i need to aproximate a circle with Ogl. and i need to draw it in any place of the screen without using the translatef command.
Well, you could use a for loop to generate your coordinates with sin and cos… Then you can simply provide the way you want to draw with glBegin( GL_LINE_STRIP ); or somtehing…
So it could be like this:
glBegin( GL_LINE_STRIP );
for( float angle = 0.0f; angle <= (2.0f*PI); angle += 0.1f )
{
x = cos(angle);
y = sin(angle);
glVertex3f( x, y , -1 ); // I dont know you coordinate system, so this -1 maybe wrong.
}
glEnd();
This should give and idea… For a better approximation, you simply increment the angle by a smaller unit…
I have not tested this 'cause i just worte it.
I hope this helps.
Originally posted by mancha:
for( float angle = 0.0f; angle <= (2.0f*PI); angle += 0.1f )
It’s better to use something like:
angle+=2.f*PI/Segments
This way all circle segments will be equal.
Also you need to have an offset varible, if you want to draw anywhere without glTranslate. I think you maybe having a problem a lot of people have using glTranslate and not also using glPop/push matrix with it, making things not translate the way you want.
Here are a few things missing from mancha’s example:
ox, oy, oz = location of center of circle anywhere on screen.
r = radius of circle.
x = r * cos(angle);
y = r * sin(angle);
glVertex3f( ox - x, oy - y , oz ); // you may need to change ether ox or oy to plus instead of minus.
Originally posted by mancha:
[b]Well, you could use a for loop to generate your coordinates with sin and cos… Then you can simply provide the way you want to draw with glBegin( GL_LINE_STRIP ); or somtehing…
So it could be like this:
glBegin( GL_LINE_STRIP );
for( float angle = 0.0f; angle <= (2.0f*PI); angle += 0.1f )
{
x = cos(angle);
y = sin(angle);
glVertex3f( x, y , -1 ); // I dont know you coordinate system, so this -1 maybe wrong.
}
glEnd();
This should give and idea… For a better approximation, you simply increment the angle by a smaller unit…
I have not tested this 'cause i just worte it.
I hope this helps.[/b]