A friend of mine and I have a disagreement about this - If you zero your rifle on level ground at 100 yards and then shoot uphill or downhill the same distance 100 yards my friend says you will shoot high uphill and low downhill. I thought I read that you would shoot high both ways. Does anyone know which is right and why?
uphill - downhill shooting
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ShaiVong
Member
The bullets will drop less shooting uphill or downhill. No difference.
goalie
Member
Gravity affects the bullet the same, it is just that when shooting uphill or downhill the bullet is not actually going as far. Think about a slope of 45 degrees. If you measure out 100 yards up the slope then tunneled straight down into the hill until you were actually at the same elevation as the shot was taken you would be less than 100 yards from where the shot was taken. Gravity pulls straight down, not on the slope. Gravity is a constant. Therefore a 100 yard shot up or downhill may be the equivilant of a 70 yard shot (or even less if it is a really steep slope) because that is the true horizontal distance that the bullet traveled.
It will shoot high in both cases.
Here's why.
When you shoot a level shot, the bullet actually leaves the barrel in a slightly upward direction. It needs to be going up a bit so that it can make it to the target. If you had the barrel perfectly level, the bullet would start falling from the instant it left the barrel and you would never be able to hit anything level with your barrel unless it was just a few feet in front of you.
So, what happens is, we use a small amount of the muzzle velocity to arc the bullet trajectory to counteract the effects of gravity.
In other words, even when you make a "level" shot, the barrel is actually pointed slightly upwards to get the arc.
When you shoot uphill, now you're pointing the barrel even more upward which means MORE of an arc. More of the muzzle velocity is counteracting gravity and the bullet will hit higher.
What about downhill?
Not quite as easy to see, but it's still all about the small amount of arc that's always there that we don't think about.
Picture this. Let's shoot almost straight downward. Where will the bullet hit?
Due to the slight upward angle of the barrel with relation to the sights (which is there to fight gravity when shooting on the level), the bullet will always go upwards somewhat--In relation to the sights. In this case we're pointing downwards so the bullet never actually gets going upwards from the level, only upward in relation to the sight. But since we're pointing almost straight down, gravity isn't fighting the arc (that little bit of upward angle to the barrel and bullet trajectory) as much as it normally would. That means the bullet will hit higher than we expect it to.
Here's why.
When you shoot a level shot, the bullet actually leaves the barrel in a slightly upward direction. It needs to be going up a bit so that it can make it to the target. If you had the barrel perfectly level, the bullet would start falling from the instant it left the barrel and you would never be able to hit anything level with your barrel unless it was just a few feet in front of you.
So, what happens is, we use a small amount of the muzzle velocity to arc the bullet trajectory to counteract the effects of gravity.
In other words, even when you make a "level" shot, the barrel is actually pointed slightly upwards to get the arc.
When you shoot uphill, now you're pointing the barrel even more upward which means MORE of an arc. More of the muzzle velocity is counteracting gravity and the bullet will hit higher.
What about downhill?
Not quite as easy to see, but it's still all about the small amount of arc that's always there that we don't think about.
Picture this. Let's shoot almost straight downward. Where will the bullet hit?
Due to the slight upward angle of the barrel with relation to the sights (which is there to fight gravity when shooting on the level), the bullet will always go upwards somewhat--In relation to the sights. In this case we're pointing downwards so the bullet never actually gets going upwards from the level, only upward in relation to the sight. But since we're pointing almost straight down, gravity isn't fighting the arc (that little bit of upward angle to the barrel and bullet trajectory) as much as it normally would. That means the bullet will hit higher than we expect it to.
Chuck Dye
Member
Not exactly!
