Reputation: 988
As you can see on the image, I have a p1 and p2 objects with (x,y) coordinates which I know the values, and I know radius of all these circle objects.
However, I want to calculate new position x,y which would be p3 center point. Basically, as you can see it's p2 position + radius.
I am doing this for java game which is based on libgdx. I would appreciate any math or java language directions/examples.
Upvotes: 0
Views: 299
Reputation: 1455
Now using libgdx for the graphics. Thus no need for polar co-ordinates, on the outside.
I am not doing frame rate relative animation. Therefore, this is no perfect match to your code.
Using the following calculation (if (theta >= 360) { theta = 0.0f; }
) at the end of the render method will let the animation restart with its original value.
package org.demo;
import com.badlogic.gdx.ApplicationAdapter;
import com.badlogic.gdx.math.Vector2;
import com.badlogic.gdx.utils.ScreenUtils;
import com.badlogic.gdx.Gdx;
import com.badlogic.gdx.graphics.glutils.ShapeRenderer;
public class CircleDemo extends ApplicationAdapter {
ShapeRenderer shapeRenderer;
float theta = 0.0f;
@Override
public void create () {
shapeRenderer = new ShapeRenderer();
}
@Override
public void render () {
ScreenUtils.clear(0, 0.4f, 0.4f, 1);
Vector2 p1 = new Vector2( Gdx.graphics.getWidth() / 2.0f , Gdx.graphics.getHeight() / 2.0f);
Vector2 smallCircleCenter = new Vector2(150.0f, 0.0f);
smallCircleCenter.add(p1); // translate center by p1
shapeRenderer.begin(ShapeRenderer.ShapeType.Line);
// static lines and circles
for (int angle = 0; angle < 360; angle += 30) {
Vector2 lineEnd = new Vector2(smallCircleCenter);
lineEnd.rotateAroundDeg(p1, angle);
shapeRenderer.line(p1, lineEnd);
shapeRenderer.circle(lineEnd.x, lineEnd.y, 20);
}
// animated line and circle in red
shapeRenderer.setColor(0.75f, 0, 0, 1);
Vector2 movingCircleCenter = new Vector2(smallCircleCenter);
movingCircleCenter.rotateAroundDeg(p1, theta);
shapeRenderer.line(p1, movingCircleCenter);
shapeRenderer.circle(movingCircleCenter.x, movingCircleCenter.y, 20);
shapeRenderer.setColor(1, 1, 1, 1);
shapeRenderer.end();
theta++;
// for the screenshot stop at 90 degrees
if (theta >= 90) {
theta = 90.0f;
}
}
@Override
public void dispose () {
shapeRenderer.dispose();
}
}
Upvotes: 1
Reputation: 988
So I wrote a test in my project, based on your approach:
package com.bigbang.test.impl;
import com.badlogic.gdx.graphics.glutils.ShapeRenderer;
import com.badlogic.gdx.math.Vector2;
import com.badlogic.gdx.utils.Array;
import com.bigbang.Game;
import com.bigbang.graphics.g2d.shapes.impl.Ellipse;
import com.bigbang.graphics.g2d.shapes.impl.Line;
import com.bigbang.graphics.gl.Color;
import com.bigbang.math.BBMath;
public class PolarToCartesianTest extends AbstractTest {
private Array<GraphicalObject> graphicalObjectArray;
private GraphicalObject dynamicGraphicalObject;
private float radius, smallCircleRadius;
private float centerX, centerY;
public PolarToCartesianTest(Game game) {
super(game);
}
@Override
public void create() {
radius = 200f;
centerX = game.getScreenController().getScreenWidth() / 2;
centerY = game.getScreenController().getScreenHeight() / 2;
smallCircleRadius = radius / 4;
graphicalObjectArray = new Array<>();
for (int angle = 0; angle < 360; angle += 30) {
GraphicalObject graphicalObject = new GraphicalObject();
graphicalObject.angle = angle;
graphicalObjectArray.add(graphicalObject);
}
dynamicGraphicalObject = new GraphicalObject();
game.getCameraController().getCamera().position.x = game.getScreenController().getScreenWidth() / 2;
game.getCameraController().getCamera().position.y = game.getScreenController().getScreenHeight() / 2;
}
@Override
public void update(float deltaTime) {
for (GraphicalObject graphicalObject : graphicalObjectArray) {
Vector2 polarToCartesianPosition = BBMath.polarToCartesian(radius, graphicalObject.angle);
graphicalObject.line.x1 = centerX + 0;
graphicalObject.line.y1 = centerY + 0;
graphicalObject.line.x2 = centerX + polarToCartesianPosition.x;
graphicalObject.line.y2 = centerY + polarToCartesianPosition.y;
graphicalObject.line.color = Color.WHITE_COLOR;
graphicalObject.ellipse.x = centerX + polarToCartesianPosition.x;
