Earlier we discussed how to start a counterweight for a dumbbell station, starting with Janhunen's dumbbell as requiring the least starting mass. Suppose Janhunen has shuttled this off to some larger rock, for settlement.
Janhunen assumes both weights are spheres. I, for one, do not wish to live in a sphere-in-rotation. First up, it concentrates the pressure on a point, the bottom "pole" if you will [UPDATE 9/5: Maths!]. Also if humans are here, in the bottom, they have to take the stairs to go up and meet anyone. Gravity fluctutes from level-to-level, as well. So does Coriolis.
Instead I am thinking, a trough. The port-starboard width at fore - "fore" is the direction of rotation - is the same back to aft. As for the bottom curvature port/starboard: sure, there is curvature, can hardly avoid it. Basically we've stumbled into the torus-section as against the classical torus.
But does this have to be a true torus-section i.e., circle - does it even have to be a semicircle, or any section of a circle? How about a parabola?
As we start, I am afraid the trough of dirt is still going to be blockey. But we'll work with that after we move in.
I can get an ugly block with 400m × 400m × 200m (height) for 3.2e7 cubic meters. I reserve the side-to-side 400m. Right now I am reserving also fore/aft 400m - right now this is an authorial fiction. Now let us handle the dimensions of the 80000m2 cross-section specifically, height.
The parabola equation is a-bx2=0, flying above the barycentre. So integrate a-bx2 dx, side-to-side, for final 80000. Units are in meters; a is the height at the centre. As a symmetry, we can integrate this from 0 to the starboard (or port) side, for 40000 m.
Now: a-bx2 = 0 where x = +/- the half-width. x boundaries: +/-root[a/b]. The integral from zero to either edge is ax - bx3/3 = 40000. eek! Tritic equation!! Luckily,
ax - (a/x2) x3/3 = 40000 a(x-x/3)=40000
For height a to fill out half of 80000, the halfwidth x = 40000 * 1.5 / a. If width is 400m, central / maximum height be 300m.
To return to the curvature of the fore-aft axis: if it's negligible, we just multiply that 80000 crosssection by 400m for total volume. In practice I expect the circumference of that "400m" to end up more than 400m at the bottom; less at the top. More so at the beginning when they're not spinning the wheel as hard; shorter radius of spin.
And they'll start at the top. 300m is a lot of height for the initial colonists to support over their heads. This can be ameliorated by settling the upper parts. There they'll suffer low-gravity and high-Coriolis.
Eventually they're working toward 150m central height, 800m fore/aft length.
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