I have always accepted the "no diff, uphill or downhill." When I began playing with the idea, I decided that gravity will actually improve performance on a downhill shot compared to an uphill shot in that the rate at which the downhill shot loses velocity will be less than the rate at which the uphill shot loses velocity. Sooo, I fired up the Oehler Ballistic Explorer, selected my .30-06’s favorite round from the ammo library, set standard atmospheric conditions, set a 200 yard zero on a 0 degree slope and then, maintaining the same sight settings, set one trace to +45 degrees, the other to -45 degrees. Once the calculations were done, I checked the traces expecting to see the uphill shot hit slightly lower than the downhill shot. The results will not alter my point of aim on anything I hunt, but they are there, as expected.
range 200 250 300
level 0.00 -2.85 -7.19
+45° 2.69 1.44 -0.86
-45° 2.70 1.48 -0.79
range 350 400 450
level -13.13 -20.78 -30.29
+45° -4.29 -8.95 -14.94
-45° -4.19 -8.79 -14.68
I have always accepted the "no diff, uphill or downhill." When I began playing with the idea, I decided that gravity will actually improve performance on a downhill shot compared to an uphill shot in that the rate at which the downhill shot loses velocity will be less than the rate at which the uphill shot loses velocity. Sooo, I fired up the Oehler Ballistic Explorer, selected my .30-06’s favorite round from the ammo library, set standard atmospheric conditions, set a 200 yard zero on a 0 degree slope and then, maintaining the same sight settings, set one trace to +45 degrees, the other to -45 degrees. Once the calculations were done, I checked the traces expecting to see the uphill shot hit slightly lower than the downhill shot. The results will not alter my point of aim on anything I hunt, but they are there, as expected.
range 200 250 300
level 0.00 -2.85 -7.19
+45° 2.69 1.44 -0.86
-45° 2.70 1.48 -0.79
range 350 400 450
level -13.13 -20.78 -30.29
+45° -4.29 -8.95 -14.94
-45° -4.19 -8.79 -14.68
Last edited:
Art Eatman
Moderator In Memoriam
Goalie is correct as to "why".
The effect is trivial for angles of less than 30 to 35 degrees, and for hunting shots within 300 yards the uphill/downhill aspect can be ignored. (At least that's the implications from looking at Mr. Sierra's handbook.)
Art
The effect is trivial for angles of less than 30 to 35 degrees, and for hunting shots within 300 yards the uphill/downhill aspect can be ignored. (At least that's the implications from looking at Mr. Sierra's handbook.)
Art
ShaiVong
Member
Huck: If you shoot uphill the x,y vectors of F(g) will retard the forward velocity of the bullet, causing it to slow faster. If you shoot downhill the F(g) will work in concert with the bullets velocity causing it to accelerate... But 9.81m/s^2 is so trivial compaired to the bullets velocity, and the shorter length from a high theta that it is usually ignored. Maybe if you lobbed a slow bullet over 800 yards it would make enough of a difference to be worth consideration.. But doesnt physics rest on the idea of ignoring everything you dont want to think about? 
goalie
Member
One of the people shown wearing the Ghillie suit in that book used to own a company that trained police here in Minnesota. IIRC, he was a little pissed off that a few pictures were misrepresented as pictures of Plaster when they were actually of him. It is a good book though.
Hkmp5sd
Member
The angle of the dangle....
The ABC's of Ballistics: Why You Hold Low To Hit Your Target When Shooting Uphill or Downhill
Thus far in our discussion, we have explored the antics of a projectile launched horizontally where, basically, the only forces in evidence are the imparted muzzle velocity and gravity, but what happens when an angle from the horizontal is injected into the equation?
Fig 2 illustrates just such a scenario.
The projectile is launched from a bow or rifle towards a target. In this example, I have chosen an angle of 45 degrees, simply because the math is easier. The actual adjustments will vary depending on the angle of the shot, either uphill or down.
In this scenario, the shooter is shooting downhill at an angle of 45 degrees, called the angle of declination. the range along the slope, as measured in one's rangefinder, is 50 yds. When illustrated thus, by simple triangulation, dropping a vertical line from the launch point to intersect the horizontal line extended from the target point, the horizontal distance is determined by multiplying the (Trigonometry function) Sine of the angle 45 deg (.707) times the Slope distance (50 yds). The product of this action shows us that the horizontal distance is 35.35 yards, or very close to 35 yds 1 ft.
