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 Determining if a circle is in your field of view in 2D "without" trig functions.  (Read 3122 times) 0 Members and 1 Guest are viewing this topic.
ClickerMonkey

JGO Coder

Medals: 20

Game Engineer

 « Posted 2013-06-28 13:27:38 »

http://pastebin.com/yeESVkhZ

Given your direction vector is normalized, and you store FOV as cos/sin, you can pretty quickly determine if a circle is within your view (either partially or entirely).

This works by creating vectors that extend from the origin to the sides of their view port - rotating them by the view's direction. Using these vectors it can rotate the vector between the circle's center and the object down to the x-axis. Once you do this, the y-component of that vector is the signed distance of the circle to that extent.

This also works where FOV >= 90 (and by FOV, I'm talking half of your actual field of view, in radians).

This is some code from a library I'm working on called [Steerio](https://github.com/ClickerMonkey/steerio).

If someone can figure out a more efficient method, I challenge thee!

Peace!

RobinB

JGO Ninja

Medals: 44
Projects: 1
Exp: 3 years

Spacegame in progress

 « Reply #1 - Posted 2013-06-28 15:24:14 »

Why is this function better then:

circle.x - circle.width/2 <= rect.x + rect.width && circle.x + circle.width >= rect.x (same with y)
ClickerMonkey

JGO Coder

Medals: 20

Game Engineer

 « Reply #2 - Posted 2013-06-28 15:31:39 »

By "your" FOV, I mean something in your game like a unit.

Imagine that on the head of a unit of yours, and it want's to know everything it can see based on it's direction, location, and FOV.

 Games published by our own members! Check 'em out!
RobinB

JGO Ninja

Medals: 44
Projects: 1
Exp: 3 years

Spacegame in progress

 « Reply #3 - Posted 2013-06-28 15:36:28 »

Oh awesome i was looking for that a while ago!
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