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View Full Version : OT: Does shooting from high ground make the bullet go high?

winchman
12-20-2009, 10:01 AM
I just caught a bit of a show on the Military Channel about the Revolutionary War. In talking about one of the battles, they said the "Loyalists soon learned the disadvantage of having the high ground as their bullets sailed over the heads of the Rebels."

Does shooting from high ground really make the shot go high?

Roger

Lew Hartswick
12-20-2009, 10:06 AM
The amount of "drop" depends on the HORIZONTAL distance not
total distance.
...lew...

lazlo
12-20-2009, 10:09 AM
Mythbusters has a current episode where they replicate the classic freshman physics question: which bullet hits the ground first: the one fired out of a gun, or the same bullet dropped on the ground ;)

Extremely well-done experiment, even for the Mythbusters.

bewards
12-20-2009, 10:11 AM
I think it has to do with gravitational pull on the bullet. Shooting down has less pull than shooting flat or up. You would think it would have to be a long shot though. There is an article in one of my books on it but it will take a while to find it again.

BE

lazlo
12-20-2009, 10:20 AM
You get the extra distance from the vertical displacement: the time the bullet takes to drop the additional height of the hill.

Black_Moons
12-20-2009, 10:32 AM
It won't make a bullet 'shoot high' but it might screw with your sights calibration and how you compensate for distance.

davidwdyer
12-20-2009, 10:37 AM
Shooting downhill OR uphill will cause the bullet to go higher than a flat shot.

You can count on it!:eek:

JoeCB
12-20-2009, 10:40 AM
To make this short and sweet... Yes. For example if you are sighted to zero at say 200 yds on the level, if you now shoot down hill OR UPHILL the bullet will impact higher. How much depends on the angle of incline or decline and of course the range.
Think of it this way, in the extreme... say you are shoothin straight up in the air, there is no gravitational component pulling the bullet off it'c course, same if downward as shooting into a well.
Joe B

Rustybolt
12-20-2009, 10:42 AM
aiming high makes the bullet go higer. There is a tendancy for a shooter who is above the target to shoot over the target. It has more to do with the shooter than the weapon. As soon as the bullet leaves the barrel gravity is acting on it.

rkepler
12-20-2009, 11:08 AM
The moment the bullet exits the barrel it starts falling due to gravity. The distance it falls from a straight trajectory is a function of the distance travelled at a right angle to the gravitational field.. If you shoot at a target 100 yards from the muzzle at 45 degree angle the bullet the bullet travels 70 yards horizontally for every 100 yards of straight line distance, and so the bullet drop would be the same as for a 70 yard horizontal shot. If the rifle was sighted in for a 100 yard zero the bullet would hit high relative to that zero in the same spot as if you were shooting a target horizontally at 70 yards.

bob308
12-20-2009, 11:15 AM
this may not sound right buy up hill or down hill you will hit low. a person in a fire fight will tend to shoot high.

J Tiers
12-20-2009, 12:03 PM
The moment the bullet exits the barrel it starts falling due to gravity. The distance it falls from a straight trajectory is a function of the distance travelled at a right angle to the gravitational field.. If you shoot at a target 100 yards from the muzzle at 45 degree angle the bullet the bullet travels 70 yards horizontally for every 100 yards of straight line distance, and so the bullet drop would be the same as for a 70 yard horizontal shot. If the rifle was sighted in for a 100 yard zero the bullet would hit high relative to that zero in the same spot as if you were shooting a target horizontally at 70 yards.

That sounds good, but has a fatal flaw. TIME OF FLIGHT.

The bullet falls a certain amount in a given time.

The traveled distance is what primarily determines time of flight

The angle makes the bullet travel horizontally *effectively* a bit slower than it would otherwise, although it is on its aimed trajectory at normal muzzle velocity. That tends to compensate the shorter linear horizontal distance. But there are other effects

obviously shooting straight up, the bullet velocity and resulting range "up" is reduced by gravitational pull, where horizontally velocity is not directly affected. The steeper the angle up the more the effect, which causes the bullet to fall short using normal sights.

Shooting DOWN, that is reversed, and gravity 'speeds up" the bullet, reducing the time of flight, and thus the total distance it falls, making it go high if aimed using the normal sights.

Evan
12-20-2009, 12:09 PM
If you shoot down hill using the same point of aim over the same distance the shot will be high. If you shoot uphill with the same point of aim the shot will be low.

The reason is that when shooting down the bullet is already heading down. It's velocity vector and the gravitational vector added together produce less drop for a given distance than when shooting perpendicular to gravity normal.

The opposite holds true for shooting up.

IdahoJim
12-20-2009, 12:31 PM
this ain't rocket science. If you shoot up or downhill, at a 45* angle, the force of gravity, on the bullet's path, is reduced by 50%. It's vector algebra. Going uphill, half the force of gravity is slowing the bullet, but the effect of that slowing is less than the reduced force of gravity on the bullet's path, so you shoot high. When shooting downhill at a 45* angle half the force of gravity is speeding the bullet up, and half is pulling the bullet toward the ground. The extra velocity is insignificant, but reducing the pull downward is a big deal, so you shoot high. Anybody who uses guns knows this.
Jim

lazlo
12-20-2009, 12:34 PM
this ain't rocket science. If you shoot up or downhill, at a 45* angle, the force of gravity, on the bullet's path, is reduced by 50%

You're missing the point of his question: does the bullet travel farther if you shoot from high ground. Not does the bullet go farther if you aim high.

Jpfalt
12-20-2009, 12:38 PM
This most likely has nothing to do with gravity and ballistics.

During the Revolutionary War, the usual firearm was the musket. Troops were trained to fire, reload, fire as quickly as possible and aimed fire wasn't part of the cirriculum.

One feature of the musket was that the ball was typically .01" smaller than the bore and if you tilted the barrel down after loading, there was a good chance that the ball would roll out of the barrel or at least separate from the charge. Given the emphasis on fast loading and not aiming down, it's no wonder the shots from high ground would go high.

Carld
12-20-2009, 12:42 PM
The opening post:

I just caught a bit of a show on the Military Channel about the Revolutionary War. In talking about one of the battles, they said the "Loyalists soon learned the disadvantage of having the high ground as their bullets sailed over the heads of the Rebels."

Does shooting from high ground really make the shot go high?

Roger

Lazlo, nowhere does he ask if the bullet travels farther but he wants to know it if the shot will go high and I agree with IdahoJim. From things I have read it was common for Civil War combatants to shoot high on flat, uphill or downhill.

lazlo
12-20-2009, 12:45 PM
One feature of the musket was that the ball was typically .01" smaller than the bore and if you tilted the barrel down after loading, there was a good chance that the ball would roll out of the barrel or at least separate from the charge. Given the emphasis on fast loading and not aiming down, it's no wonder the shots from high ground would go high.

LOL! Cool story!

When I took the NRA DCM (M1 Garand) class, they told us that because the powder in the cartridges was nearly 100 years old, that we were supposed to tilt the barrel up and slowly lower it before firing.

They could have been messing with us (we were a bunch of engineering geeks at Virginia Tech), but it seemed plausible at the time.

J Tiers
12-20-2009, 12:57 PM
But, since the ball was rammed down surrounded by a patch (or oughtta have been) it very likely would NOT "roll" out..... Not sure I buy that one.

Carld
12-20-2009, 12:57 PM
Lazlo, the lifting of the barrel is to put the powder in front of the orifice from the cap. If there is a space between the powder and the flash and the cap is weak and the powder weak it may not fire with full force. Some if not all target shooters that load their own shells use a filler on top of the powder charge to keep the powder against the cap orifice. The seating of the bullet packs the charge so it can't move.

As J Tiers said, the ball is loaded with a patch, either paper or cloth. I think most the musket charges were paper covered with the charge and ball inside and they were torn open, the power dumped in and the ball rammed in with the paper as a wrapping. I guess in the heat of battle they would be doing good to get the powder and ball in and to hell with the paper.

JoeCB
12-20-2009, 01:25 PM
Interisting how this thread has wandered around, bunch of must have time on our hands on this snowy Sunday morning...
As for high / low issue.. Jim / Idaho has it exactely right.
As for the Revolutionary war musket.. they did not use a patched ball.The ball pretty much rolled down the un-rifled barrel and the paper wrap, known as the cartridge was stuffed down on top of the ball with the ram rod. Any muzzel loader guy will tell you that a sure way to blow up any gun is to fire it with the ball not seated down on the powder... acts just like a barrel obstruction.
Joe B

Yankee1
12-20-2009, 01:33 PM
The bullet will impact high. Think of it like a right angle. The bullet flight is along the hypotenuse but the gravity effect is along the adjacent side.
Yankee1

davidwdyer
12-20-2009, 01:36 PM
I don't think the kind of gun matters.

I cannot explain the science of it. BUT,

from experience I can tell that both uphill and downhill shots shoot high.

IdahoJim
12-20-2009, 01:41 PM
You're missing the point of his question: does the bullet travel farther if you shoot from high ground. Not does the bullet go farther if you aim high.

That isn't what I read. I thought they were aiming at the rebels, but missing high because they were shooting downhill.
Jim

Teenage_Machinist
12-20-2009, 02:14 PM
If you shoot down from a hill at a flat surface, you have to aim right since a bullet won't have a chance to hit other targets behind your target-

The same goes for shooting at people who are on a hill.

If both you and the enemy are on level ground each salvo should eventually find a mark depending how deep the lines are.

winchman
12-20-2009, 02:21 PM
Jim's interpretation is the same as mine. They were aiming at the rebels, but the bullets went high because they were shooting downhill.

I just don't have enough experience with guns to know the answer or to know all the things that would have to be considered to figure it out.

Roger

Evan
12-20-2009, 02:44 PM
As for high / low issue.. Jim / Idaho has it exactely right.

Yes he is. It's also what I posted 20 minutes earlier.

doctor demo
12-20-2009, 03:02 PM
If you shoot down from a hill at a flat surface, you have to aim right
.

Why can't You aim left?

Steve:D

IdahoJim
12-20-2009, 03:41 PM
Yes he is. It's also what I posted 20 minutes earlier.

Excuse me, Evan, but what you posted is that the bullet impact would be high shooting downhill, and low shooting uphill. In fact, the impact is higher at any angle, up or down, than it would be shooting level.
Jim

winchman
12-20-2009, 04:21 PM
So the reasoning is as follows:

The sight on the gun is set for level ground, and the soldier makes no compensation for the uphill/downhill angle.

The soldier's aim is based on the straight-line distance to his target.

The bullet travels the distance, but drops less than expected because __?___

I don't understand how the effect of gravity is altered by the angle. The bullet is in flight for the same period of time, so the drop caused by gravity should be the same. Right?

Roger

beanbag
12-20-2009, 04:41 PM
this ain't rocket science. If you shoot up or downhill, at a 45* angle, the force of gravity, on the bullet's path, is reduced by 50%. It's vector algebra. Going uphill, half the force of gravity is slowing the bullet, but the effect of that slowing is less than the reduced force of gravity on the bullet's path, so you shoot high. When shooting downhill at a 45* angle half the force of gravity is speeding the bullet up, and half is pulling the bullet toward the ground. The extra velocity is insignificant, but reducing the pull downward is a big deal, so you shoot high. Anybody who uses guns knows this.
Jim

replace half by 0.71, and that's vector algebra :)

beanbag
12-20-2009, 04:45 PM
So the reasoning is as follows:

The sight on the gun is set for level ground, and the soldier makes no compensation for the uphill/downhill angle.

The soldier's aim is based on the straight-line distance to his target.

The bullet travels the distance, but drops less than expected because __?___

I don't understand how the effect of gravity is altered by the angle. The bullet is in flight for the same period of time, so the drop caused by gravity should be the same. Right?

Roger

when shooting horizontally, ALL of the force of gravity is used to pull the bullet away from it's path, so you need the most aim compensation. When shooting upwards or downwards, SOME of the force of gravity pulls it off it's path, and SOME goes into speeding it up or slowing it down, but that's not a big deal. The bullet deflects less than you expected and therefore you have overcompensated too high. What Jim said is qualitatively correct.

Edit: Would anybody who actually shoots a gun like to share some numbers?

IdahoJim
12-20-2009, 05:02 PM
replace half by 0.71, and that's vector algebra :)

I'm not much of a mathematician, but why is it .707, rather than .50 at 45*? At some angle, half the gravity is effecting bullet path, and half is effecting the velocity. What angle is it?
Thanks,
Jim

winchman
12-20-2009, 05:14 PM
Thanks, beanbag. Now I see what's going on.

Roger

beanbag
12-20-2009, 05:24 PM
Because to say something like "what fraction of gravity is used for [this effect]?" means to take the gravity vector and project it along the direction of interest. The definition of projection (or "what fraction of this vector is in common with this other vector") is cos(theta) where theta is the angle between the two directions.

For example, if you shoot a gun at 45°, the projection of gravity in the same direction as the path (i.e. slowing it down) is cos 45 = .71. The projection of gravity trying to deflect the bullet is also cos 45 because the "deflection" vector points perpendicular to the "bullet path" vector" i.e. 45° also. IT would be correct to say that gravity "acts equally" to both deflect and slow down the bullet.

If you shoot a gun at 60° up from the horizon, then the deflection vector is 30° and the gravity vector is 90°, so the deflection component is cos 60 = 1/2. However, the "slowing down" component is cos 30 =.86something.

There is no angle at which you can aim the gun such that gravity will project to 1/2 in both the deflection and slowing down directions.

Hope that makes things clear as mud.

IdahoJim
12-20-2009, 06:04 PM
Interesting...so it's an equal effect, but not 50%. I was sure it was equal at 45*, but had no idea that the sum of the two forces could total more than 1.
Thanks,
Jim

Evan
12-20-2009, 06:09 PM
In fact, the impact is higher at any angle, up or down, than it would be shooting level.

That is incorrect. It will happen as I described it. I used to teach marksmanship and also have the Expert Medal in Marksmanship from the US Army with one of the highest scores recorded.

edit: I will draw a diagram if I have time.

John Stevenson
12-20-2009, 06:14 PM
That is incorrect. It will happen as I described it. I used to teach marksmanship and also have the Expert Medal in Marksmanship from the US Army with one of the highest scores recorded.

Is this whilst going in reverse or pop riviting at 1,000 per hour ?

.

IdahoJim
12-20-2009, 06:34 PM
That is incorrect. It will happen as I described it. I used to teach marksmanship and also have the Expert Medal in Marksmanship from the US Army with one of the highest scores recorded.

edit: I will draw a diagram if I have time.

Think about it like this: The angle of departure of the bullet is 45* (as an example). In level flight, the effect of gravity on the bullet is at roughly a 90* angle to the bullet path. When the angle of departure is either up, or down, the effect of gravity on the bullet path is the cosine of the angle of departure x the level bullet drop. So, if your bullet drops 2" at 100 yards, you compensate to hit the target by adjusting your sights so the bullet is lofted, or slightly above horizontal to be dead on at 100 yards. In effect, you're raising the barrel to compensate for the bullet drop. Now if you shoot steeply, up or down, at say that 45* angle, the drop is .707 (the cosine of the angle of departure of the bullet x the level drop of 2", or 1.414". Using the same sight setting you are high by .586". There are some other minor factors in there, but that's roughly what happens. All of this is explained very well in Sierra bullet's reloading manual...the portion on Exterior Ballistics.
Jim

jdunmyer
12-20-2009, 07:00 PM
Here's another vote for "you'll shoot high when shooting uphill or downhill"

http://www.chuckhawks.com/shooting_uphill.htm

he explains why in the article.

