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Topic: Can you parametrize this helix?
Replies: 9   Last Post: Dec 22, 2010 2:18 PM

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Posts: 12
Registered: 12/17/10
Re: Can you parametrize this helix?
Posted: Dec 21, 2010 12:30 AM
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Thank you. That's the one. I appreciate it.

Here's some things, I've realized, later, in thinking about this whole thing.

So, say: s=r(theta)

As s increases, the initial line emanating from the 11th division of the circle begins. For instance when the circle has gone 15 degrees clockwise, the first line has reached a minimum in its decent at the value 1. And as the circle reaches 30 degrees the value of the spiral is e. At 45 degrees, I believe, the spiral (the first one I mean, not the circular spiral) has a value of pi. And so on. As the circle continues along its path, the logarithmic spiral begins to turn in on itself faster and faster.

Thus, as the circle approaches the ninth division, the spirals are spiraling very fast. And then it all quits at the ninth division. So, although this first spiral isn't a logarithmic spiral, per say (I don't know what it is exactly), it might have an exponential kind of growth, of some kind, to a power of (theta), in some manner.

Just something to ponder, if anyone's interested.

I think this particular rant I'm going on now, better explains what I meant initially by the "relationship" between the circle, its growth, and the creation of the first spiral. The second spiral is also related to the growth of the circle, as well as the first spiral. So with s=r(theta), and the equations for the second spiral, the question of what these drawings mean might be answered, as long as the initial spiral can be put into terms.

If anyone is interested, I've got some more drawings that show how uniform the logarithmic spiral is in relation to the circle. I'll post them when I get a chance. But, in fact, with the circle laying flat on a plane (or at a slight angle) its possible to see how the initial spiral generates itself quite nicely in two dimensions.

Thank you again for the equations, Avni.

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