How do bullet weights and velocity affect point of impact?

[1911 Jim]

I want to clarify something before I go forward, as I've been somewhat less than explicit in encompassing all forces involved here, and might have given a misconception in err.

The bullet has a force acting on it which is equal to the force acting on the frame. Since we know this will accelerate the bullet much faster than the frame, the bullet will clear the bore before the frame moves all that much. When this happens, the remaining charge causes a much greater force rearward as the gasses are now not only trying to move a projectile and slide, but they're pushing against a large volume of air, and that air is now pushing back as well. We know static air can do a lot of work, as it's how we fly (everyone's aware that pushing air against the ground is only part of the forces inducing flight right? The wings suck the aircraft up, they don't push it).

"Exhibit A" to support this assertion is how "limp wristing" will cause misfeeds in semi autos. If the bullet's force was the sole means of actuating the mechanism, then there wouldn't be any problem with how loosely you held the weapon as it would contain within itself the force needed to fully cycle the action.

"Exhibit B" is the problem of a plugged bore. If the equal and opposite force of the bullet against the gun was solely responsible for the cycling of the action, there wouldn't be any blown up barrels from plugs. You'd simply launch the frame rearward and violently eject the casing. We know this isn't the case.

In response to the introduction of the revolver to this discussion, I only have one point to make - you have an open system between the cylinder and the barrel via the gap. You're actually inducing force from the charge to the atmosphere prior to the bullet exiting the barrel. You're pre-recoiling the gun, and as such must be compensated for.

Lastly, if bullet acceleration was so heavily responsible for muzzle rise, it would be far more evident in longer barreled weapons. You don't have any issue with high powered rifles rising up as the bullet makes way down the bore. We know this isn't the case either.

There's also the force of the air being compressed in front of the bullet pushing back against the bullet/gun which hasn't been mentioned. I don't think it's of much influence, but it is present none the less.

There's a lot of dynamics involved in such a simple act as shooting a gun.

Now if you want to get really nasty, we can start talking about compensated barrels and how they don't need to direct gasses vertically to completely eliminate muzzle rise. If it's explicitly the bullet acting against the gun, these wouldn't work. I don't have to say it do I?

 

[1911 Jim]

And John, you can clearly see that the muzzle hasn't flipped even though it has started moving until after the bullet has cleared. 2:16 ish

 

[Double Naught Spy]

Gravity is always present and always pulls down at an acceleration of 9.8 meters per second, per second. This means that for the first second, the bullet is going to drop 9.8 meters, but for the second second, it's going to drop 19.6, and the third it would drop 29.4. So for 3 seconds of travel, the bullet has dropped 58.8 meters and is accelerating towards earth.

This only applies if the bullet is fired from an exactly level bullet or dropped from a stationary position. Bullets fired from an upward tilted barrel do not drop at 9.8 meters per second for the first second because they are traveling upward initially.

 

[brickeyee]

And John, you can clearly see that the muzzle hasn't flipped even though it has started moving until after the bullet has cleared. 2:16 ish

The amount of movement required to affect the point of aim is far smaller than the large flip.

There is not enough resolution to see if the gun has moved a few thousandth to hundredths of an inch, all it takes to start affecting point of aim.

The bullet moving gives it momentum, and the momentum affects the position of the gun instantly.

If the bullet is moving the gun is starting to move.

Even the benchrest guys try to make sure that all recoil is exactly the same in every shot, since it affects point of impact.

 

[kle]

but they're pushing against a large volume of air, and that air is now pushing back as well. We know static air can do a lot of work, as it's how we fly (everyone's aware that pushing air against the ground is only part of the forces inducing flight right? The wings suck the aircraft up, they don't push it).

Wings pushing on air? It's Physics--Newton's 3rd Law of Motion: For every action there is an equal and opposite reaction. With forward thrust, the wings are pushed through a fluid (air). The wings direct that air downwards (or push it downwards, if you like)--the opposite reaction is that the wings are then pushed upwards. Since the wings are attached to an airplane, the airplane also rises.

