# Is this a logarithmic spiral?

I'm trying to draw a logarithmic spiral by hand (actually I need to use a plotter to cut a spiral on wood, but that is another story) and I saw this method:

http://www.wikihow.com/Draw-A-Perfect-Spiral

it seems to me like a log spiral, because, if precisely done, the distance to the center is shortened geometrically.

Let me know if I'm correct!.

Thanks!.

• I don't know of an easy mechanical way to draw a logarithmic spiral. I would just calculate and plot points, either on graph paper or in some program. Excel will plot a scatter plot, but it is hard to maintain vertical vs horizontal scale. – Ross Millikan Apr 2 '14 at 22:34

## 1 Answer

I believe that that is an Archimedian Spiral. Each time it goes around, its shortens by the circumference of the pencil. This arithmetic decreasing is not logarithmic at all.

For more info about drawing and recognizing spirals, see http://en.wikipedia.org/wiki/Spiral

• The drawings on that website are not to scale. Think about it. Each time the pencil goes around, it shortens the thread to center by the circumference of the pencil. Therefore The distance decreased by equal amounts (the circumference of the pencil) and it is Archimedian – Asimov Apr 2 '14 at 22:27
• Actually it gets shorten by the circumference of the green thumbtack, but, yeah, it is the same. So, then my question is, how much should I shorten the string to achieve a log spiral effect?. – Artemix Apr 2 '14 at 22:39
• a golden spiral aka a logarithmic spiral with growth factor e, is easy enough to draw if you can draw rectangles and do fibonacci numbers. wikihow.com/Draw-the-Golden-Spiral – Asimov Apr 2 '14 at 22:41
• Yeah, the thing is, I need to know, based on the method previously linked (the string, thumbtack and pencil) how much should I shorten the string to achieve a logarithmic effect. This is because I need to plot on wood using a plotter. – Artemix Apr 2 '14 at 22:47
• I don't think you'll have a tidy way to do this "mechanically". A logarithmic (equiangular) spiral is self-similar: for any choice of points one "turn" (2 $\pi$ radians) apart in angle, the arclength must increase by a constant factor. It would be pretty tricky to "pay out" the right amount of string from the pencil everywhere along the path, since the amount of string would have to grow (or decrease, if you're "winding in" toward the limit point) "geometrically". [This is vaguely related to why Jacob Bernoulli got the wrong spiral engraved on his tombstone...] – colormegone Apr 2 '14 at 23:46