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I have this polar function:

r = A / log(B * tan(t / 2 * N)

where A, B, N are arbitary parameters and t is the angle theta in the polar coordinate system.

Example graph for A=8, B=0.5, N=4

Sample graph

How can I plot this function onto a cartesian coordinate grid so I get an image like the one above?



migration rejected from Apr 4 '15 at 4:44

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closed as off-topic by Jonas Meyer, Sally, Newb, David K, Zev Chonoles Apr 4 '15 at 4:44

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-1. Seriously look at and find out how to convert (r,t) into (x,y). If you have a programming question then post it again at SO. – ja72 Jan 1 '11 at 21:39
This is not a simple matter of converting polar to cartesian so this is in fact a programming question. If you would have bothered to use your brain you could've probably figured that out by reading my answer to this question which I posted about 43 minutes before your unnecessary comment and downvote. Jerk. – Raoul Duke Jan 1 '11 at 21:58
Whether it is a valid description doesn't change the fact that it is unprofessional. I also don't understand why this question was migrated. – Qiaochu Yuan Jan 2 '11 at 15:03

Incomplete pseudocode sample, but you should get the idea:

for t in [0, 2pi):
    r = /* whatever you got depending on t */
    x = r * cos(t)
    y = r * sin(t)
    draw line to (x,y)
Thanks for your answer but I'm sorry to say this is not very helpful. This won't produce a smooth curve. How would you do that exactly with this pseudocode? Also please elaborate on how to iterate from 0 to 2pi because this is a continuous range of real numbers this isn't as easy as your pseudocode makes it out to be. – Raoul Duke Jan 1 '11 at 20:10
The answer exactly matches the question. Never in the original question metions that the answer has to produce a smooth curve (to what criteria?) – ja72 Jan 1 '11 at 21:34
@ jalexiou I wrote "so I get an image like the one above" – Raoul Duke Jan 1 '11 at 22:00

Ok, I figured it out. Some example Java code:

import static java.lang.Math.*;

import java.awt.Color;
import java.awt.Graphics;
import java.awt.Point;
import java.awt.image.BufferedImage;

import javax.swing.ImageIcon;
import javax.swing.JFrame;
import javax.swing.JLabel;

public class TestPolarPlot {
    public static void main(String[] args) {
    final int width = 512;
    final int height = 512;
    BufferedImage img = new BufferedImage(width, height, BufferedImage.TYPE_4BYTE_ABGR);
    Graphics g = img.getGraphics();
    g.fillRect(0, 0, width, height);
    final double A = 8;
    final double B = 0.5;
    final double N = 4;
    final double scale = 128;
    final double zoom = 50;
    final double step = 1 / scale;
    Point last = null;
    final Point origin = new Point(width/2, height/2);

    for (double t = 0; t <= 2*PI; t+= step) {
        final double r = zoom * polarFunction(t, A, B, N);
        final int x = (int)round(r * cos(t));
        final int y = (int)round(r * sin(t));
        Point next = new Point(x, y);
        if (last != null) {
            g.drawLine(origin.x + last.x, origin.y + last.y,
                origin.x + next.x, origin.y + next.y);
        last = next;

    JFrame frame = new JFrame("testit");
    frame.getContentPane().add(new JLabel(new ImageIcon(img)));

    public static double polarFunction(double t, double A, double B, double N) {
        return A / log(B * tan(t / (2 * N)));

I didn't expect this to create smooth curves but it works pretty well.

alt text alt text


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