Calculating the position of points in a circle

Question

I'm having a bit of a mind blank on this at the moment. I've got a problem where I need to calculate the position of points around a central point, assuming they're all equidistant from the center and from each other.

The number of points is variable so it's DrawCirclePoints(int x) I'm sure there's a simple solution, but for the life of me, I just can't see it :)

Solution 1

A point at angle theta on the circle whose centre is (x0,y0) and whose radius is r is (x0 + r cos theta, y0 + r sin theta). Now choose theta values evenly spaced between 0 and 2pi.

Solution 2

Given a radius length r and an angle t in radians and a circle's center (h,k), you can calculate the coordinates of a point on the circumference as follows (this is pseudo-code, you'll have to adapt it to your language):

float x = r*cos(t) + h;
float y = r*sin(t) + k;

Solution 3

Here's a solution using C#:

void DrawCirclePoints(int points, double radius, Point center)
{
    double slice = 2 * Math.PI / points;
    for (int i = 0; i < points; i++)
    {
        double angle = slice * i;
        int newX = (int)(center.X + radius * Math.Cos(angle));
        int newY = (int)(center.Y + radius * Math.Sin(angle));
        Point p = new Point(newX, newY);
        Console.WriteLine(p);
    }
}

Sample output from DrawCirclePoints(8, 10, new Point(0,0));:

{X=10,Y=0}
{X=7,Y=7}
{X=0,Y=10}
{X=-7,Y=7}
{X=-10,Y=0}
{X=-7,Y=-7}
{X=0,Y=-10}
{X=7,Y=-7}

Good luck!

Solution 4

Placing a number in a circular path

// variable

let number = 12; // how many number to be placed
let size = 260; // size of circle i.e. w = h = 260
let cx= size/2; // center of x(in a circle)
let cy = size/2; // center of y(in a circle)
let r = size/2; // radius of a circle

for(let i=1; i<=number; i++) {
  let ang = i*(Math.PI/(number/2));
  let left = cx + (r*Math.cos(ang));
  let top = cy + (r*Math.sin(ang));
  console.log("top: ", top, ", left: ", left);
}

Solution 5

Using one of the above answers as a base, here's the Java/Android example:

protected void onDraw(Canvas canvas) {
    super.onDraw(canvas);

    RectF bounds = new RectF(canvas.getClipBounds());
    float centerX = bounds.centerX();
    float centerY = bounds.centerY();

    float angleDeg = 90f;
    float radius = 20f

    float xPos = radius * (float)Math.cos(Math.toRadians(angleDeg)) + centerX;
    float yPos = radius * (float)Math.sin(Math.toRadians(angleDeg)) + centerY;

    //draw my point at xPos/yPos
}

View more solutions

Comments

[Melsi]

Excellent! Worked great for me, I already translated it to php-cairo and works great!

[Andreas]

You have flipped cos and sin functions should be sin for x and cos for y. Not the other way around.

[Brian Driscoll]

My degree in mathematics, as well as every other answer here, say you are incorrect.

[Andreas]

Hm.. well on the swedish wikipedia it says sin is x axis I know this is not secure source but then I used sin on x and cos on y my cube started moving in the right direction. Even my math teacher pointed out that I flipped them. Can you think of any other reason why my cube would move in a strange pattern away from the target location and then I flipped them it moves to it's position?

[Andreas]

This is the code I wrote maybe you could tell why it works with them flipped? jsfiddle.net/Lf5sZ

[Brian Driscoll]

@Andreas Without looking at your code I would guess that you have flipped something around somewhere, or some user input is not behaving as you expect.

[Andreas]

Looks like since the cord is moving from top left to bottom right instead of bottom left to top right things got flipped.

[Brian Driscoll]

In screen coordinates the positive y-axis is reversed, so that makes sense.

[Ruyonga Dan]

am looking to do the same kind of task, however mine is depedent on the Triggertrap/SeekArc · GitHub , when a user moves the thumb , i want to place an image to indicate that selected progress of the person....all i have tried give me the points a bit off and the not a perfec

[nirvanaswap]

The classic question- is value of pi 3.14 or 180? (i.e is the angle in deg or radian?)

[Gareth McCaughan]

Definitely radians. If you use degrees you need angles between 0 and 360 instead.

[Gareth McCaughan]

(The value of pi is 3.14ish regardless of how you prefer to write angles, of course. It is what it is.)

[Jason]

I believe the parenthesis is incorrect for $newx and $newy, putting the coordinates way outside the circle radius. Try $newx = (int)($center->getX() + ($radius * cos($angle))); and similar for $newy.

[Patrick Hund]

Perfect, thanks! Just what I was looking for.