Same with recoil in firearms--action/reaction. The force necessary to drive the bullet forwards also drives the firearm backwards. Since the bullet has less mass, it moves faster and exits the barrel before the muzzle has risen too much (but the muzzle does rise a little before the bullet has left the barrel). Once the bullet has left the barrel, the gun continues to recoil because of inertia (Newton's first law of motion). There is a little bit of gas pressure left acting on the firearm, yes, but it doesn't act on the air. At this point it's just like a rocket--the expanding gas acts on the chamber. If it acted on the air, then rockets wouldn't work in space, since there isn't any air up there to act against.

Then the gun stops recoiling as it is slowed and stopped by the larger mass of the shooter (Newton's 2nd law).

 

[JohnKSa]

Since we know this will accelerate the bullet much faster than the frame, the bullet will clear the bore before the frame moves all that much.

Correct. But it doesn't have to "move all that much" to affect the point of impact on the target. As pointed out, 1mm (less than a 25th of an inch) of flip will make 6" of difference at 25 yards from a 6" bbl.

And John, you can clearly see that the muzzle hasn't flipped even though it has started moving until after the bullet has cleared. 2:16 ish

If you watch the other clips in that compilation you will see that the shooter is forcibly holding the guns in the video clips against a solid rest. He is preventing any movement of the gun in order to get a clear video. The point of that video is that you can CLEARLY see that the slide and barrel are moving in recoil "long" before the bullet exits. That contradicts your claim that recoil is created by the jet effect and doesn't start until the bullet exits the muzzle.

If you look at the second video (a gun shot from a more conventional hold) you can see muzzle rise before the bullet exits.

When this happens, the remaining charge causes a much greater force rearward as the gasses are now not only trying to move a projectile and slide, but they're pushing against a large volume of air, and that air is now pushing back as well.

You are correct in the sense that there is more momentum imparted to the gun than to the bullet. That is due to the fact that part of the ejecta (the gases and unburned powder) is not the bullet but still has momentum. In other words, the total momentum of the ejecta at the moment it leaves the muzzle is equal to the total momentum of the firearm as it recoils. The bullet is only PART of the ejecta and therefore the bullet gets only part of the momentum (the largest part) imparted to the ejecta as a whole.

It is not difficult to calculate the recoil due to the bullet momentum and the recoil due to the rest of the ejecta's momentum and find that the bullet is the biggest contributor.

The idea that the "remaining charge causes a much greater force rearward" is inconsistent with reality.

"Exhibit A" to support this assertion is how "limp wristing" will cause misfeeds in semi autos. If the bullet's force was the sole means of actuating the mechanism, then there wouldn't be any problem with how loosely you held the weapon as it would contain within itself the force needed to fully cycle the action.

It does contain within itself the force needed to fully cycle the action. The recoil velocity due to the conservation of momentum is imparted to the slide & barrel. The problem is that they must move independently of the frame in order for the gun to function. If that is to happen, something must restrain the frame or the recoil velocity of the slide and barrel will couple to the frame via the recoil spring (and the hammer/hammer spring in a hammer fired gun) and the entire gun will try to move as a unit.

If it moves as a unit, or even if the frame just moves too much, the slide & barrel won't keep enough of their "share" of the recoil velocity to move independently of the gun and function the action properly.

"Exhibit B" is the problem of a plugged bore. If the equal and opposite force of the bullet against the gun was solely responsible for the cycling of the action, there wouldn't be any blown up barrels from plugs. You'd simply launch the frame rearward and violently eject the casing. We know this isn't the case.

The bullet comes to an abrupt stop at the plug. When the gases can't vent through the muzzle as normal the excess pressure is what blows up the barrel.