graphicalObject.ellipse.y = centerY + polarToCartesianPosition.y;
graphicalObject.ellipse.width = 2 * smallCircleRadius;
graphicalObject.ellipse.height = 2 * smallCircleRadius;
graphicalObject.ellipse.color = Color.WHITE_COLOR;
}
float shift = 0;
float theta = (shift * smallCircleRadius) * (centerY / centerX);
Vector2 pos = BBMath.polarToCartesian(radius, theta);
dynamicGraphicalObject.line.color = new Color(Color.RED);
dynamicGraphicalObject.line.x1 = centerX + 0;
dynamicGraphicalObject.line.y1 = centerY + 0;
dynamicGraphicalObject.line.x2 = centerX + pos.x;
dynamicGraphicalObject.line.y2 = centerY + pos.y;
dynamicGraphicalObject.ellipse.x = centerX + pos.x;
dynamicGraphicalObject.ellipse.y = centerY + pos.y;
dynamicGraphicalObject.ellipse.width = 2 * smallCircleRadius;
dynamicGraphicalObject.ellipse.height = 2 * smallCircleRadius;
dynamicGraphicalObject.ellipse.color = new Color(Color.RED);
}
@Override
public void draw() {
game.getShapeRenderer().begin(ShapeRenderer.ShapeType.Line);
for (GraphicalObject graphicalObject : graphicalObjectArray) {
graphicalObject.line.draw();
graphicalObject.ellipse.draw();
}
dynamicGraphicalObject.line.draw();
dynamicGraphicalObject.ellipse.draw();
game.getShapeRenderer().end();
}
class GraphicalObject {
Ellipse ellipse;
Line line;
float angle;
public GraphicalObject() {
this.ellipse = new Ellipse(game);
this.line = new Line(game);
}
}
}
Which is same math like in your example, with some modifications:
However, you can notice I have this dynamicGraphicalObject (red circle), which I want to shift position around circle by using theta value calculated as (shift * smallCircleRadius) * (centerY / centerX);
. This works perfect for shift=0 value. It's properly positioned/overlapping white. But if I would change shift variable to 1, 2, 3, or 11, you can see that it's not precisely aligned with white circles. Is this floating point issue or am I missing something in calculation of theta ?
shift values used: 2,6 and 11 in order by images
--
SOLUTION:
float fixPrecision = 1.1f;
float theta = (shift * fixPrecision) + ((shift * smallCircleRadius) * (centerY / centerX));
Upvotes: 0
Reputation: 1455
See code comments for explanation.
import java.awt.*;
import java.awt.geom.Ellipse2D;
import java.awt.geom.Line2D;
import java.awt.geom.Point2D;
import javax.swing.*;
class CenteredCircle extends Ellipse2D.Double {
CenteredCircle(Point2D.Double p, double radius) {
super(p.x - radius, p.y - radius, 2 * radius, 2 * radius);
}
}
public class CircleDemo extends JFrame {
public CircleDemo() {
int width = 640; int height = 480;
setSize(new Dimension(width, height));
setLocationRelativeTo(null);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
setVisible(true);
JPanel p = new JPanel() {
@Override
public void paintComponent(Graphics g) {
Graphics2D g2d = (Graphics2D) g;
// center p1
Point2D.Double p1 = new Point2D.Double(getSize().width/2, getSize().height/2);
double radius = 130.0;
// big circle
Shape circle2 = new CenteredCircle(p1, radius);
g2d.draw(circle2);
// 12 small circles
for (int angle = 0; angle < 360; angle += 30) {
// this is the magic part
// a polar co-ordinate has a length and an angle
// by changing the angle we rotate
// the transformed co-ordinate is the center of the small circle
Point2D.Double newCenter = polarToCartesian(radius, angle);
// draw line just for visualization
Line2D line = new Line2D.Double(p1.x, p1.y, p1.x + newCenter.x, p1.y+ newCenter.y);
g2d.draw(line);
// draw the small circle
Shape circle = new CenteredCircle(
new Point2D.Double(p1.x + newCenter.x, p1.y + newCenter.y),
radius/4);
g2d.draw(circle);
}
}
};
setTitle("Circle Demo");
getContentPane().add(p);
}
public static void main(String arg[]) {
SwingUtilities.invokeLater(new Runnable() {
@Override
public void run() {
new CircleDemo();
}
});
}
static Point2D.Double polarToCartesian(double r, double theta) {
theta = (theta * Math.PI) / 180.0; // multiply first, then divide to keep error small
return new Point2D.Double(r * Math.cos(theta), r * Math.sin(theta));
}
// not needed, just for completeness
public static Point2D.Double cartesianToPolar(double x, double y) {
return new Point2D.Double(Math.sqrt(x * x + y * y), (Math.atan2(y, x) * 180) / Math.PI);
}
}
Upvotes: 1