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Can anyone provide the name of a type of spiral described in the following, and a link to any description of the math that describes the spiral:

It is like an Archimedean spiral but with one difference.

An Archimedean spiral is the locus of points on a plane corresponding to the locations over time of a point that moves away from a fixed point with a constant speed along a line that rotates with constant angular velocity. There is a very good animated illustration on a website that shows the movement of the point in an Archimedean spiral both along the line and in rotation. The site is http://www.mathematische-basteleien.de/spiral.htm.

The spiral that I am trying to identify is the locus of points corresponding to the locations over time of a point that (a) moves away from a fixed point along a line that rotates with a constant angular velocity, and (b) moves with a constant speed relative to the plane (rather than a constant speed relative to the rotating line).

Because the point moves at a constant speed relative to the plane, its speed along the rotating line decreases as the point moves farther from the fixed center. Ultimately the point might move far enough outward on the rotating line that its tangential velocity equals its constant speed. Then the point would stop moving outward along the line and move in a circle (although I don't know whether this would occur, or whether the point would approach the circle asymptotically). In any event, before then the point moves outward in a spiral.

Thanks.

AnoNY
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  • What do you mean by constant speed "relative to the plane"? What plane? The plane it's in? What does that mean? Isn't the point moving within a fixed plane to begin with? – MPW Jan 18 '16 at 00:54
  • It is not a spiral, it is a circle. – achille hui Jan 18 '16 at 01:00
  • By "relative to the plane" yes, I mean the plane that it is in -- the speed of the point on its curved path. The Archimedean spiral is explained here, where you will see that the point moves at a constant speed relative to the line and at a constant angular velocity, so its speed in the plane increases as it moves outward on the line. http://www.mathematische-basteleien.de/spiral.htm ( – AnoNY Jan 18 '16 at 01:49
  • It is a spiral, not a circle, until the point moves far enough out along the rotating line that its tangential velocity equals its constant speed. Prior to that, the point's speed exceeds its tangential velocity -- and therefore the point continues to move outward along the line. The difference from the illustration in the link above is that the point moves outward on the line at a constant speed in the link above (in the Archimedean spiral), but in the case I describe the point moves outward on the line at a decreasing speed relative to the line as its distance from the center increases. – AnoNY Jan 18 '16 at 02:49

2 Answers2

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Your spiral can be described by the equations $$\dot \theta=k\\ \sqrt{\dot r^2+r^2k^2}=v$$ where $k$ is the angular velocity and $v$ is the velocity of the point in the plane. This gives $$\frac {dr}{\sqrt{v^2-r^2k^2}}=dt$$ which Alpha integrates to $$\frac 1k \arctan \left(\frac {kr}{\sqrt{v^2-r^2k^2}}\right)=t+c\\\frac {kr}{\sqrt{v^2-r^2k^2}}=\tan(\theta +c')$$ If you start at $r=0, \theta=0$ then $c'=0$

Ross Millikan
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There are two components, circumferential and radial. Speed along arc is constant.

In terms of $\theta$ rotation

$$ \sqrt { r^2 + {(\frac{dr}{d\theta}})^2 }= a $$ $$ \frac {dr}{d \theta} = \sqrt { a^2 - r^2 } $$

which integrates with boundary condition ($\theta =0, r=0 $ ) to

$$ r = a \sin \theta$$

It is a circle of diameter $a$ passing through origin or the fixed point.

There is no radial velocity component at $ r=a, r=0. $

Narasimham
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