finding the sin/cos of a point on a circle
Author
I need to know how, if you had a circle as x,y,r and a point in 2d space of x,y which will reside somewhere on the perimeter of the circle - how can I get the sin/cos values of that point relative to the centre?
That probably doesn''t make much sense, so here''s what I''m doing:
I''m making a 2D game and the baddies in the game will have spherical shield systems. I want to make these look reeel purdy by making it so when a projectile strikes the shield, I get the point of impact and then generate multiple particles at that point. these particles will then spread out across the shield sphere to make a nice shield hit effect. I have figured out how to make the particles move around the sphere''s surface using sin/cos but I can''t figure out how to get the initial sin/cos values once I have established the point of impact. Anyone understand me and can help please?
Also, I only need to get the sin/cos for the x/y of the impact. Being a 2D game we can chuck the z vector and assume all impacts occur at the equator line of the shield.
It''ll try to draw a picture to make this easier to understand:
(x1,y1) being the center of your circle, (x2,y2) being the point on the perimeter...
So, assuming you want to find the sin/cos/tan of the angle at (x1,y1) You would use the following forumulas sin(angle)=opposite/hypotenus cos(angle)=adjacent/hypotenus and tan(angle)=opposite/adjacent
For the angle at (x1,y1), "adjacent" or the length of the adjacent size, is (x2-x1), opposite is (y2-y1) and hyp. is "r"
So sin(angle) = (y2-y1)/r cos(angle)=(x2-x1)/r and tan(angle)=(y2-y1)/(x2-x1)
Finding the angle itself is as easy as taking the inverse sin/cos/tan of the result
o (x2,y2) /| len=r / | len=(y2-y1) / | (x1,y1)O---o (x2,y1) len=(x2-x1) (x1,y1) being the center of your circle, (x2,y2) being the point on the perimeter...
So, assuming you want to find the sin/cos/tan of the angle at (x1,y1) You would use the following forumulas sin(angle)=opposite/hypotenus cos(angle)=adjacent/hypotenus and tan(angle)=opposite/adjacent
For the angle at (x1,y1), "adjacent" or the length of the adjacent size, is (x2-x1), opposite is (y2-y1) and hyp. is "r"
So sin(angle) = (y2-y1)/r cos(angle)=(x2-x1)/r and tan(angle)=(y2-y1)/(x2-x1)
Finding the angle itself is as easy as taking the inverse sin/cos/tan of the result
Author
Wow, fast response. Thanks, I think I understood it, I''ll have to do some tests and see if I got it right 
Here''s a routine to return an angle, if (x1, y1) is the origin of your circle and (x2, y2) is a point on the circle.
double getAng(double x1, double y1, double x2, double y2){ double x = x2 - x1; double y = y2 - y1; double dist = sqrt(x * x + y * y); double ang = acos(x / dist); if (y >= 0) return ang; else return -ang + 2 * PI;} This topic is closed to new replies.
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