If you want to look at the recoil from a plugged barrel you have to look at the forces. A large force was applied to accelerate the bullet down the bore. That resulted in the gun beginning to recoil backward. When the bullet hit the plug it applied a large force against the plug. That resulted in the gun now moving forward due to the large forward force applied by the bullet against the plug and therefore against the barrel. The net effect (in terms of the motion of the gun) isn't significant in either direction as a result of the opposite forces applied almost simultaneously.

In response to the introduction of the revolver to this discussion, I only have one point to make - you have an open system between the cylinder and the barrel via the gap. You're actually inducing force from the charge to the atmosphere prior to the bullet exiting the barrel. You're pre-recoiling the gun, and as such must be compensated for.

Any force due to the gas escaping from the barrel/cylinder gap would be in the opposite direction of the movement of the escaping gases. Since they escape radially and roughly equally in all directions the net effect is zero. None of the escaping gases move significantly forward or rearward so the effect on forward/rearward motion of the firearm is zero.

It doesn't "pre-recoil" the gun, it has no measurable effect on the gun's motion at all and therefore there is no need to compensate for it.

Lastly, if bullet acceleration was so heavily responsible for muzzle rise, it would be far more evident in longer barreled weapons. You don't have any issue with high powered rifles rising up as the bullet makes way down the bore. We know this isn't the case either.

Shoulder fired weapons tend to have less muzzle flip than handguns because the bore axis is much more in line with the stock axis. I don't have a lot of videos of rifles, however the video of the shotgun muzzles I provided definitely shows motion and some muzzle rise due to recoil prior to the bullet exiting.

Your ideas about recoil are completely inconsistent with the well-accepted laws of physics and contrary to what can be observed in high-speed videos.

 

[Glockman]

Muzzle rise is real

What this discussion needs is Physics teacher that is also an NRA certified Pistol instructor. Well here I am. Yes Muzzle rise is real and is at its worst in tall revolvers. In a semi-auto (Glock) with the barrel lower down and closer to the supporting hand this problem is greatly reduced. Go to you-tube and look at all the powerful revolvers shot by novice shooters that flip around and put the front sight right into their forehead. Heavier bullets go down the bore more slowly so they exit at a higher angle.

 

[rodfac]

Glockman is correct. Heavier bullets out of a hand gun, impact higher on target than lighter ones, given identical sight settings and grip strength. Revolvers, due to the high barrel location relative to the hand grip position are more prone to show large differences than automatics.

Jullian Hatcher wrote about the cause and effect in, Hatcher's Notebook. Col Chas. Askins wrote about it and had pictures taken with high speed cameras as early as the late '30's in his eminently readable, "The Art of Pistol Shooting" which clearly showed "muzzle flip" minute tho it may be. Finally, Keith wrote about it as well in his book, "Sixguns". All three were not physics professors, but were keen handgunners, and reliable witnesses. All three attributed the phenomena to the longer barrel time incurred by a slower heavy bullet. Too, its heavier recoil increases the amt that the barrel rises, and it does rise as recoil begins with bullet movement..."action - reaciton" in physics parlance.

We can all talk till we're blue in the face, but it happens to a greater or lesser degree to us all. If the OP observes the opposite effect then a new variable has been introduced or our collective chains are being pulled.

JMHO, and no offense to anyone who disagrees, Rodfac

 

[JohnKSa]

After working out the math my grayed out comments in this post are incorrect. rodfac is correct. tipoc's assessment below is correct, the POI difference observed by the shooter is due to the shooter anticipating the recoil, it is not consistent with the POI difference expected due to loading differences alone.

Heavier bullets out of a hand gun, impact higher on target than lighter ones, given identical sight settings and grip strength.

And given reasonably similar recoil.

If the muzzle is flipping upwards under the force of reasonably similar recoil then the slower (heavier) bullet will exit when the muzzle has flipped higher and therefore will impact higher on the target.

If the OP observes the opposite effect then a new variable has been introduced or our collective chains are being pulled.

