trig - Printable Version +- iDevGames Forums (http://www.idevgames.com/forums) +-- Forum: Development Zone (/forum-3.html) +--- Forum: Graphics & Audio Programming (/forum-9.html) +--- Thread: trig (/thread-8633.html)

trig - tesil - Feb 12, 2011 07:24 PM This is done in 2d only. So ultimately what I'm trying to achieve is... remainder = shadow_height - length(p,n); (where remainder should be equal to the distance between m and p I believe) ...the only problem is that I can't figure out how to find n (or m for that matter). p.s. my math is terrible. RE: trig - Applewood - Feb 13, 2011 03:44 AM Whoa, I'm struggling with this given your diagram looks 3D. Assuming m,n,p an etc are all 2D vectors (they have an x and y component): The length of the line between them is the sum of the componentwise differences, all squared, added up and then square rooted again. ie: diffx = p.x-n.x diffy = p.y-n.y length = sqrt(diffx*diffx + diffy*diffy) Note that the multiplies effectively remove any minus signs so it doesn't matter which order you subtract - could be n minus p for the same answer. I hope I haven't made a silly mistake there, I got up literally 20 seconds ago and am still waiting on the kettle for my first coffee...