# sighting in slightly downhill

grafsk8er

July 17, 2007, 10:45 PM

in the back of my grandparents house i can set up targets 100 yards away and do my sighting in there. but their back yard is slightly downhill, i'd say a gradient of 5-7 degrees. will this in any way affect POI or just accuracy in general if i were to shoot at a target that is on a level plane?

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trueblue1776

July 17, 2007, 10:51 PM

Depending upon the platform you should sight in just a tad higher when shooting down a gentle slope.

Some of the more physics minded guys on here have a better knowledge of this than me. I would love to see hard numbers on more severe slopes (i.e. 30, 45,60).

browningguy

July 17, 2007, 10:55 PM

Point of impact will be very slihtly lower with that difference in levation, probably not enough to worry about at 100 yards, but I can't calculate tonight. The thing to remember is your geometry, gravity is acting over the horizontal route. So at 100 yards measured distance, with a 5-7 degree downslope, carry the 3, round up to... never mind. Probably not more than an inch at this distance, the horizontal distance is probably 95-97 yards if you want to calculate, a sq. + b sq. = c sq. A is horizontal distance, b is the vertical drop, c is the line of sight distance.

grafsk8er

July 17, 2007, 10:57 PM

thanks browningguy, simple physics, i didn't even think of that. and to think i took the class this year, haha

W.E.G.

July 17, 2007, 10:59 PM

http://forums.cabelas.com/archive/index.php/t-183.html

Whether shooting uphill, or downhill, you always need to shoot lower than if you were on a level field. This is due to the actual flight path of the bullet being shorter, in vertical distance, than linear distance. In essence, gravity affects the bullets as it travels in a horizontal distance, not linear.

Now, to be realistic about it, its doubtful you will find many situations where you are shooting at a severe enough angle, or long enough distance to warrant holding your point of aim anywhere other than dead center. Especially if you are in a hunting situation.

See also http://www.gunsandammomag.com/gun_columns/notes/gn0409/

trueblue1776

July 17, 2007, 11:04 PM

Wow, I retract my previous statement, you guys are great!

AndyC

July 17, 2007, 11:16 PM

You're looking for the cosine value of the angle, whether uphill or downhill.

To calculate it, you'd need a Cos table (easy to find online, print it out for easy reference) and you'd need to know two things:

1. Distance to your target (laser rangefinder or similar)

2. Angle of the shot ( http://www.snipertools.com/aci.htm )

Assume a measured 600 yard shot and 40 degree upward (or downward) angle.

Look up 40 degrees in your Cos table and you'll see the value is .766

Simply multiply your measured 600 yards by .766 (the Cos value of 40 degrees) and you'll get the horizontal distance: 459.6 yards.

Edit: In short, if you're shooting at an upward or downward angle, the target is actually closer (in terms of horizontal distance) than your rangefinder indicates - which is why people typically shoot over the target; their sights are set too high, so you need to actually shoot lower.

Cosine table for easy reference:

http://img168.imageshack.us/img168/1306/tablcoseb8.jpg

30Cal

July 18, 2007, 09:57 AM

It generally doesn't make a difference until you go past 20 degrees.

Ty

SaMx

July 18, 2007, 11:17 AM

the drop will be over the horizontal distance, so if it's 100 yards measured down the slope, the actual horizontal distance the bullet travels is less than 100 yards, and the bullet will hit a little high. I actually did some physics problems like this in class last year.

cameron.personal

July 18, 2007, 11:30 AM

Why not just set the target stand the same height as the shooter position then a 5-7 degree grade will have no effect.

Cameron

martinc64

July 18, 2007, 11:51 AM

Why not just set the target stand the same height as the shooter position then a 5-7 degree grade will have no effect.

Cameron

I tried that once but the damn deer refused to climb the ladder I gave him.

:neener:

45crittergitter

July 29, 2007, 07:42 PM

As was posted in another thread:

Just to be painfully technical, the horizontal distance is irrelevant to the discussion. If you are reading for entertainment, you may want to stop here.

Please bear with me as I explain without the benefit of the necessary diagrams and way too much time elapsed since high school physics and engineering school.

