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Spiraling Leg equation


Unity+

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Well, this is something that isn't really useful, but more fun.

 

Well, here is the idea. The idea is there is a line of size x which then another line is placed at the other line's midpoint, or any given point for generalization, where one end of the new line is positioned at this particular coordinate. This is to repeat onto infinity.

 

The idea is to find the distance between the end of the first line to the end of the last line that is placed in the particular fashion told about above.

 

diagram1.png

 

The distance of this is equal to the following function, given x as the length of the first line and F being the fraction at which to place the new line at on the original line.

 

[math]\eta (x,F)=\frac{Fx - x-\sum_{n=1}^{\infty }\frac{Fx}{F^{2n+1}}}{F}[/math]

The way this was found was first to add up all the lengths of the lines that added distance to the line that is being analyzed. Since the length of the lines are being halved their parents, the way to find the distance is to simply subtract the sum of these lines from the length x of the original line. Also, the beginning must be taken into account when subtracting the sum because the original line's height must also be accounted for to find the distance.

 

[math]\eta (x,F)=x-\frac{x}{F}-\sum_{n=1}^{\infty }\frac{x}{F^{2n+1}}[/math]

 

[math]\eta (x,F)=\frac{Fx - x}{F}-\sum_{n=1}^{\infty }\frac{x}{F^{2n+1}}[/math]

 

[math]\eta (x,F)=\frac{Fx - x-F\sum_{n=1}^{\infty }\frac{x}{F^{2n+1}}}{F}[/math]

 

[math]\eta (x,F)=\frac{Fx - x-\sum_{n=1}^{\infty }\frac{Fx}{F^{2n+1}}}{F}[/math]

 

I'm going to be adding more to this later on, but that is what I have so far. There are some other interesting patterns I am finding when I modify the rules. Tell me if I have some incorrect math here if I do.

Edited by Unity+
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Here is a more generalized equation that both takes into account the fraction of the line the midpoint is and the size of each line after each new line.

 

[math]\eta (x,F)=\frac{\kappa x-x}{\kappa }-\sum_{n=1}^{\infty }\frac{1}{F^{2n}\kappa}[/math]

 

Where F is the new fraction size after each time while κ is the positioning of the newline based on what particular fraction the new point will be on.

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Unity - you are obviously a far better mathematician than I, but permit me to make a few observations on the style and presentation of your idea. I don't think I can help you on the maths - but I think that your communication of your idea may be hindering those who could help from engaging.

Well, this is something that isn't really useful, but more fun.

Well, here is the idea. The idea is there is a line of size x which then another line is placed at the other line's midpoint, or any given point for generalization, where one end of the new line is positioned at this particular coordinate. This is to repeat onto infinity.

Take a line L_0 of length x_0 starting at the origin in the positive y_hat direction. A point p_0 on L_0 is defined as the point dividing L_0 in the ratio given by the fraction F. At point p_0 a new line L_1 of length ? (I cannot see where you specify this length - guessing x_0.F surely not x_0/F) is drawn; this line L_1 is orthogonal to L_0 and rotated by pi/2 radians clockwise. A similar process is followed on line L_1 to create line L_2 although (for some unspoken reason) L_2 is rotated by -pi/2. This process is continued by iteration with lines alternating between positive y_hat direction and positive x_hat. The distance mu is defined as the distance between the origin and the end of line L_n as n-> infinity.

 

It is more verbose - but if you specify everything you start to realise what you have assumed internally and not spelled out in your description. For example - your diagram has M, alpha, and beta - none of which are defined. It also appears at first glance that you have defined x_hat to be up the page which just smells wrong (even if upon investigation it becomes clear you have not


For the simplest form where F=1/2 I would do it this way (I hope this is right):

 

1. set length of L_0 to 2 makes things easier - and it must be a simple multiplier that can be removed from calcs

2. start at origin - makes life easier

3. calculated distance in x and y separately - they are orthogonal and cannot mix

4. L_even go in y, L_odd go in x

5. Distance moved is away from origin is half length of line. And is halved for each line - and quartered for each line in same direction

6. Distance moved in y is 1 + 1/4 + 1/16 ...+ 1/2(2n-2)

7. Distance moved in x is 1/2 + 1/8 +1/32 ... + 1/2^(2n-1)

8. The two infinite sums solved easily to y distance being 4/3 and x distance being 2/3

9. Pythagoras gives distance mu as sqrt (20/9)

 

This is easily generalisable as the 2 in the 1/2(2n-2) or 1/2^(2n-1) is the division of the line - if you move a third of the way along the line that will be a 3; ie it is 1/F. You can then just work out the new sums and use pythagoras.

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Unity - you are obviously a far better mathematician than I, but permit me to make a few observations on the style and presentation of your idea. I don't think I can help you on the maths - but I think that your communication of your idea may be hindering those who could help from engaging.

 

Take a line L_0 of length x_0 starting at the origin in the positive y_hat direction. A point p_0 on L_0 is defined as the point dividing L_0 in the ratio given by the fraction F. At point p_0 a new line L_1 of length ? (I cannot see where you specify this length - guessing x_0.F surely not x_0/F) is drawn; this line L_1 is orthogonal to L_0 and rotated by pi/2 radians clockwise. A similar process is followed on line L_1 to create line L_2 although (for some unspoken reason) L_2 is rotated by -pi/2. This process is continued by iteration with lines alternating between positive y_hat direction and positive x_hat. The distance mu is defined as the distance between the origin and the end of line L_n as n-> infinity.

 

It is more verbose - but if you specify everything you start to realise what you have assumed internally and not spelled out in your description. For example - your diagram has M, alpha, and beta - none of which are defined. It also appears at first glance that you have defined x_hat to be up the page which just smells wrong (even if upon investigation it becomes clear you have not

 

 

For the simplest form where F=1/2 I would do it this way (I hope this is right):

 

1. set length of L_0 to 2 makes things easier - and it must be a simple multiplier that can be removed from calcs

2. start at origin - makes life easier

3. calculated distance in x and y separately - they are orthogonal and cannot mix

4. L_even go in y, L_odd go in x

5. Distance moved is away from origin is half length of line. And is halved for each line - and quartered for each line in same direction

6. Distance moved in y is 1 + 1/4 + 1/16 ...+ 1/2(2n-2)

7. Distance moved in x is 1/2 + 1/8 +1/32 ... + 1/2^(2n-1)

8. The two infinite sums solved easily to y distance being 4/3 and x distance being 2/3

9. Pythagoras gives distance mu as sqrt (20/9)

 

This is easily generalisable as the 2 in the 1/2(2n-2) or 1/2^(2n-1) is the division of the line - if you move a third of the way along the line that will be a 3; ie it is 1/F. You can then just work out the new sums and use pythagoras.

The equations that I worked with mainly work with the y direction. You added on to the idea, which I appreciate greatly.

 

Yes, I have difficulty presenting ideas. It has always been my weakest points. Thank you for taking the time to allow other members to understand it. I will try to improve my skills in presentation.

 

EDIT: I need to make a correction in my generalized equation:

 

[math]\eta(x,F)=\frac{\kappa x-x}{\kappa }-\sum_{n=1}^{\infty }\frac{x}{F^{2n}\kappa}[/math]

Edited by Unity+
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I will look out some better references for you.

 

Wiki is a good place to start, but often rather abrupt.

 

A good book for a mathematician to start with is

 

Fractal Geometry

 

Kenneth Falconer (Wiley)

 

(He has written several others as well)

Edited by studiot
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