!! Please help !! Java2d geometry question.

[Lappa]

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:

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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]

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

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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]

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:

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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.

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[ryanm]

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.