If you want all the math, go here: http://en.wikipedia.org/wiki/Trajectory

Yankee1
12-20-2009, 07:03 PM
If you are 300 yards up on a cliff and your target is out 50 yards from the bottom of the cliff directly below the edge of the cliff. You best have your sights set for 50 yards or you will go high over the target.
The 50 yard distance is the distance that gravity will affect your shot.
It makes no difference Werther or not your shooting uphill or downhill the gravity vector is the same. And yes Evan the Army says I'm an expert too.
I also have a few national records.
Yankee1

terry_g
12-20-2009, 07:52 PM
If you fire a bullet horizontally the bullet will be affected by gravity and will travel in a curve, dropping as it travels through its trajectory the sights will be calibrated to correct this.
If you shoot a bullet straight up it will travel in a straight line, if you shoot a bullet straight down it will again travel in a straight line as gravity will only affect its speed . So if you shoot at an angle would the the bullets trajectory not be less than the horizontal trajectory that the sights are calibrated for?
If you shoot up or down at a forty five degree angle would the bullet drop trajectory not be half of the horizontal trajectory and therefore the bullets would travel in a straighter line going over their targets ?????

Evan
12-20-2009, 09:55 PM
Here is how it works.

http://ixian.ca/pics7/shoothill.gif

The bullet shot up never falls so it loses energy all the way to the target.

The bullet shot level falls part way through the trajectory and hits the point of aim.

The bullet shot down falls all the way to the target and gains energy from gravity all the way. The further the bullet falls on the way to the target the greater the gain in energy and the higher the point of impact.

Rich Carlstedt
12-20-2009, 10:05 PM
Man what a bunch of bad shots here..
Do you guys shoot ?
I Go with Yankee 1

Rich

Evan
12-20-2009, 10:14 PM
If you want all the math, go here

That math does not describe the situation correctly. It assumes the range to the uphill target is different than to the level target. That is comparing apples to oranges and doesn't give a meaningful answer. Note in my diagram all targets are at the same range so only the effect of energy loss or gain is a factor, not differing ranges.

If you want to answer the question asked then the range to the target must be considered the same in each instance. There is no valid reason to presume the range changes for a down hill or uphill shot.

Example: I am holding a hill. The guy at the bottom is taking it. We both have the same range and the same weapons. I am shooting down, he is shooting up. For his bullet to go high at my position would require more energy than for my bullet to go on target at his position.

Impossible, it violates the laws of physics.

lazlo
12-20-2009, 10:35 PM
I used to teach marksmanship and also have the Expert Medal in Marksmanship from the US Army with one of the highest scores recorded.

Evan, you said you were in the Army for 2 years, and your MOS was 68G20, which my Dad (Huey pilot, three tours in Vietnam) tells me is a guy who patches bullet holes in aircraft.

So how did you have time to be a firearms instructor?

JCHannum
12-20-2009, 10:51 PM
It is Ballistics 101. The bullet is affected by the pull of gravity only by that part of the distance of its flight which is parallel to the surface of the earth (if you assume for the moment that the earth is flat for that short distance). If you fire uphill or downhill, the length of the bullet's flight affected by gravity is the base of a right triangle while the line of sight is the hypotenuse.

Thus, whether uphill or down, you will shoot high unless you compensate.

Evan
12-20-2009, 10:57 PM
So how did you have time to be a firearms instructor?

Five years teaching military cadets how to shoot.

Thus, whether uphill or down, you will shoot high unless you compensate

Impossible, unless you can figure out how to give the bullet going up more energy than the bullet going down.

lazlo
12-20-2009, 11:03 PM
Five years teaching military cadets how to shoot.

But if you were in the Army for 2 years -- how did you spend 5 years teaching military cadets how to shoot?

JoeCB
12-20-2009, 11:06 PM
Strange, what topics generate hot discussion on a "shop" board. I know that the answer is "high" for both up hill and down... Idaho has the best explaniation. As for Evan's theory on the up hill scenairo.. he has a point of truth I think but the effect is so minor as to not overcome the governing (Idaho) principle. Evan's point is that AT THE END of the trajectory path the gravity vector on an up hill shot is greater than the down hill shot. This (I think) is true because the bullet path is nearer to horizontal on the former than the latter. So given two identical shots, one at say 30* up hill and one at 30* down hill the point of impact for both will be high, however the up hill shot will be SLIGHTLY less high than the other.
Joe B

JCHannum
12-20-2009, 11:08 PM
So given two identical shots, one at say 30* up hill and one at 30* down hill the point of impact for both will be high, however the up hill shot will be SLIGHTLY less high than the other.
Joe B

Quite true. That is how it works when the myriad outside influences of ballistic coefficient, drag, velocity and on and on are added into the equation. The basic concept is quite simple.

IdahoJim
12-20-2009, 11:10 PM
Here is how it works.

http://ixian.ca/pics7/shoothill.gif

The bullet shot up never falls so it loses energy all the way to the target.

The bullet shot level falls part way through the trajectory and hits the point of aim.

The bullet shot down falls all the way to the target and gains energy from gravity all the way. The further the bullet falls on the way to the target the greater the gain in energy and the higher the point of impact.

The velocity/energy lost and gained due to gravity is negligible....the acceleration/deceleration due to gravity, at a 90* angle is 32ft/sec./sec. A bullet, in this case from a muzzle loader, is going 1,800fps...that 32ft/sec doesn't do much. In the first full second, the ML bullet is already 500 yards downrange. A modern rifle bullet is at least 70% faster. You draw pretty pictures, buy they don't mean a thing.
Jim

lazlo
12-20-2009, 11:10 PM
Strange, what topics generate hot discussion on a "shop" board. I know that the answer is "high" for both up hill and down...

Yeah, but it's funny as Hell to watch! I'm sitting here with a glass of scotch -- much better than what's on TV! ;)

Evan
12-20-2009, 11:18 PM
You sure don't read well Robert. I have explained this before in threads where you participated. Both me and my wife were civilian instructors for the Local Royal Canadian Sea Cadet Corp, RCSCC 202 Chilcotin. They are a branch of the Canadian Military with direct command at each unit by a reserve officer of the military. All funding is by the military including official training at millitary bases as well as the opportunity for actual terms of service on board military vessels.

Evan
12-20-2009, 11:20 PM
The velocity/energy lost and gained due to gravity is negligible....the acceleration/deceleration due to gravity, at a 90* angle is 32ft/sec./sec. A bullet, in this case from a muzzle loader, is going 1,800fps...that 32ft/sec doesn't do much. In the first full second, the ML bullet is already 500 yards downrange. A modern rifle bullet is at least 70% faster. You draw pretty pictures, buy they don't mean a thing.

WHAT??? Try telling me that after you bore sight your rifle on the bull and set the sights the same. Where do you think the round will end up?

IdahoJim
12-20-2009, 11:21 PM
Strange, what topics generate hot discussion on a "shop" board. I know that the answer is "high" for both up hill and down... Idaho has the best explaniation. As for Evan's theory on the up hill scenairo.. he has a point of truth I think but the effect is so minor as to not overcome the governing (Idaho) principle. Evan's point is that AT THE END of the trajectory path the gravity vector on an up hill shot is greater than the down hill shot. This (I think) is true because the bullet path is nearer to horizontal on the former than the latter. So given two identical shots, one at say 30* up hill and one at 30* down hill the point of impact for both will be high, however the up hill shot will be SLIGHTLY less high than the other.
Joe B

That's correct, but if you look at Sierra's info, the difference is really minute.
The difference is because the velocity loss from air drag is very slightly decreased by gravity when going downhill, and very slightly increased by gravity when going uphill. According to Sierra's info the difference due to the speeding/slowing of the bullet from gravity is in the area of .03" at 200 yards for a bullet at 2200fps. That for a 45* angle. For lesser angles it would be less.
Jim

lazlo
12-20-2009, 11:22 PM
You sure don't read well Robert. I have explained this before in threads where you participated.

Sorry, haven't seen this before Evan.

Both me and my wife were civilian instructors for the Local Royal Canadian Sea Cadet Corp

Ah, that's a very different scenario than what I read here:

I used to teach marksmanship and also have the Expert Medal in Marksmanship from the US Army with one of the highest scores recorded.

beanbag
12-20-2009, 11:32 PM
Evan,

I think your mistake is believing that if you shoot uphill and the bullet goes high, it somehow "gained" kinetic energy. This is not true. This magical gain in energy (actually potential energy) comes out of the kinetic energy of forwards motion, e.g. the bullet is moving very slightly slower because gravity has a component of backwards pull on it.

It's a fair comparison to say that the distance to the target must be the same in all instances. Thus the time for the bullet to reach the target is about the same time.

The equations of motion minus friction are

X=Vox*t
Y=Voy*t -.5g*t^2

Vox and Voy are the initial x and y component of velocity depending on the angle at which you shoot.

Vox*t and Voy*t are just the straight line velocity paths in the absense of gravity. Only the y component in position feels the drag from gravity. Since the time to reach the target is approx the same in all circumstances, there is the same y droop in all cases.

HOWEVER, the y droop is perpendicular (maximal) to the bullet path in the horizontal shot, thus this shot has the most droop, as measured as an angle from the line of sight of the shooter.

At an inclination or declination shot, the y droop is at an angle to the bullet path, and thus the angular error as viewed from the shooter is smaller.

J Tiers
12-20-2009, 11:42 PM
At an inclination or declination shot, the y droop is at an angle to the bullet path, and thus the angular error as viewed from the shooter is smaller.

Bingo.........

"as viewed from the shooter" is the key element......

The effect quoted by Evan (and I also mentioned it) is actually there, but the "apparent correction" for droop is in a different direction and not perpendicular to the path. That is dominant, even though the deceleration and acceleration are factual, real effects. It would take a lot more deceleration, etc to compensate than is generally present.

I was pretty sure there was a real explanation, and that's the one.

Evan
12-21-2009, 12:25 AM
I think your mistake is believing that if you shoot uphill and the bullet goes high, it somehow "gained" kinetic energy

edit: Yes, that is what would have to be the case. see below.

I can't believe that everybody is having such a hard time with this.

Here is a plot produced by a ballistic simulator. The straight lines are the points of aim. The curves are the actual trajectory. The circles are the points of impact for each trajectory at equal distance from the firing position. This is the exact equivalent of the scenario I gave when somebody is shooting up hill and I am shooting back down. We all agree that the downhill shot will be high but there is no way in that scenario that the up hill shot can also be high. It's impossible. The difference isn't minor either.

http://ixian.ca/pics7/ballsim.gif

lynnl
12-21-2009, 12:35 AM
this ain't rocket science. If you shoot up or downhill, at a 45* angle, the force of gravity, on the bullet's path, is reduced by 50%. It's vector algebra. Going uphill, half the force of gravity is slowing the bullet, but the effect of that slowing is less than the reduced force of gravity on the bullet's path, so you shoot high. When shooting downhill at a 45* angle half the force of gravity is speeding the bullet up, and half is pulling the bullet toward the ground. The extra velocity is insignificant, but reducing the pull downward is a big deal, so you shoot high. Anybody who uses guns knows this.
Jim

Where are you getting the 50% factor.
It isn't a question of half, or some other fraction of gravity, it's a question of the full force of gravity slowing only the vertical component of the resultant velocity vector (which is the initial speed along the initial path).

Viewed on an X-Y graph, the velocity vector 'V' of a shot fired upward at a 45deg angle can be broken down into a horizontal component of 'X' and vertical component of 'Y'. Both X & Y will have a value of .707 times V. (.707 is the sine and cosine of 45deg)
Gravity will not be acting on X at all, but solely on Y. However, since X (horizontal motion) is diminished (same as Y) the bullet will be in the air longer before traveling any given horizontal distance (i.e. longer than if shot level), therefore gravity will have longer to act on the bullet, which means more drop.
But/however (again), in this scenario the bullet starts out with a much greater vertical component than when it was originally sighted in at, say 200 yds. But that's neither here nor there, Ol' man gravity is still accelerating that component of travel downward at 32ft/sec/sec. And in this case, unlike when it was being sighted in, that vertical component Y represents .707 x V of the velocity toward the target, i.e. it's being slowed toward the target.

J Tiers
12-21-2009, 12:51 AM
Evan, it isn't anything to DO with your issue, which I agree with and which IS there....... it just isn't BIG enough...

Here's teh deal

Totally forget about velocity etc, it isn't the issue.

http://img.photobucket.com/albums/0803/jstanley/droop.jpg

The distance "d' is the droop, which is IDENTICAL in both cases within the small error we agree about......

But the EFFECTIVE d, AS SEEN, is smaller, as shown by "d eff" which is smaller than the actual "d" because of the angle. The smaller effective 'd" leads to a smaller angle to the endpoint of the droop, again as seen by the shooter.
The "effective theta", or angle to the droop, is smaller than the angle seen in horizontal shooting.

What we assert DOES happen, it's just not enough to overcome the angular factor.

So, if you were to aim per horizontal shooting, you'd shoot high. Same effect works for downhill.

Evan
12-21-2009, 01:25 AM
But the EFFECTIVE d, AS SEEN, is smaller, as shown by "d eff" which is smaller than the actual "d" because of the angle. The smaller effective 'd" leads to a smaller angle to the endpoint of the droop, again as seen by the shooter.

That makes no sense. It doesn't matter if I look at a target from straight on or at an angle. That doesn't change the point of aim at all. It will compress the image of the target but that makes no difference to the point of aim. The parameters are no change in the sight setting, no change in sight picture, no change in the muzzle energy or type of firearm and no change in the distance to target.

That describes the up hill / down hill situation I described which is extremely common in warfare and was a part of my training with half the company using live fire taking a hill and the other half keeping score of hits on targets waved above a trench at the top.

Yankee1
12-21-2009, 03:40 AM
Evan
The data used to calculate gravity effect on a shot is the cosine of the angle. If the shot is a specific distance that line of sight distance is not used to calculate gravity effect if the shot is an angle up or down hill.
When calculating the gravity effect use cosine tables for that specific distance to get the distance of the cosine to figure sight setting for that cosine distance.

John Stevenson
12-21-2009, 04:31 AM
Oh oh,
I now see the usual signs where our resident ex-spurt is again becoming a drip under pressure.

.

beanbag
12-21-2009, 05:52 AM
http://ixian.ca/pics7/ballsim.gif

I see what you did there. And depending on the situation, that is either really cleaver or really dirty. You are choosing very ballistic paths, with some combination of relatively low muzzle velocity, and long flight times such that the angle of aim is significantly different from the angle to the object. In these situations, at very high angles of compensation required, gravity has enough time to act on the projectile to slow it down, thus increasing the time it takes to reach the target, which increases the droop even more than I initially suggested.

Here's an example of what I mean. Suppose you had a M16 rifle with muzzle velocity of 1000m/s. Let's even say you wanted to shoot at something outside it's usual range - 1000m to make the math easy.
It takes about 1s to reach the target, at which point gravity caused the projectile to droop by 5m. That is an error of approx 0.3°, which is what you'd set your sight at.

Next, assume you are shooting at an insurgent on a cliff high above you, say 45° even. In the time it takes the bullet to reach him, the bullet has climbed 707 meters. Using conservation of energy equations

.5mVi^2=.5mVf^2 +mgh
Vf^2=Vi-2gh
Vf=sqrt(10^6-1400)=993m/s

That's the final velocity, so the average velocity was 996m/s. Thus, it actually took 1.004 seconds to reach him instead, and in the meantime, the bullet drooped 5.04 meters instead. You aimed your gun .3° above him, which is 5 m above his head, but unfortunately for you, it's the WRONG 5 METERS, and that is 5 meters above AND IN FRONT of his head. See attached diagram:
http://i139.photobucket.com/albums/q286/beanbag137/bullet.png

Now instead, if you had a laser sight that was coaxial with the muzzle, and you knew that for 1000m you are supposed to aim 5 meters above, and you take into account that you actually aim 5 meters directly above the target, (because he is also wearing a 5 meter vertical flagpole that you can aim the laser at), then yes, the bullet will hit 4cm low, due to the Evan Effect.
I don't shoot guns, so I don't know if people aim compensate by degrees, or by guessing at an actual height above an object

Getting back to the cleaver/dirty trick you pulled, if you shoot really really long ranges, like 50,000 meters, then the flight time is 50 s, at which point the bullet drops 12500 meters, or about 25-ish degrees of aim compensation required at horizontal. If you shoot up the cliff, the bullet is down to nearly half the initial velocity by the time it reaches the top. It travelled about a third longer than usual, and thus dropped 1.3*1.3 times more, or ALMOST TWICE MORE THAN IN THE HORIZONTAL CASE. The Evan Effect is very large here. No wonder your trajectory plot doesn't show the effect that everybody else is talking about.

