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Any Mathmeticians out there?

OK here's my problem This my second year bowhunting and I Shoot an Onieda Black Eagle. My sights are simple fiber-optic pegs but to hit the vitals of my deer target from my shooting platform I have to aim at the point where the deer's leg attaches to the body. I have no problems getting tight groups this way BUT!! My shooting platform is 17 ft tall while my stand is 27 ft off the ground. The distance from the bottom of my tree to the trail I'm hunting and the bottom of the platform to the target is the same. I must keep my stand elevation because of thick under brush. Will this change in elevation affect my shot placement any? If so by about how much? Any help is apprecited!

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[ This Message was edited by: bnow0707 on 2003-10-13 10:20 ]

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I'm not a bow hunter but I'll take a stab at helping you figure this out. Assuming the tree and trail are at a right angle to each other the equation you need is:

(a^2 + b^2)^1/2 = c

where:

a = distance from trail (deer) to tree
b = distance from ground to you
c = distance from deer to you (hypotenuse of the right triangle)

If "a" is large, the increase in "c" from a 10 ft increase in "b" is minimal. Just running some numbers, if the distance from the deer to the tree is 20 yards, the increase to 27 feet of your stand only increases the distance between you and the deer by 3.5 feet. If the deer is 30 yards from the tree (90 feet) the increase is just 2.5 feet.

So if the deer is between 20 and 30 yards from the tree assume between an extra 2.5 and 3.5 feet of shot distance.

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Bitmaster
Very good but I think the formula you are looking for is this [Sq.Rt(a^2 + b^2)]=c ?
I could be wrong here but I do not remember ever getting ^1/2 in any math/geomitry classes, or is that the same as Sq. Rt?
In bow hunting tho elivated distances are figgured from the base of the tree to the target and hight is not part of the equation at all. A gravity factor or something, someone other than myself might be able to explain the physics behind this?

bnow0707, back to the question. Some bows just shoot high from an elivated position. I have had 3 bows that I have hunted with. Two shot the same no matter the angle, but 1 a PSE Thunderflight shot consistantly high from above. The owner of my local archery shop gave me 2 options. 1- sight in (with hunting heads) from the height you will be hunting at. 2- a pendulum sight. I chose the new sight because I hunted from multiple stands all at different heights. The pendulum worked great for that bow from level ground to high in a tree.

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Something to the half power is the same as square root. Thanks for pointing that out though, because half power is rather confusing. I kind of just threw it up there without thinking how it looked. A few examples:

(a^2)^1/2 = a
(a^3)^1/3 = a
(a^3)^1/2 = a^3/2
(a^c)^d = a^(c*d)
(a^c)*(a^d) = a^(c+d)

Your right about the rise of the arrow (shooting high) as the angle gets tighter (arrow pointing more toward the ground) it has to do with the angle between the arrows path and the vertical (the direction in which gravity pulls). Although quantifying that is difficult without knowing anything about how the bow is shooting when at the perpendicular.

I was going to suggest maybe holding just a little bit lower, but it really depends on how the bow is sighted in, as you pointed out Adkbear, plus the shot difference between the two is so small in the examples above, i figured it might not make much difference.

How does a pendulum sight work? That sounds interesting. I assume it has a mark on it that you sight in on (for various distances) and as it swings away from the bow (as you tilt toward the ground) the swing outward compensates for the rise in the arrow.

[ This Message was edited by: bitmasher on 2003-10-14 22:20 ]

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Bitmaster, yes once I took/got some time to think about it to the power of one half would be the same as Sq.rt. I just cannot ever remember using it. I like it, what do they call that? "Out of the box thinking"?

The pendulum sight (usualy 1 pin) the pin is hinged at the top of the sight window. Thus the pin is self leveling due to gravity. As you aim the bow down the pin swings out away from you and compensates for arrow lift.

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My first year of college, i had to take a "bonehead" (that is exactly what they called it) pre-calculus class. My highschool math was not quite up to snuff, thus the remedial course before jumping into the deep side of the pool.