In order to hit our target as thusly illustrated, one would need to hold as if he were shooting 35 yds, not 50 yds. Or, on your bow, use your 35 yard pin, not your 50 yard pin.
This ratio holds true no matter what the slope distance is. If you were antelope hunting and had measured the distance down the hill to be 600 yds, you would hold as if you were shooting 425 yds (.707X600).
As this angle of declination changes, the horizontal distance will vary as well, but it is always the Sine of the angle X the slope distance. For example, if one were shooting downhill at a very steep angle of 30 degrees, the horizontal distance would be 1/2 the slope distance, or 25 yds. (Sine 30 deg = .5) Conversely, if one were shooting at a shallower angle of 60 deg, the horizontal distance would be 37.5 yards (Sine 60 deg = .75).
As one can see from this discussion, it is very important to practice shooting from various ranges, and various angles to determine for oneself where your projectile is going to hit, relative to the slope distance. If one fails to take this factor into consideration, you will overshoot your target every time.
Now, understanding the effects of shooting downhill, how does one compensate for shooting at these same angles uphill? Figure 3 is an illustration of this effect. As one can readily see, the horizontal distance when shooting uphill is precisely the same as it is when shooting downhill at the same angle. Hence, we would adjust our point of aim in exactly the same manner as previously described. I know that this flies in the face of what we'd believe to be true, but, nonetheless, nothing has changed. The horizontal distance is still 35 yds as it was before, and it is over this 35 yards that gravity acts on the projectile. As Gravity is the ONLY force pulling the projectile to the ground, and it remains constant, the actual range remains constant.
The ABC's of Ballistics: Why You Hold Low To Hit Your Target When Shooting Uphill or Downhill
45crittergitter
Member
wrong wrong wrong
As an engineer, perhaps I can help clear this up.
First, the effect of gravity pulling back (down) on an upward-fired bullet or forward (down) on a downward-fired bullet is truly negligible, for all practical purposes.
Second, the deal about the horizontal distance is false. Horizontal distance makes no difference whatsoever unless the shot is horizontal. However, the geometry makes that fallacy work out correctly mathematically, which is probably why it is perpetuated. What does make a difference is the actual distance of the shot. 50 yards to the target, for instance, is 50 yards, regardless of the angle of the shot. (Vertical) Drop is totally dependent upon time of flight (and gravity). TOF is dependent upon distance (shot distance, NOT horizontal distance), ballistic coefficient, and velocity. In other words, the TOF, and therefore the (vertical) drop, is the same for a given load, regardless of the angle of inclination (see paragraph #1). The reason that the POI is higher on angled shots is that the DIRECTION of gravity (or drop) is no longer perpendicular to the line of flight, therefore, the pull of gravity on the bullet is proportionately less relative to the line of sight because gravity is not acting in a perpendicular direction to the line of sight. The actual vertical drop is the same, but it looks like less on the target.
As an engineer, perhaps I can help clear this up.First, the effect of gravity pulling back (down) on an upward-fired bullet or forward (down) on a downward-fired bullet is truly negligible, for all practical purposes.
Second, the deal about the horizontal distance is false. Horizontal distance makes no difference whatsoever unless the shot is horizontal. However, the geometry makes that fallacy work out correctly mathematically, which is probably why it is perpetuated. What does make a difference is the actual distance of the shot. 50 yards to the target, for instance, is 50 yards, regardless of the angle of the shot. (Vertical) Drop is totally dependent upon time of flight (and gravity). TOF is dependent upon distance (shot distance, NOT horizontal distance), ballistic coefficient, and velocity. In other words, the TOF, and therefore the (vertical) drop, is the same for a given load, regardless of the angle of inclination (see paragraph #1). The reason that the POI is higher on angled shots is that the DIRECTION of gravity (or drop) is no longer perpendicular to the line of flight, therefore, the pull of gravity on the bullet is proportionately less relative to the line of sight because gravity is not acting in a perpendicular direction to the line of sight. The actual vertical drop is the same, but it looks like less on the target.