When you guys were shooting at each other on a hillside, were you using cannon balls and mortars? Catapults? Was there an archery brigade?;)

Evan
12-21-2009, 06:19 AM
So you are saying the ballistic simulation is wrong? I don't think so and neither do people whose business it is to know about these things.

Are you actually reading what I posted about the conditions? No change in point of aim other than elevation to target, no change in range to target, no change in sight picture and no change in load or weapon.

The data used to calculate gravity effect on a shot is the cosine of the angle. If the shot is a specific distance that line of sight distance is not used to calculate gravity effect if the shot is an angle up or down hill.

Why are you making the assumption that the range to target changes with elevation? There is no reason to do so. In the example I described the range is the same for both shooters. It is no more than very basic ballistics which was something I actually paid attention to in 8th grade.

It makes absolutely no sense to apply a cosine function to the effect of gravity on the trajectory. That isn't how ballistics works. Gravity has the same influence on the bullet regardless of the angle of the trajectory. It produces a downward acceleration of ~10 meters per second per second in all cases.

The formula for the height of the ballistic arc is all we need to compare the actual trajectory to the point of aim. To simplify we leave out the effect of air resistance. If we include it it only makes the shot lower still. To further simplify we calculate for the highest point reached in the arc since that makes it easy to compare to the point of aim.

The point of aim is somewhere on the circumference of a circle that has a radius equal to the distance to the target. That point is a on a direct line of sight from the shooter to the circumference of the circle and is determined by the angle of inclination. The height above ground of the point of aim is simple the sine times the distance to the target.

The height achieved by the projectile is (velocity squared) times (sine squared of the angle of inclination) divided by 2 times the acceleration of gravity.

Note that the term for gravity stays the same and that the only function is a sine function. You can plug in any values you wish other than straight up and the resultant is always the the height achieved by the projectile is below the point of aim.

beanbag
12-21-2009, 07:00 AM
Forget semantics, here's a question:

Suppose you shoot horizontally at a target at 100 yards away, and you find out that the bullet drops by 1".

Now put a ball of diameter 2" on a 45 degree hillside, and 100 yards away from you, as measured with a string.

Where do you aim at? (At what point of the ball is your barrel pointed at?) (Or how do you set up your rifle to aim?)

http://i139.photobucket.com/albums/q286/beanbag137/ball.png

philbur
12-21-2009, 07:04 AM
But using this graph the down-hill shoot also impacts low. All the graph does is shows that a projectile will impact below the line of aim, but then everybody already knew that. The only two cases where this is not true is directly up or directly down.

Interestingly the difference between aim and impact get less as the angle increases which tells me that if I had set my sights for a low angle shot then used that setting on a high angle shot the shot would be high.

Phil:)

edit: Yes, that is what would have to be the case. see below.

I can't believe that everybody is having such a hard time with this.

Here is a plot produced by a ballistic simulator. The straight lines are the points of aim. The curves are the actual trajectory. The circles are the points of impact for each trajectory at equal distance from the firing position. This is the exact equivalent of the scenario I gave when somebody is shooting up hill and I am shooting back down. We all agree that the downhill shot will be high but there is no way in that scenario that the up hill shot can also be high. It's impossible. The difference isn't minor either.

http://ixian.ca/pics7/ballsim.gif

Evan
12-21-2009, 07:21 AM
Here is the actual situation when the shot is elevated. The line of sight is directly on the target and the bullet follows a ballistic arc to the target. That arc changes only lower under the effect of gravity regardless of how the sight must be set for the distance.

When you set the sights you set the line of sight below the trajectory that the bullet will follow. The distance makes no difference to the kind of path the bullet takes. It will always describe the same arc first rising then falling to hit the target. The only thing that varies with range is how much rise must be added to compensate for the fall. The end result is always that the rise and fall must be equal to hit the target.

When the elevation is above horizontal the only difference to the trajectory is that the fall will exceed the rise over the optical path which is your sight picture.

http://ixian.ca/pics7/shootup2.gif

Philbur,

If you shoot down in the table I posted the range increases therefor the point of impact rises.

beanbag
12-21-2009, 07:30 AM
http://ixian.ca/pics7/shootup2.gif

Is the initial angle between the green line and red trajectory the same in both instances?

Evan
12-21-2009, 07:34 AM
Yes, it is the same. The assumption is that no change will be made to the sights for an elevated shot. If you do change the sights then the answer is meaningless in the context of the question.

beanbag
12-21-2009, 07:40 AM
Yes, it is the same..

Then I assert that for relatively flat trajectories, your picture is wrong.

Could you please go to my post #68, and say where you would aim the barrel of the gun. Never mind the sights, assume you had a laser pointer down the center of the barrel.

philbur
12-21-2009, 07:54 AM
We are now entering into the smoke-screen stage, and for gods' sake try not to use the word relative.:D

Phil:)

JCHannum
12-21-2009, 08:25 AM
Forget all about droop and sag and the ballistic tables drawn for a different situation. Almost all effects of velocity, projectile weight and design will cancel out as long as the same round and rifle is used. The concept is very simple and is nothing more than that the projectile is acted on only during that part of its flight which covers the the distance to the target relative to the pull of gravity.

The projectile will be more or less high in one situation than the other due to various influences, but it will always be high relative to point of aim. The longer the line of sight and the greater the angle, the greater the error will be.

It has always been so and always will be so. With the rifles in use in muzzle loading times and the rainbow trajectories they demonstrated, it was much more of a factor than it is today with the high velocity, relatively flat shooting ammunition currently in use. Over short distances and minor elevations, it is not much of a factor, but long ranges and steep angles will very definitely affect the point of impact.

J Tiers
12-21-2009, 08:59 AM
That makes no sense. It doesn't matter if I look at a target from straight on or at an angle. That doesn't change the point of aim at all. It will compress the image of the target but that makes no difference to the point of aim.

Don't be ridiculous.........

The "d" hasn't anything to do with the target. it is target independent.

The distance "d' in my drawing is THE BULLET FALL DISTANCE AS SEEN FROM THE SHOOTER'S POSITION.

Because the bullet FALLS APPARENTLY LESS AS VIEWED FROM THE SHOOTER'S POSITION (actually it falls about the same distance regardless, the line of sight angle causes the apparent difference*) THE SHOOTER MUST NOT AIM AT AS HIGH A COMPENSATING ANGLE AS ON THE FLAT.

The angular deal works for the bullet also. The target is foreshortened, and the bullet fall distance is foreshortened. By aiming at the foreshortened angle, instead of teh same angle as on the flat, the bullet will then hit the same point on the target as it would on the flat.

Aiming at the ORIGINAL angle same as on flat, the bullet will pass high, because the inverse of teh foreshortening means the fall distance ASSUMED by the original angle corresponds to a larger fall distance than actually occurs.

it is a sort of "geometrical illusion" which must be compensated out

* There IS a definite but very small effect due to the varying time of flight shooting up vs down. It's "THERE", it just is not big enough to make a difference because the difference in time of flight is small.

firbikrhd1
12-21-2009, 09:05 AM
This is why I love this site. There are some brilliant people here all demonstrating their brilliance and willing to put their credibility on the line to prove their point. I'll be retiring soon and one of the things I plan to do is attend some Engineering classes, not to become an Engineer but so I can figure out problems such as this one when they arise. Reading this thread gives me some apprehension though, the people with great experience and engineering backgrounds can't seem to agree. I hope this doesn't turn out like the Science of Global Warming. No one seems to agree on that either so I've had to come to my own conclusions. I guess to learn the truth regarding this issue I'll have to do the same, go out and shoot some targets uphill & downhill and draw my own conclusions.
It makes me wonder if I have what it takes to learn this stuff.
Carry on!!!

Rustybolt
12-21-2009, 09:06 AM
You're missing the point of his question: does the bullet travel farther if you shoot from high ground. Not does the bullet go farther if you aim high.

Well. Yes. kinda. The bulet will fall further because the gun is higher, but if it is shot at the same angle as a gun on the ground, the ground the bullet covers will be the same.

JCHannum
12-21-2009, 09:14 AM
Don't be ridiculous.........

The "d" hasn't anything to do with the target. it is target independent.

The distance "d' in my drawing is THE BULLET FALL DISTANCE AS SEEN FROM THE SHOOTER'S POSITION.

Because the bullet FALLS APPARENTLY LESS AS VIEWED FROM THE SHOOTER'S POSITION (actually it falls about the same distance regardless, the line of sight angle causes the apparent difference*) THE SHOOTER MUST NOT AIM AT AS HIGH A COMPENSATING ANGLE AS ON THE FLAT.

The angular deal works for the bullet also. The target is foreshortened, and the bullet fall distance is foreshortened. By aiming at the foreshortened angle, instead of teh same angle as on the flat, the bullet will then hit the same point on the target as it would on the flat.

Aiming at the ORIGINAL angle same as on flat, the bullet will pass high, because the inverse of teh foreshortening means the fall distance ASSUMED by the original angle corresponds to a larger fall distance than actually occurs.

it is a sort of "geometrical illusion" which must be compensated out

* There IS a definite but very small effect due to the varying time of flight shooting up vs down. It's "THERE", it just is not big enough to make a difference because the difference in time of flight is small.

It is not an illusion nor does it have anything to do with the apparent size of the target. Your aiming point is a dot the diameter of the projectile.

The fall of the bullet is not apparent, but is a set distance determined by the time in flight over the distance traveled which is affected by the pull of gravity. As such, it is directly related to the velocity of the projectile.

Rustybolt
12-21-2009, 09:18 AM
This is why I love this site. There are some brilliant people here all demonstrating their brilliance and willing to put their credibility on the line to prove their point. I'll be retiring soon and one of the things I plan to do is attend some Engineering classes, not to become an Engineer but so I can figure out problems such as this one when they arise. Reading this thread gives me some apprehension though, the people with great experience and engineering backgrounds can't seem to agree. I hope this doesn't turn out like the Science of Global Warming. No one seems to agree on that either so I've had to come to my own conclusions. I guess to learn the truth regarding this issue I'll have to do the same, go out and shoot some targets uphill & downhill and draw my own conclusions.
It makes me wonder if I have what it takes to learn this stuff.
Carry on!!!

People that shoot a lot know something about ballistics. telescopic sites are graduated in moment of angle. The shooter compensates for things like distance, wind, hight or lowness of target to shooter.
There are some very grisly but impressive videos of Canadian snipers taking out targets at insanely long distances with a Barret .50.

lazlo
12-21-2009, 09:30 AM
Reading this thread gives me some apprehension though, the people with great experience and engineering backgrounds can't seem to agree.

Everyone has been in agreement since page 2. To whom are you referring?

I think taking engineering fundamentals classes is an excellent idea. Statics and dynamics would be especially good, and very appropos to many machinist projects -- they're usually the very first class a engineering candidate takes. There's a considerable amount of math involved, so you might consider a calculus and vector math refresher.

IdahoJim
12-21-2009, 09:36 AM
Where are you getting the 50% factor.
It isn't a question of half, or some other fraction of gravity, it's a question of the full force of gravity slowing only the vertical component of the resultant velocity vector (which is the initial speed along the initial path).

Viewed on an X-Y graph, the velocity vector 'V' of a shot fired upward at a 45deg angle can be broken down into a horizontal component of 'X' and vertical component of 'Y'. Both X & Y will have a value of .707 times V. (.707 is the sine and cosine of 45deg)
Gravity will not be acting on X at all, but solely on Y. However, since X (horizontal motion) is diminished (same as Y) the bullet will be in the air longer before traveling any given horizontal distance (i.e. longer than if shot level), therefore gravity will have longer to act on the bullet, which means more drop.
But/however (again), in this scenario the bullet starts out with a much greater vertical component than when it was originally sighted in at, say 200 yds. But that's neither here nor there, Ol' man gravity is still accelerating that component of travel downward at 32ft/sec/sec. And in this case, unlike when it was being sighted in, that vertical component Y represents .707 x V of the velocity toward the target, i.e. it's being slowed toward the target.
Yes, I erred when I said .50. The true value is .707. However, the bullet slowing factor due to gravity is very minor, compared to the slowing factor caused by air drag...really, really minor. The initial velocity is so high compared to the accelerating/decelerating effects of gravity, that those effects can be ignored for practical purposes. If you assume the .707 X the 32fps/ps, you can see that. In the first second after firing at an uphill angle of 45* gravity is slowing the bullet by 22.6fps....that's almost meaningless compared to the muzzle velocity. In fact, the normal shot -shot variations in velocity are often more than that. As I mentionjed in an earlier posting Sierra's engineers give a value of drop caused by the slowing effect of gravity, at an angle of 45*, of .03" at 200 yards...hardly worth worrying about.
Jim

lazlo
12-21-2009, 09:40 AM
There are some very grisly but impressive videos of Canadian snipers taking out targets at insanely long distances with a Barret .50.

There's a show on the Military Channel running about the Top 10 sniper rifles. Quite a few start-up company proposals that I hadn't seen, in addition to the M24 and the Barrett M82.

They mentioned that a Canadian sniper set the record for the longest confirmed kill -- over a mile and a half in Afghanistan with a Barrett.

JCHannum
12-21-2009, 09:53 AM
An even more remarkable feat was performed by Billy Dixon in 1874 when he killed an indian at a distance approaching one mile with a Sharps 45-90 rifle equipped with iron sights.

lazlo
12-21-2009, 09:56 AM
An even more remarkable feat was performed by Billy Dixon in 1874 when he killed an indian at a distance approaching one mile with a Sharps 45-90 rifle equipped with iron sights.

Holy Cow! How can you even see that far?

Did he teach marksmanship to Sea Cadets?

Evan
12-21-2009, 10:34 AM
Then I assert that for relatively flat trajectories, your picture is wrong.

Could you please go to my post #68, and say where you would aim the barrel of the gun. Never mind the sights, assume you had a laser pointer down the center of the barrel.

Wrong how? Let's assume that I bore sight directly on the target in both cases. What do you think will happen?

BTW, we get a LOT of practise with this particular exercise on a regular basis shooting squirrels in the trees. We use a 500 fps .22 air pistol with a red dot finder that is sighted to shoot dead on at 50 feet level, which is about how high the squirrels are in the trees. To hit them in the trees one must always aim a bit high. My wife is a deadly shot on the tree rats.

dp
12-21-2009, 10:50 AM
This stuff is really simpler than is being made of it. Gravity pulls the bullet down at 1g per second per second over the flight time of the bullet. You can calculate the possible drop time simply by calculating the flight time and the initial altitude of the bullet when it leaves the barrel. This is standardized around a barrel angle tangent to the surface.

If there is a vertical element, + or -, that is calculated into the possible flight time. Simple vectors, so far. In a vacuum on a non-curving surface of a non-rotating plain, we're done calculating.

