Wireless Electricity Project

Earlier this year I represented my school at Spectrum 2012, a science fair. We were tasked with making a project which demonstrated a new technology which would be prevalent in the future. Inspired by a TED talk I saw, I decided to make the project on wireless electricity. The talk was delivered by Eric Giler who works Dr. Solijacic at the 'Witricity' corporation.

I read up on the physics which made the efficient energy transmission work. I learned about how resonant inductive coupling works. I was fascinated by how I could related the concept to that of a swing, where a push at just the right moment yields the most efficient usage of energy. My project consisted of two coils each with a capacitor of the same value connected in parallel. This meant that each coil would have the same resonant frequency. { Frequency (F) in Hertz of an LC circuit, where L=inductance in Henries and C=capcitance in Farads,  F = 1/(2Pi*sqrt(LC))} So that both coils had the same inductance I ensured both coils had the same number of turns and the same radius.

I hooked up an oscilloscope to the output coil so that we could see the voltage induced. We needed a function generator  for the driver coil but I couldn't find one and they were VERY expensive so buying was out of the question. So I built one using some parts I had, however it couldn't meet the power requirements and I couldn't fine tune it either, and fine tuning was important if I wanted to ensure that the driver coil hit the resonant frequency. After fruitlessly searching far and wide for a function generator I could borrow my eyes caught some knobs on the oscilloscope we were using. The scope had a built in function generator! It was perfect, I could tune it exactly to the frequency we needed.

Here are some pictures:

Demonstrating how power can be transmitted through concrete/stone

Different angle

Demonstrating that there is no ionizing radiation and that we can transmit energy through water easily
(No fishes were harmed in the making of this project :) )

A view of the coils

top left you can see the rectifier I built to power an LED with the output coil

The way I understand it,  this is how resonant inductive coupling works:

First, the driver coil is energized when current passes through it, this produces a magnetic field around the coil. As a result of the magnetic field a current is induced in the output/secondary coil. The 'resonant' part comes in here. When the current in the driver coil changes direction (An  AC current is obviously driving it) the magnetic field breaks down and reverses direction. In the secondary coil, due to the current induced, will also have its own magnetic field. This magnetic field too breaks down at this point. What would now usually happen if the secondary coil did not have the same resonant frequency is that its field would break down at a different time than that of the driver coil. However as the driver frequency matches the resonant frequency of the secondary coil, the field induced will reverse direction at the same time as the one in the secondary breaks down. This means that the driver coil will actually 'aid' the breaking down of the field in the secondary coil and simultaneously induce a current in the opposite direction (producing a field in the secondary in the same direction as that of the primary). I imagine the driver coil as a sort of flexible rod attached to a pendulum (the secondary coil) which pushes and tugs at the same resonant frequency of the pendulum to make it oscillate. 

Thus the advantage of having the driver coil's frequency and resonant frequency of secondary coil the same is that the magnetic fields induced do not oppose each other and thus do not waste energy cancelling each other out. By the same token, they also actually reinforce each other and overall make the energy transfer much more efficient.

( If you have any questions or if I have left anything out or have got something wrong please email me: tuniomurtaza@gmail.com )



During the demonstration we showed how energy can be transmitted through various substances at distances significantly farther than those which can be achieved through normal inductance. ( We showed this by varying the frequency of the driver coil and using coils with different turns and capacitances.)  Ultimately the project won us the 1st prize at the fair. What I enjoyed most out of the experience was the realization that I could 'do science' myself. I've always wanted to be on the frontier where I could make new things and move mankind forward. It excited me to think that I was able to replicate the results of something real scientists were working on. It made me feel like I was on that frontier.



My team on this project without whom I could not have built the project was: Mustafa Rashid, Manish Kumar, Minaam Abbass and Saad Hirani.

- Murtaza Tunio

Project Euler Problem #2

Here is my attempt at the second problem, I got it right first shot, but again I think the way I used the while loop and if statement is not very elegant. If i make any revisions I'll update. Until till then, here:

Problem 2: 
"Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms."

Code:

//Fibonacci Problem 2

int[] nums = new int[3];

int total = 2;

nums[0] = 1;

nums[1] = 2;

while (nums[2] < 4000000) {

if (nums[2]%2 == 0) {

total += nums[2];

}

nums[2] = nums[0] + nums[1];

nums[0] = nums[1];

nums[1] = nums[2];

}

text(total,0,50);

Project Euler Problem #1

I just found this cool new website which has a bunch of interesting math/computer programming related problems. Heres my go at the first problem using Processing. It was quite trivial to solve in the inefficient way I did but some of the problems ahead are pretty intense!

Problem 1: "If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000."

Here is the code:

int total = 0;




for (int i = 0; i<1000 code="code" i="i">




  if (i%3 == 0 || i%5 == 0) {




    total += i;




  }




}







fill(0);






text(total,0,70);