# Homework Help: Geometric expressions for a spandrel cut at an arbitrary point

1. Jul 9, 2012

### Engineering01

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. Jul 9, 2012

### LCKurtz

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

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$$\bar x_g =\frac{\int_0^c\int_{\sqrt{r^2-y^2}}^r x\, dxdy}{\int_0^c\int_{\sqrt{r^2-y^2}}^r 1\, dxdy} \ \bar y_g =\frac{\int_0^c\int_{\sqrt{r^2-y^2}}^r y \, dxdy}{\int_0^c\int_{\sqrt{r^2-y^2}}^r 1\, dxdy}$$

 Corrected typos in the last two denominators

Last edited: Jul 9, 2012
3. Jul 9, 2012

### Engineering01

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.

4. Jul 9, 2012

### LCKurtz

$r$ 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
5. Jul 10, 2012

### Engineering01

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

6. Jul 10, 2012

### LCKurtz

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.

7. Jul 10, 2012

### Ray Vickson

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

RGV

8. Jul 11, 2012

### LCKurtz

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.

9. Jul 11, 2012

### Ray Vickson

Thank you for that.

RGV

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