Correct, the new variable is the amount of recoil.

If you have two loadings that are VERY different in recoil then the one with the most recoil may result in a higher impact on the target even if it involves a lighter bullet.

In other words, a light target .44spl load with a relatively heavy bullet might impact lower on the target then a full house magnum with a lighter bullet out of the same revolver. Even though the heavy bullet spends more time in the barrel, the overall muzzle flip with the full house load will occur much more rapidly and will be much more pronounced. That might be enough to result in the lighter bullet from the much more powerful loading hitting higher on the target than the heavy bullet from the powderpuff loading.

The OP was shooting a 10mm, and to my knowledge, all of the 135gr factory loadings for the 10mm are loaded to maximum performance for the caliber. On the other hand, a good bit of the practice ammo in 200gr for that caliber is pretty lightly loaded. Barely hotter than a .40S&W if that.

That could get into the region where the recoil in the two loadings is very dissimilar and things could get more confusing.

 

[rodfac]

JohnK...."If you have two loadings that are VERY different in recoil then the one with the most recoil may result in a higher impact on the target even if it involves a lighter bullet."

I've never observed that effect, John, but then maybe I've never had the speed high enough on the heavy bullet....44, .41 or .357 Magnum would be the probable candidates for what you're commenting on. As I rarely push even the magnums to their full potential in that regard, it's possible that there is another outcome that I've never encountered. Have you pushed one to that level, where the heavier bullet shot lower than the lighter one as you're describing?

I've been working with one of the 50th Anniversay Ruger .357 Magnums off and on for the past six months...using 125 JHP's and 158 gr cast lead SWC's. The 125's are going a chrono'd 1318 fps vs. a chrono'd 1050 fps for the 158's. Comparing impact pts for these two examples, shows a 2" difference at 25 yds...the heavier bullet impacting higher. I use a moderately heavy hand grip hold on the gun, and test my loads, sitting with a back rest position...basically the same position I'd use hunting from a stand...it allows me to shoot 1-1/4" gps with good loads on most days...the limit of my eye sight over iron sights.

It's been an interesting discussion....always something to learn. And I'd welcome anyone who's seen "the light bullet hitting higher" effect to share their loads and the type of gun used.

As an aside...I wonder if the automatic, with it's lower barrel position, vs the higher bore line of the revolver minimizes the upward motion? Measureing the difference between one of my Rugers vs a Colt 1911A1 this morning, I found 1-1/4" for the Colt vs 1-3/8" for the Ruger...measuring from the bore centerline to the top of the web between trigger finger and thumb...not much really! The Colt with its grip safety blocks upward movement of the piece in recoil, however, whereas the Ruger rolls upward in my hand with no fixture to stop the movement, and in my observation, it's recoil that accounts for any POI difference. I've never worked with a Glock...is the barrel mounted closer to the web of the hand than say a Colt 1911?

Just some thoughts here on a cold KY morning...Regards, Rodfac

 

[crghss]

AMP 44 has it right.

There is no recoil till the bullet leaves the barrel. If this wasn't true then why would we port barrels to reduce recoil? What does porting do? It releases pressure in the barrel so that when the bullet leaves the barrel there is less backward pressure on the gun.



kle

The wing of a plane uses lift not air redirection to fly. The top side of the wing spreads the air (pressure) over a larger area then the bottom side. This produces lift. The flaps and rudder steer the plane by pointing the plane in a certain direction but to fly the wings produce lift. Sails (Sailboats, Kite surfers) work in the same manner. You have more pressure on one side then the other.

 

[tipoc]

This is an interesting discussion and to help sort some of it out I picked up my copy of Robert Rinker's "Understanding Firearm Ballistics". This is a useful book which every shooter should have.

First off it does not take a physics professor to understand this stuff. All that it takes is a fella picking up a book and reading.

