I have a bunch of vectors normal to window surfaces in a 3D modelling software. Projected to the XY-Plane, I would like to know in which direction they are facing, translated to the 8 compass coordinates (North, North-East, East, South-East, South, South-West, West and North-West).
The vectors work like this:
Given a vector (x, y) I am looking for the closest of the 8 compass coordinates. Any ideas on how to do this elegantly?
This works in Java, computing a value 0...7 for the eight directions:
import static java.lang.Math.*;
int compass = (((int) round(atan2(y, x) / (2 * PI / 8))) + 8) % 8;
The result maps to the compass as follows:
0 => E
1 => NE
2 => N
3 => NW
4 => W
5 => SW
6 => S
7 => SE
I would probably just do a call to atan2() to figure out the heading angle ("yaw"), and then use either a sequence of if
:s or some maths to "snap" it to a multiple of 90 degrees.
no need to do an atan function.
if you do: y/x you'll get the slope of the line. Judging by the number you get you can determine the angle/octant.
for positive x's (x>0)
and a similar list for the negative x's
and finally the exception cases:
addendum: I only report this method for when calculating an atan is a no go (for instance on an embedded system))
I had to dig a bit. Here's a highly optimized routine I use (used in mobile games).
input: x1,y1 = startpoint of vector x2,y2 = endpoint of vector output (0-7) = 0=north, 1=northwest, 2=west,...etc
int CalcDir( int x1, int y1, int x2, int y2 )
{
int dx = x2 - x1, dy = y2 - y1;
int adx = (dx<0)?-dx:dx, ady = (dy<0)?-dy:dy, r;
r=(dy>0?4:0)+(dx>0?2:0)+(adx>ady?1:0);
r=(int []){2,3,1,0,5,4,6,7}[r];
return r;
}
void CalcDirTest(){
int t = CalcDir(0, 0, 10, 1);
printf("t = %d",t);
t = CalcDir(0, 0, 9, 10);
printf("t = %d",t);
t = CalcDir(0, 0, -1, 10);
printf("t = %d",t);
t = CalcDir(0, 0, -10, 9);
printf("t = %d",t);
t = CalcDir(0, 0, -10, -1);
printf("t = %d",t);
t = CalcDir(0, 0, -9, -10);
printf("t = %d",t);
t = CalcDir(0, 0, 1, -10);
printf("t = %d",t);
t = CalcDir(0, 0, 10, -9);
printf("t = %d",t);
}
This will result in the following output:
t = 7
t = 6
t = 5
t = 4
t = 3
t = 2
t = 1
t = 0
(The vectors for the test might look oddly chosen, but I tweaked them all a bit to be clearly in one octant and not on the exact border)
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