First I get the angle of the coordinate.
angle=tan(OLD_Y/OLD_X)
Then I add 90 degrees.
angle+=90
I find the radius, which is OLD_Y squared times OLD_X all square rooted, then the new coordinates as followed:
NEW_X=radius*cos(angle)
NEW_Y=radius*sin(angle)
...or maybe I switched the cos and sin.
First I get the angle of the coordinate.
angle=tan(OLD_Y/OLD_X)
Then I add 90 degrees.
angle+=90
I find the radius, which is OLD_Y squared times OLD_X squared and all square rooted, then the new coordinates as followed:
NEW_X=radius*cos(angle)
NEW_Y=radius*sin(angle)
...or maybe I switched the cos and sin.
Then flip it vertically over the x,y axis
~ 1 ~ ~ ~ ~ 5 ~ ~ ~
~ 2 ~ ~ ~ ~ 4 ~ ~ ~
~ 3 ~ ~ ~ becomes ~ 3 ~ ~ ~
~ 4 ~ ~ ~ ~ 2 ~ ~ ~
~ 5 ~ ~ ~ ~ 1 ~ ~ ~
Did I do something incorrectly?
I think this would work for curved lines to.
------------------
Tell the truth and you will never fear someone will figure out you lied.
<<I'm sure I'm quoting someone out there, wish I knew who!>>
David Abresch
Otherwise, if you will be dealing with variable size arrays, just use the trig ;(
[This message has been edited by Splat (edited October 08, 1999).]
B = A [ j , 6 - i ]<P> <IMG SRC="http://www.gamedev.net/community/forums/ubb/smile.gif">
Question:
Say you have a 5x5 grid, and perform a 90 degree clockwise rotation around the middle cell (3,3). Is there an equation to calculate the new row and col based on the cell's pre-rotation row,col?
I know there is an equation but I can't remember plus I sold all my college math books back. 
Thanks in advance for any help.
Wesley
B = A [ j , 6 - i ]<P><BR><P>——————<BR>–Ali Seyedof (It's all dark !)
