NB: Sorry, this post has been rewritten. Talk about burying the lead! Facts and results unchanged, so need to read again if you got them the first time....using the center of mass of a semicircle and equating the levers from them to the position of the center of mass, x, one gets a general solution for a radius R and a ratio of densities ρ2/ρ1:
x = 4/(3π) (ρ2/ρ1-1)/(ρ2/ρ1+1) R
Turns out that ta0's solution for a circular "half-and-half" rotor generalizes nicely to a wide range of rotor "shapes" as seen in plan view.
General formula: You have a half-and-half rotor like the one
ortwin envisioned in Reply #45. The rotor has a certain "shape" in plan view with a sharp density contrast along a straight "boundary" passing through the shape's center and dividing it into equal "halves". For shapes like circles, ellipses, rectangles, rhombi, squares, reqular hexagons, regular octagons, and many others sharing certain key properties, the perpendicular distance
x from the boundary to the rotor's CM is then
x = (
γ - 1)
xcentroid / (
γ + 1),
where
γ > 1 is the ratio of
areal densities across the boundary, and
xcentroid is the perpendicular distance from the boundary to the centroid of the chosen half-shape.
Wikipedia's list of centroids is a good resource here.
Octagonal example: In the half-and-half top below, the rotor shape is an octagon of side
S = 32 mm = 4 studs. The chosen boundary between halves is clear. The centroid location comes out to
xcentroid = 17.2 mm = 0.543
S on both halves. (You don't wanna see the math.) With
γ = 5 in this example, the overall CM is then on the thickened half at distance
x = 11.6 mm = 1.45 studs from the boundary.
Unfortunately, wobble is barely tolerable. The yellow stem and tip mounts at
xactual = 1.50 studs were the closest I could get to
x = 1.45 studs, and the difference matters.
Which shapes qualify? Applicable rotor shapes must have a center of symmetry crossed by at least one pair of perpendicular "mirrors" (lines of mirror symmetry at right angles to each other). All of the shapes mentioned above fit that description. But stars are out because they're not convex. The regular pentagon is out because it has no
perpendicular mirror pairs. And the isoceles triangle is out because it has only one mirror and no center of symmetry.
If the rotor shape has more than one perpendicular mirror pair through its center (e.g., a regular hexagon), pick a pair, call one of its mirrors the "boundary", and divide the shape into halves accordingly. For either half,
xcentroid is the perpendicular distance from the boundary to that half's centroid.
For example, a rectange with sides
a >
b has a single perpendicular mirror pair through its center of symmetry. If you pick the mirror parallel to
b as the boundary, then
xcentroid =
a / 4. A circle of radius
R, on the other hand, has an infinite number perpendicular mirror pairs. Just pick a pair and make one of its mirrors the boundary. Result: 2 semicircles, each with
xcentroid = 4
R / 3
π.
Why areal densities? Because they're more general than mass densities measured in kg/m³. Recall that if half-rotor
i has uniform uniform thickness
Li and mass density
ρi in kg/m³, then its areal density is
αi =
ρi Li in kg/m². As long as you watch out for couple unbalance, doesn't matter then whether you change
ρi or
Li to adjust
αi. All that counts here is the final value.
In these terms,
γ =
α1 /
α2 > 1, where the subscript "1" points to the denser half.