But we have to add in drag from the atmosphere. If the barrel is rifled then we have to consider gyroscopic effects and precession. The bullet does not want to keep the point in the direction of flight - it is a gyro. It wants to continue pointing in line with the barrel. The longer it flies on a rotating body (earth) the more off line it becomes with the direction of flight (which is curved). The shape of the bullet keeps it pointing in the direction of least resistance but precession is not to be trifled with, so the direction of least resistance is not the least drag angle through the air. Imagine the gyroscopic forces on a 2,000 pound round flying a 20 mile arc.

Then there is the problem of shooting on a spinning planet. Drop an object from a 1,000 foot tower and it will land further away from the tower's base than the distance it was from the tower's top. In the case of a bullet fired, the direction of flight changes the accuracy and flight time. Fire a bullet parallel to the current longitudinal line across either pole of the earth and watch that bullet miss the opposite longitudinal line it was fired from.

And then there's that pesky "aim" that is bandied about. In a well made rifle the sights are arranged to compensate for a number of things. In conversation, and in this conversation it was implied, that point of aim is the sight picture, not where the barrel is pointing. In the case of artillery there are no sights - everything is based on where the barrel is pointing. That has to be considered when analyzing the flight path where sights are used and which are not calibrated for shots with a large vertical component. In other words, the sights are calibrated for targets that are at or nearly at the same elevation as the shooter.

I've fired my 45-70 Marlin often in mountainous country, and it is a round you can follow with your eye. At the best of times this round drops like a stone, and when shooting up hill it will fall well short of the sight point for the same distance to target as when shooting on level ground.

See? Really quite simple :)

winchman
12-21-2009, 10:51 AM
I mentioned this thread to a friend, and he suggested an easy way to compensate for shooting high on a slope would be to tilt the gun (gangsta-style):
http://www.istockphoto.com/file_thumbview_approve/5812953/2/istockphoto_5812953-man-aiming-gun.jpg

The tilt angle of the gun would match the angle of the uphill/downhill slope.

His theory was that tilting the gun would cancel the effect of the sights being set for level ground.

How would that work?

Roger

dp
12-21-2009, 10:59 AM
Holding the gun sideways and rapid firing it will cause horizontal recoil so later shots will not be over the heads of people in the room. Otherwise it's a very inaccurate way to form a sight picture. There was a long article about it on Slate some time back.

Evan
12-21-2009, 11:21 AM
Dennis,

That's all pretty obvious stuff but for the purpose of making the answer at least somewhat clear it is necessary to reduce the complications as much as possible. The place to start is to first determine what the ballistic arc will be for different elevations of fire with all other variables held fixed. If you don't do that then you can't calculate anything else.

Incidentally, the ballistic arc isn't really a parabola although that is a useful approximation. The arc is a very short portion of an ellipse which is what all bodies in free fall orbit follow.

For the purpose of answering the question a valid result can be had by placing the experiment on the moon or just by assuming an airless Earth, non rotating.

JCHannum
12-21-2009, 11:42 AM
For the purpose of answering the question a valid result can be had by placing the experiment on the moon or just by assuming an airless Earth, non rotating.

That will serve to eliminate many of the side issues, but the outcome will still be the same, the projectile will go high whether shooting uphill or down.

Billy Dixon did admit that luck played a large part in his feat, but it remained for over 100 years as the longest confirmed kill by a rifle. I was not aware that it had been outdone, but it took a purpose built firearm and some fancy optics to accomplish it. Billy was using a Sharps.

lazlo
12-21-2009, 12:04 PM
Billy Dixon did admit that luck played a large part in his feat, but it remained for over 100 years as the longest confirmed kill by a rifle. I was not aware that it had been outdone, but it took a purpose built firearm and some fancy optics to accomplish it.

...and a spotter, a precision anemometer and a ballistics computer.

Both are amazing feats. You have to wonder if there have been any black ops that have have surpassed the Afghanistan shot...
Where's Rigger? :)

Evan
12-21-2009, 01:40 PM
That will serve to eliminate many of the side issues, but the outcome will still be the same, the projectile will go high whether shooting uphill or down.

I have not yet seen any explanation that follows known Newtonian principles of ballistics that confirms your assertion.

dp
12-21-2009, 01:52 PM
Dennis,

That's all pretty obvious stuff but for the purpose of making the answer at least somewhat clear it is necessary to reduce the complications as much as possible. The place to start is to first determine what the ballistic arc will be for different elevations of fire with all other variables held fixed. If you don't do that then you can't calculate anything else.

Incidentally, the ballistic arc isn't really a parabola although that is a useful approximation. The arc is a very short portion of an ellipse which is what all bodies in free fall orbit follow.

For the purpose of answering the question a valid result can be had by placing the experiment on the moon or just by assuming an airless Earth, non rotating.

In any discussion like this, though, it's important to discuss the impact point in terms of sight point or barrel point orientation. In every case you hit where you aim but it's important to know that where you think you aim and where you actually aim can be different. The conversation started out speaking in terms of sight point vs impact point for changes in elevation of shooter and target, and drifted off to barrel control. The site point and impact point will change if the only variable is elevation differences between the shooter and target (range and environment being the same). The sight and impact points do not merge when shooting perpendicular to the surface with a sight point, but will using barrel point.

dp
12-21-2009, 01:55 PM
I have not yet seen any explanation that follows known Newtonian principles of ballistics that confirms your assertion.

It depends on how many times the projectile crosses the sight line on the way to the target.

camdigger
12-21-2009, 02:32 PM
It depends on how many times the projectile crosses the sight line on the way to the target.

Doesn't the bullet cross the sight path twice? (At least for telescopic or iron sights)? Once on the way up, and once on the way down. The higher the muzzle velocity, the flatter the trajectory, and generally speaking, the further these two points are apart.
IF SO, the bullet is above the line of sight for a portion of it's path whether shooting up, down, or flat so the point of impact will be higher than the line of sight for a portion of all three situations. In this case plain old trig hints that the situation will be aggravated whether uphill or downhill

Most sight systems (telescopic, iron, laser) I'm aware of are aligned with the barrel viewed from above, but are not aligned with the barrel viewed from the side, they're either above (optical, iron) or below (laser). When the sights are set so the flight path of the bullet crosses the line of sight at, say 25 m ( a standard shooting range distance), how far is it until the bullet drops back across the line of sight? How far if the gun is sighted in for 75 yds or 100?

How is the distance affected (if at all) if the line of sight is tilted (shooting uphill or downhill)?

Allan Waterfall
12-21-2009, 02:54 PM
They mentioned that a Canadian sniper set the record for the longest confirmed kill -- over a mile and a half in Afghanistan with a Barrett.

Evan probably trained him in marksmanship ?:D

Allan

JCHannum
12-21-2009, 02:57 PM
I have not yet seen any explanation that follows known Newtonian principles of ballistics that confirms your assertion.

It is not "my assertion", it is a fact of ballistics that has been explained by myself and several others in this thread. I cannot account for your inability to grasp the simple interraction of gravity and trigonometry involved.

Camdigger gets it.

IdahoJim
12-21-2009, 03:12 PM
Doesn't the bullet cross the sight path twice? (At least for telescopic or iron sights)? Once on the way up, and once on the way down. The higher the muzzle velocity, the flatter the trajectory, and generally speaking, the further these two points are apart.
IF SO, the bullet is above the line of sight for a portion of it's path whether shooting up, down, or flat so the point of impact will be higher than the line of sight for a portion of all three situations. In this case plain old trig hints that the situation will be aggravated whether uphill or downhill

Most sight systems (telescopic, iron, laser) I'm aware of are aligned with the barrel viewed from above, but are not aligned with the barrel viewed from the side, they're either above (optical, iron) or below (laser). When the sights are set so the flight path of the bullet crosses the line of sight at, say 25 m ( a standard shooting range distance), how far is it until the bullet drops back across the line of sight? How far if the gun is sighted in for 75 yds or 100?
How is the distance affected (if at all) if the line of sight is tilted (shooting uphill or downhill)?

It depends on the muzzle velocity. For modern rifle cartridges, in the 3,000fps range, and with a telescopic sight mounted about 1 1/2" above the bore, and sighted in for zero at 200 yards, the LOS is crossed close to the muzzle at about 30 yards +/-. It crosses back at the zero range of 200 yards. With faster cartridges the first crossing point will be farther from the muzzle. Also, with bigger objective lenses on modern scopes, the scope to bore distance will be larger, which means the crossing point will be farther from the muzzle. Also, the farther out the zero range, the closer to the muzzle the first crossing point will be. There are lots of variables in all this, which is why serious shooters spend a lot of time at the range...LOL.
Since the sights are rigidly mounted to the barrel, you're not changing the relationship between the LOS and the line of departure of the bullet, by changing the angle at which you aim.
Jim

J.Ramsey
12-21-2009, 03:32 PM
camdigger, for all practical purpose figure the the trajectory crossing the line of sight at 25 yards and 100 yards for the "average" center fire cartridge.

Here's an example of my model 70 Winchester in .243 Winchester driving a 100 grain bullet at 3,300 fps muzzle velocity, zeroed at 100 yards.

Scope line of sight 1.800 above the bore.
25 yards - 1/2"
200 yards -2.1"
300 yards -8.1"
We shot at 600 yards for giggles and because of the wind group size really opened up, but estimated bullet drop between 55- 60 inches.

dp
12-21-2009, 03:33 PM
It depends on the muzzle velocity. For modern rifle cartridges, in the 3,000fps range, and with a telescopic sight mounted about 1 1/2" above the bore, and sighted in for zero at 200 yards, the LOS is crossed close to the muzzle at about 30 yards +/-. It crosses back at the zero range of 200 yards. With faster cartridges the first crossing point will be farther from the muzzle. Also, with bigger objective lenses on modern scopes, the scope to bore distance will be larger, which means the crossing point will be farther from the muzzle. Also, the farther out the zero range, the closer to the muzzle the first crossing point will be. There are lots of variables in all this, which is why serious shooters spend a lot of time at the range...LOL.
Since the sights are rigidly mounted to the barrel, you're not changing the relationship between the LOS and the line of departure of the bullet, by changing the angle at which you aim.
Jim

The number of times the projectile can cross the sight line is zero, one, or two, depending on the target range and sight adjustment. For short distances the projectile will reach the target at the first crossing. A consequence is that at some further range it will cross the line again if it is still flying. For that sight setting, hitting a closer target will prevent the round from crossing the sight line, and the shot will be low.

If the barrel and sight are parallel the projectile will never cross the sight line and all shots will be low.

If the sight line is set sufficiently high there is one and only one point down range where the site line and projectile meet. At all other distances, plus or minus, the shot will be low.

If the sight line is set such that the projectile crosses the sight line twice you have a very elongated kill zone and two bullseye points with a high-velocity round. With a low velocity round like the 45-70 you will have two kill zones separated by a space with a lofting round. If the target is large enough you can still get a hit in the gap. Mostly I just feel good if I hit the world. :)

Jimno2506
12-21-2009, 03:44 PM
Evan,

In your diagram, can you see the horizontal distance to the target is less for the high and low shots? ALL bullets drop at 32ft/sec/sec.

You keep asking why the distance has to change for the high and low shots. I would add that the Revolutionary soldiers measured distance by line of site as very few of them would concern themselves with trig.

They simply did not compensate for the horizontal distance being less than the line of sight distance, and therefore time in the d=v*t equation. The bullet was in flight shorter time than they expected, had less time to fall, and went high.

Granted your assertion that bullets will be <less> high shooting up hill, the numbers are so close that it won't matter in virtually any shooting scenario.

Nikon and several other rangefinders have an elevation compensator. It's not for the added energy of bullet going down...it's for the angle.

Check one out.

Jim

Evan
12-21-2009, 05:58 PM
In your diagram, can you see the horizontal distance to the target is less for the high and low shots? ALL bullets drop at 32ft/sec/sec.

The projected horizontal distance isn't important. The drop of the bullet is controlled by the time it is in free fall and only that variable. The direction in which it is fired also makes no difference to the acceleration of gravity. All objects that have some horizontal velocity vector will describe a ballistic arc with only the maximum altitude and the range as dependent variables. The higher the angle of fire the greater the difference is between the maximum altitude reached by the bullet and the point of aim of the barrel where it crosses the plane of impact. It's a tangent function which increases asymptotically with angle. The the point of aim becomes infinitely higher than impact at 90 degrees.

As for the number of times the path of the bullet crosses the line of sight, that is irrelevant. In this problem it should be granted that the weapon is properly sighted for the range in question when horizontal. The distance of the sights above the barrel is a simple offset and require no special consideration. The bullet will travel in a ballistic arc that will coincide exactly with the point of aim when fired horizontally. When fired up the point of aim will be still be exactly on the target but the ballistic arc will have a lower endpoint.

The sight picture and the relationship of the sights to the trajectory do not change because the weapon is pointed in a different direction. Neither does the effect of gravity. If we also grant that the range stays the same then the time of flight has very little change as well. The only significant change at all is the loss or gain of energy due to the gain or loss of altitude in the gravity well.

The trigonometry of the situation doesn't change when the point of aim is higher unless the assumption is made that the target is the same horizontal distance away in which case the situation has no equivalence and cannot be compared since the range to the target will change. The only valid comparison is that which I described where one person shoots down the hill and the other shoots up the same hill.

Rustybolt
12-21-2009, 06:41 PM
An even more remarkable feat was performed by Billy Dixon in 1874 when he killed an indian at a distance approaching one mile with a Sharps 45-90 rifle equipped with iron sights.

7/8ths of a mile. slightly uphill.
An experiment was conducted at Ft Sill OK back in the 70s or early eighties. Three rifles, chambered for 45-70 45-90 and 50-120 were tested. They wanted to find out the maximum leathal range of those calibers. Turned out that if you could hit a man with a 45-70 at two miles, it had enough retained energy to kill him.

Rustybolt
12-21-2009, 06:45 PM
There's a show on the Military Channel running about the Top 10 sniper rifles. Quite a few start-up company proposals that I hadn't seen, in addition to the M24 and the Barrett M82.

They mentioned that a Canadian sniper set the record for the longest confirmed kill -- over a mile and a half in Afghanistan with a Barrett.

Judging from the videos, when the human body is hit by a .50 cal round body parts fly in all directions.

philbur
12-21-2009, 06:57 PM
Using the following simple equations shows that for the example given the projectile passes 2.3m over the target (not under) if you do not correct the sight for inclination compared to sight setting for a level shot.

Vm= muzzle velocity
vertical velocity component Vx= Vm.cos(muzzle angle)
Horizontal velocity component Vy= Vm.sin(muzzle angle)
Vertical distance travel Y= Vy.t + g.t^2/2
Horizontal distance travel X = Vx.t

The equations require iteration to obtain the correct answer.

The results below are for the following three shots.
A= Level shot
B= Uphill shot at correct sight setting - setting + 0.21 degrees
C= Uphill shot at same sight setting as level shot - sight setting + 0.30 degrees

http://i186.photobucket.com/albums/x36/philbur/Image2.jpg

Phil:)

JoeCB
12-21-2009, 07:56 PM
I've always been on the "high" - "high" side of this discussion because I know from 40 years of competitiving shooting that that's the way it is.
Pillbur has the definative argument.
But for those that still doubt try going thru this thought process... same rifle , same sight settings, you shoot straight up 90* , there is (should be ) no doubt that the bullet will hit the target high, correct. Now tilt the barrel down,, just say 89* and do it again, no one would doubt that the shot will again be high. keep going... untill you get to horizontal 0* the shot will now be on target. Nothing magical happens at some point when you progress from 90* down to 0 The influenct of gravity on the verticle position of the point of impact gradually changes as you move closer to horizontal where it is at a maximum.
Joe B

Yankee1
12-21-2009, 08:23 PM
The rifle used by the Canadian sniper was a Mc Millan Tac 50 to make that 1.5 mile kill. He took two shots to see where he was hitting and made the hit on the third shot using the information he got from the first two shots.
Barrett .50 calibers are used by snipers but in this case it was a McMillan Tac 50.

philbur
12-21-2009, 08:31 PM
The same example also occurred to me. It's a very good, subjective way of understanding what's going on without having to do the math.

