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Calculating the dimensions of an arc

  1. Jun 20, 2011 #1
    Dear all,

    I feel this should be a simple problem but I can't solve it. Could you give me a hand?

    Imagine if an arc is bounded by a rectangle of dimensions width and height. The arc starts in the bottom left corner of the rectangle, and ends in the bottom right corner. The apex of the arc is the center of the top side of the rectangle. A picture of this is shown half way down this page:
    http://www.mathopenref.com/chordsintersecting.html" [Broken]

    Assuming that the arc is less than a semi circle, if I know the length of the arc and the height of the rectangle, how can I calculate it's width?

    Any help gratefully received,

    kenneth
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 20, 2011 #2

    micromass

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    Hi kennethgilpin! :smile:

    So you know some calculus? In my opinion that's the easiest way to solve this.

    First we will have to find the equation of the circle given the coordinates of the rectangle.
    Say that the vertices of the rectangle have coordinates (0,L),(0,-L),(H,L),(H,-L).

    The general equation of a circle is

    [tex](x-x_0)^2+(y-y_0)^2=R^2[/tex]

    with (x0,y0) the center of the circle and R the radius. We see easily that the center of the circle must lie on the y-axis, and thus x0=0.

    Now, what you have to do is to find y0 and R such that (0,L) and (H,0) lie on the circle

    [tex]x^2+(y-y_0)^2=R^2[/tex]
     
  4. Jun 20, 2011 #3

    I like Serena

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    @MM: I'm afraid this won't work.
    It's the arc length that is given, and the width that is asked.

    I believe that the resulting equations can not be solved algebraically.
    Of course it can be approximated numerically.
     
  5. Jun 20, 2011 #4

    micromass

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    Yes, I know. What I was attempting is to find an equation which calculated the arc length if you have the length and width given. Then I would use this equation to find an equation calculating the width... I think this ought to work, given that the equations are not too difficult...
     
  6. Jun 20, 2011 #5

    I like Serena

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    I created the set of equations:

    [tex]
    \begin{eqnarray}
    S &=& 2 R \alpha \\
    W &=& 2 R \sin \alpha \\
    R &=& H + R \cos \alpha
    \end{eqnarray}
    [/tex]

    where S is the arc length, and [itex]\alpha[/itex] is half of the angle of the arc.

    If you try to solve it for W, with given S and H, you'll quickly see that you're left with an equation that is afaik not solvable algebraically.
     
  7. Jun 21, 2011 #6
    thank you for your replies. @likeSerena - i got similar equations to you which I could not solve.

    This is for use within an iterative model so I will try to solve it iteratively.

    kenneth
     
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