# Math Help - Concentric Circle In Square for Music

1. ## Concentric Circle In Square for Music

Hi,

I'm a Masters student of Music and I'm currently working on a project that involves writing music for a picture of concentric circles inside a square.

I came to this forum because I'm studying music and not maths so i thought someone might be able to help me with this.

I'm trying to find a way of calculating this:

If the whole picture is 100%, how much of that 100% is the circle in the middle ? (photo: File:Archery Target 80cm.svg - Wikipedia, the free encyclopedia)

Or how much percent is the first red circle ?

I hope I've explained this well.

Thanks for your time,

Fred

2. Originally Posted by freddyG
If the whole picture is 100%, how much of that 100% is the circle in the middle ? (photo: File:Archery Target 80cm.svg - Wikipedia, the free encyclopedia)
Or how much percent is the first red circle ?
I sorry to tell you that it probably cannot be done with so little information. If we knew that each circle is uniform as far as its radius is concerned then maybe. There are eleven circles on that target.
Let's say the overall target has radius R. That means the innermost circle must have radius $\frac{R}{11}$.
However, the picture in the link suggest that none of that is true.
If it were true, then we can find the area of the first red ring and divide by the total.

3. As Plato says, you can't really give an answer to this problem unless you assume that all the rings have the same width. If that is the case, then the central yellow circle will have approx 0.8% of the total area of the whole target. The inner red ring will have about 5.8% of the total area.

The percentage of the total area for each of the 11 rings in the target, starting with the central one, will be approximately

$0.8\%\quad 2.5\%\quad 4.1\%\quad 5.8\%\quad 7.5\%\quad 9.1\%\quad 10.7\%\quad 12.4\%\quad 14.0\%\quad 15.7\%\quad 17.4\%$

4. Hi,

Thank you for the quick reply. I tried what you said but it didn't really add up because I'm sure I've made a few mistakes.

I'll try to explain this better. I've attached the picture that I'm working on at the moment. What is the best mathematical way of calculating what percent each of the rings take on each of the squares ?

Is it possible ?

I'm looking for approximate values and not exact values.

Thanks again

5. Hello, freddyG!

If the whole picture is 100%, how much of that 100% is the circle in the middle?

(photo: File:Archery Target 80cm.svg - Wikipedia, the free encyclopedia)

Or how much percent is the first red circle?

I assume that the radius is divided into ten equal parts,
. . and the innermost segment is bisected.

Using a radius of 10, we have these facts:

. . $\begin{array}{cccccccc} \text{Color} & \text{Area} & \% \\ \hline \\[-4mm] & \frac{1}{4}\pi & \frac{1}{4}\%\\ \\[-4mm] \text{Yellow} & \frac{3}{4}\pi & \frac{3}{4}\%\\ & 3\pi & 3\%\\ \hline \text{Red} & 5\pi & 5\% \\ & 7\pi & 7\% \\ \hline \text{Blue} & 9\pi & 9\% \\ & 11\pi & 11\% \\ \hline \text{Black} & 13\pi & 13\%\\ & 15\pi & 15\% \\ \hline \text{White} & 17\pi & 17\%\\ & 19\pi & 19\% \\ \hline \hline \text{Total} & 100\pi & 100\%\end{array}$

6. I basically want to calculate how much space every ring takes in every square approximately. For example, on the top left square, if the whole square is 100%, how much of that percent is the blue circle in the middle ? Or the orange ring around ?

The painting is by Wassily Kandinsky and it's called Farbstudie Quadrate, http://www.globalgallery.com/prod_im...l-ad-ea078.jpg

None of these squares or circles are perfect and I'm looking for approximate values.

Is there a way of doing this ?

Thanks again

7. What you are asking about comes down to

$\frac{(\mbox{area of washer})}{(\mbox{area of square})}$

(Using the word washer instead of ring due to personal preference.)
If you assume that the square is circumscribed about the disk (the whole, that is) and you assume that each of the washers are of equal width (and that the inner disk will have radius equal to this width), then you can designate the radius of the whole disk as R and the length of a side of the square as 2R (area 4R^2).

If you want N washers (not including the inner disk), you have to divide R into N+1 pieces.

If you designate the disk as position 0, the jth washer (counted from the center) will have area

$\pi * (\frac{R}{N+1})^{2}((j+1)^2 - j^{2})$

Then the percentage (as a decimal) you are looking for is

$\frac{\pi * (\frac{R}{N+1})^{2}((j+1)^2 - j^{2})}{4R^{2}}$

$=\frac{\pi * ((j+1)^2 - j^{2})}{4(N+1)^{2}}$

Multiply this by 100 and append a percentage sign and you've got your answer.

So basically, in the first panel of that Kandinsky the black(ish?) washer is number 2 (so j=2), and I count 4 distinct washers (N = 4).
Then the black washer is

$\frac{\pi * ((3)^2 - 2^{2})}{4(4+1)^{2}}=\frac{5\pi}{100}\approx 15.7$ %

of the square's area.

If you try to check your math, note that the whole disk (the washers plus the inner disk) is

$\frac{\pi}{4}\approx 78.5$%

of the square.