Prevent dragged circle from overlapping

Question

As mentioned in the title, i have two Circle 's the first is draggable and the second is fixed, I would rotate (with the drag) the first one around the second without overlapping them but my Circle reacts oddly, I'm sure the error comes from the drag condition but I don't know how to solve it, that's why I need your help, here is a minimal and testable code :

import javafx.application.Application;
import javafx.scene.Scene;
import javafx.scene.layout.Pane;
import javafx.scene.paint.Color;
import javafx.scene.shape.Circle;
import javafx.stage.Stage;

public class Collision extends Application{

private Pane root = new Pane();
private Scene scene;
private Circle CA = new Circle(20);
private Circle CB = new Circle(20);
private double xOffset = 0;
private double yOffset = 0;

@Override
public void start(Stage stage) throws Exception{

    initCircles();
    scene = new Scene(root,500,500);
    stage.setScene(scene);
    stage.show();

}

private void initCircles(){

    CA.setCenterX(100);
    CA.setCenterY(100);
    CA.setFill(Color.rgb(255, 0, 0,0.2));
    CA.setStroke(Color.BLACK);

    CB.setCenterX(250);
    CB.setCenterY(200); 
    CB.setFill(Color.rgb(255, 0, 0,0.2)); 
    CB.setStroke(Color.BLACK);

    CA.setOnMousePressed(evt->{

        xOffset = CA.getCenterX() - evt.getSceneX();
        yOffset = CA.getCenterY() - evt.getSceneY();

    });

    CA.setOnMouseDragged(evt->{

        //get Scene coordinate from MouseEvent
        drag(evt.getSceneX(),evt.getSceneY()); 

    });

    root.getChildren().addAll(CA,CB);

}

private void drag(double x, double y){

    /* calculate the distance between 
     * the center of the first and the second circle
     */
    double distance = Math.sqrt (Math.pow(CA.getCenterX() - CB.getCenterX(),2) + Math.pow(CA.getCenterY() - CB.getCenterY(),2));

    if (!(distance < (CA.getRadius() + CB.getRadius()))){

        CA.setCenterX(x + xOffset);
        CA.setCenterY(y + yOffset);

    }else{


        /**************THE PROBLEM :Condition to drag************/

        CA.setCenterX(CA.getCenterX() - (CB.getCenterX()-CA.getCenterX()));
        CA.setCenterY(CA.getCenterY() - (CB.getCenterY()-CA.getCenterY()));

        /*What condition must be established for the 
         * circle to behave correctly
         */
        /********************************************************/

    }

}    
public static void main(String[] args) {
    launch(args); 
}

}

Here is a brief overview :

Note: for my defense, i searched and found several subject close to mine but which have no precise or exact solution, among which:

-The circle remains blocked at the time of the collision

-Two circle that push each other

-JavaScript, Difficult to understand and convert to java

Thank you for your help !

James_D · Accepted Answer · 2017-01-22T18:12:51.540

Point2D can be interpreted as a 2D vector, and has useful methods for creating new vectors from it, etc. You can do:

private void drag(double x, double y){

    // place drag wants to move circle to:
    Point2D newCenter = new Point2D(x + xOffset, y+yOffset);

    // center of fixed circle:
    Point2D fixedCenter = new Point2D(CB.getCenterX(), CB.getCenterY());

    // minimum distance between circles:
    double minDistance = CA.getRadius() + CB.getRadius() ;

    // if they overlap, adjust newCenter:
    if (newCenter.distance(fixedCenter) < minDistance) {

        // vector between fixedCenter and newCenter:
        Point2D newDelta = newCenter.subtract(fixedCenter);

        // adjust so that length of delta is distance between two centers:
        Point2D adjustedDelta = newDelta.normalize().multiply(minDistance);

        // move newCenter to match adjusted delta:
        newCenter = fixedCenter.add(adjustedDelta);
    }

    CA.setCenterX(newCenter.getX());
    CA.setCenterY(newCenter.getY());

}

Obviously, you could do all this without using Point2D and just doing the computation, but I think the API calls make the code easier to understand.

Prevent dragged circle from overlapping

1 Answers1