My Math Forum Overlapping Circles & Lattice Efficiencies

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February 2nd, 2018, 09:20 AM   #1
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Overlapping Circles & Lattice Efficiencies

Hello,

I was wondering if anyone could help me with an overlapping circle problem... I'm trying to prove that the efficiency of a triangular lattice is higher than that for a square lattice.

Given the figure below:

The blue dots represent LEDs and the black lines show the lattice. The red circles represent the radiance of the LEDs. To simplify the problem, I'm assuming the radiance of each LED is the same regardless of distance from the source, for a given radius.

The left hand side of the figure shows a triangular lattice and the right hand side shows a square lattice. It can be seen that there are 4 unique levels of radiance. Let 1 be the weakest and 4 be the strongest. This gives the triangular lattice 2 radiance levels of 3 and 4 and the square lattice 3 radiance levels of 2, 3 and 4.

Finally, assume the radius of each circle is 1 and therefore the length of a side for either the triangle or square in each respective lattice will be 1.

So the question is, how do I calculate the percentage area for each radiance level for either the square or the triangular lattice?

I've successfully answered this for the triangular lattice...

The area of a triangular cell is simply: $\displaystyle \frac{\sqrt{3}}{4}a^2$

As $\displaystyle a = 1$ we get: $\displaystyle \frac{\sqrt{3}}{4}\approx 0.433$ (Eqn. 1)

The area of a circle is: $\displaystyle \pi r^2$

As $\displaystyle r = 1$ we get: $\displaystyle \pi$

Given the arc is $\displaystyle \frac{1}{6}$ of the circle, we get the area of the arc as: $\displaystyle \frac{\pi}{6}$ (Eqn. 2)

Subtracting (1) from (2) we get: $\displaystyle \frac{\pi}{6}-\frac{\sqrt{3}}{4}$

As there are three such segments in one cell, this gives: $\displaystyle 3\left ( \frac{\pi}{6}-\frac{\sqrt{3}}{4} \right )\approx 0.272$ (Eqn. 3)

To find the percentage of the arc segments in the triangular cell, simply divide (3) by (1) which gives $\displaystyle \approx 62.8\%$

So, this solves the triangular lattice, but I really can't get my head around the square lattice?

I know that each arc segment in the square lattice is $\displaystyle \frac{\pi}{4}$ and that the area of a cell is $\displaystyle 1^2=1$ but don't know how to separate each individual segment.

I look forward to hearing from you. If you have any questions, or I've missed anything out, please do let me know and I'll do my best to answer them or fill in the gaps.

---------------------------------

Kind Regards

Edward
Attached Images
 Lattice Efficiencies.jpg (22.3 KB, 26 views)

Last edited by edwardholmes91; February 2nd, 2018 at 09:24 AM.

 March 6th, 2018, 07:42 AM #2 Newbie   Joined: Mar 2018 From: Europe Posts: 5 Thanks: 0 A relatively easy part is level 2. Divide the square in half, and you get it divided in 3 pieces: a triangle, a sector and a piece of level 2.
April 9th, 2018, 07:13 AM   #3
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Joined: Mar 2016
From: UK

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Hi Everyone,

I initially tried to solve this problem using the attached scan:

I realised that I needed to create 3 unique equations in order to solve for A, B and C. As can be seen from my working out, I wasn't able to achieve this and it proved to be more difficult than I'd originally anticipated.

The two figures at the top of the scan were drawn on the computer and so I was able to overlay a grid with 20 divisions, this allowed me to create the approximations.

Quote:
 Originally Posted by snorkack A relatively easy part is level 2. Divide the square in half, and you get it divided in 3 pieces: a triangle, a sector and a piece of level 2.
Snorkack, sorry, I'm not sure as I understand what you mean, however I did find this: https://www.mathalino.com/reviewer/p...quarter-circle which sounds the same as what you mentioned?

By combining this and also the following two examples:

https://www.mathalino.com/reviewer/p...uarter-circles

https://www.mathalino.com/reviewer/p...uarter-circles

I should be able to come to a solution.

Thanks to everyone for their assistance.
Attached Images
 h002048a_09-04-2018_15-43-48.jpg (92.3 KB, 3 views)

 April 9th, 2018, 08:04 AM #4 Newbie   Joined: Mar 2016 From: UK Posts: 23 Thanks: 1 Further to my last reply, Mathalino get the following values for the different areas: $\displaystyle A=126.04\\B=17.36\\C=51.12$ Given their total area is: $\displaystyle 20^2=400$ This gives the following: $\displaystyle Total\ Area = A+4\left(B+C\right)\\=126.04 +4\left(17.36+51.12\right)\\=399.96$ With some slight rounding error. Calculated as percentages: $\displaystyle Total\ Area=\frac{126.04}{400}+4\left( \frac{17.36}{400}\right) + 4 \left( \frac{51.12}{400} \right)\\=31.51\%+17.36\%+51.12\%\\=99.99\%$ The values that I approximated for the different areas were: $\displaystyle \color{red}{A=120\\B=16\\C=50}$ As I used a 20 division grid for my approximations, I can use the same area: $\displaystyle \color{red}{20^2=400}$ Which gives the following percentages: $\displaystyle \color{red}{Total\ Area=\frac{120}{400}+4\left( \frac{16}{400}\right) + 4 \left( \frac{50}{400} \right)\\=30\%+16\%+50\%\\=96\%}$ Based on the approximations that I carried out, this would indicate that the values which Mathalino calculated are correct.

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