Usbrth

Archimedean spiral $oplus$ sawtooth = circles

But what about the involute of the circle instead of the Archimedean spiral?

Let the unit circle be given by

$$gamma_textcirc(t) = (x_textcirc(t),y_textcirc(t)) = (cos( t),sin( t))$$

and consider the linear function $f(t) = t$.
_{(For the sake of simplicity of the following formulas please assume that $2pi = 1$ 🙂)}

Let the Archimedean spiral be given by $gamma_textarc = gamma_textcirc oplus f$ with

$$x_textarc(t) = x_textcirc(t) + f(t)cos( t) = cos( t) + tcos( t) = (1+ t)cos( t)$$

$$y_textarc(t) = y_textcirc(t) + f(t)sin( t) = sin( t) + tsin( t) = (1+ t)sin( t)$$

The Archimedean spiral may be considered as modulating the unit circle (blue) by the linear function $f(t) = t$:

Now do the reverse:

Modulate the Archimedean spiral by the sawtooth wave $f'(t) = lfloor trfloor - t$, i.e. perform $gamma_textarc' = gamma_textarc oplus f'$ with

$$x_textarc'(t) = x_textarc(t) + f'(t)cos( t) = (1 + t + (lfloor trfloor) - t)cos( t) = (1+ lfloor trfloor)cos( t)$$

$$y_textarc'(t) = y_textarc(t) + f'(t)sin( t) = (1 + t + (lfloor trfloor) - t)sin( t) = (1+ lfloor trfloor)sin( t)$$

What one gets for $gamma_textarc'$ are obviously circles with equally spaced radii, i.e. $gamma_textarc' = bigcup_k=0^infty (1 + k)gamma_textcirc$:

You may do something similar with the involute of the circle given by

$$x_textinv(t) = cos( t) + tsin( t)$$

$$y_textinv(t) = sin( t) - tcos( t) $$

compared to the Archimedean spiral (gray) given by

$$x_textarc(t) = cos( t) + tcos( t)$$

$$y_textarc(t) = sin( t) + tsin( t) $$

The Archimedean spiral (gray) and the involute of the circle (black) are more different than one might expect by just looking at their drawings, especially by different arc lengths $mathrmds_textarc = sqrt1+t^2 mathrmdt$ as opposed to $mathrmds_textinv = t mathrmdt$.

When modulating the involute by the sawtooth wave $f'(t)$ as above by $gamma_textinv' = gamma_textinv oplus f'$, i.e.

$$x_textinv'(t) = x_textinv(t) + f'(t)cos( t)$$

$$y_textinv'(t) = y_textinv(t) + f'(t)sin( t)$$

the different arc lengths result in a quite different picture:

i.e. not distinguished circles anymore, but another but cut-off spiral.

I wonder how this can be "rescued", i.e.

1. By which other function $g(t)$ does one have to modulate the involute of the circle to get circles as for the Archimedean spiral?




2. Alternatively: By which other operation $oplus$ would $f'(t)$ yield circles for the involute of the circle?