Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
| |
|
|
You are not logged in. #1 2005-12-04 22:00:12
calculate position c based on position a and bHi there, #2 2005-12-04 22:49:54
Re: calculate position c based on position a and bIt is actually easy if you know more about the position of A and B. You probably won't need sine or cosine. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman #3 2005-12-05 00:34:02
Re: calculate position c based on position a and bThanks a lot for responding so quick and great forum (it took me about 5 secs of googling to find it)! #4 2005-12-05 01:06:08
Re: calculate position c based on position a and bYou have a points A and B, (ax, ay) and (bx, by). The vector between these two points is <ax - bx, ay - by>, going from B to A. Now what you need is the unit vector. To get this, we divide that vector by it's length. So let L = √( (ax - bx)² + (ay - by)² ), which is the length. <(ax - bx) / L, (ay - by) / L> is the unit vector. Then multiply this by the distance you want the camera, D: <D*(ax - bx) / L, D*(ay - by) / L>. Last edited by Ricky (2005-12-05 01:39:49) "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." #5 2005-12-05 03:00:49
Re: calculate position c based on position a and bThanks Ricky, I think I got it to work. #6 2005-12-05 06:42:33
Re: calculate position c based on position a and bAh, sorry, I misread your post. But yes, that is precisely right. By multiplying the vector by -1, you are reversing it's direction, so in front becomes behind. "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." |