Archimedean spiral $\oplus$ sawtooth = circles

Question

Archimedean spiral $\oplus$ sawtooth = circles

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

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Let the unit circle be given by

$$\gamma_\text{circ}(t) = (x_\text{circ}(t),y_\text{circ}(t)) = (\cos( t),\sin( t))$$

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

Let the Archimedean spiral be given by $\gamma_\text{arc} = \gamma_\text{circ} \oplus f$ with

$$x_\text{arc}(t) = x_\text{circ}(t) + f(t)\cos( t) = \cos( t) + t\cos( t) = (1+ t)\cos( t)$$

$$y_\text{arc}(t) = y_\text{circ}(t) + f(t)\sin( t) = \sin( t) + t\sin( t) = (1+ t)\sin( t)$$

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 $f'(t) = \lfloor t\rfloor - t$, i.e. perform $\gamma_\text{arc'} = \gamma_\text{arc} \oplus f'$ with

$$x_\text{arc'}(t) = x_\text{arc}(t) + f'(t)\cos( t) = (1 + t + (\lfloor t\rfloor) - t)\cos( t) = (1+ \lfloor t\rfloor)\cos( t)$$

$$y_\text{arc'}(t) = y_\text{arc}(t) + f'(t)\sin( t) = (1 + t + (\lfloor t\rfloor) - t)\sin( t) = (1+ \lfloor t\rfloor)\sin( t)$$

What one gets for $\gamma_\text{arc'}$ are obviously circles with equally spaced radii, i.e. $\gamma_\text{arc'} = \bigcup_{k=0}^\infty (1 + k)\gamma_\text{circ}$:

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

$$x_\text{inv}(t) = \cos( t) + t\sin( t)$$

$$y_\text{inv}(t) = \sin( t) - t\cos( t) $$

compared to the Archimedean spiral (gray) given by

$$x_\text{arc}(t) = \cos( t) + t\cos( t)$$

$$y_\text{arc}(t) = \sin( t) + t\sin( t) $$

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 $\mathrm{d}s_\text{arc} = \sqrt{1+t^2}\ \mathrm{d}t$ as opposed to $\mathrm{d}s_\text{inv} = t\ \mathrm{d}t$.

When modulating the involute by the sawtooth wave $f'(t)$ as above by $\gamma_\text{inv'} = \gamma_\text{inv} \oplus f'$, i.e.

$$x_\text{inv'}(t) = x_\text{inv}(t) + f'(t)\cos( t)$$

$$y_\text{inv'}(t) = y_\text{inv}(t) + f'(t)\sin( t)$$

the different arc lengths result in a quite different picture:

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

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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?