JohnBT
Member
I'm having trouble visualizing how TOF and actual shot distance relates to bullet drop in a shot directly overhead(such as shooting up a tree or cliff.) You still have to aim low, right? Horizontal distance to target equals zero and TOF wouldn't seem to matter.
John
John
30Cal
Member
You need to start considering uphill/downhill when you are talking about 30+ degrees. You'll need to start holding low.
The bullet will drop most when aimed horizontally. This is the case where gravity is pulling the bullet the most away from the line of sight. As you get farther from that situation, the affect of gravity pulling away from the line of sight becomes less and less. When you reach vertical, the bullet will come up and cross the line of sight on it's way out and will continue to diverge instead of falling back down to the line of sight.
The bullet will drop most when aimed horizontally. This is the case where gravity is pulling the bullet the most away from the line of sight. As you get farther from that situation, the affect of gravity pulling away from the line of sight becomes less and less. When you reach vertical, the bullet will come up and cross the line of sight on it's way out and will continue to diverge instead of falling back down to the line of sight.
P95Carry
Moderator Emeritus
I will go with 45crittergitter on this .. I am also an engineer tho from a math POV probably gettin rusty.
However .... I agree - it is still only actual shot distance that counts. The parabola described with a horizontal shot is close to being ''normal'' to the acceleration due to G ... (at 90º) and thus G exerts max effect ...... once the trajectory deviates from normal then the parabola will flatten (marginally) ... simply because the vector represented by G force is producing a resultant at a shallower angle.
As said ... at shorter ranges and with high vel rifle rounds, the effect can be all but ignored but .. at long ranges should be taken into consideration much as we would windage. Shots will tend just high.
However .... I agree - it is still only actual shot distance that counts. The parabola described with a horizontal shot is close to being ''normal'' to the acceleration due to G ... (at 90º) and thus G exerts max effect ...... once the trajectory deviates from normal then the parabola will flatten (marginally) ... simply because the vector represented by G force is producing a resultant at a shallower angle.
As said ... at shorter ranges and with high vel rifle rounds, the effect can be all but ignored but .. at long ranges should be taken into consideration much as we would windage. Shots will tend just high.
HKmp5sd's graphic shows it well.
And when you start talking about shots of 300 or more yards, you have to figure your elevation on the horizontal range but your windage on the slope range.
We did shooting from a three-story tower at the Badlands Tactical Advanced Phase I course.
Even with three stories, only the shots up close to the tower and at a very steep angle showed enough difference in POI to matter.
From the top of the three-story tower, the 400 yard target was at an angle of only about 15 degrees....so it was put on your 400 yard dope and zap it.
The 25 yard target might have been at an angle something like 60 degrees from the top of the tower, and thus required more compensation.
Just to help visualization...the 400 yard target was a "crazy Ivan" half silhouette---chuck of steel about 18 inches wide by 18 20 inches tall.
The 25 yard target was a sheet of paper with a small dot about .30 inches in diameter drawn on it with a magic marker.
hillbilly
And when you start talking about shots of 300 or more yards, you have to figure your elevation on the horizontal range but your windage on the slope range.
We did shooting from a three-story tower at the Badlands Tactical Advanced Phase I course.
Even with three stories, only the shots up close to the tower and at a very steep angle showed enough difference in POI to matter.
From the top of the three-story tower, the 400 yard target was at an angle of only about 15 degrees....so it was put on your 400 yard dope and zap it.
The 25 yard target might have been at an angle something like 60 degrees from the top of the tower, and thus required more compensation.
