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How to rotate a polygon?

  1. Aug 14, 2015 #1
    Let's say I have a 5 vertex polygon drawn on a x/y graph. Each of those vertices have an x/y coordinate.

    Now let's say I want to rotate the polygon 25 degrees clockwise and calculate the new coordinates of each vertex.

    How can I do that so that by knowing the original coordinate I can do a math transformation and know the resulting coordinate?

    Is matrix algebra involved? Is there a simple way that a math challenged guy like me can understand?

  2. jcsd
  3. Aug 14, 2015 #2


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    Homework Helper

    Shift everything so that the center is your origin.
    Counterclockwise rotation of a point (x;y) is the same as multiplication by a 2x2 matrix.
  4. Aug 14, 2015 #3
    Assume the center is the origin.

    A rotation of ##\theta## radians clockwise is given by the matrix

    [tex]\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{array}\right)[/tex]

    Thus if you have the original coordinates ##(x,y)##, then after the rotation, you have the coordinates (written in a column)

    [tex]\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right)[/tex]
  5. Aug 14, 2015 #4
    I have no experience or education in matrix math but I"m willing to learn.

    Do you mean to put the center of the polygon at the x/y origin? How do I determine where that is if it is a complex poly?
    And do I multiply the 2x2 matrix by the 1x2 xy matrix?

  6. Aug 14, 2015 #5
    Are you not working with a regular polygon? If not: which point is the center of rotation.

    If the center of rotation is ##(0,0)## and ##(x,y)## are the old coordinates, then the new coordinates are
    [tex](x\cos(\theta) - y\sin(\theta), x\sin(\theta) + y\cos(\theta))[/tex]
  7. Aug 15, 2015 #6


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    Science Advisor

    What do you mean by a "complex poly"?
  8. Aug 15, 2015 #7
    I mean a polygon that has many sides and is irregular in shape. Im working on an algorithm that will determine which points of the polygon are the outermost. By that I mean if I were to redraw a fence aroung the outer most points it would encompass all point including the ones inside of that. Anyway...my way of doing determining which points those are is to rotate the polygon twice and remeasure the extreme points.

    So I just need to know a good method (that I can understand) to do the rotation. I want to take the polygon and measure it in its original position, then rotate it 120 degrees and measure it again, and do that one more time.

    So it seems that all I have to do is to put the point of rotation within the polygon, note the coordinates of each vertex in that orientation then rotate about that center point 120 degrees. I can do that. Im just wondering if there is a mathematical way (such as matrix math, on which I will need some help) that will mathematically rotate the polygon and produce the new coordinates.

  9. Aug 15, 2015 #8
    For Euclidean geometry, rotation is an isometric transformation. The distances between points is preserved. I do not understand what "measurement" of the polygon means. Based on isometry, I expect that you would get the same result for each rotation.
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