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Lappa

Junior Newbie

 « Posted 2008-04-08 13:55:39 »

Hi! I have a java2d geometry problem that I am trying to solve for a project I'm currently working on. My maths isn't great by any stretch of the imagination, so it might be true that this is a lot simpler than I am imagining. Please forgive my ignorance. I have made a quick visual representation of what I am trying to achieve (see). The scenario is as follows:

My (circular) bot has a surrounding Circle representing it's sensor range. When this circle intercepts an obstacle (also a circle) my checkCollision() method returns true. The Bot class holds the bearing (in radians) which the bot is currently travelling towards. When the bot collides with the Obstacle, I want it to call the rotate() method - this rotates the bot by a small increment each game loop - until the bot is facing in the opposite direction to the Obstacle. I create a line to join the obstacle with the bot, representing the bearing of the Obstacle. But what I really need to know is theta - the angle between the current bearing and the new bearing.

My first idea was to call rotate() until the bearing line was parallel with the obstBearing line, however I can't work out how to do that and it might cause a problem if the bot rotates clockwise instead of anti-clockwise. So it looks like I'm in for some complicated trigonometry. Any ideas(?) much appreciated. Thanks for reading this.

Here's some of my code to give an idea of how my program works:

 1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18 `public double calculateBearing() {      double newBearing = 0;      Line hypotenuse;      Vector2f intersection;      double m, c, x2, x1, y2, y1, r;            x1 = botCentre.x;      y1 = botCentre.y;      x2 = obstLoc.x;      y2 = obstLoc.y;      r = sensCircle.getRadius();      m = (botCentre.y - obstLoc.y)/(botCentre.x - obstLoc.y); // gradient (or slope) of line      c = y1 - (m * x1); // y intersection //calculations here            return newBearing;   }`

ryanm

Senior Devvie

Projects: 1
Exp: 15 years

Used to be bleb

 « Reply #1 - Posted 2008-04-08 14:27:48 »

I'm assuming you're using vecmath, based on the names in the diagram:

 1  2  3  4  5 `Vector2f desiredDir = new Vector2f( x1 - x2, y1 - y2 );Vector2f botDir = new Vector2f( Math.cos( botBearing ), Math.sin( botBearing) );double theta = desiredDir.angle( botDir );theta *= Line2D.relativeCCW( 0, 0, botDir.x, botDir.y, desiredDirx, desiredDir.y );`

Untested, so that final multiplication might need to be negated but it'll be easy to spot if that's the case.

edit: premature posting
Lappa

Junior Newbie

 « Reply #2 - Posted 2008-04-09 13:10:46 »

Thanks for your help. I found out that it is in fact a lot simpler than I thought. By using arctan on the change in x and y I get the correct result:
 1  2  3  4  5  6  7  8  9  10  11  12  13  14  15 `   public void calculateBearing() {            updateObstacleLine();            float dx = (botCentre.x - obstLoc.x);      float dy = (botCentre.y - obstLoc.y);            double thetaRadians = (StrictMath.atan2(dy,dx));       double thetaDegrees = StrictMath.toDegrees(thetaRadians);             targetAngle = thetaRadians;            targetBearingVec = new Vector2f(thetaDegrees);   }`

However, I now need to work out how to decide whether to rotate CW or CCW to get to this new bearing. Many thanks for your input.
ryanm

Senior Devvie

Projects: 1
Exp: 15 years

Used to be bleb

 « Reply #3 - Posted 2008-04-09 16:04:45 »

However, I now need to work out how to decide whether to rotate CW or CCW to get to this new bearing. Many thanks for your input.

Check out Line2D.relativeCCW for this.
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