The problem you have is the number of variables. You could use excel to calculate the distances for every variable but what you are after is a projection for every arrow speed and every initial trajectory. You could probably then just extend the curves using a french curve approximation but it's a lot of work. And I'm not sure why you need it. I don't know how they came up with 150yds. What arrow speed was used and what initial trajectory was used and why?Heehe beautiful in its simplicity. heehee
I worded my post badly, sorry. I wonder if there is a chart of some sort, with results such as drop in level x requires extra overshoot=Y
Or simpler still, every drop of 1 foot requires extra x feet overshoot.
They might, or must have. I hope they didn't just guess , but they may well have the above equations punched in already, then all they need to do is set initial trajectory and velocity and slope angle. Punching that stuff into Excel is not difficult but probably a 3 cup of tea job at leastThanks for that.
I wondered if assessors of archery fields used a chart that would show how much needs adding for a drop of let's say 1m in 20m. Or some other simple to apply formula.
Okay... because the base of the trench is at a given depth the solution to the distance traveled is just the addition of an initial height to the starting equations for a flat field.Imagine a level field and a 150y line is marked on it. An arrow lands on the line.
Imagine, next that the ground either side of that line is excavated and a wide 1foot deep trench is cut out parallel to the line.
An arrow that would have landed on the line will now travel further. How much further depends on the flight path of the arrow.
Assuming the compound is faster than the longbowAnd a compound arrow's parabola will be flatter than a longbower's.
I'm guessing you must be using a coefficient for drag, which I don't use.I have a code that will calculate arrow trajectories and here are the results for my bow:
1 - Launch velocity 285 fps, 330 grn arrow.
2 - 6.16 degrees above horizontal will allow my arrow to land at 150 yds on the flat.
Yes, 2.5*pi/4.0 as suggested byI'm guessing you must be using a coefficient for drag, which I don't use.
Ahhh... okay, I would use 2x 9.8m/s/s resistance giving me an average arrow speed of 90m/s gives me 148m at 6.16 degrees elevation....cool