Just to help visualization...the 400 yard target was a "crazy Ivan" half silhouette---chuck of steel about 18 inches wide by 18 20 inches tall.
The 25 yard target was a sheet of paper with a small dot about .30 inches in diameter drawn on it with a magic marker.
hillbilly
45 is right. It is more than just using simple geometery.
In the theory that Hkmp posted, where do you aim when you shoot (flight path of the projectile) 90° up or down at a target 50 yds away? At the 0 pin? You will surely miss the target. The projectile will have a laser straight path and you would be aiming too low.
In the theory that Hkmp posted, where do you aim when you shoot (flight path of the projectile) 90° up or down at a target 50 yds away? At the 0 pin? You will surely miss the target. The projectile will have a laser straight path and you would be aiming too low.
Art Eatman
Moderator In Memoriam
Sorry, but HKmp5sd is another fella who agrees with Mr. Sierra's handbook.
What kinda injunears we turning out, these days?
P.E. registration 1968,
Art
What kinda injunears we turning out, these days?
P.E. registration 1968,
Art
It's not too hard to come up with easy formulas that give answers that are close enough to work practically.
That doesn't mean that they explain the physical phenomenon behind the problem.
In other words, just because the Sierra formula gives you the right answer (or close enough so that it doesn't make a difference) doesn't mean that their little picture is telling the whole story about WHY it gives you the right answer.
That doesn't mean that they explain the physical phenomenon behind the problem.
In other words, just because the Sierra formula gives you the right answer (or close enough so that it doesn't make a difference) doesn't mean that their little picture is telling the whole story about WHY it gives you the right answer.
P95Carry
Moderator Emeritus
I don't find the Sierra explanation ideal.... tho it is trying to explain in one sense ''along the right lines''. From a sighting POV then the shot might ''seem'' as if the shorter distance (35) and could strike high ... in that respect I can see its approach.
My earlier mention should have one thing made very clear ... all forces in this analysis are based on the long axis of a bullet ..... it's flight path. We might imagine the bullet having a small hole thru the middle and travelling along a parabolic thin wire... and 'seeing'' the ''pull'' of gravity. That is under normal ''horizontal'' flight acting as I said ... ''normal'' to the trajectory and so exerts max effect.
Leaving math etc out of it again for now ... try this small experiment ... anyone. Take a long and thin but springy length of wire, rod etc (say 6 feet long) .. . and place a substantial weight on the end. Now, with the weight on the ground and rod vertical .... pick up the free end and move the whole thing thru 180º until the weight is now at the top.
At low and high positions the rod is effectively straight. But thoughout the 180º, notice that the bend in the rod increases to max at mid point ... 90º ... when gravity is exerting max ''pull'' ... it is ''normal'' to the rod's axis. As you progress thru towards top position the bend is once more decreasing. The effect is of course most obvious thru the first and last 15º let's say ... and either side of horizontal there does not appear so much obvious change.
So, whilst the effect is less obvious either side of horizontal it is still there .... and the ''bend'' or parabola - or trajectory curve .. is slightly flattened/reduced ... so for a sight setting done for level shooting ... (the maximized parabola for a given distance) .. this will be giving the bullet that much more lift and so a slightly higher impact point will result unless corrected for.
This could benefit from a diagram ... and some trig' ... hmmm
My earlier mention should have one thing made very clear ... all forces in this analysis are based on the long axis of a bullet ..... it's flight path. We might imagine the bullet having a small hole thru the middle and travelling along a parabolic thin wire... and 'seeing'' the ''pull'' of gravity. That is under normal ''horizontal'' flight acting as I said ... ''normal'' to the trajectory and so exerts max effect.
Leaving math etc out of it again for now ... try this small experiment ... anyone. Take a long and thin but springy length of wire, rod etc (say 6 feet long) .. . and place a substantial weight on the end. Now, with the weight on the ground and rod vertical .... pick up the free end and move the whole thing thru 180º until the weight is now at the top.
