Geometric expressions for a spandrel cut at an arbitrary point

In summary: GIMP?In summary, Kurtz is trying to find general geometric expressions for a quarter-circular spandrel that is split into two segments along either its domain or range (they are equal). He is stuck on this problem and needs the equation of the spandrel to even begin solving it. He has 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 him in the right direction, Kurtz would be grateful. This problem is not directed at homework.
  • #1
Engineering01
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Homework Statement



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

Homework Equations



Area of circle = ∏r2

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

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.
 
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  • #2
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
 
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  • #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.
 
  • #4
Engineering01 said:
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.

##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.
 
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  • #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
 
  • #6
Engineering01 said:
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

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
LCKurtz said:
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

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

RGV
 
  • #8
Ray Vickson said:
Nice figure (especially the shading of non-regular shapes). What package did you use to draw it?

RGV

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
LCKurtz said:
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.

Thank you for that.

RGV
 

FAQ: Geometric expressions for a spandrel cut at an arbitrary point

1. What is a spandrel cut and why is it important in geometry?

A spandrel cut is a type of geometric expression that involves creating a shape by cutting a portion of a geometric figure at an arbitrary point. This is important in geometry because it allows for the creation of new shapes and the exploration of mathematical concepts.

2. How is a spandrel cut different from other geometric expressions?

A spandrel cut is different from other geometric expressions because it involves cutting a portion of a figure at an arbitrary point, rather than following a specific geometric rule or formula.

3. Can a spandrel cut be used to create any shape?

No, a spandrel cut can only create shapes that are possible within the original figure. For example, if the original figure is a rectangle, a spandrel cut can create a parallelogram or a trapezoid, but it cannot create a circle.

4. How do you determine the measurements of a spandrel cut?

The measurements of a spandrel cut can be determined by using the geometric properties of the original figure. For example, if the original figure is a triangle, you can use the Pythagorean theorem to find the length of the spandrel cut.

5. What real-world applications use spandrel cuts?

Spandrel cuts can be used in architecture and engineering to create unique and visually appealing designs. They are also used in mathematics education to teach geometric concepts and problem-solving skills.

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