Light Amusement Brooming Tubs

Question

If you've ever brushed the sides of a tub with a wide upright broom you'd have noticed that only the ends touch the tub wall. Consider the tub sides elliptical in shape so the broom forms a chord that is deeper at major axis ends of the ellipse than the minor.

I) Is there a condition where the handle (midpoint of brush chord) traces a circle?

Obvious limiting cases are: (i) a zero chord width broom path matching the ellipse itself when circular, and then (ii) a tub Diameter width(ed) broom chord, whose mid-point is fixed to a Center of a then also circular tub.

II) If it ever does traverse a circle between those limits, then is it (i) a unique configuration (of major, minor axes and chord lengths) or is it (ii) a many solution (functional or manifold) in R^2?

For extra credit, III) If we admit Complex fields can we always find a circular path for the now imaginary broom?

Heh heh, on the complex field (light amusement types please?). — MKhomo, Apr 18 '21 at 11:06
Related: "What is the path equation that is created with the middle point of a fixed length line segment that touching both ends to an ellipse." ... In particular, my answer shows that, in the real case, the midpoint path is typically described by a degree-$4$ polynomial in $x$ and $y$. I believe it's not difficult to show that this reduces to a circle (and a point) when and only when $a=b$. — Blue, Apr 18 '21 at 16:10

Paul Sinclair · Accepted Answer · 2021-04-18T14:15:09.567

1

Suppose that the midpoints did trace a circle.

By symmetry, the chords bisected by the major and minor axes of the ellipse have to be tangent to the circle. But if $M, E$ are the midpoint and an endpoint of a chord bisected by the major axis, and $m, e$ are the same for the minor axis, and $C$ is the center of the circle, then $CM = Cm$ because $M, m$ are both points on the circle, and $ME = me$ since the chords are all the same length. Further $\triangle CME$ and $\triangle Cme$ are both right triangles. Thus by Pythagorus, we must have $CE = Ce$. But that is only true when the ellipse is a circle.

I.e., the only time the midpoints of the chords trace out a circle is when the original ellipse is a circle.

edited Apr 18 '21 at 14:15

answered Apr 18 '21 at 14:01

Paul Sinclair

43,643

The real broom shall apparently never trace a circle as @Blue earlier demonstrated in his detailed answer (see comment) to this very problem type. When paddling the broom one felt it tracing a circle when in fact a near figure of eight, especially when the chord equated to the minor axis span for a fattest possible '8' path. The intermediate paths have apparently inspired the Wankel engine for example. – MKhomo Apr 18 '21 at 16:35

Light Amusement Brooming Tubs

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