Orrery Designs using LEGO® Elements
Since 2007 I have worked on Orrery designs using LEGO elements. This began when I looked at the NASA Kepler Mission website and found their educational material including a miniature tabletop "solar system" designed (in 2004) to demonstrate how the Kepler mission will work.
I liked their model so much I decided to work on a few improvements, then began developing my own completely new designs.
As this "page" has gotten rather long, I have broken up the material into a few pages, organized chronologically (oldest first).
Table of Contents
Constrained models (using parts from only two brandnew sets)
Model F: "Constrained" 5Planet Design
G: Adding Moons and Refinements (OffRoad Truck rev. 2)
H: Finishing the Inner Planets (OffRoad Truck rev. 3)
More Fancy Ideas
The Hailfire Model: Extreme Coplanar Gears
Robust and Modular Designs
Stacked Turntables: TwoModule Prototype
Stacked Turntables: FourModule Prototype
Appendices
Script for model D YouTube video
Script for Geartrain Module video
History: Previous Versions of the Orrery
20040915
My 2004 designs are descended from versions by Allan Ayres and Dave Koch. The first two were 2planet and 3planet designs by Ayres. I know little of the 2planet version, but I suspect it was like its successor but without the middle planet.
Here is a picture of the 3planet design; the building instructions for it are here: transitdemo.pdf.
A1. Ayres' 3planet orrery
A large Technic turntable, whose base remains fixed, is made to turn through epicyclic gearing. 4 rotations of the inner shaft makes the turntable top rotate once. The 3^{rd} planet is attached to the turntable, and the 1^{st} planet to the central shaft. Another gear on the turntable makes the 2^{nd} planet turn at a rate twice that of the 3^{rd}.
The orbit period ratios can be measured easily just by turning the crank a few times. Here are the periods; I have also shown a "radius" computed by the formulas discussed here.

These orbital periods might seem a little artificial, but it does actually happen in real life — Jupiter's moons Io, Europa and Ganymede have exactly this ratio of periods, due to orbital resonance. You can approximate their orbits by setting up the model so that the 1^{st} and 3^{rd} planets are aligned and the 2^{nd} is all the way around on the other side:

20050729
You may notice in the above 3planet designs, there is a gear on the upper part of the differential housing that doesn't come into contact with any other gear. That gear provides an opportunity to add a fourth planet, and Dave Koch did exactly that. He took Ayres' 3planet design and modified it to look roughly like this:
B1. My Rendering of the Ayres/Koch 4Planet Orrery (without moon)
(Here, I have added halfstud adjustments to make the planet distances more closely conform with Kepler's 3^{rd} law (discussed below).)
Then he added a moon:
Koch's version (with moon on 2^{nd} planet)

These 4planet designs incorporate two modifications over the 3planet design:
 A new planet is added, carried by an arm with epicyclic gears that compute a weighted average of the rotations of its neighbors (details below). What were the 2^{nd} and 3^{rd} planets are now the 3^{rd} and 4^{th}, respectively.
 The differential housing is turned upside down, and gearing changed so that its rotation is slower (but still faster than that of the 4^{th} planet).
The instructions to build this model are here: The Kepler Transit Demonstration
{Updated instructions for the Kepler project's 4planet orrery are here: LEGOorrery2011.pdf RPM 20121223}
Kepler's Third Law
To make the model more accurate, one should adjust the lengths of the arms so that the planets' distances are related to their orbital periods by the formula
(^{P}/_{2π})^{2} = ^{a3}/_{(G(M+m))}
where
P is the orbital period,
a is the semimajor axis of the orbit (the radius, for a circle)
G is the universal gravitational constant
M and m are the masses of the sun and planet
for our purposes we can simplify it to
a = K P^{2/3}
where K is an arbitrary constant, which can be 1 if you express P and a as multiples of one planet's period and radius. So for example, knowing that Mars has an orbital period 1.88 times that of Earth, the formula tells us that its semimajor axis is 1.88^{2/3} {~=} 1.523 times that of Earth.
All of the models above (A and my versions of B) are built with the proper distances to within ±2mm.
20071212
B's orbit periods are a bit more complex than A's. The turntable turns the same way as before, through epicyclic gearing. 12 turns of the crank make it go around once. This is one orbit of the 4^{th} planet. 36 turns of the crank is the time it takes the 4^{th} planet to orbit 3 times. In this period the 3^{rd} planet orbits 4 times, the 2^{nd} planet orbits 7 times, and the innermost planet orbits 12 times. I call this a syzygy cycle because if you start the orrery with all planets aligned, it takes this long for them all to become aligned again.
Here is a table showing the planet orbital rates and periods, and the radii used in Koch's model (K) and my two models (M_{1} and M_{2}):