Phil:)

But for those that still doubt try going thru this thought process... same rifle , same sight settings, you shoot straight up 90* , there is (should be ) no doubt that the bullet will hit the target high, correct. Now tilt the barrel down,, just say 89* and do it again, no one would doubt that the shot will again be high. keep going... untill you get to horizontal 0* the shot will now be on target.
Joe B

lazlo
12-21-2009, 08:45 PM
Barret .50 calibers are used by snipers but in this case it was a McMillan Tac 50.

Ah, a bolt-action .50 -thanks for the clarification! The Military Channel episode mentioned several new Wildcat rounds that were designed for even longer range than the .50 BMG -- any truth to that?

beanbag
12-21-2009, 08:55 PM
Wrong how? Let's assume that I bore sight directly on the target in both cases. What do you think will happen?

BTW, we get a LOT of practise with this particular exercise on a regular basis shooting squirrels in the trees. We use a 500 fps .22 air pistol with a red dot finder that is sighted to shoot dead on at 50 feet level, which is about how high the squirrels are in the trees. To hit them in the trees one must always aim a bit high. My wife is a deadly shot on the tree rats.

The chart you drew is wrong for the following reasons(s):
I don't have any problem with the green and red line you drew for the horizontal shot. However, in this case, gravity is puling on the bullet perpendicular to its direction of travel, therefore causing maximal deflection.

You then copy pasted and rotated this set of lines and put it at an angle. The true bullet path is now deflected LESS by gravity because it is being pulled at some angle, instead of perpendicular to the path. (In the extreme case where you shoot straight up, gravity does not deflect the bullet path, it only shows it down.) If the true bullet path is deflected less than before, you should take the red line and unbend it some, while keeping the same initial angle to the green line, which means the bullet hits high.

So repeating what some others have said, the bullet hits high not because of some strange gain in kinetic or potential energy, but due to an artifact of aim compensation in degrees, instead of distance directly above the target. See my diagram in post 66 as an example. Philbur's calculations basically confirm this effect.

(by the way, my attempt to use energy conservation equations to add further corrections to the bullet path is a dirty hack, and should be ignored, if only because it confuses the issue)

I don't know why you have to aim high to hit the squirrels, though. It could have to do with other factors such as how far off axis the sight is mounted to the barrel, or how far you stand from the base of the tree.

BTW, the correct answer as to where to aim the barrel on the ball is the orange line. If you used your sights calibrated for a horizontal shot, you'd end up aiming along the blue line, which is too high.

J Tiers
12-21-2009, 09:22 PM
It is not an illusion nor does it have anything to do with the apparent size of the target. Your aiming point is a dot the diameter of the projectile.

The fall of the bullet is not apparent, but is a set distance determined by the time in flight over the distance traveled which is affected by the pull of gravity. As such, it is directly related to the velocity of the projectile.

Since you seem to AGREE with the majority here (at least you disagree with Evan) , I am puzzled by your inability to grasp the fact that there IS an "apparent" drop, which is the drop AS SEEN FROM THE SHOOTERS POSITION WHEN FIRING UPHILL.

THAT IS WHAT THE SHOOTER AIMS OFF OF

If that aim-off, is not adjusted by 'foreshortening" it, it will be inappropriate, and will cause a high shot. It's extremely simple.

MickeyD
12-21-2009, 09:32 PM
Here is a simple way of looking at it. Say you have gun that fires a bullet at a relatively low velocity but the bullet has no drag. Assume the gun is sighted for 100 yards. Because the velocity on the bullet is very slow but with no deceleration due to drag, the gun is actually pointed 1 yard over the target (you are "lobbing" the bullets at the target in a perfect arc - if you shoot at a target at 50 yards you will be shooting 1 yard high) . If you happen to glance up and see a duck hovering 100 yards directly over your head and decide that it will make a fine dinner, you will have to aim 2 yards low to actually hit it. This is an extreme example, but shooting both down and uphill requires adjustments that might not be apparent at first.

dp
12-21-2009, 10:07 PM
But for those that still doubt try going thru this thought process... same rifle , same sight settings, you shoot straight up 90* , there is (should be ) no doubt that the bullet will hit the target high, correct.

True if you shoot straight up. But if you aim straight up the bullet will land behind you. Again, it's important to speak of line of delivery vs line of sight here.

beanbag
12-21-2009, 10:23 PM
http://i186.photobucket.com/albums/x36/philbur/Image2.jpg

Phil:)

Thanks for running the numbers. As long as you have the simulation set up, could you do another run, where you have to aim at something really far away, such that the barrel has to be pointed significantly higher than the line of sight to the object? Evan's picture of his ballistic simulations doesn't really show the "bullet flies high when aimed at high objects" effect, so I wonder if other effects start to come into play at these long ranges.

Jimno2506
12-21-2009, 11:02 PM
The drop of the bullet is controlled by the time it is in free fall and only that variable.

Exactly, and if a shot is measured at 100 yds horizontal, it travels for time 't'. If it is measured at 100 yards at a 45 degree angle, its time is .707t.

The time of flight is reduced, the drop by gravity is reduced, and you will hit high.

Happens every time.

Jim

Yankee1
12-21-2009, 11:07 PM
Hi Lazio
I do not Know of any more powerful than the .50 caliber. I understand that the Ball2 ammo is rated at about 4.25 miles. To shoot at longer distances Barrett makes special scope rings that allow the rear of the scope to be tilted up to allow for more elevation adjustment. Barrett says 2000 yards maximum effective range. They use a 750 grain Hornady A-MAX bullet.
The marines use a ballistic computer to calculate wind, drop,distance formulas. That is far advanced from tho old way. With the special Barrett scope rings and the ballistic computer it would depend on the skill of the shooter as to how far he could extend the range and of course the conditions.
There is a .338 lapua sniper rifle but thats not even close.
There are also plenty of 30 caliber sniper rifles but nothing that I know of
better than the .50 caliber.
Just imagine the wind conditions changing 3 or 4 times between you and the target and you will get an idea of the difficulties..That heavy 750 grain bullet helps a whole lot under those long range conditions.

JoeCB
12-21-2009, 11:33 PM
One last time... here is the definative answer, and yes it's high in both cases. Just a little bit less high in the up hill case. The attached chart is from the Sierra Reloading Manual page 1050. If you can read the attached ... example the 7MM Rem in a 30* incline at 300 yards prints 2.68 in. high on 30* incline and prints 2.73in. high on 30* decline.
Joe B
http://i201.photobucket.com/albums/aa106/JoeCB/ballistic.jpg

dp
12-21-2009, 11:53 PM
So you register your sights and have an impact point at 300 yards shooting horizontally - by that I mean the bull is at the same elevation as the barrel of the gun, 0º angle. Then as you elevate your weapon incrementally, say 5º vertically using the sights aimed at a magic compass out there at 300yds, you will strike the target at incrementally higher points above the the degree mark. Still shooting at 300 yds, sights remain fixed. And the further up the compass you go the higher above your degree mark you will hit.

By the time you are aiming straight up your rounds are landing behind you because they are not going straight up. Your sights are straight up but your barrel is past the vertical. You have to aim less than 90º to hit the 90º mark.

Now do the reverse - start shooting lower incrementally. The same thing happens until you are aiming straight down - you are still hitting high except this time you're not shooting behind you. You have to aim past -90º to hit the -90º mark.

Evan
12-22-2009, 01:14 AM
First I will acknowledge that Philbur is correct. Allthough the drop is still the same amount it is no longer perpendicular to the path traveled.

However, where I have been having problems is with some of the other explanations such as Jim's.

Exactly, and if a shot is measured at 100 yds horizontal, it travels for time 't'. If it is measured at 100 yards at a 45 degree angle, its time is .707t.

Not possible unless the bullet for unaccountable reasons travels faster. It still has to travel the same distance which will take slightly more time since it will travel slightly slower going up against gravity. The amount of drop is exactly the same for the time traveled plus whatever amount of slowing takes place. Gravity always acts straight down so the drop is always straight down.

The example of shooting with a point of aim at 90 degrees also has some problems. It is a limit case and if the range happened to be at the extreme limit the drop will ensure that the bullet never reaches the target.

gzig5
12-22-2009, 02:34 AM
Ah, a bolt-action .50 -thanks for the clarification! The Military Channel episode mentioned several new Wildcat rounds that were designed for even longer range than the .50 BMG -- any truth to that?

Yes, there are several that are using lathe turned bullets with very high ballistic coefficients. The .408 CheyTac is one. There are also several rounds that were developed to get around the California ban on .50 caliber rifles. Basically smaller caliber on a shortened or necked down 50 BMG case. Leave it to the Liberals to further the development of modern small arms. :D

Yankee1
12-22-2009, 02:40 AM
Hello Again Evan
Gravity affects the horizontal shot with maximum pull downward. As the angle of the shot increases the affect of gravity becomes proportionally less.
Using the right angle to demonstrate the hypotenuse is the path of the bullet and the adjacent side represents cosine of the angle which represents the horizontal distance that gravity will affect the pull on the bullet.
Let us say that the angle of the hypotenuse is 45 degrees. The cosine of 45 degrees is.707 and the shot distance is 300 yards. 300 yds.* .707 =212.1yds
If you are sighted for 300 yards and the gravity only affects it for 212.1 yds.
it becomes easy to see the shot will go high because gravity is affecting almost 30% less than the horizontal shot. The difference between shooting this shot uphill or downhill is so slight as to not become of any importance .
Both shots will go high. I hope this explanation is clear enough.

dp
12-22-2009, 02:43 AM
Hello Again Evan
Gravity affects the horizontal shot with maximum pull downward. As the angle of the shot increases the affect of gravity becomes proportionally less.

STOP!

This is nonsense. Gravity has no idea which way a bullet is going and doesn't care. It tugs relentlessly at 1G, period. The projectile can be going in any direction imaginable and the effect is the same. One G, straight to the center of the world. No exceptions.

Yankee1
12-22-2009, 02:56 AM
Dennis
Gravity places its full pull on the horizontal shot and less pull on the shot that angles up or downward. The length of the cosine of the angle is used to determine how much gravity will affect the angled shot. Look up the cosine of the angle and multiply it times the line of sight distance and you have the distance the shot will be affected by gravity.
If you do not understand this maybe reading all the posts here will help.

Paul Alciatore
12-22-2009, 03:15 AM
With the same aim offset, as seen in the shooter's sights, the bullets will go high at any angle above or below horizontal. The explanation does not require any analysis of the bullet's speed. In fact, the amount of speed added or subtracted from the bullet is very small and can be neglected in the analysis. Likewise, the difference in flight time is also quite small and can be neglected. My simple analysis only assumes that:

1. The distance from shooter to target remains the same.

and

2. The sight offset, as seen by the shooter, remains the same.

Simple geometry explains the high trajectory of the bullets.

http://img.photobucket.com/albums/v55/EPAIII/BulletStrike.jpg

In the drawing, ALL three shots were drawn with a 1" VERTICAL bullet drop. A well trained shooter will compensate for this drop with an offset in his sights. This is actually an angular offset in his sight picture.

So, in the horizontal shot, the sight offest AND the bullet drop are both perpendicular to the line of the shot and they are EQUAL both in absolute magnitude and in their vector magnitued. The center shot hits the target (shown by a small oval).

But in both the upper and lower shots, firing uphill and downhill, the drop of the bullet, while it is still the same 1" in my drawing, IT IS NOT perpendicular to the flight of the bullet and so it subtends a SMALLER angle as viewed by the shooter. But the shooter has made that SAME, offset in his sights. This offset is shown by the lines that are perpendicular to the lines of flight: the same angular offset he used for the horizontal shot. Hence, the offset is simply too much. And the bullet flies high in both cases.

In my drawing the bullet drops are all vertical as they will be in the actual firing. But the sighting offsets are perpendicular to the lines of flight. This is where the difference comes from, the difference in the angles of the bullet drop and the sighting offest.

philbur
12-22-2009, 03:58 AM
Jim's answer is correct but I think his reasoning is wrong. As you point out, for the same linear distance the time to target of an uphill shot is longer than the time to target of a horizontal shot. The drop (due to gravity) is of course always straight down but for the angled shot part of this drop is down the line of the trajectory so it has no influence on the "direction" of the shot, only on the time to get to the target. This is where the cosine/sine rule comes into play.

A shot has a limit on range at any muzzle angle. So I think that as long as you stay within that range there is no discontinuity in the math, therefore the vertical example is a valid way to visualise what is happening.

Phil:)

Not possible unless the bullet for unaccountable reasons travels faster. It still has to travel the same distance which will take slightly more time since it will travel slightly slower going up against gravity. The amount of drop is exactly the same for the time traveled plus whatever amount of slowing takes place. Gravity always acts straight down so the drop is always straight down.

The example of shooting with a point of aim at 90 degrees also has some problems. It is a limit case and if the range happened to be at the extreme limit the drop will ensure that the bullet never reaches the target.

philbur
12-22-2009, 04:11 AM
Its all in the wording, he should have said the effect "on the trajectory" is less. This can be seen from the vertical shot, where gravity has no influence on the trajectory at all.

This is why engineers use equations to explain things and not words, there is less scope for misunderstanding.

Phil:)

STOP!

This is nonsense. Gravity has no idea which way a bullet is going and doesn't care. It tugs relentlessly at 1G, period. The projectile can be going in any direction imaginable and the effect is the same. One G, straight to the center of the world. No exceptions.

JCHannum
12-22-2009, 09:05 AM
The time of flight over a given distance in the horizontal plane is what determines the ballistic arc, nothing more, nothing less.

The degradation of velocity due to resistance to the atmosphere and other factors play a secondary role in the shape of the arc, as velocity decreases, the arc becomes more pronounced, creating the elliptical shape. When considering the performance of identical cartridges and firearms, these factors cancel out and can be ignored.

A high velocity projectile will have a much flatter trajectory than a low velocity one. A projectile with zero velocity will drop straight down. A projectile that is fired perfectly straight up will drop straight down when its velocity is spent (discounting side effects such as the earth's rotation and wind). These effects are due to the length of time the projectile is exposed to the pull of gravity and the distance traveled in the horizontal plane in that time. Change either of these two variables and the ballistic arc will be changed.

Evan
12-22-2009, 09:16 AM
A shot has a limit on range at any muzzle angle. So I think that as long as you stay within that range there is no discontinuity in the math, therefore the vertical example is a valid way to visualise what is happening.

90 degrees up is a limit case in that it has zero horizontal range regardless of velocity. It is also in the limit case because the aiming compensation required is zero.

A shot has a limit on range at any muzzle angle

You of course mean any positive angle. All downward shots may result in a discontinuity.
Using the equations of motion downward shots have a strictly theoretical problem of including infinite solutions using Newtonian math.
This is so because there in no mathematical limit on how high the shooter may be above the target and still be able to reach it.

J Tiers
12-22-2009, 09:19 AM
I am not sure who agrees and who DISagrees any more..... it's really mixed up.

But the answer is SO simple....

Assume a target that is a vertical surface...... paint a line on it that is the bullet drop for that straight line range.

To hit the bottom of that line, you aim at the top. ALWAYS*.

So when you look uphill at it the line LOOKS shorter, but you STILL aim at the top. Your sights have to be set differently or else you will shoot high, you would be aiming above the top of the line if you didn't adjust.

Same downhill.