I've read that heavier bullets tend to have a higher point of impact, but shooting my Glock 29 I've found the opposite to be true. 200gr bullets shoot to point of aim, while 135gr bullets shoot 4" high at only 25 feet.

At 8 yards the difference between the point of aim and the point of impact of the 135 gr. versus the 200 gr. will be negligable, maybe an inch if that. But the felt recoil between the two will be quite a bit and in this case the difference in a 6" point of impact at 8 yards is due to the shooter. It is the shooter who is causing the lighter bullet to hit 6" higher than the heavier at 8 yards and not the gun or the load.

All else being equal (and it rarely is) A handgun will shoot lower on the target with light loads and higher with heavy loads no matter the velocity.

There are two reasons for this: One, recoil begins the moment the bullet begins to leave the case. In response the barrel begins to jump up as the gun is fired. This happens as the bullet moves down the barrel. Sights are set so that the center line of the bore is pointed below the intended point of impact at the time of firing. The gun moves rearward and the muzzle up. The bullet leaves the barrel as the muzzle comes up to the proper point.

Now all manufacturers have the formulas for the height of their front sight from the axis of the bore for their fixed sight guns and they are based on bullet weight and length of the barrel. These formulas anticipate the amount of recoil and muzzle rise for bullets of particular weights (158 gr. for the 38/357, 230 gr. for the .45acp, etc.) shooting at usually 25 yards for usually a 6 o'clock hold on a 6" bull.

The second reason is: If two bullets are fired at the same velocity but one is considerably heavier than the other, actual recoil will be heavier for the heavier bullet. At the same velocities the two bullets will reach the muzzle at the same time. But the heavier bullet will have the greater recoil and kick and cause the muzzle to rise higher and leave the barrel at a higher angle and it will hit the target higher than the lighter round.

Now if we add more powder to the case of the lighter bullet, so to increase it's recoil to the same as the heavier round, we also increase it's velocity. This means that it will leave the barrel sooner than the heavier bullet. So if both the heavy and the light bullet have the same recoil but the light is faster it will leave the barrel before the heavier and also before the muzzle has risen to the proper angle.

This is why from my BHP at 25 yards from a rest a 147gr bullet hits about 2-3" higher than a 115 gr.

tipoc

 

[kle]

There is no recoil till the bullet leaves the barrel. If this wasn't true then why would we port barrels to reduce recoil? What does porting do? It releases pressure in the barrel so that when the bullet leaves the barrel there is less backward pressure on the gun.

If you notice, the porting on gun barrels is generally on the top side (maybe pointing off to one side on rifles like AR15s). This directs the still-expanding gas upwards in that last fraction of a second where the tail-end of the bullet is still in the barrel and the gas is still 'contained'. Directing the gas upwards has an equal and opposite effect on the barrel, directing it downwards (or at least retarding/braking its upward climb, Newton's 3rd Law) and against the recoil already acting on the entire firearm.

Get a gunsmith to cut a port in the bottom of a handgun barrel and see if the 'pressure release' there still helps to reduce recoil.

The wing of a plane uses lift not air redirection to fly. The top side of the wing spreads the air (pressure) over a larger area then the bottom side. This produces lift. The flaps and rudder steer the plane by pointing the plane in a certain direction but to fly the wings produce lift. Sails (Sailboats, Kite surfers) work in the same manner. You have more pressure on one side then the other.

How can an airplane with a symmetrical wing airfoil fly, then? How can airplanes fly upside down? How can sailboats and windsurfers (whose sails are the same area on both sides of the sail, being sheets of fabric) travel into the wind, then? If the air pressure explanation (i.e. the Bernoulli Principle, where a fluid's (air) pressure decreases as its speed increases) were true, then the Blue Angels program at an airshow would be very boring.

The force we call "lift" is generated by the wings using flow redirection--airflow is directed downwards (relative to the path of the airfoil through the air) and "lift" is generated as a result.