20071213
The number 7 in the above ratios might come as a bit of a surprise, if you consider that the gears (with 8, 16, 24 and 40 teeth) cannot by themselves produce any ratios involving the prime number 7.
The 7 (and other prime factors in the more complicated ratios below) comes from epicyclic gearing formulas that involve addition and subtraction, and division of ratios.
Consider first the drive of the 4^{th} planet:
B2's Base.
This is a planetary gear mechanism (a type of epicyclic gearing). The base (black) is the annulus, and has 24 teeth. The planetcarrier (clear, and gray) carries the planet gear (blue) with it. The planet gear has 8 teeth. The sun gear (yellow) has 8 teeth. In this particular setup, the base is fixed, the sun is the input, and the planetcarrier is the output.
Let R_{p} be the rate of rotation of the planetcarrier, and R_{s} be the rate of rotation of the sun gear. R_{b}, the rate of rotation of the base, is fixed at zero.
It is easiest to compute the gear ratio from within the rotating reference frame of the planet. Relative to the planet, the base is rotating at R_{p} and the sun is rotating at R_{s}  R_{p}. The base has 24 teeth, the sun has 8, and the planet acts as an idler gear. Because the base's gear is insideout (teeth facing in) a direction reversal is added. The gear ratio is 1×24/8, or 3/1. Therefore, the motion of the planetcarrier and sun will fit the following formula:
R_{s}  R_{p} = 3 (R_{p})
and therefore (by algebra)
R_{s} = R_{p} + 3 R_{p} = 4 R_{p}
This shows that the 3^{rd} planet in model A will revolve once for each 4 times the 1^{st} planet revolves.
Notice that we get a ratio of 4, even though the gears we used (24 and 8) have a ratio of 3.
Now consider the transfer of motion to the 3^{rd} planet:
B2 Transfer from 4^{th} to 3^{rd}
We will use R_{4} now to refer to the rotation rate of the 4^{th} planet arm (R_{p} above), R_{1} instead of R_{s}, and R_{3} for the 3^{rd} planet arm. Viewed from the rotating reference of the 4^{th} planet's arm, this is an ordinary gear train. The sun gear (yellow) with 8 teeth drives a 24tooth gear, which shares an axle with an 8tooth gear (gray) which drives the output, a 24tooth gear (dark gray). The input is R_{1}R_{4}, the output is R_{3}R_{4}, and the ratio is (24/8)×(24/8) = 9/1. Therefore:
R_{1}R_{4} = 9 (R_{3}R_{4})
we already know from above that R_{1} = 4 R_{4}. So we have
3 R_{4} = 9 R_{3}  9 R_{4}
which gives
R_{3} = 4 R_{4} / 3
Still a fairly simple ratio, no strange prime numbers here. But let's see what happens when we go up to the 2^{nd} planet:
B2's 2^{nd} Planet Between 1^{st} and 3^{rd}
The 2^{nd} planet's arm rotates at rate R_{2}, and has gears mounted on it. The dark gray gear is turning at R_{3} — but from the 2^{nd} planet's point of view, it rotates at R_{3}R_{2} (which is negative, hence backwards). The yellow gear rotates at R_{1}, which is R_{1}R_{2} from the 2^{nd} planet's point of view. An extra 16tooth idler ensures that the gear train has only one direction reversal. The gear train will obey the equation
R_{1}R_{2} = (40/24)×(16/16)×(R_{3}R_{2})
Which gear is the input? We already know that R_{1} and R_{3} are determined as just described. Therefore, they are both inputs, and R_{2}, the 2^{nd} planet's arm, is the output. The equation reduces as follows:
R_{1}R_{2} = 5 (R_{3}R_{2}) / 3
4R_{4}  R_{2} = 5 (4R_{4}/3  R_{2}) / 3
R_{2}  4R_{4} = 20R_{4}/9  15R_{2}/9
24 R_{2} / 9 = 56 R_{4} / 9
R_{2} = 7 R_{4} / 3
There's that surprise prime number 7. Every time the 4^{th} planet completes 3 orbits, the 2^{nd} planet completes 7 orbits.
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