The cube is flooded with deuterium gas plasma and all the positively charged ions have less potential energy as they move towards the center. Hopefully with a high enough voltage you could get them to fuse.

## Friday, June 24, 2011

### In 3d

Continuing that last thing , I found out that if you have a cube, the total of the distances to the 8 corners squared generates a spherical gradient much like the square made a circular one. So I thought my first attempt at designing a charge concentrator could look like this (sort of):

## Tuesday, June 21, 2011

### Circle from Square

I got thinking of a thought experiment where you are in a square room and thinking about the four distances to each of the four corners all added together. I wanted to know what that would look like if you graphed it at different points around the room. I was having trouble visualizing it though so I used Maple (a math program) to graph it for me.

The formula up top is adding the four distances to the corners up... I was surprised at the graph, it's like a black hole! Looking at it for a couple seconds starts tricking my eyes like an optical illusion. The darkest part is where the total distance from the four corners is the least, 2*(square root of 2) and it's the most at each corner 2+(square root of 2) . I was suprised at how you can make a circle by saying where this function is equal to say, 3. That's different than the way you usually define a circle as a certain distance from the center.

*EDIT*

I worked some more on the math behind it but if you just want to hear what I think might be practical applications you can skip to the *APPLICATIONS* below....

A professor of mine, Dr. Rose, was asking me how I knew that that where the above graph is the same color creates circles, not some other shape that just appears circular at a glance. Well, the way I defined it above where the four distances to the corners are added up ends up being pretty hard for me to prove. But I did prove that adding the

In polar coordinates, the distances A^2+B^2+C^2+D^2 from a point somewhere within the square to the four corners is:

Which at first doesn't look very simple or nice, but through the magic of my math program Maple an equivalent formula or simplification to that above is:

Now that is much nicer. In polar coordinates the fact that the theta cancels out means the distances squared to the four corners added together only depend on how far you are from the center of the square, not the angle from the center. In other words, you can move around the center of the square in a circle and not change the sum of the squares of the distances to the corners. The graph still looks pretty much the same as the one in the beginning of this post but the dark area is a bit tighter towards the center.

*APPLICATIONS*

So my formula above says that if you have four equal forces pulling towards something from the corners of a square, the place where it should go to get as close as it can to all four is the very center. And it's like a funnel (circular) that will move anything affected by the forces to the center.

So, in fusion experiments, what I've seen is they try to use a bunch of magnets around the middle in a circle to "push" what they are trying to concentrate in the middle, but I've shown that you can concentrate it with just four magnets flipped the other way "pulling" in the four opposite directions. It seems kind of counter intuitive at first, but the math shows that it would work. And with a circle of magnets pushing in, there will usually be dead spots in between the magnets that allow the material to escape, but with just these four magnets pulling you get a perfect funnel type field to concentrate the particles.

The formula up top is adding the four distances to the corners up... I was surprised at the graph, it's like a black hole! Looking at it for a couple seconds starts tricking my eyes like an optical illusion. The darkest part is where the total distance from the four corners is the least, 2*(square root of 2) and it's the most at each corner 2+(square root of 2) . I was suprised at how you can make a circle by saying where this function is equal to say, 3. That's different than the way you usually define a circle as a certain distance from the center.

*EDIT*

I worked some more on the math behind it but if you just want to hear what I think might be practical applications you can skip to the *APPLICATIONS* below....

A professor of mine, Dr. Rose, was asking me how I knew that that where the above graph is the same color creates circles, not some other shape that just appears circular at a glance. Well, the way I defined it above where the four distances to the corners are added up ends up being pretty hard for me to prove. But I did prove that adding the

*square*of each distance together results in a circle. It goes like this:In polar coordinates, the distances A^2+B^2+C^2+D^2 from a point somewhere within the square to the four corners is:

Which at first doesn't look very simple or nice, but through the magic of my math program Maple an equivalent formula or simplification to that above is:

Now that is much nicer. In polar coordinates the fact that the theta cancels out means the distances squared to the four corners added together only depend on how far you are from the center of the square, not the angle from the center. In other words, you can move around the center of the square in a circle and not change the sum of the squares of the distances to the corners. The graph still looks pretty much the same as the one in the beginning of this post but the dark area is a bit tighter towards the center.

