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Stupid V3 geometry question Displaying 1-7 of 7 total.
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Gayo
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This is kind of embarrassing, but although I *should* be able to figure out how to do this on my own, it's stymied me long enough to ask. I'm probably not going to be able to describe it too well, so bear with me.
I have a circle. Given the coordinates of the circle, I want to be able to make a square that bounds the circle -- that is to say, each side of the square would touch the circle at a point, and that the slope of that side would be the tangent of the circle at that point. Of course you can just work this out by hand if you need it for a specific angle of square, but I want to be able to draw the square at any angle, which means I need a specific formula to determine the four lines of the square from the lines of the circle and an angle measurement.
I've poked around with this for a couple hours, but I won't bore with the minutiae of what I've done so far since it's probably fruitless. Any ideas?
Posted on 2004-04-06 00:30:41
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mcgrue
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Well, do you have the radius of the circle? And do you have the centerpoint of the circle?
It should be a trivial task if you do. otherwise it becomes more intresting.
Posted on 2004-04-06 00:43:28
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Gayo
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Yes.
Posted on 2004-04-06 01:03:12
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mcgrue
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okay, let's make this simple
assuming you have:
int center_x, int center_y, radius_length;
you want to find:
int topleft_x, topleft_y, botright_x, botright_y;
topleft_x = center_x - radius_length;
topleft_y = center_y - radius_length;
botright_x = topleft_x + (radius_length*2);
botright_y = topleft_y + (radius_length*2);
and then to draw the rectangle:
Rect(topleft_x, topleft_y, botright_x, botright_y, myColor, screen);
That should be right, but I'm not testing it or anything.
Posted on 2004-04-06 01:33:11
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Gayo
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That only gives me the square in that one position, though. If I want, say, a square where each point is rotated 17 degrees from that position, it's trickier.
Edit: Well, I found a messy way to do it, but it only works with squares and isn't generally applicable to drawing the tangent line intersecting a circle at a given point.
Posted on 2004-04-06 01:35:45 (last edited on 2004-04-06 01:52:39)
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vecna
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How about:
Draw a circle. Find the endpoints of the 'easy' rectangle.
Take your angle and rotate each of those four points by that angle. Simple sin/cos.
Use linedraws to draw the rectangle. If you want it filled, you can use Triangle().
Posted on 2004-04-06 02:46:10
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Gayo
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That would work! What I ended up doing was finding the square root of 2, and multiplying that by the radius of the circle. That then gave me the length of the lines from the centre of the circle to the corners of the square, and I could find those with sin/cos and then join them.
Posted on 2004-04-07 02:43:48
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Displaying 1-7 of 7 total.
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