 

[crghss]

airflow is directed downwards

A symetric wing creates lift because it is moving through the air at an angle of attack that creates high and low pressure areas around itself. This results in displacement of air.

porting on gun barrels is generally on the top side

Some pistols are most are not. In rifles porting is done uniformly around the barrel. So why is this? Why wouldn't they all be only ported out the top?

Because what is important is that gas is released not in what direction.

sails are the same area on both sides of the sail, being sheets of fabric

Like a wing one side of the sail is bowed. So the sail has a side that has more surf area exposed to the wind. Wind isn't actually passing over the inside of the surface area of the sail. This spreads the pressure across a larger area thus having more pressure on one side then the other.

If you willing to come to the beach in south Florida I can show you. The kitesurfing is excellent right now, it would be fun.

How can sailboats and windsurfers (whose sails are the same area on both sides of the sail, being sheets of fabric) travel into the wind, then?

Sailboats and Kit-surfers don't and can't travel into the wind. They tack or travel 45 degrees to the left of or to the right of the head wind.

 

[stevieboy]

Here's a personal example. I have my sights on my 6" 686 zeroed for 158gr. magnum. At 10 and 25 yards with this ammo my gun is dead on if I center the front sight on the bullseye. If I switch to 125 gr. magnums, the gun shoots 2" low at 10 yards and about 4" low at 25 yards. The groups remain tiny but they consistently fall below the point of aim. My 158 gr. magnums are rated at about 1200 feet per second. The 125s, close to 1500 feet per second. So, yes, faster = lower.

 

[kle]

Some pistols are most are not. In rifles porting is done uniformly around the barrel. So why is this? Why wouldn't they all be only ported out the top?

I've yet to see a pistol barrel ported on just the sides or just the bottom. Magna-Porting and Weigand porting are done on the top of the barrel for a reason--to direct gases up to counteract muzzle flip. Porting done on the bottom of a barrel would increase muzzle flip, and porting done on the sides would not impact muzzle flip at all.

Which rifles are ported omni-directionally? I thought that was called a flash-hider--burning gases are allowed to dissipate more quickly into the environment and not be so concentrated when exiting the barrel, hiding the flash. Or there's the muzzle-brake on the Barrett 50-cal rifles--you'll notice the vents on that one are directed backwards. This directs gas backwards to counteract the backwards-recoil of the rifle. Again, action/reaction.

Because what is important is that gas is released not in what direction.

Wouldn't the muzzle be a sufficient gas port then? It's a pretty big hole, compared to the gas ports. If the idea was to simply vent gas, then why isn't porting cut into the barrels just after the chambers (or as "early" as possible in a bullet's travel down the barrel)? Why is it done at the end of the barrel? Why not have a barrel that looks like Swiss cheese?

To get bullets up to sufficient speed before venting, is one answer. Another is that gases escaping through vent at the end of the barrel has the greatest effect (longest moment arm on the pistol's fixed point--the shooter's hand) on muzzle flip.

Some revolvers are ported (and porting can be cut into revolvers aftermarket), yet they have a gap between the barrel and cylinder (Nagant notwithstanding). Wouldn't this gap be sufficient to reduce muzzle flip? Why, then, when porting is cut into revolvers' barrels, it is cut into the top of the barrel?

Or why even cut holes at all? The bullet leaves the barrel so quickly that the sights have barely moved and the shooter hasn't even begun to register the recoil; why bother?

A symetric wing creates lift because it is moving through the air at an angle of attack that creates high and low pressure areas around itself. This results in displacement of air.

Ah yes, Angle of Attack. A wing at a positive AoA would direct air downwards, thereby producing lift upwards. Yes, pressures are generated, and yes they play a role in aerodynamics, but I contend that the bulk of the lift is generated by directing air downwards.

Why are wings curved on top, then, and flat(er) on the bottom? That has to do with maintaining laminar (smooth) flow. Smoothly directing the air downwards via the Coanda effect (the airflow will 'stick' to the top of the wing) is better than abruptly doing so because it creates less drag.