*APPLICATIONS*

So my formula above says that if you have four equal forces pulling towards something from the corners of a square, the place where it should go to get as close as it can to all four is the very center. And it's like a funnel (circular) that will move anything affected by the forces to the center.

So, in fusion experiments, what I've seen is they try to use a bunch of magnets around the middle in a circle to "push" what they are trying to concentrate in the middle, but I've shown that you can concentrate it with just four magnets flipped the other way "pulling" in the four opposite directions. It seems kind of counter intuitive at first, but the math shows that it would work. And with a circle of magnets pushing in, there will usually be dead spots in between the magnets that allow the material to escape, but with just these four magnets pulling you get a perfect funnel type field to concentrate the particles.

## Wednesday, June 8, 2011

### A form of polynomial

I was looking at a function f(x) that passes through the points (0, a), (1, b), (2, c), (3, d). This is what I came up with:

(1-x)*(1/2)(2-x)*(1/3)*(3-x)*a+

(x) *(2-x)*(1/2)*(3-x)*b +

(1/2)*x *(x-1) *(3-x)*c +

(1/3)*x *(1/2)*(x-1)*(x-2)*d

If I had considered 5 terms instead the fifth would be:

(1/4)*x *(1/2)*(1/3)*(x-1)*(x-2)*(x-3) *(5-x) * e

So it generalizes pretty well.

The reason the formula I found for going through four points works is if you look at the four lines of the formula, at x=0 all that preceeds A is equal to 1, and at least one of those is 0 at every x after that so A is being multiplied by 0. The same is true for each line for x=0,1,2,3 respectively A,B,C,D are being multiplied by 1 at x=i and 0 elsewhere so the ith term is all that doesn't cancel out.

Anyway a useful feature of this type of interpolation is that if you know the f(x) values at x=0,1,... you can easily put it into this form which is a very nice and simple form (practically factored) to look at f(x) in. .

(1-x)*(1/2)(2-x)*(1/3)*(3-x)*a+

(x) *(2-x)*(1/2)*(3-x)*b +

(1/2)*x *(x-1) *(3-x)*c +

(1/3)*x *(1/2)*(x-1)*(x-2)*d

If I had considered 5 terms instead the fifth would be:

(1/4)*x *(1/2)*(1/3)*(x-1)*(x-2)*(x-3) *(5-x) * e

So it generalizes pretty well.

The reason the formula I found for going through four points works is if you look at the four lines of the formula, at x=0 all that preceeds A is equal to 1, and at least one of those is 0 at every x after that so A is being multiplied by 0. The same is true for each line for x=0,1,2,3 respectively A,B,C,D are being multiplied by 1 at x=i and 0 elsewhere so the ith term is all that doesn't cancel out.

Anyway a useful feature of this type of interpolation is that if you know the f(x) values at x=0,1,... you can easily put it into this form which is a very nice and simple form (practically factored) to look at f(x) in. .

## Wednesday, June 1, 2011

### Lateral Doppler-Like shift

So it's well known that if you are moving away or towards an emitter of electromagnetic radiation you will perceive what is called a red shift or a blue shift depending on whether you are moving away or towards respectively. The red shift means the frequency of the wave seems to decrease as you move quickly away and the blue shift means the frequency seems to increase as you move quickly towards. Think if you were in a boat at the ocean and moving into the waves each wave would reach the boat more quickly than if you were riding away from the source of the waves.

So anyway today I was reading about relativity and I had this thought experiment:

A spaceship launched from the planet and accelerated to a significant fraction of the speed of light. Here a wave is being sent from the line A to B. Let's say it's a radio signal and he has a radio on his ship so he can hear some music playing. At a certain velocity it would seem to him because of relativity that he could hear the signal on his radio for 2 seconds but to us it looked like it took him four seconds to pass through the area where the waves were... this is known as the Twin's Paradox or Time dilation if you want to read on Wikipedia.