At low and high positions the rod is effectively straight. But thoughout the 180º, notice that the bend in the rod increases to max at mid point ... 90º ... when gravity is exerting max ''pull'' ... it is ''normal'' to the rod's axis. As you progress thru towards top position the bend is once more decreasing. The effect is of course most obvious thru the first and last 15º let's say ... and either side of horizontal there does not appear so much obvious change.
So, whilst the effect is less obvious either side of horizontal it is still there .... and the ''bend'' or parabola - or trajectory curve .. is slightly flattened/reduced ... so for a sight setting done for level shooting ... (the maximized parabola for a given distance) .. this will be giving the bullet that much more lift and so a slightly higher impact point will result unless corrected for.
This could benefit from a diagram ... and some trig' ... hmmm
P95Carry
Moderator Emeritus
Ok, I'm a stubborn ole curmudgeon ... let me throw this diagram at you ...... severely simplified but I hope it makes a point.
Here we represent a downward trajectory (as has been said the ''offset'' will be same either side of horizontal.) For pure convenience I have set it up so we have (imagine) a classic ''3,4,5'' triangle ... that gives us an angle of declination of 36.87º. This is now the value for theta. Notice that this is showing twice.. of course!
I am calling the gravity forces G(n) for the normal effect of gravity as it would be at 90º ..... and the other value G(t) for trajectory ''effective value'' .. as gravity acts on this non horizontal line (shown straight BTW for simplicity's sake!).
The G(t) then is expressed by {G(n) cos theta}. In this case using simple figures and forgetting that 9.81 is the acceleration due to gravity .. lets just give it a simple ''mass value'' .. it's more convenient .. call it simply ''9''. I said I was keeping it simple so excuse.
Then G(t) becomes {9 Cos 36.87} ....... which is {9 x 0.8} in this case .. giving G(t) a value of 7.2. This is a reduced downward force relative to this trajectory at any fixed given velocity at the point shown..... when compared with a simple near horizontal flight line.
So ---- drop will reduce for a given velocity and distance relative to horizontal ... ergo the shot will hit high. QED I hope.
Ok .. it's over simplified but .. this is my working premise. If I'm wrong entirely then .. I'm off for a beer, and all my rusty math is ........ just that ... corroded to hell.
Here we represent a downward trajectory (as has been said the ''offset'' will be same either side of horizontal.) For pure convenience I have set it up so we have (imagine) a classic ''3,4,5'' triangle ... that gives us an angle of declination of 36.87º. This is now the value for theta. Notice that this is showing twice.. of course!
I am calling the gravity forces G(n) for the normal effect of gravity as it would be at 90º ..... and the other value G(t) for trajectory ''effective value'' .. as gravity acts on this non horizontal line (shown straight BTW for simplicity's sake!).
The G(t) then is expressed by {G(n) cos theta}. In this case using simple figures and forgetting that 9.81 is the acceleration due to gravity .. lets just give it a simple ''mass value'' .. it's more convenient .. call it simply ''9''. I said I was keeping it simple so excuse.
Then G(t) becomes {9 Cos 36.87} ....... which is {9 x 0.8} in this case .. giving G(t) a value of 7.2. This is a reduced downward force relative to this trajectory at any fixed given velocity at the point shown..... when compared with a simple near horizontal flight line.
So ---- drop will reduce for a given velocity and distance relative to horizontal ... ergo the shot will hit high. QED I hope.
Ok .. it's over simplified but .. this is my working premise. If I'm wrong entirely then .. I'm off for a beer, and all my rusty math is ........ just that ... corroded to hell.
Hkmp5sd
Member
Nah, that's just the easiest way to explain it.
If you wanna do the math, check out Applications of Kinematics and Dynamics - Projectile Motion .
P.E. Third Period - Coach Brown
Cosmoline
Member
I'm picturing the TFL crew arguing about this while the zombie horde climbs up the hill.
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