* the pull of gravity affects the uphill and downhill shots differently because in one case it is pulling the bullet "ahead" and the other case it is slowing it. With a high velocity round the time is short enough that the effect is maybe 1/10 that of the "target line" difference above, and fairly insignificant.

dp
12-22-2009, 09:35 AM
Dennis
Gravity places its full pull on the horizontal shot and less pull on the shot that angles up or downward. The length of the cosine of the angle is used to determine how much gravity will affect the angled shot. Look up the cosine of the angle and multiply it times the line of sight distance and you have the distance the shot will be affected by gravity.
If you do not understand this maybe reading all the posts here will help.

I think you've worded this badly. Again, the angles mean nothing to gravity. Unless the angle affects the flight time of the projectile. If that is your point then let's rework it to say anything that affects flight time of the round will give gravity more or less time to act on the projectile. Gravity's full pull is never more and never less than 1G over the distances we're assuming here.

dp
12-22-2009, 09:40 AM
90 degrees up is a limit case in that it has zero horizontal range regardless of velocity. It is also in the limit case because the aiming compensation required is zero.

You of course mean any positive angle. All downward shots may result in a discontinuity.
Using the equations of motion downward shots have a strictly theoretical problem of including infinite solutions using Newtonian math.
This is so because there in no mathematical limit on how high the shooter may be above the target and still be able to reach it.

I've presumed, because it's just one more variable that has not been established, that the sight line range to target for all test angles is the same. The shooter is suspended in the center of a large bubble and the bubble itself is the target. For my purposes the bubble also has convenient graduations on it oriented with gravitational "up" and marked in degrees :).

Some of my comments in this thread slyly recognize this critical variable has not been specified as when I wondered how many times the bullet crosses the LOS.

Paul Alciatore
12-22-2009, 10:24 AM
Read mypost above. You guys are making this WAY too difficult and it is simple.

dp
12-22-2009, 10:32 AM
Read mypost above. You guys are making this WAY too difficult and it is simple.

You've made it too simple - and wrong. Your model does not allow for +-90º.

JoeCB
12-22-2009, 10:36 AM
Please stop the pro and con... just read the ballistic chart in post #118.
Although I have to say it has been fun!

Joe B

dp
12-22-2009, 10:54 AM
Please stop the pro and con... just read the ballistic chart in post #118.
Although I have to say it has been fun!

Joe B

The chart doesn't say why, just what. I don't know that the "what" has been disputed all that much but there's a lot of "why's" :)

lazlo
12-22-2009, 11:04 AM
Please stop the pro and con... just read the ballistic chart in post #118.

Not only are the bullet drop tables (with negative adjustment for inclination) in every handloading book, but anyone who's fired a rifle off-range (i.e., hunting, trap and skeet...) has experienced it first hand :)

MickeyD had a very apt description a couple of pages back.

Ironically, David (and then you) answered the question correctly on the first page, followed by a more detailed answer by Yankee on the second page, but it's taken 12 more pages...

Jimno2506
12-22-2009, 11:04 AM
Perhaps my 't' equation is not the best way to look at it. Maybe if you consider the vector for gravity pulling (almost) perpendicular to the flight path for a horizontal shot (value of 1g) you will see the effect of gravity pulling down is a value of .707g for a 45* angle shot. I believe this is the 'normal force' vector.

Jim

IdahoJim
12-22-2009, 12:41 PM
Here's something to consider...the angle of departure between the LOS and the bullet are fixed. When we elevate the LOS to an uphill angle...lets use that 45* as the angle. We know that the force of gravity's effect on the bullet path is only 70.7% of the effect at the horizontal. So what has happened is we've extended the ballistic trajectory. But, we didn't flatten it as far as how high the highest point is relative to the LOS, we might even have caused the high point to be higher relative to the LOS. By extending the trajectory what we did was move the highest point closer to the target. That alone would make the bullet impact higher than at the horizontal. Compounding that is that the effective range we're shooting, compared to the horizontal, is only 70.7% as far.
Both of these tend to bring the highest point of the ballistic arch closer to the target. That combined effect is why the amount we shoot high is more than just a minor factor.
Jim

Evan
12-22-2009, 12:48 PM
The effect of gravity is to produce a certain amount of drop. The amount of drop depends only on the time the bullet is in flight and the amount of drop is the same no matter what direction the bullet is aimed.

If the bullet is in flight for 1 second the drop is ~16 feet straight down no matter what the velocity or the point of aim.

The only time that the "shoots high" rule comes into play is when the sights are adjusted incorrectly to compensate for this drop. If the weapon is boresighted on the target it will always shoot low no matter what the angle above zero is except at 90 degrees.

Compounding that is that the effective range we're shooting, compared to the horizontal, is only 70.7% as far.

How do you figure that?

Optics Curmudgeon
12-22-2009, 01:23 PM
Here's a suggestion. We're talking about a statement made on a TV channel, right? Maybe they are just a little confused and don't have things just right. For instance, what if they heard about a "shots going over the head" problem associated with the people firing being on high ground. The assumption is that these people are firing muskets. An entirely different problem presents itself if it's artillery. Even in those days the sighting process for artillery consisted of determining the target range, determining the correct angle of elevation, and setting this with a gunner's quadrant. Direct fire, as with small arms, was the exception. There is a factor in determining the correct elevation known as "angle of site", which is the angle from horizontal of a line between the gun and the target. Disregard angle of site when the gun is higher than the target and you will overshoot. For direct fire, such as with a musket, the gunner automatically corrects for this. This would have been known to professional artillerists, but these were militia, and may have been less well trained, only on level ground for instance. On the other hand, maybe the story is just apocryphal and seeking a reason is a wild goose chase.

Joe

BWS
12-22-2009, 02:33 PM
Shooting large bore lead bullets.............if you can get the Sun behind you at juuust the right angle,the reflection off slow moving bullet base is akin to shooting tracers.For S&Gs we set up clays @100m and send .45ACPs....you can walk them right into the target.carry on,BW

IdahoJim
12-22-2009, 03:19 PM
Shooting large bore lead bullets.............if you can get the Sun behind you at juuust the right angle,the reflection off slow moving bullet base is akin to shooting tracers.For S&Gs we set up clays @100m and send .45ACPs....you can walk them right into the target.carry on,BW

Interesting....in some atmospheric conditions we've noticed that you can see a vapor trail of high velocity bullets. You have to be standing behind the shooter and looking right down the LOS to see it. We've actually walked bullets into rock chucks at long range doing that.
Jim

IdahoJim
12-22-2009, 03:54 PM
The effect of gravity is to produce a certain amount of drop. The amount of drop depends only on the time the bullet is in flight and the amount of drop is the same no matter what direction the bullet is aimed.

If the bullet is in flight for 1 second the drop is ~16 feet straight down no matter what the velocity or the point of aim.

The only time that the "shoots high" rule comes into play is when the sights are adjusted incorrectly to compensate for this drop. If the weapon is boresighted on the target it will always shoot low no matter what the angle above zero is except at 90 degrees.

How do you figure that?

Sure, the drop is always constant, relative to the horizontal. What changes is the bullet path and LOS. And you're right...if you bore sight on the target, and shoot horizontal, the impact point will be below the target. You've made no adjustment in the sights to allow for gravity's pull on the bullet. Of course, nobody does that in the real world. They pick a range at which they want their weapon zero'd, and adjust the sights until the impact point and the sights agree. What they've done is angle the barrel up relative to the LOS. No big deal. Now, if you leave that rifle as sighted, and raise the LOS to 45*, and set the same target up the hill, at the same distance you were shooting at it when horizontal...let's say 100 yards. Two targets, one at 100 yards horizontal and one at 100 yards up the hill at a 45* angle. Distances are the same, Right?. But the horizontal distance to the target up the hill is only 70.7 yards. When we sighted the rifle to be on the money at 100 yards, we caused the bullet to go above the LOS about 30 yards in front of the muzzle. The bullet stays above the LOS until the ZERO range, in this case 100 Yards, where both the bullet path, and the LOS coincide. Now back to our target on the hill. Since the horizontal distance to that target is only 70.7 yards, the bullet impact point is above the LOS.
Even if you use bore-sighting, the same thing occurs. The bullet path is an ellipse, or curve, call it what you will. The bullet is higher at 70.7 yards than it is at 100 yards. Can't be anything else. The only difference in the two shots is that gravity is pulling the bullet directly away from the bullet path when shooting horizontal, and at an angle away from the bullet path when shooting up or down hill. If gravity pulls the horizontal bullet 10" away from the path on the horizontal, it will pull it 10" vertically, but not at a 90* angle, from the path up the hill. But since we're shooting at an angle the amount of 90* drop from the bullet's path is the cosine of the angle X that 10" vertical drop.
Jim

IdahoJim
12-22-2009, 04:04 PM
You shoot you're target at 100 yards horizontal. Now put a target up the hill at 100 yards at a 45* angle. Lets say the bullet hit right on the money on your horizontal target. Now lets build a platform straight up in the air above the shooters position. Lets build it so high that the shooter is now horizontal with that target up the hill. How far is it from the shooter's position to the target?
That's right, 70.7 yards. Now he's shooting at the same target, but shorter range. But the target is still 100 yards from the shooter on the ground. If they both shoot, they both hit the target in the same place. The horizontal distance to the target is the same for both shooters. Gravity effects both bullets exactly the same, but one shooter was shooting 100 yards, and the other was only shooting 70.7 yards.
Jim

Evan
12-22-2009, 05:09 PM
Even if you use bore-sighting, the same thing occurs. The bullet path is an ellipse, or curve, call it what you will.

No, it doesn't. The range is still 100 yards and the drop is still the same amount straight down . Even in zero gee with no drop at all the bullet won't be high since the bore is aimed directly at the target. Add in ANY drop and it will be low at any angle but 90. The ballistic curve starts dropping from the moment the bullet leaves the barrel. The bullet never rises at all compared to the LOS of the barrel, only compared to the sights.

This is what I have been getting at all along although I did neglect the issue of the sight compensation. Shooting high is only related to how the sights are set, nothing else. If you look at Philbur's math you will see that what he calculated is the difference between the horizontal sight setting and the proper setting for a 45 degree shot over a distance of almost 2/3rds mile. That of course isn't a realistic shot wih most weapons but that isn't what we are considering.

JoeCB
12-22-2009, 07:06 PM
Boy I hate to have to do this, but here I go again... Evan's last post did it. "bullet drop is only a function of time of flight" NOT TRUE ...it's also dependent on the upward force vector imparted by elevating the line of departure ( bore line). A rifle zeroed at 100 yds has no bullet drop measurable at the 100 yard target because the projectile was launched with some component of vertical force vector. So, X time of flight = zero bullet drop AT THE TARGET. After all, this entire discussion pertains ONLY to the question of point of impact at the target.
Joe B

philbur
12-22-2009, 07:22 PM
I ran several simulations with a lot of interations to obtain each result. For every shot there are two successful trajectories. the direct (rifle) shot and the lobbed (artillary) shot. The rifle shot is always high/high. The artillary shot has no meaning in the high/high Low/high discussion because the tageting metod is different, that is you don't use sights for targeting in an artillary shot.

Phil:)

.
Thanks for running the numbers. As long as you have the simulation set up, could you do another run, where you have to aim at something really far away, such that the barrel has to be pointed significantly higher than the line of sight to the object? Evan's picture of his ballistic simulations doesn't really show the "bullet flies high when aimed at high objects" effect, so I wonder if other effects start to come into play at these long ranges.

Evan
12-22-2009, 07:23 PM
Joe,

Sorry but that is incorrect. Anything in free fall does exactly that, fall. Any upward vector is merely compensating for the fall but it still accelerates downward by the exact same amount regardless. If you fire a bullet 1000 yards in one second and have the sights set to compensate for the drop over that distance the bullet would hit 16 feet high if there was no gravity. However, it drops 16 feet during that one second which completely cancels out the rise of 16 feet and puts it on target.

dp
12-22-2009, 07:25 PM
You shoot you're target at 100 yards horizontal. Now put a target up the hill at 100 yards at a 45* angle. Lets say the bullet hit right on the money on your horizontal target. Now lets build a platform straight up in the air above the shooters position. Lets build it so high that the shooter is now horizontal with that target up the hill. How far is it from the shooter's position to the target?
That's right, 70.7 yards. Now he's shooting at the same target, but shorter range. But the target is still 100 yards from the shooter on the ground. If they both shoot, they both hit the target in the same place. The horizontal distance to the target is the same for both shooters. Gravity effects both bullets exactly the same, but one shooter was shooting 100 yards, and the other was only shooting 70.7 yards.
Jim

Let's simplify your target placement. Pay out 100 yds of chain (the near end is anchored) and attach it to the target which is level with the shooter. Shoot the target. Now raise the target 45 degrees - it is magic chain that has no weight so don't let that distract you; this is science after all :)

Now the target is 100 yds away but elevated 45 degrees. Shoot it. How far does the bullet go? 100 yards. The flight time of the bullet is immeasurably close to the flight time of the shot at the horizontal target. The effect of gravity over time is for all practical purposes the identical for both shots.

JoeCB
12-22-2009, 07:42 PM
Evan ( #149) you are exactly correct in that statement, gravity is unrelenting. You however are missing the point of this discussion, please re-read ... it's all about POINT OF IMPACT on the target not about trajectory and inflight bullet drop etc etc. Again I refer you to the Seirra ballistics table... that tells the story.
Joe B

philbur
12-22-2009, 07:52 PM
Most of the current posts are getting bogged down in poor word usage. You guys could dance around this topic for another 500 posts if you are not careful.

For example the word "drop" seems to have different meaning to different people.

definition 1 - Drop = moving closer to the earth's surface.
definition 2 - Drop = moving further below the line of sight to the target.

As in any language if you don't have a common definition for a word you are going to be missunderstood.

Phil:)

Evan
12-22-2009, 08:35 PM
Evan ( #149) you are exactly correct in that statement, gravity is unrelenting. You however are missing the point of this discussion, please re-read ... it's all about POINT OF IMPACT on the target not about trajectory and inflight bullet drop etc etc. .

The bullet is falling the entire time it is in flight. That is why the condition is called "free fall". The end result after one second is that the trajectory of the bullet has dropped by 16 feet no matter where it was aimed. The word drop means exactly that, it's what happens when you let go of something and let it fall. If you lob a baseball to somebody it immediately begins falling the moment it leaves your hand. If there was no gravity then it wouldn't fall.

All objects in a ballistic trajectory are falling toward the Earth. A ballistic trajectory is a portion of an orbit. If it has a minor axis less than the radius of the earth it will impact the earth before completing an orbit.

Again I refer you to the Seirra ballistics table... that tells the story

Yes it does and everything I mention is incorporated in those tables. The difficulty is in separating the factors and considering them independently. Bullet drop and bullet rise are two factors that when added together result in the observed tracjectory. Each has a value that depends on separate factors but when combined to produce a single result such as point of impact those separate values are not obvious.

dp
12-22-2009, 08:48 PM
The bullet is falling the entire time it is in flight.

General info for nobody in particular (and ignoring atmospheric drag): This is true even for a projectile fired straight up. Its rate of ascent is declining at 1G until it stops, and then it accelerates downward at 1G. But it is considered to be falling even as it is rising because nothing is holding it up. Just as objects in orbit are falling and a baseball lofted straight up begins falling the instant it leaves your hand.

IdahoJim
12-22-2009, 09:01 PM
No, it doesn't. The range is still 100 yards and the drop is still the same amount straight down . Even in zero gee with no drop at all the bullet won't be high since the bore is aimed directly at the target. Add in ANY drop and it will be low at any angle but 90. The ballistic curve starts dropping from the moment the bullet leaves the barrel. The bullet never rises at all compared to the LOS of the barrel, only compared to the sights.

This is what I have been getting at all along although I did neglect the issue of the sight compensation. Shooting high is only related to how the sights are set, nothing else. If you look at Philbur's math you will see that what he calculated is the difference between the horizontal sight setting and the proper setting for a 45 degree shot over a distance of almost 2/3rds mile. That of course isn't a realistic shot wih most weapons but that isn't what we are considering.

