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Re: Some simple 2d game-related maths questions

by moritz (Cardinal)
on Aug 27, 2008 at 21:29 UTC ( #707321=note: print w/replies, xml ) Need Help??

in reply to Some simple 2d game-related maths questions

First of all the game is quite nice, but there's a simple method with which you can win every level pretty fast, which kind of kills the fun as it is now.
How can I tell whether the circular sprit intersects with an arbitrary line.

I don't know how familiar you are with linear algebra, it really helps ;-)

(Update: When I talk about "points", I really mean "position vectors to points". Thanks FunkyMonk)

I'll call the end points of your line A and B, the center of your sprite M, the point on the line which is the closest to M is called C (all those are vectors), the radius of the sprite is r (a scalar value).

Since C is on the line between A and B, it must be expressible in the form A + x * (B-A). Since it's the closest point to M, the connection line from C to M is perpendicular to the original line, or as a formula:

(B-A) . (M-C) = 0 or (B-A) . (M - A - x * (B-A)) = 0 # (that's a scalar product)

The only unknown part in this equation is x, which you can calculate from that equation, but to which I'm too tired at the moment.

If 0 <= x <= 1 and |M - C| < r, then it circle touches or intersects the line.

Likewise if |M-A| <= r or |M-B| <= r it also touches or intersects. In all other case it doesn't.

To get a realistic bouncing behaviour, the angle between (dx, dy) and (M-C) after the bounce has to be negative of what it was before the bounce.

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[davido]: And remember; time is hard. :) As soon as you think, I've got this, you'll discover that time is harder than that. :)
[1nickt]: TCLion are you parsing the dates with DateTime or another tool?
[1nickt]: I note DateTime::Format:: Flexible, announced immediately below this box in the CPAN feed
[Corion]: 1nickt: That one sounds sensible as it will tell you when it found an error
[1nickt]: Throws on error, I believe
[1nickt]: But should handle month as number or word

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