Draw arc from point?

[FerrisWheel]

This is more a maths question, but I'm sure many people here can answer it.

I wish to draw an arc. That is, mark a series of points equidistant from a set point going from 0 degrees to 180 degress (so really, half a circle).

How would I do this using screen points?

[ThePuska]

(x0, y0) is the center of the circle, r is its radius

(x - x0)^2 + (y - y0)^2 = r^2

Solving for y, since you probably want the half-circle to be horizontal (0-180 is generally from right to left)

y^2 - 2*y*y0 + y0^2- r^2 + (x - x0)^2 = 0

y = ( -(-2*y0) ± sqrt( (-2*y0)^2 - 4*(y0^2 - r^2 + (x - x0)^2) ) )/2

Give x any values between x0-r and x0+r to get corresponding values on the y-axis. Choose either + or - for the ± part. Draw lines between those points or something, I don't know about that part.

[Mattz333]

Another method would involve trigonometry.

y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);

Where t is your degree point. In this setup (t=0) is the very top of the circle, (t++) will move clockwise.
To draw your half-circle you could do:

for(int t = 0; t < 180; t++)
{
y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);
}

If you wanted to draw an arc going clockwise from an arbitary point:

// a is your arbitary point
for(int t = a; t < (a+180); t++)
{
y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);
}

In these examples the amount you add to t every step is dependant. If you are using a large radius then I would make t a float and decrease the step t makes (maybe there is a relation to it and the radius).

[ThePuska]

[QUOTE=Mattz333;20154890]Another method would involve trigonometry.

y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);

Where t is your degree point. In this setup (t=0) is the very top of the circle, (t++) will move clockwise.
To draw your half-circle you could do:

for(int t = 0; t < 180; t++)
{
y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);
}

If you wanted to draw an arc going clockwise from an arbitary point:

// a is your arbitary point
for(int t = a; t < (a+180); t++)
{
y = r * cos( t * Pi / 180);
x = r * sin ( t * Pi/ 180);
}

In these examples the amount you add to t every step is dependant. If you are using a large radius then I would make t a float and decrease the step t makes (maybe there is a relation to it and the radius).[/QUOTE]

Also notice that these will indeed be clockwise even though they might appear counter-clockwise on your screen. That's because the y-axis in graphics is usually mirrored.

[Mattz333]

[QUOTE=ThePuska;20156112]Also notice that these will indeed be clockwise even though they might appear counter-clockwise on your screen. That's because the y-axis in graphics is usually mirrored.[/QUOTE]

Ah yes, I was thinking of the standard Cartesian system when I wrote this out. To correct the code you just need to negate the y-coordinate calculation.

[FerrisWheel]

Using Mattz code (yes i'm lazy, and chose it because I understand it better), I've been able to get the arc.

Changing the code to

for(int t = 90; t < 270; t++)

Should make it an arc going from/to two horizontal points?

[ThePuska]

[QUOTE=FerrisWheel;20160294]Using Mattz code (yes i'm lazy, and chose it because I understand it better), I've been able to get the arc.

Changing the code to

for(int t = 90; t < 270; t++)

Should make it an arc going from/to two horizontal points?[/QUOTE]

Yes, you can do that, or simply swap the sin and cos.