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View Full Version : [SOLVED] Game Design. Circular Surface Deflection



Dr Belka
June 12th, 2012, 11:29 PM
I'm making a simple 2D game. It's coded in Java.

In the game, there is a planet and objects that move around the planet according the force of gravity. I've got the gravity working but I just can't wrap my mind around how to deal with the object to planet collisions.

When one of the moving objects strikes the surface of the planet, I want it to properly deflect based on the angle at which it struck. I just can't wrap my brain around how to do this.

Something to work with:




class Planet {

public int getX();
public int getY();
public int getRadius();

}

class MovingObject {

public int posX;
public int posY;
public float velX;
public float velY;

}

class Physics {

public static float getDistane(int x1, int y1, int x2, int y2);
public static float getAngle(int x1, int y1, int x2, int y2);

}

Planet p;
MovingObject s;
.....;
// if collision
if (Physics.getDistance(s.posX, s.posY, p.getX(), p.getY()) < p.getRadius) {
// deflection occurs
}





Any help would be much appreciated. Thank you.

Vaphell
June 12th, 2012, 11:54 PM
arctan delta(y)/delta(x) -> angle of movement vector

arctan (yi-y0)/(xi-x0) -> angle created by r anchored in x0,y0 in the point of impact xi,yi

calculate the difference and mirror that angle on the other side of radius. Use this newly created angle to calculate new movement vector

... or something

Dr Belka
June 13th, 2012, 12:30 AM
Thank you so much for your reply.


arctan delta(y)/delta(x) -> angle of movement vector

Yup, that's what Physics.getDistance() does.


arctan (yi-y0)/(xi-x0) -> angle created by r anchored in x0,y0 in the point of impact xi,yi

And that's what Physics.getAngle() does.


calculate the difference and mirror that angle on the other side of radius. Use this newly created angle to calculate new movement vector

... or something

And this is what I'm trying to figure out how to do.

Dr Belka
June 13th, 2012, 12:33 AM
Here's a picture showing what I would like to happen.

219598

Vaphell
June 13th, 2012, 01:44 AM
a - movement angle
b - radius angle
----------------
c = (180+a)-b
n = b-c = 2b-180-a

let's check it: from the picture, a is roughly -80deg (object moving down), b 30deg
c = 70deg
n = 30deg-70deg = -40deg which is what you approximately see

Dr Belka
June 14th, 2012, 07:55 PM
a - movement angle
b - radius angle
----------------
c = (180+a)-b
n = b-c = 2b-180-a

let's check it: from the picture, a is roughly -80deg (object moving down), b 30deg
c = 70deg
n = 30deg-70deg = -40deg which is what you approximately see

Thank you so much. I'm going to try this out

muteXe
June 15th, 2012, 11:53 AM
Isn't this an object bounding off a flat surface?

Vaphell
June 15th, 2012, 12:32 PM
doesn't matter (unless you are writing some serious physics simulator). You can assume that deflection off the circular surface = deflection off the tangent plane in the point of contact (dashed line on picture)

muteXe
June 15th, 2012, 01:59 PM
yea exactly. not sure what the confusing picture's all about then. Or even the mention of a sphere come to think of it.