the muzzle flip is due to escaping gases from the muzzle itself so the projectile would have left the barrel prior to the gas release.  Since you more or less grip a firearm below the bore axis, the hot gases force the muzzle in an upward direction due the fulcrum point of the grip rather than straight back along the axis.

[Pecos Bill]

Gentlemen and/or Ladies, I think I'd like to call a halt to this discussion but of course I can't. I have read through the various posts and seen the explanations and theories. Some of you give what sounds like very good theories about recoil and when it starts. I have had some of that (recoil doesn't start till the hot gases exit the muzzle) my question to that is didn't some guy named Newton come up with a rule that said that for every action there is an equal and opposite reaction? If the bullet is forced down the barrel of a gun would there be a reaction the that action and be the start of recoil? If that is so then the longer the bullet is in the barrel the higher the muzzle rises before the bullet exits. Therefore at some reasonable range that bullet will strike the target higher than a bullet moving faster and having a shorter time in the barrel and thus the muzzle rises less before bullet exit would strike lower on the same target. This assumes that all variables remain constant except the speed of the bullet and that the bullet hits the target "on the rise" not as it is dropping. If it were dropping the the faster bullet will indeed strike the target at the higher point. I thought I mentioned that in my second post.

To do this with two different calibers with two different bullets would never be a valid comparison.

I don't know the mathematics or the physics of this I just know what I've seen. Thanks for all the help and responses so unless y'all need me for something I'm outta here.

Pecos

[Timothy]

Bill,

My contention is that muzzle rise is a product of the projectile exiting the barrel because of the hot escaping combustion gases and has no affect on the bullets rise.  A firearm is aimed with a sight that is somewhere above the bore axis.  It needs to be zeroed at a distance from the muzzle regardless of what type of firearm is used.  Using that zero point with two identical calibers with two different velocities will cause the faster bullet to get to the aim point sooner thereby hitting the target at a point above the normal zero of the sight line.

If you have a free floating barrel with nothing to impede it's rearward motion it would not rise at all vertically.  It would move straight back along it's bore axis.  That is Newtons 3rd law of motion "Third law: The mutual forces of action and reaction between two bodies are equal, opposite and collinear."

If you bothered to watch all five of the Mythbuster video's posted previously, you'll clearly see the bullet exiting the barrel well before the 1911 began to cycle.  The muzzle flip had no effect on the bullet what so ever.  It's only in the barrel for about 1/2 of a millisecond or .0005 seconds.

850fps/1000 = .85' per ms x 12 = 10.2" of travel per millisecond

[tombogan03884]

Muzzle rise has nothing to do the bullet or gases exiting the barrel.
if you used a universal receiver bolted securely you could fire a 16 inch gun and there would be no muzzle rise.
muzzle rise is caused by the recoil impulse trying to force the gun backward, the gun however is braced against the shoulder, (hand in a pistol ) which causes the rifle to pivot on the point of resistance. Since it is also braced at the bottom by the off hand the motion is always upwards.

[Timothy]

Muzzle rise has nothing to do the bullet or gases exiting the barrel.
if you used a universal receiver bolted securely you could fire a 16 inch gun and there would be no muzzle rise.
muzzle rise is caused by the recoil impulse trying to force the gun backward, the gun however is braced against the shoulder, (hand in a pistol ) which causes the rifle to pivot on the point of resistance. Since it is also braced at the bottom by the off hand the motion is always upwards.

Which is exactly what I just said...

Recoil impulse is directly related to escaping hot gases.