Anyway, that bonehead class was one of the best i ever took, because the prof was particular about laying a good geometry/algebra foundation. That's when i learned a "root" was simply a "fractional" exponent, which was really quite nice because you can then break apart equations more easily, prior to that a "root" was some sort of mystical function to me (which becomes even more mystical with cube roots, quad roots, etc). Why they don't teach it that way from the start, i don't know, would make things easier in my opinion.

Any way, that pendilum sight idea is clever. Good suggestion, had never heard of or seen such a thing (like i said, i'm not a bow hunter, but have been around archers), sounds like it is just the ticket to clear up the elevation change issue.

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cube roots?.......quad roots?....???
yall need to hush up cause you are making me feel like I am unedumacated. Grin

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That's about the same thing I said when I first heard of them too JTapia. Wink

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It will take a lot of calculating, but you can find the mathematical difference between the 2. You'll need a scientific calculator (one with sin, cos and tan functions).

For purposes of this example we'll say it's 60ft (20 yds).

Divide 27 (height of your stand) by 60. On your calculator hit: "2 FNCT" (2nd function), "tan", input whatever that number is (27/60) and then =. That will give you the actual mathematical angle you're hunting from.

Then do the same thing except divide 17 by 60. That will give you the angle you're practicing from (it will be less than the first one).

Subtract the second angle from the first. We'll say that it's 10deg for example purposes (I'm at my in-laws and don't have a calculator).

Go back to your calculator, hit "tan 10". Multiply that number by 60 and that will give you the difference in impact point between the 2 stands in FEET.

(Multiply by 12 to find out how many inches that is).

Like others said, that won't take into account the performance of your bow/arrow though, but it's something to work from.

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bnow,
As an electrician I have used the formulas that bitmasher, adkbear and saskie have given here many times to get the length of guide wires supporting poles, but all that will get you is a distance from your target at a given hight and angle.
What you are experiencing are the effects of Gravity on trajectory.
Anything that you wish to propel thru the air is effected by gravity trying to pull it to the ground. This includes Arrows, Bullets, and baseballs(how bout them Marlins).:smile: To compensate for this effect we must Lob, for lack of a better word, an object(arrow in your case) at an arch, at a speed to overcome gravity and make it land at a point that we call "on target". this is why your 30 yard pin is below your 20 yard pin so that you will have to tilt the arrow head upwards in order to align your sight pin with the target at your other sight, if you use a peep sight.Your object (arrow) will actually cross your aiming line of sight twice on its way to the target. This is why you have bullet drop listings in ballistics reports. If you sight your rifle in at 200 yards then thats where your bullet will cross your aiming sight line the second time and at 300 yards it is well below your aiming sight line. It is this distance below your aiming sight line that becomes "bullet drop" and at 100 yards your bullet is still higher that your aiming sight line at that point having crossed it only once, hence you have a positive distance above your aiming sight line.
Now, when you elevate yourself you are still set up on your bow sights as if you were shooting on flat ground so your arrow will leave your bow at a predetermined arch in order to hit the "target" at the yardage that the particular sight pin you aimed with is set for. Problem is you are already shooting toward the ground, in effect helping gravity do its work so that you dont need that arch anymore but its still there cause you are set up to shoot flat and your arrow will never cross your aiming sight line but once, (when it leaves the bow), before you reach your target giving you a shot that lands higher than your aiming sight line. The closer that your target gets to your stand the higher that your shot will land and the lower that you must aim.
I have never had this problem but I did overhear once that if that became a problem then you should try aiming with your 20 yard pin for 10 yard shots, 30 yard pin for 20 yard shots...etc.
Hope this helps some and I never had to use a Cube or Quad root. Grin

[ This Message was edited by: JTapia on 2003-10-21 16:19 ]

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Completey off topic, but I gotta respond to that Marlins comment. While I'm no fish fan, I'd like to seem them beat the Yanks. FL has their work cut out for them though....