Percentage area of circles overlapping a triangle

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February 21st 2008, 07:20 PM
Jemma Lovell

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Percentage area of circles overlapping a triangle

Hi all,

I have an interesting problem regarding the area coverage of a triangle, in relation to the diameter of three equal circles centred at each of its three vertices (please see attached image).

Is it possible to formulate a function that will exact the required circle diameter for a given percentage-coverage of the triangular area? For example, what would be the correct circle diameter for an 89% coverage of the triangular area (i.e. 11% of the triangular area is not obscured).

As the circles will need to overlap in order to accommodate the complete 0% - 100% range, is it possible to create a unified formula or will the overlapped and non-overlapped percentages need to be addressed separately?

Any help or insight would be most appreciated.

Best regards, Gemma.
February 21st 2008, 10:47 PM
earboth

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Quote:

Originally Posted by Jemma Lovell

I have an interesting problem regarding the area coverage of a triangle, in relation to the diameter of three equal circles centred at each of its three vertices (please see attached image).

Is it possible to formulate a function that will exact the required circle diameter for a given percentage-coverage of the triangular area? For example, what would be the correct circle diameter for an 89% coverage of the triangular area (i.e. 11% of the triangular area is not obscured).

...

The 3 sectors of the 3 circles which are in the area of the equilateral triangle form a half circle. Let s denote the side of the triangle then the not obscured area can be calculated by:

$A_{n.o} = A_{triangle} - \frac12 \cdot \pi r^2~,~0 \leq r \leq \frac23 s$

$A_{n.o} = \frac14 \cdot s^2 \cdot \sqrt{3} - \frac12 \cdot \pi r^2~,~0 \leq r \leq \frac23 s$

That means: If you know the length of s and the value of $A_{n.o}$ you can calculate the value of r.
February 22nd 2008, 03:52 PM
Jemma Lovell

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Hi Earboth,

Thank you for the reply.

I think I understand what is going on here, but I still don't see how this works as the coverage approaches 100% (please see attached plot).

Sorry for being a little dense and thank you for your patience.

Best regards, Gemma.
February 22nd 2008, 09:31 PM
earboth

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Quote:

Originally Posted by Jemma Lovell

Hi Earboth,

Thank you for the reply.

I think I understand what is going on here, but I still don't see how this works as the coverage approaches 100% (please see attached plot).

Sorry for being a little dense and thank you for your patience.

Best regards, Gemma.

I'm sorry that I didn't take into account that there will be overlapping areas if the radius is greater than $\frac12 s$ .

If the radius is $0 \leq r \leq \frac12 s$ the formula of the previous post works fine. For those cases that $\frac12 s < r \leq \frac23 s$ you have to perform a little bit extra work:

The complete area which covers the triangle and which is produced by the 3 circles consists of
6 right triangles (2 of them are painted in orange)
3 small sectors

1. Calculate first the length of h: $h = \sqrt{r^2 - \frac14 s^2}$

2. Calculate the angle $\beta$ : $\tan (\beta) = \frac h{\frac12 s}=\frac{\sqrt{r^2 - \frac14 s^2}}{\frac12 s}~,~0^\circ \leq \beta \leq 30^\circ$

3. Then $\alpha = 60^\circ - 2 \beta$

4. Then the area of the triangle which is not covered by any of the circles consists of:

$A_{not\ covered}= A_{triangle} - 6 \cdot A_{right\ triangles} - 3 \cdot A_{small\ sectors}$ . That means:

$A_{nc} = \frac14 \cdot s^2 \cdot \sqrt{3} - 6 \cdot \frac12 \cdot \frac12 s \cdot \sqrt{r^2 - \frac14 s^2} - 3 \cdot \frac{\alpha}{360^\circ}\cdot \pi r^2~,~\frac12s < r \leq \frac23 s$

5. This is the rough version of the formula. But I believe it is easier to see what's going on and how I got this result if I leave it in this form. It is up to you to simplify this equation a little bit (if possible).