Figuring yardage from treestand

[PhillyB]

On Saturday, l was able to put up my treestand. Unfortunately I was unable to take my bow with me and practice from the stand in the field.

I figured i would be able to use the Pythagorean Theorem, a^2+b^2=c^2, to figure out the distance. I have both the distance from the ground and to my mark, essentially the A & B of the formula.

When I ran the numbers through the formula, the distances came out the same as if the shot was being taken on flat ground.

From the ground to bow height is 14ft. I had trees marked at 20 yds, 25 yds, 30 yds, etc... but according to what I figured there is no difference between using my pins from the ground or tree stand.

Any thoughts, or is it that, at 14ft, there is not enough of a change in degrees and angle, to affect arrow flight?

[BOHNTR]

14' up is not that high up and there for probably not much of a degree at those distances. Although the longer you shoot it WILL eventually have a difference in inclination yardage vs. actual yardage.

[firefighterbraun]

You'd be suprised at how much it changes. I have the Chuck Adams Bushnell Rangefinder so it does all the calculating for me! My spot that I got this year is on a hillside with a blind that I built. For about a 15 yd shot I end up shooting as for 14 yds due to the slight slope that I am on. Its probably only about 6 feet difference. So depending on how far you're shooting it can change. Being 14 ft up I would imagine you would be taking out 5 yds off your shot. If you want get me your distances to me from the bottom of your tree to your target area and I can set it up at my house and get you the distance with the range finder. Also depending on your bow, your shot might not have to b changed much. I have my first pin set for 20 yd, and it works for any shot 20 yds or less.

[PhillyB]

firefighterbraun- l have two shooting lanes setup one for 20 yds and the next for 30 yds...

I appreciate you taking the time to figure that out for me. I am interested to see what you find. Tomorrow after work l am planning on setting up a makeshift-redneck stand (ladders/boards/bricks) on my side yard to see what l find...

['Ike']

Also remember, bend at the waist...And practice shooting from the sitting position!

[cwp]

If you are using the pathagorean therum the distance would be figured as if you were standing on the ground, it does not take into consideration the angles, just the lengths.

[firefighterbraun]

Set it up today at my house and kinda suprised at the results. My rangefinder said at your 20 yd shot from 14 ft you are at a 15 degree angle and only take 1 yd off your shot. (So shoot as if 19 yd) For your 30 yd shot from 14 ft you are at a 12 degree angle and only take one yd off your shot. (Shoot as if to 29 yd) I thought it would be more than that but guess not! Hope that helps ya zero in on your sweet spot when the moment comes!

[PhillyB]

Set it up today at my house and kinda suprised at the results. My rangefinder said at your 20 yd shot from 14 ft you are at a 15 degree angle and only take 1 yd off your shot. (So shoot as if 19 yd) For your 30 yd shot from 14 ft you are at a 12 degree angle and only take one yd off your shot. (Shoot as if to 29 yd) I thought it would be more than that but guess not! Hope that helps ya zero in on your sweet spot when the moment comes!

Very cool... thanks for taking the time to help me out. It is good to know the exact distances.

If my scouting trips pay off and all goes "as planned" I may just get a crack at a good buck

[johnyutah5]

I think the issue is confused because when you cut yardage(Like a cut chart or range finder does) using the formula the cut chart/rangefinder is solving for B not C. If you were in the treestand with a rangefinder(That doesn't have angle compensation) you would be given C when ranging an object. What you would then need to know is the yardage you should actually shoot for which would be B. If I understood you correctly you already have 'B'(from the base of the tree to the target). Gravity effects the arrow as though it is traveling the distance of B even though it is traveling a further distance (Along the path of C). Hope that makes sense, makes sense in my mind anyway.