The major error commonly made is stating that the horizontal distance to the target affects the bullet's drop. In fact, the horizontal distance is irrelevant. For our purposes, keeping miscellaneous variables such as atmospheric conditions, etc. equal, there is only one force causing bullets to drop - gravity. The only other variable in amount of drop is the time that gravity has to act upon the bullet in flight, i.e., time of flight (TOF). That is because gravity is a vector, acting vertically downward with an acceleration of about 32.2 feet per second per second.

Here is an example. Suppose a pair of shots, both equal distances from the muzzle, say 100 yards. The first is a perfectly horizontal shot. The second is at the mathematically convenient angle of 45 degrees upward (could be downward as well) from horizontal. The horizontal distance of the second shot is less than the first, measuring only 70.7 yards, but this is irrelevant.

Now here is where one must grasp the physics. Both shots are 100 yard shots; the first is a horizontal 100 yards, and the second is 100 yards at an angle of 45 degrees, but still 100 yards from muzzle to target. Since both bullets travel 100 yards at the same average velocity, the times of flight of both shots are equal. Since the TOF are equal for both shots, gravity has an equal time to act upon both bullets, meaning the drop must be the same, right?

Wrong. Bear with me.

While it is true that gravity has the same time to act upon both bullets, the acceleration of gravity does not act equally on them relevant to the direction of drop. Remember, gravity is a vector. In this example, the bullet is a vector also. Vectors have both a magnitude (in the case of gravity, an acceleration of 32 fps/s) and a direction (vertically downward). Vector mechanics involves both the angles and the magnitudes of the involved vectors.

Now back to the horizontal bullet fired in shot one. By definition, horizontal is perpendicular to gravity. Therefore, gravity acts in a downward direction that happens to be perpendicular to the original path of this particular bullet. Therefore, the drop of the bullet is related directly to the TOF and the effect of the full magnitude of gravity (32 fps/s) because gravity in this case acts in exactly the same direction as drop. Simple concept; everybody understands this.

For the bullet fired in shot two it gets more complicated. You see that bullet has a vector direction of 45 degrees to horizontal, and to vertical. While gravity still acts vertically upon the bullet, the bullet is not traveling horizontally or perpendicular to gravity, so gravity no longer acts perpendicularly to the original path of the bullet, but at the 45 degree angle. (This is easier with a diagram.) In other words, gravity pulls the bullet vertically downward, but vertically downward in this case is not the same as drop. Drop, as I understand the definition, is the amount the bullet falls perpendicularly (but not necessarily vertically) away from the line of the original path (barrel). Since the barrel in this case is not horizontal, drop is not vertical.

Now let's do the math. Picture shot two, upward at an angle of 45 degrees from horizontal, with a muzzle-to-target distance of 100 yards.

Gravity is acting at an angle of 45 degrees from the path of the bullet, rather than perpendicular to it. The component of gravity that is acting in the direction of drop, or perpendicular to the path of the bullet, is described by the equation: (Gravity) x (sine 45 degrees). You will notice that due to the angle this amount is only 0.707 times the horizontal shot's gravity effect, or about 23 fps/s. The other component of gravity is acting perpendicularly to the first. It acts exactly parallel, but opposite (for an uphill shot) in direction to the path of the bullet and the magnitude is described by the equation: (Gravity) x (cosine 45 degrees). This component can be legitimately ignored as insignificant relative to the velocity and TOF of the bullet.

Just for fun, let's consider a third shot. Vertical. Up or down, I don't care. There is zero drop; the bullet stays on its original path with no drop from gravity, because all of gravity is acting exactly parallel to the path of the bullet, and not at any angle to pull the bullet off that path.

In summary, the drop depends upon TOF and angle of bullet path from vertical (gravity). For two equidistant shots with the same load, the TOF are equal, leaving only the different angles to determine the difference in drop.

I hope this has clarified rather than confused.

trueblue1776

July 30, 2007, 10:00 AM

Excellent, knew it was more complicated than triangle trig, but I'll leave it to the PE to put good words to it! (unless he went to Ole Miss, then I know that explanation was all a lie :D)

AndyC

July 30, 2007, 12:46 PM

Ok, so let's assume a shot at a lasered 300 yards, upwards angle of 30 degrees. Explain how you would set your sights.

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