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Geometric expressions for a spandrel cut at an arbitrary point

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I am after finding general geometric expressions for a quarter-circular spandrel that is split into two segments along either its domain or range (they are equal). I.e. Taking the geometry provided in the sketch below (Figure 1) I am after expressions for area and centroids x1, y1 and x2, y2 (Figure 2).

    http://oi47.tinypic.com/95y4ig.jpg
    Figure 1

    http://oi46.tinypic.com/35i3pc7.jpg
    Figure 2

    2. Relevant equations

    Area of circle = ∏r2

    Equation of quarter circle: y(x) = √(x2-r2)

    3. The attempt at a solution

    I can define the area under the quarter circle by some simple geometry, chiefly:

    Area of rectangle = r2

    Area of quarter circle = ∏r2/4

    Subtracting the two leaves me with the area of the spandrel: A = r2(1-∏/4)

    Apart from stating the obvious I am stuck on this problem.

    I think I need to find the equation of the spandrel rather than the circle to even begin solving this problem (something I haven't been able to find nor derive). I have tried to approximate the spandrel by using the equations of a spandrel of nth degree by back calculating the value of n to no avail.

    Breaking the resultant two shapes into two composite shapes (rectangle and spandrel) seems to make things even more complicated.

    If anyone can point me in the right direction I would be grateful as I have spent a while thinking about it. This is a problem directed at self study, not homework.
     
    Last edited by a moderator: Jul 9, 2012
  2. jcsd
  3. Jul 9, 2012 #2

    LCKurtz

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    Here's a re-oriented picture with what I hope are the two spandrel regions colored:

    spandrel.jpg

    The formulas for ##\bar x## and ##\bar y## for the blue region is$$
    \bar x_b =\frac{\int_c^r\int_{\sqrt{r^2-y^2}}^r x\, dxdy}{\int_c^r\int_{\sqrt{r^2-y^2}}^r 1\, dxdy}
    \ \bar y_b =\frac{\int_c^r\int_{\sqrt{r^2-y^2}}^r y
    \, dxdy}{\int_c^r\int_{\sqrt{r^2-y^2}}^r 1\, dxdy}$$

    For the green region they are$$
    \bar x_g =\frac{\int_0^c\int_{\sqrt{r^2-y^2}}^r x\, dxdy}{\int_0^c\int_{\sqrt{r^2-y^2}}^r 1\, dxdy}
    \ \bar y_g =\frac{\int_0^c\int_{\sqrt{r^2-y^2}}^r y
    \, dxdy}{\int_0^c\int_{\sqrt{r^2-y^2}}^r 1\, dxdy}$$

    [Edit] Corrected typos in the last two denominators
     
    Last edited: Jul 9, 2012
  4. Jul 9, 2012 #3
    Thank you for the reply Kurtz. Should I be taking the value of τ as the radius? I'm interpreting it to mean the value at the top/tangent.
     
  5. Jul 9, 2012 #4

    LCKurtz

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    ##r## is the radius of the circle. It is also the same ##r## in the equations of the horizontal and vertical tangent lines. Also note I corrected a couple of typos in the outer limits of the last two denominators.
     
    Last edited: Jul 9, 2012
  6. Jul 10, 2012 #5
    Thanks for the clarification. I have been playing with these equations for a short while but am going wrong somewhere. Say, for example, my radius was 10 and I cut the spandrel at c = 4. I have used autocad to produce the exact solution of Ygreen = 3.00853937. No matter what I try I am getting wildly incorrect answers.

    I should mention that I am using wxmaxima to solve this integral which leads me to believe that I am messing up my bounds. For the example I have used above, what bounds should I be using to solve for Ygreen?

    Cheers
     
  7. Jul 10, 2012 #6

    LCKurtz

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    You should use ##r=10## and ##c=4##. Maple gives the following results using those values in the integrals$$
    \bar x_b = 7.659701567,\ \bar y_b = 8.021856234,\ \bar x_g=9.751445529,
    \ \bar y_g = 3.008539377$$the last of which agrees with your value.
     
  8. Jul 10, 2012 #7

    Ray Vickson

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    Nice figure (especially the shading of non-regular shapes). What package did you use to draw it?

    RGV
     
  9. Jul 11, 2012 #8

    LCKurtz

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    A shareware program called Mayura Draw. See www.mayura.com. I did the shading by coloring the two rectangles and placing an opaque white circle in front of them.

    I used that drawing program for many years for exam figures because the nice vector graphics don't have the ugly rasterization. I highly recommend it.
     
  10. Jul 11, 2012 #9

    Ray Vickson

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    Thank you for that.

    RGV
     
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