Fitting a circular heliostat field in a square plot

Diagram of an oversized, circular heliostat field in a square plot of land.

A heliostat field wants to be circular, but plots of land are square.  Determining how much of a square plot to cover with heliostats would require a complex optimization, and it seems unlikely that filling the square all the way out to its corners would be the result.

An initial, plausible solution is an oversized circle with energy loss (land in the corners not covered by heliostats) that is twice as big as brightness loss (portions of the oversized circle that will be truncated by the square.) The reason to prefer energy loss, is that it is cheaper: we've already paid for the lamp to handle the radiation that brightness loss causes to go missing. By comparison, un-built-upon land is cheap.

By symmetry, this problem can be solved by looking at a single octant of the square and circle. From the diagram, the area of the full square exceeds the area of the full circle by 8*A. Assuming the circle is a unit circle, area A is:

A = (π*θ/2π)  - (sinθ cosθ)/2

= θ/2 - (sinθ cosθ)/2

Since the square, having side length s, exceeds the area of the circle by 8*A,

π + 8*A = s²,

π + 4θ - 4sinθ cosθ = s².

Also, s = 2cosθ, so

π + 4θ - 4sinθ cosθ = 4cos²θ

which is solved by

θ ≈ 0.421 radians, or 24.1° .

The circle diameter, d, is s/cosθ = 1.096 s .

Nine heliostat fields on square plots with d/s = 1.096.

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Getting a handle on the value of land is difficult because its value changes when we build on it; also, once a property is hemmed in by other claims, land is the one resource of which no more can be trucked in. The figure above, intuitively, looks a little under-filled to me. One factor to consider is that the diameter of the circle (being visible only where it touches already-purchased, use-it-or-lose-it land in the corners of the plot) should be larger than might be expected from the height of the lamp: it should the diameter of a theoretical system built on free land.

If unused land is equally as valuable as unused lamp, the calculation is simpler. Then the tangerine and cyan areas in the top diagram are equal, so the full area of the square is equal to the full area of the circle.

s² = π/4 * d²


So d/s = √(4/π) = 1.128

Nine heliostat fields on square plots with d/s = 1.128.

As above, the area truncated from the circle is:

8 * A = 8 * ( θ/2 - (sinθ cosθ)/2 ) = 4θ - 4sinθ cosθ

since the area of the full unit circle, π,  is also the area of the full square, the covering factor is

1 - (4θ - 4sinθ cosθ)/π

s = d/1.128 = 2cosθ 

d = 2.256 cosθ

but d = 2, so cosθ = 0.887, θ = 0.48 or 27.6°, giving a covering factor of 0.91.