Finding an AngleDate: 08/09/2002 at 08:43:00 From: Timo Petmanson Subject: Sine and cosine maths Hello, I need to know how to calculate an angle between two 2D points. I know how to calculate a new point by angle, but I can't figure out how otherwise. Calculating by angle: a = 143 // this is the angle x = 1 // 2d point x coord y = 1 // 2d point y coord u = 3 // how many units we change coords So, if we want to change point's coordinates by angle, then we must do: x = x + SIN(a * (pi / 180)) * u // (pi / 180)radians to degrees y = y + COS(a * (pi / 180)) * u // I hope I didn't change sin and cos but what must I do to calculate the angle? x1 = 1 y1 = 1 x2 = (-1) y2 = (-5) 0 | (x2,y2) | / |angle/ | / | / | / | / (x1,y1)------------ 90 | | | | | | | 180 I hope you understood my problem. Thanks. Timo Date: 08/09/2002 at 09:48:14 From: Doctor Rick Subject: Re: Sine and cosine maths Hi, Timo. Yes, I understand: you know how to find the coordinates of a point given the starting point, angle and distance, but you don't know how to do the reverse - calculate the angle given the starting point and ending point. (You probably do know how to calculate the distance.) We can take your equations as a starting point. You did have sin and cos interchanged, if the angle is measured counterclockwise from the positive x axis as we usually do. I'll call the two points (x1,y1) and (x2,y2). x2 = x1 + cos(a * (pi / 180)) * u y2 = y1 + sin(a * (pi / 180)) * u Subtract the starting point coordinates: x2 - x1 = cos(a * (pi / 180)) * u y2 - y1 = sin(a * (pi / 180)) * u Divide the second equation by the first: (y2-y1)/(x2-x1) = (sin(a*pi/180)*u)/(cos(a*pi/180)*u) What is sin/cos? It's the tangent. (y2-y1)/(x2-x1) = tan(a*pi/180) We can find the angle by taking the inverse tangent (arctan) of both sides: a*pi/180 = arctan((y2-y1)/(x2-x1)) a = 180/pi * arctan((y2-y1)/(x2-x1)) There's your formula. However, there are some tricky things to add. One is that x2 may equal x1, and you'll get a divide-by-zero error. Another is that the arctan won't distinguish between the angle from point 1 to point 2 and the angle from point 2 to point 1. These should be 180 degrees apart, but they will come out identical. In summary, to do it right, you need to consider separate cases for x2-x1 negative, zero or positive. It looks as if you're writing a program in C++. If so, check out the function atan2(y,x). It computes the arctangent of y/x, and it takes care of those special cases too, because it is designed for exactly the sort of thing you are doing. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/20/2002 at 14:10:09 From: Timo Petmanson Subject: Thank you (sine and cosine maths) Thanks, that helped me. I was working on a computer game and I needed that formula for the movements. |
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