You missed the point again. If you bore sight level. And shoot level. The bullet is higher at 70 yards than it is at 100 yards. Period. The bullet path curves downward from the force of gravity. So, if you bore sight uphill at a 45* angle the same 100 Yards, but the horizontal, or level distance is only 70.7 yards, the bullet impact point will be higher. There is no refuting that.
Jim

JoeCB
12-22-2009, 09:02 PM
We have gotten off track... the original question was

Does shooting from high ground ( ie. down hill) really make the shot go high?

The answer the gentleman was after is YES ...(assuming the same firearm, range, load, point of aim, etc)

Oh and coincedently shooting up hill also makes the shot go high. (given same assumptions)

Between there and page 16 we have had some very interisting and enlightening conversation... thanks to all that contributed, it was fun.

Joe B

Evan
12-22-2009, 09:30 PM
You missed the point again. If you bore sight level. And shoot level. The bullet is higher at 70 yards than it is at 100 yards. Period. The bullet path curves downward from the force of gravity. So, if you bore sight uphill at a 45* angle the same 100 Yards, but the horizontal, or level distance is only 70.7 yards, the bullet impact point will be higher. There is no refuting that.

Yes, it will be higher but not higher than the point of aim. Also, while your explanation for the difference does arrive at the right answer it does so for the wrong reasons. The difference is accounted for by the change in gravitational force vector because of the change in angle. It just happens that this change is also proportional to the change in X distance but that is incidental to the actual cause. What your explanation does not account for is where the energy is going. When the gravity force vector is not perpendicular to the path of the bullet it then either adds to or subtracts from the velocity of the bullet. A proper calculation of the ballistics will take this into account.

J Tiers
12-22-2009, 09:55 PM
Evan, refer to post 130, AMONG OTHERS.

Then forget about gravitational details for a moment and think.

JCHannum
12-22-2009, 10:17 PM
True bullet drop explained.

Evan
12-22-2009, 10:44 PM
From that site:

The true vertical bullet drop is the same for level fire and uphill or downhill shooting for the same range. See Figure 1. The vertical drop, do, is the same for all three methods of shooting over the same range.
Correct

The bullet velocity is the same whether shooting over a level range or shooting uphill or downhill. In other words the bullet does not slow down faster in uphill shooting than with level shooting and the bullet velocity does not increase when shooting downhill.
Incorrect, the bullet MUST lose some energy going up. There is no free ride. Likewise, it gains energy going down.

A rifle zeroed in at level range will shoot higher when shooting uphill or downhill.
Correct

For a given angle of fire the bullet will shoot high by the same amount weather shooting uphill or down hill.
Almost correct but not quite. The energy loss going up increases the drop slightly and vice versa going down

Jimno2506
12-22-2009, 10:48 PM
One way to think of it is this: As soon as the bullet leaves the muzzle it begins to fall (drop) away from the boreline. Consequently, when bullet arrives at target, it will impact at a point 1 below the point 2 where the boreline intersects the target. We call the distance between points 1 and 2: Drop.

To compensate for Drop so that the point of impact will correspond with the point of aim, the boreline is elevated by the unique angle that raises its point of intersection exactly 1 Drop above the point of aim.

The crucial point is that, because Drop is independent of inclination, elevation angle isn't.

That's because as the inclination angle gets steeper and steeper (whether up or down), the apparent separation of points 1 and 2 gets smaller and smaller, making elevation angle smaller and smaller. In other words, increasing inclination angle tends to make 1 and 2 merge, shrinking the required elevation angle along with the shrinking apparent separation of the points.

Short version: The greater the inclination the less the elevation.

http://i45.tinypic.com/308e7q1.jpg

authored by Steve in NC

Regards,
Jim

Jimno2506
12-22-2009, 10:59 PM
...are we disagreeing on?

Evan says:
A rifle zeroed in at level range will shoot higher when shooting uphill or downhill.
Correct

If we have succumbed to arguing the .015moa difference, I give up.

Regards,
Jim

dp
12-22-2009, 11:07 PM
From that site:

The bullet velocity is the same whether shooting over a level range or shooting uphill or downhill. In other words the bullet does not slow down faster in uphill shooting than with level shooting and the bullet velocity does not increase when shooting downhill.

Incorrect, the bullet MUST lose some energy going up. There is no free ride. Likewise, it gains energy going down.

This one is funny. Funny because any round that did not lose energy going uphill would by default have escape velocity when shot straight up from either pole - it would never slow down except by declining drag through the air, and would leave the earth never to be seen again.

ulav8r
12-22-2009, 11:17 PM
Not arguing with anyone but it could be stated that shooting up or down hill does not cause the bullet to hit higher. The lack of making the necessary sighting correction when shooting up or downhill makes the bullet hit higher than the aiming point.

dp
12-22-2009, 11:24 PM
Short version: The greater the inclination the less the elevation.
authored by Steve in NC

Regards,
Jim

In order for this to be the case you have to adjust the sights between shots. It was implied in the OP that the sights were not adjusted - the range did not change, only the elevation of the target relative to the shooter changed. If you don't adjust the sights and aim using those unadjusted sights pointed at the elevated target you will miss high.

We gotta all be talking about the same thing here for this to make sense.

dp
12-22-2009, 11:25 PM
Not arguing with anyone but it could be stated that shooting up or down hill does not cause the bullet to hit higher. The lack of making the necessary sighting correction when shooting up or downhill makes the bullet hit higher than the aiming point.

True, but that is not how the question was posed in the OP. And that has led to a great deal of confusion.

Jimno2506
12-22-2009, 11:41 PM
Short version: The greater the inclination the less the elevation.

In order for this to be the case you have to adjust the sights between shots. It was implied in the OP that the sights were not adjusted - the range did not change, only the elevation of the target relative to the shooter changed. If you don't adjust the sights and aim using those unadjusted sights pointed at the elevated target you will miss high.

We gotta all be talking about the same thing here for this to make sense.

What that means is the higher the inclination, the less elevation must be dialed into a sight to make the poa=poi. ie, the bullet drops less at the greater inclination. I agree with everything you said about the OP's question.

Jim

dp
12-23-2009, 01:30 AM
We can make it even simpler, conceptually.

We are given a fixed sight rifle. No matter what kind of sights. What matters is we don't adjust them in this test because we can't.

We fire the weapon and observe the range at which the projectile crosses the sight line the second time when shooting horizontally. We use our line of sight crossing detector - a miracle of engineering made entirely of unobtanium billet and hi-speed digital electronics with +- 1 angst. of accuracy. It records exactly 300 yds.

Now we raise our rifle +10º and fire again. Our LineOSightOmatic says the projectile has traveled farther before crossing the line than for the previous shot.

Repeat at +20 and the LOSOmatic shows that again, the projectile has traveled farther before crossing the LOS a second time than for the previous shot.

As we continue elevating our rifle we decide to throw some analysis at the problem and we determine that at some point our projectile will no longer cross the LOS a second time and that that occurs when our barrel (not our sights) is pointed straight up. Our sights are still pointed somewhat less than perpendicular.

The readers who are still awake probably realize I'm fudging a bit here, because there is a second crossing but only when the projectile falls straight down through the LOS at the same point it crossed it on the way up.

Then we repeat this test aiming incrementally lower and we find that again, the range to the second crossing increases as we increase the angle of fire. We conjecture then, that the shortest possible distance to the second crossing happens at exactly 0º. The one exception to the experience with the elevation raised vs lowered is that there actually is a point at which the projectile crosses the LOS once and that is when the barrel is pointed at -90º. Again, the sights will be at somewhat less than perpendicular because they have passed through -90º.

By now I think we know just about all that can be known about this subject. That can only mean it's time for a knurling debate! Pitch and ratio deniers on the left, plunge and take what you get deniers on the right - Begin!

JCHannum
12-23-2009, 07:29 AM
The site in the link gives the clearest explanation of any that I found. (BTW, I only looked at three or four, it is not that obscure.) While it is not correct to the finest point, there is a slight energy loss uphill, and commensurate gain downhill, the concept is the same, gravity affects only that portion of the projectile's flight which is in the horizontal plane, and point of impact is what it would be using a rifle sighted for the distance to target at the true horizontal distance.

This is the point that we have been trying to get across and why the point of impact is high regardless of inclination or declination no matter how hard some would try to prove otherwise.

Evan
12-23-2009, 07:46 AM
gravity affects only that portion of the projectile's flight which is in the horizontal plane,

Incorrect. Gravity affects all portions of a projectile's flight. If fired straight up it is gravity that stops the projectile from leaving the Earth.

camdigger
12-23-2009, 08:03 AM
Incorrect. Gravity affects all portions of a projectile's flight. If fired straight up it is gravity that stops the projectile from leaving the Earth.

Hmmm, unless gravity acts in all directions now, it should only act on the vertical component of the projectile's velocity.

Resolving the total velocity into vertical and horizontal components was how I was taught to solve ballistic problems in school.

The horizontal component is unaffected after leaving the muzzle with the exception of air resistance. The vertical component might have an initial component of muzzle velocity which is changed by the acceleration of gravity and air resistance. That's apparently what gives the projectile path it's parabolic shape. There are relatively minor affects from other stuff like gyroscopic stabilization from the spin of the rifling and wind drift effects and, and, and, that affect accuracy, but the bullet's path is basically the same second order type curve.

JCHannum
12-23-2009, 08:39 AM
Incorrect. Gravity affects all portions of a projectile's flight. If fired straight up it is gravity that stops the projectile from leaving the Earth.

Captain Pedantic strikes again, quoting out of context as usual. My statement is correct in the context of this discussion.

Evan
12-23-2009, 10:55 AM
What quote out of context? Your post is directly above mine.

Hmmm, unless gravity acts in all directions now, it should only act on the vertical component of the projectile's velocity.

It does act in all directions. The vector sum is toward the centre of the Earth.

If the Earth were a hollow sphere with a thin layer of densium plus a door so you could enter you would be weightless anywhere within the sphere.

The vertical component might have an initial component of muzzle velocity which is changed by the acceleration of gravity and air resistance. That's apparently what gives the projectile path it's parabolic shape.

The shape isn't precisely parabolic with or without air resistance figured in. It is a very small section of an elliptical orbit which approximates a parabola. For calculations of bullet trajectories a parabola is close enough. For calculation of theatre class ballistic missiles it isn't close enough for good accuracy and for calculation of ICBM flight paths it isn't applicable at all.

IdahoJim
12-23-2009, 11:02 AM
Yes, it will be higher but not higher than the point of aim. Also, while your explanation for the difference does arrive at the right answer it does so for the wrong reasons. The difference is accounted for by the change in gravitational force vector because of the change in angle. It just happens that this change is also proportional to the change in X distance but that is incidental to the actual cause. What your explanation does not account for is where the energy is going. When the gravity force vector is not perpendicular to the path of the bullet it then either adds to or subtracts from the velocity of the bullet. A proper calculation of the ballistics will take this into account.

Yeah...I've been trying to come up with a way to explain it that you would understand, since what you originally stated was that I was wrong when I said the bullet impact would be high when shooting either uphill, or downhill, relative to the impact point when shooting horizontally. You posted that the impact would be high when shooting downhill, and low when shooting uphill. I have yet to see a statement from you that changed that position, so that encouraged me to keep trying other ways of explanation to help you understand what's going on.
Also, the added, or subtracted, velocity change from the force of gravity when shooting downhill/uphill is so minor that it can be ignored for all practical purposes. The added/subtracted velocity is less than the normal standard muzzle velocity deviation for most rifle cartridges.
Jim

Evan
12-23-2009, 11:04 AM
I guess you missed the post way back when I allowed that Philbur was correct.

JCHannum
12-23-2009, 11:22 AM
[QUOTE=Evan]What quote out of context? Your post is directly above mine./QUOTE]

It is out of context in that you extracted only a few words from a much longer statement that is correct in the context of the ongoing discussion.

IdahoJim
12-23-2009, 01:19 PM
I guess you missed the post way back when I allowed that Philbur was correct.

If I did, I apologize.
Jim

winchman
12-23-2009, 02:57 PM
OK, I think I've figured out how to think about the question by considering three cases:

First consider the soldier aiming horizontally:

http://images.clipartof.com/small/5268-Male-Military-Union-Soldier-Aiming-Rifle-Clipart-Illustration.jpg

The bullet passes through the sight line, and then drops back to the sight line at the target distance because of the effect of gravity on the trajectory.

Second, consider the soldier aiming straight up. The bullet passes through the sight line, arcs over slightly behind the soldier until it runs out of momentum, and then falls behind him. The major effect of gravity is to slow the bullet down; the minor effect is to make it arc over behind the soldier. If the bullet reaches the target distance, it will be "high" with respect to the aim point.

Lastly, consider the soldier standing at the edge of a cliff and aiming the gun straight down. The bullet passes through the sight line, but never crosses it again. The major effect of gravity is to increase the speed of the bullet. The minor effect of gravity is to cause the bullet to arc toward the sight line, but it can never cross it. When the bullet reaches the target distance, it will be "high" with respect to the aim point.

The bullet goes high when the gun is aimed vertically up or down, and it hits the target when fired horizontally. Therefore, it MUST go high to a varying degree at any angle other than horizontal.

Roger

Yankee1
12-23-2009, 03:18 PM
Jc Hannum, Idaho Jim you both have been correct all along. Now I believe Evan is on board except for semantics. In earlier posts there was a question about my statement on the affect of gravity on projectile and I was told to STOP. The statement was about the AFFECT of gravity being less because as you said JC "the horizontal fliight is what is considered in computation of gravity's affect on the projectile" I know that may not be verbatim. I have to watch my semantics. This has been interesting seeing all the different conceptions of the problem.

IdahoJim
12-23-2009, 04:15 PM
Jc Hannum, Idaho Jim you both have been correct all along. Now I believe Evan is on board except for semantics. In earlier posts there was a question about my statement on the affect of gravity on projectile and I was told to STOP. The statement was about the AFFECT of gravity being less because as you said JC "the horizontal fliight is what is considered in computation of gravity's affect on the projectile" I know that may not be verbatim. I have to watch my semantics. This has been interesting seeing all the different conceptions of the problem.

many thanks. The thing we have to remember is that gravity's effect is always vertical to the earth. Any angle of bullet flight, other than horizontal, will not feel the full effects of gravity.
And I agree...it's been a very interesting thread, and of value to me as I learned a thing, or two.
Jim

12-23-2009, 05:12 PM
Here you go. Technology to the rescue.

http://www.brownells.com/.aspx/pid=19637/Product/COSINE_INDICATOR

dp
12-23-2009, 06:17 PM
Any angle of bullet flight, other than horizontal, will not feel the full effects of gravity.

Just when I thought the science was settled... :(

beanbag
12-23-2009, 06:27 PM
Just when I thought the science was settled... :(

append "in terms of flight path deflection". Yes it's settled already except for semantic issues.

SteveF
12-23-2009, 06:45 PM
Here you go. Technology to the rescue.

http://www.brownells.com/.aspx/pid=19637/Product/COSINE_INDICATOR

Too complicated. Use this.

http://www.chuckhawks.com/slope_doper.htm

Damn, you guys got to almost 200 posts arguing about something that they make simple tools for?

Steve.

Evan
12-23-2009, 06:50 PM
append "in terms of flight path deflection". Yes it's settled already except for semantic issues.

