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Finding coordinates along an arc, given arc length

  1. Apr 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find coordinates of a piont on an arc from begining coordinates at the arc distance of 6.6821. Given Radius of 25 at (100,100), Begining coordinates at (125,100), Ending Coordinates at (115.6994,119.6301), with an overall arc length of 22.4472.

    2. Relevant equations

    Newton's method, maybe?


    3. The attempt at a solution

    I can think of a way to do it with trig, finding the chords between the end points and the length to the desired point then using the cosign law to find my angles, but I tried drawing this up in CAD as a check and my answer didn't match exactly. I can't figure out how to set it up using calculus methods.
     
  2. jcsd
  3. Apr 13, 2011 #2

    Stephen Tashi

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    Science Advisor

    I don't understand the significance of the "Ending Coordinates" data.

    You can convert arc length to a radian measure of an angle theta between two radii that go to the ends of the arc.

    If a circle has center at point (Cx,Cy) and point (Px,Py) is on it then the radius drawn to (Px,Py) makes some angle phi with x_axis ( and also with a horizontal line drawn through (Cx,Cy)) You can find phi. The end of the arc is on a radius that makes angle (phi + theta) with that line. So you can find it's coordinates.

    The problem will be simpler to visualize if you convert to a coordinate system where the center of the origin is at (0,0). This amounts to subtracting the coordinates of the circle from the given coordinate data and then adding them back after you find the answer.
     
  4. Apr 13, 2011 #3

    LCKurtz

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    You don't need much calculus. You know the radius and arc length to the second point so you can calculate the angle between by s = rθ. Then you can use the fact that the arc length is proportional to the angle to calculate the angle β to your other point.

    Then just use polar coordinates at (100,100) to get the x and y coordinates.
     
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