Like a wing one side of the sail is bowed. So the sail has a side that has more surf area exposed to the wind. Wind isn't actually passing over the inside of the surface area of the sail. This spreads the pressure across a larger area thus having more pressure on one side then the other.

But a sail is two sided, and the inside is just as bowed as the outside. Plus, since the sail is flexible, it won't create pressures like a rigid wing. How, then, does a sailboat go forward? The answer is that the sail redirects the wind/airflow. By directing this airflow, say, to the left/port and to the rear/stern, an equal/opposite force is generated to the right/starboard and to the front/bow. This would tend to make the boat go right, except that there's a keel in the water which resists this rightward tendency (because water is much more viscous than air). The resulting net force is forward. Windsurfers (sailboats without the 'boat', just the sail/wing, a board to stand on, and a fin/keel in the water) do the same thing.

If you willing to come to the beach in south Florida I can show you. The kitesurfing is excellent right now, it would be fun.

You got a plane ticket ready for me? Dulles International to Miami International, any carrier, any time is fine with me

It's interesting to note that in my research on this topic, both theories (the Bernoulli Principle/Equal-Transit-Time theory and the Newtonian theory) are presented, and a lot of sources debunk the Bernoulli Principle (as applied to the generation of lift). I'm more inclined to accept the Newtonian theory because it adequately handles scenarios that the Bernoulli Principle theory would have a hard time with (i.e. flat wings, symmetrical wings, flying upside down). I also used to live and breathe this stuff (as a hobby) before I got into firearms.

We're also quite off-topic here.

 

[JohnKSa]

Some pistols are most are not. In rifles porting is done uniformly around the barrel. So why is this? Why wouldn't they all be only ported out the top?

Because what is important is that gas is released not in what direction.

Porting on top of the barrel is to reduce muzzle flip, porting distributed around the barrel does little for muzzle flip but helps to reduce recoil.

By venting the very high pressure gases in a direction that tends to counteract the movement of the gun due to recoil you can reduce the recoil somewhat.

AMP 44 has it right.

--------------------------------------------------------------------------------

There is no recoil till the bullet leaves the barrel.

I believe you misread what 44 AMP said in his post. He does not state that recoil begins when the bullet leaves the barrel, he clearly says that recoil begins when the cartridge goes off.

By the way, I posted links to high-speed videos on the first page of this thread demonstrating conclusively that recoil begins before the bullet exits the muzzle.

 

[tipoc]

From stevieboy,

My 158 gr. magnums are rated at about 1200 feet per second. The 125s, close to 1500 feet per second. So, yes, faster = lower.

What you are seeing is not the effects of velocity directly but the different bullet weights. The higher velocity of the lighter bullet is causing it to leave the muzzle, during recoil as the barrel rises, before the point at which the gun is sighted in for and so it hits lower on the target.

If both bullets were traveling at the same velocity the lighter would still hit lower. This time though it would be due to the greater recoil of the 158 gr. pill.

tipoc

 

[Dubs]

First semester physics... Gravity is always present and always pulls down at an acceleration of 9.8 meters per second, per second. This means that for the first second, the bullet is going to drop 9.8 meters, but for the second second, it's going to drop 19.6, and the third it would drop 29.4. So for 3 seconds of travel, the bullet has dropped 58.8 meters and is accelerating towards earth.

After 3 seconds, it dropped 44.1 meters. distance in meters = 4.9(time in seconds)^2 !!! Integrate!

 

[darkgael]

Bullets fired from an upward tilted barrel do not drop at 9.8 meters per second for the first second because they are traveling upward initially.

Yeah, they do. Gravity acts the same way on the object regardless of its angle of flight and velocity. As soon as it exits the barrel, gravity starts to pull it toward the center of the Earth. A bullet fired upward will still rise though the amount/rate of that rise will be affected by the pull downward effected by gravity.
Pete