Not quite. It isn't a semantic issue. Gravity affects the flight path in full no matter what angle it is fired at. It always produces the exact same amount of drop toward the Earth in a given period of time. How that affects the flight path depends on how the vectors add up. It isn't a small issue as it entirely controls where the bullet ends up. If fired straight up it is what makes the bullet fall back down. That is a pretty significant effect.

beanbag
12-23-2009, 07:11 PM
Not quite. It isn't a semantic issue. Gravity affects the flight path in full no matter what angle it is fired at. It always produces the exact same amount of drop toward the Earth in a given period of time. How that affects the flight path depends on how the vectors add up. It isn't a small issue as it entirely controls where the bullet ends up. If fired straight up it is what makes the bullet fall back down. That is a pretty significant effect.

yes it is a semantic issue which leads to all these misunderstandings. "path" means the line/curve the bullet traces out, and doesn't take into account the speed. If I took your quote and replaced all instances of "path" with "path and speed", then I would agree with everything.
As has been shown earlier, bullets fired at elevated angles (at relatively close targets) have less bend in the path.

What you and dp are taking issue with is when somebody uses some variant of "not feeling the full effects of gravity". What they means is that gravity is not perpendicular to the path, therefor not causing maximal deflection.

IdahoJim
12-23-2009, 07:16 PM
Not quite. It isn't a semantic issue. Gravity affects the flight path in full no matter what angle it is fired at. It always produces the exact same amount of drop toward the Earth in a given period of time. How that affects the flight path depends on how the vectors add up. It isn't a small issue as it entirely controls where the bullet ends up. If fired straight up it is what makes the bullet fall back down. That is a pretty significant effect.

Yup...the exact same amount of drop towards the earth, but not the same amount of drop relative to the flight path.

And, as I was having dinner, it dawned on me my last statement was incorrect, or incomplete as stated....LOL. It's easy to do when talking about technical things. I think I'll just leave it though, as an example.
Jim

IdahoJim
12-23-2009, 07:18 PM
What you and dp are taking issue with is when somebody uses some variant of "not feeling the full effects of gravity". What they means is that gravity is not perpendicular to the path, therefor not causing maximal deflection.

yup...that's an elegant way to say that, too. Wish I had.
Jim

lazlo
12-23-2009, 08:27 PM
What you and dp are taking issue with is when somebody uses some variant of "not feeling the full effects of gravity". What they means is that gravity is not perpendicular to the path, therefor not causing maximal deflection.

Very well said. Gravity affects the bullet on its horizontal travel (the vector perpendicular to gravity). So if you're shooting up or down by X°, the horizontal distance is Cosine X * Sighted Distance, (where Cosine X is less than 1.0).

dp
12-23-2009, 09:31 PM
Very well said. Gravity affects the bullet on its horizontal travel (the vector perpendicular to gravity). So if you're shooting up or down by X°, the horizontal distance is Cosine X * Sighted Distance, (where Cosine X is less than 1.0).

Possible for some combinations of firearm, rifling, bullet profile, muzzle velocity, and range at which the measurement is made. Consider the vast variables between jacketed boat tail and ball ammo, for example.

lazlo
12-23-2009, 09:40 PM
Possible for some combinations of firearm, rifling, bullet profile, muzzle velocity, and range at which the measurement is made. Consider the vast variables between jacketed boat tail and ball ammo, for example.

None of that matters. All that matters is the horizontal distance the bullet travels.

So if you're taking a 1,000 yard shot, 40° uphill, the actual horizontal distance the bullet travels is Cosine 40° * 1,000 = 766 yards. So gravity is acting on the the bullet for 766 yards. So you look up your ballistics tables for the bullet drop for your boat tail match round for 766 yards, and you hit the target.

In other words, the sighted distance (1,000 yards) the horizontal distance (766 yards) and the difference between the two comprise a right triangle.

Evan
12-23-2009, 09:58 PM
So if you're taking a 1,000 yard shot, 40° uphill, the actual horizontal distance the bullet travels is Cosine 40° * 1,000 = 766 yards. So gravity is acting on the the bullet for 766 yards. In other words, the sighted distance (1,000 yards) the horizontal distance (766 yards) and the change in bullet drop comprise a right triangle.

You are repeating the same mistaken idea that others have. The bullet time of flight to the target is almost the same (*slightly longer*) so the effect of gravity on the altitude of the bullet is exactly the same. The altitude of a bullet that travels for 1 second is reduced by 16 feet approx (compared to no gravity at all) no matter what direction it is traveling.

JCHannum
12-23-2009, 10:11 PM
Possible for some combinations of firearm, rifling, bullet profile, muzzle velocity, and range at which the measurement is made. Consider the vast variables between jacketed boat tail and ball ammo, for example.

The trajectory curve will change from load to load and bullet design to bullet design. The trajectory is based on the projectile's velocity and how quickly it sheds that velocity. There are other variables of course, but for a given firearm and load, the offset will be the same from shot to shot as all other factors will cancel out and the cosine calculation will be valid.

beanbag
12-23-2009, 10:39 PM
You are repeating the same mistaken idea that others have. The bullet time of flight to the target is almost the same (*slightly longer*) so the affect of gravity on the altitude of the bullet is exactly the same. The altitude of a bullet that travels for 1 second is reduced by 16 feet approx (compared to no gravity at all) no matter what direction it is traveling.

Lazlo is right, but he said it in a way that maximizes your chance of misinterpreting what he said.

All of gravity acts on the bullet for the whole time of 1.000x seconds. However, it acts on it at a 50 deg angle. This is equivalent to acting on it for "only" 766 yards. (more or less)
The drop of the bullet is 16 feet as you said relative to the center of the earth.
However, it is less than 16 feet as measured perpendicularly to the line of the barrel, which is the relevant distance as viewed from your scope.
You use the 766 yard mark on your scope to aim correctly.
This was stated waaaay early in the thread, that you use the horizontal distance to the target as the compensation to use on the scope.

Yankee1
12-23-2009, 10:46 PM
Evan
The bullets path is no longer perpendicular to the force exerted on it by gravity. You are ignoring the fact that the cosine of the angles length is used to compute the affect of gravity on a projectile that is angled either upward or downward. Why? Others including me have stated that only the horizontal bullet path distance is considered. To calculate the horizontal bullet path distance we must multiply the cosine of the angle by the total distance to the target.. If we were incorrect then all the ballistics manuals and books are in need of correction. You know that is not so.

dp
12-23-2009, 11:17 PM
None of that matters. All that matters is the horizontal distance the bullet travels.

So if you're taking a 1,000 yard shot, 40° uphill, the actual horizontal distance the bullet travels is Cosine 40° * 1,000 = 766 yards. So gravity is acting on the the bullet for 766 yards. So you look up your ballistics tables for the bullet drop for your boat tail match round for 766 yards, and you hit the target.

In other words, the sighted distance (1,000 yards) the horizontal distance (766 yards) and the difference between the two comprise a right triangle.

Let me know how that works for you when shooting a paintball gun. There's no general case where the math works. You have to consider the context.

Evan
12-23-2009, 11:32 PM
Evan
The bullets path is no longer perpendicular to the force exerted on it by gravity. You are ignoring the fact that the cosine of the angles length is used to compute the affect of gravity on a projectile that is angled either upward or downward. Why? Others including me have stated that only the horizontal bullet path distance is considered. To calculate the horizontal bullet path distance we must multiply the cosine of the angle by the total distance to the target.. If we were incorrect then all the ballistics manuals and books are in need of correction. You know that is not so.

You must be joking, yes? It doesn't matter what velocity or what direction anything is traveling, if it isn't supported by something then it is accelerated by gravity equally for all objects of all masses toward the center of the Earth. The acceleration is independent of anything else the object may be doing and there is no way to prevent it, shield it or cancel it. The object falls (neglecting air resistance) 16 feet in the first second from the time it is released from support. That 16 feet is subtracted from it's altitude at the end of one second unless it is stopped by something such as the ground. Galileo proved it in the 17th century and it hasn't changed.

IdahoJim
12-23-2009, 11:59 PM
None of that matters. All that matters is the horizontal distance the bullet travels.

So if you're taking a 1,000 yard shot, 40° uphill, the actual horizontal distance the bullet travels is Cosine 40° * 1,000 = 766 yards. So gravity is acting on the the bullet for 766 yards. So you look up your ballistics tables for the bullet drop for your boat tail match round for 766 yards, and you hit the target.

In other words, the sighted distance (1,000 yards) the horizontal distance (766 yards) and the difference between the two comprise a right triangle.

You got it Lazlo...Ain't it simple? It's the effective horizontal distance that matters. If I can do this here's Sierra's figure of it:
http://www.concreteraising.com/uphilldownhill shooting.jpg

http://www.concreteraising.com/uphilldownhill shooting.jpg

dp
12-24-2009, 12:03 AM
That 16 feet is subtracted from it's altitude at the end of one second unless it is stopped by something such as the ground. Galileo proved it in the 17th century and it hasn't changed.

Which also illustrates why a general case for ballistics cannot be made without a context. Galileo's experiment required similar objects in regards to atmospheric drag. Had he dropped a golf ball and a feather duster he'd have arrived at a totally different result.

When drag through the air is a large component of the forces acting on a ballistic object it will have a non-linear effect on the object's curve through space. A feather shot from a cannon will quickly reach a point in its flight where it falls straight to the earth.

From an unlikely source: http://www.worldalmanacforkids.com/WAKI-ViewArticle.aspx?pin=x-ba016800a&article_id=477&chapter_id=9&chapter_title=Numbers&article_title=Ballistics

Exterior Ballistics
In exterior ballistics, elements such as shape, caliber, weight, initial velocities, rotation, air resistance, and gravity help determine the path of a projectile from the time it leaves the gun until it reaches the target.

Yankee1
12-24-2009, 12:16 AM
Evan
Unfortunately no. Try this. Draw a vertical line and from the base of that line at a right angle draw a horizontal line that represents the path of the bullet to the target. Now draw another line of the same length as the line that represents the path of the bullet starting from the base of the vertical line up at 45 degrees. We now have two lines starting at the base of the vertical line. One horizontal and one at 45 degrees.
If the vertical line is the start time you will notice how far away the end of the horizontal line is when compared to the end of the 45 degree line from the vertical line which represents the start time. I think you will find the distance of the 45 degree lines end from the vertical will represent .707 of the horizontal line length.

Paul Alciatore
12-24-2009, 12:21 AM
You've made it too simple - and wrong. Your model does not allow for +-90º.

Of course it does. If you aim the same amount "above" the target IN YOUR SITES when shooting straight up or down, you will miss by just about the amount of that aiming offset because the bullet will go just about straight (up or down). The thing is, "above the target" becomes forward of or rearward of the target due to the vertical angle.

"Above the target" becomes toward the top of the shooter's head when he takes aim.

If the shooter were to aim at a point that was a given distance (the correct distance for the time of flight of the bullet) directly above the target, he would hit the target in all cases. But this distance would appear to him to be less and less as he changes from horizontal to vertical (up or down). On a vettical shot (your +/- 90 degrees) this point that is a distance above the target would be directly in line with the target.

Paul Alciatore
12-24-2009, 12:31 AM
True bullet drop explained.

This site has EXACTLY the same diagram and explanation that I stated about one/two pages ago. It is simply a difference between the angle of the offset in the sites and the angle at which the bullet drops. That's all. Period.

Paul Alciatore
12-24-2009, 12:33 AM
Not arguing with anyone but it could be stated that shooting up or down hill does not cause the bullet to hit higher. The lack of making the necessary sighting correction when shooting up or downhill makes the bullet hit higher than the aiming point.

Well stated and exactly correct.

Paul Alciatore
12-24-2009, 12:38 AM
You got it Lazlo...Ain't it simple? It's the effective horizontal distance that matters. If I can do this here's Sierra's figure of it:
http://www.concreteraising.com/uphilldownhill shooting.jpg

http://www.concreteraising.com/uphilldownhill shooting.jpg

Once again, that is exactly the drawing I posted about two pages ago. Plain and simple. Just a difference in angular viewpoints.

Study this drawing, it does explain the situation correctly.

dp
12-24-2009, 12:55 AM
Of course it does. If you aim the same amount "above" the target IN YOUR SITES when shooting straight up or down, you will miss by just about the amount of that aiming offset because the bullet will go just about straight (up or down). The thing is, "above the target" becomes forward of or rearward of the target due to the vertical angle.

"Above the target" becomes toward the top of the shooter's head when he takes aim.

If the shooter were to aim at a point that was a given distance (the correct distance for the time of flight of the bullet) directly above the target, he would hit the target in all cases. But this distance would appear to him to be less and less as he changes from horizontal to vertical (up or down). On a vettical shot (your +/- 90 degrees) this point that is a distance above the target would be directly in line with the target.

There are three lines in the ballistic chart and yours is missing one. The line of delivery is where the barrel is pointing and is the upper line in your diagram, the line of sight is where you are looking but is missing in your diagram, and the trajectory is where the bullet goes as shown in the lower line. Your description seems to suggest the shooter is actively compensating for elevation. The OP, in order to make any sense, disallows that. If the shooter were allowed to adjust for elevation we'd be talking about machines and projects now :)

In the OP the presumption is that the target is in the line of sight frame and the sights are fixed. The line of delivery is a fixed offset to the line of sight regardless of the angle of fire. If you aim straight up at the target you will miss high. If you shoot straight up you will hit the target but there is no 1" drop that you say exists for all angles. In fact that 1" drop fades to zero as the elevation rises.

Presumably the bullet will hit the target and carry on through (we're shooting paper so as to be civil). Your description describes what happens well enough to a point but because it fails for all conditions it does not define it. It is like describing the violent shaking of trees as the source of wind.

philbur
12-24-2009, 06:57 AM
I don't think it clear that it explains the situation correctly as there are two complicating issues.

1) The average velocity in each case is not the same therefore the time to target is not the same and "drop" is proportional to g.t^2/2.
2) Time to target also depends on the definition of distance to target. The definition in the example is the point of closest approach of the projectile to the target, which is not the same in each case, even though the target is at the same linear distance.

These two issues may cancel each other and a geometrical analysis may be correct, however it is not clear that this is so.

Phil:)

Once again, that is exactly the drawing I posted about two pages ago. Plain and simple. Just a difference in angular viewpoints.

Study this drawing, it does explain the situation correctly.

JCHannum
12-24-2009, 07:47 AM
A simple demonstration of the effect of gravity on the angle from horizontal is to imagine a very long, slender rod such as a limber fly rod. Hold it at horizontal and it will droop a certain amount. Raise it or lower it to the vertical and the droop will decrease until it is at 90* when it will become straight. The same thing happens to the projectile's flight.

lazlo
12-24-2009, 11:00 AM
Lazlo is right, but he said it in a way that maximizes your chance of misinterpreting what he said.

LOL! That's one of the nicest insults I've ever gotten Beanbag :p

Trig-related discussions like these are much easier in front of a whiteboard. IdahoJim has a good picture of it, but it would be easier to follow if they hadn't stretched the right triangle, and reflecting it to show the symmetry of inclined and declined shots makes it more confusing.

The right triangle starts at the rifle, they hypotenuse is the sighted distance, and the base is the horizontal distance traveled.

IdahoJim
12-24-2009, 11:06 AM
I should have looked at your before posting my Sierra chart. That's usually the case with me...\$2.00 short, and a day late...LOL
Jim

ulav8r
12-24-2009, 06:02 PM
I'm afraid Evan may be closer to correct than most will admit. On a 1000 yard flight that has a horizontal component of 760 yards, the gravitational affect will be more than on a 760 yard horizontal flight due to the longer flight time for the angled shot. Rather than using a 760 yard setting, it might should be closer to a 770 yard setting. For 99% of shooters, the difference will not be measurable. To get exact answers for any given bullet at a given velocity would require much more math capability than I have, and/or extensive firing results to exhaustively determine the exact differences from the accepted norm. We are all wasting too much time on small differences that most cannot determine well enough to prove to everyone else.:(

For most shooting, using the horizontal component for sight setting is completely suitable.