My Math Forum  

Go Back   My Math Forum > High School Math Forum > Geometry

Geometry Geometry Math Forum


Reply
 
LinkBack Thread Tools Display Modes
February 2nd, 2018, 10:20 AM   #1
Newbie
 
Joined: Mar 2016
From: UK

Posts: 21
Thanks: 1

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
File Type: jpg Lattice Efficiencies.jpg (22.3 KB, 12 views)

Last edited by edwardholmes91; February 2nd, 2018 at 10:24 AM.
edwardholmes91 is offline  
 
Reply

  My Math Forum > High School Math Forum > Geometry

Tags
circles, efficiencies, lattice, overlapping



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Three non overlapping circles with a triangle nikkibunny Geometry 2 November 13th, 2016 11:22 PM
How do you show a Möbius transformation maps circles to circles and lines? math93 Geometry 0 November 3rd, 2015 03:43 PM
CI of β1 of different regression models are not overlapping Scrooge2013 Probability and Statistics 0 July 11th, 2014 12:55 PM
Finding overlapping rectangles given co-ordinates DeadViper Algebra 6 September 2nd, 2013 03:03 AM
overlapping circles rockey191 Algebra 2 July 3rd, 2009 03:08 AM





Copyright © 2018 My Math Forum. All rights reserved.