ellipse algorithm
You could probably use the equation of an ellipse. I don''t have it memorized but it should be covered in any trig/geo book. Look under the conics section. There might be a faster method than the ellipse equation though.
ECKILLER
ECKILLER
You could do this...
for ( int i = 0; i < 360; i++ )
{
PutPixel(CenterX + HorizontalRadius*cos(i*(PI/180)), CenterY + VerticalRadius*sin(i*(PI/180)), ColorChoice);
}
This will draw an ellipse!
As you probably know, the GDI uses a rectangle to identify where the ellipse should be drawn. Well, here's what I would do if I wanted to code the same function.
Hope this helps!
----------------------------------------------
That's just my 200 bucks' worth!
..-=gLaDiAtOr=-..
Edited by - Gladiator on July 21, 2000 1:08:56 PM
for ( int i = 0; i < 360; i++ )
{
PutPixel(CenterX + HorizontalRadius*cos(i*(PI/180)), CenterY + VerticalRadius*sin(i*(PI/180)), ColorChoice);
}
This will draw an ellipse!
As you probably know, the GDI uses a rectangle to identify where the ellipse should be drawn. Well, here's what I would do if I wanted to code the same function.
struct RECTANGLE{ long left; long top; long right; long bottom;};void DrawEllipse(RECTANGLE &rect, byte color){ int i; int HorizontalRadius; int VerticalRadius; HorizontalRadius = ( rect.right - rect.left ) / 2; VerticalRadius = ( rect.bottom - rect.top ) / 2; for ( i = 0; i < 360; i++ ) { SetPixel( hDC, rect.left + HorizontalRadius + HorizontalRadius * (float) cos( i * ( 3.14159 / 180 ) ), rect.top + VerticalRadius + VerticalRadius * (float) sin( i * ( 3.14159 / 180 ) ), color ); }} Hope this helps!
----------------------------------------------
That's just my 200 bucks' worth!
..-=gLaDiAtOr=-..
Edited by - Gladiator on July 21, 2000 1:08:56 PM
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