But if we saw him pass through for 4 seconds and he only experienced 2 seconds of radio, it seems to me that he must have heard on his radio music at twice the frequency than was being sent out. I mean he heard what to us was 4 seconds of music but in 2 seconds so the singers must have sounded like chipmunks.

So anyway that's why I call it a lateral Doppler-like ship because he's moving perpendicular to the waves and experiences a blue shift as if he were moving rapidly towards the waves.

So anyway today I was reading about relativity and I had this thought experiment:

A spaceship launched from the planet and accelerated to a significant fraction of the speed of light. Here a wave is being sent from the line A to B. Let's say it's a radio signal and he has a radio on his ship so he can hear some music playing. At a certain velocity it would seem to him because of relativity that he could hear the signal on his radio for 2 seconds but to us it looked like it took him four seconds to pass through the area where the waves were... this is known as the Twin's Paradox or Time dilation if you want to read on Wikipedia.

But if we saw him pass through for 4 seconds and he only experienced 2 seconds of radio, it seems to me that he must have heard on his radio music at twice the frequency than was being sent out. I mean he heard what to us was 4 seconds of music but in 2 seconds so the singers must have sounded like chipmunks.

So anyway that's why I call it a lateral Doppler-like ship because he's moving perpendicular to the waves and experiences a blue shift as if he were moving rapidly towards the waves.

### A non random normallish curve

I started thinking about the idea that a normal curve can be found if you flip a large set of coins many times and count how many heads you get each time. Then plot how many times you got each possible number of heads. You'll end up getting no heads or all heads on all the coins very few times and about half heads the most. This makes a normal curve.

Well I started thinking what if you take the randomness out, and assume that you flipped every possible outcome of a series of heads and tails for say, 10 coins, exactly once. There would be 2^10 or 1024 possible outcomes, each one would show somewhere between 0 total heads or all 10 heads. I also figured that one can use the combinatoric formula called n choose r for telling how many ways there were to get each total number of heads. For instance, for 10 coins and you want to know how many ways there are to have 5 heads on the coins you can do 10 choose 5 or 10!/5!*(10-5)! = 252. Maple the math program actually took a few seconds and derived this formula:

This is a good check, it means if you add the 252 I just got for five with the formula for all the other numbers from 0 to 10 they add up to 2^10 or 1024 which should be the total number of ways to flip 10 coins.

So the next thing was to graph the combinatoric part over the 2^10 whole as you vary r from 0 to 10:

Maple has a way of estimating factorials even in between integers where factorial isn't typically thought of existing but as you can see this leads to a nice normal looking curve. The height is different than the standard normal distribution, though, here is the formula and graph one typically uses:

You can see this formula has all kinds of e's and pi's whereas mine has only factorials and 2^n. But the overall shape is very similar.

Well I started thinking what if you take the randomness out, and assume that you flipped every possible outcome of a series of heads and tails for say, 10 coins, exactly once. There would be 2^10 or 1024 possible outcomes, each one would show somewhere between 0 total heads or all 10 heads. I also figured that one can use the combinatoric formula called n choose r for telling how many ways there were to get each total number of heads. For instance, for 10 coins and you want to know how many ways there are to have 5 heads on the coins you can do 10 choose 5 or 10!/5!*(10-5)! = 252. Maple the math program actually took a few seconds and derived this formula:

This is a good check, it means if you add the 252 I just got for five with the formula for all the other numbers from 0 to 10 they add up to 2^10 or 1024 which should be the total number of ways to flip 10 coins.

So the next thing was to graph the combinatoric part over the 2^10 whole as you vary r from 0 to 10:

Maple has a way of estimating factorials even in between integers where factorial isn't typically thought of existing but as you can see this leads to a nice normal looking curve. The height is different than the standard normal distribution, though, here is the formula and graph one typically uses:

You can see this formula has all kinds of e's and pi's whereas mine has only factorials and 2^n. But the overall shape is very similar.

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