The gearset that could not be assembled

Draw a planetary gearset. A big circle for the ring, a small one in the middle for the sun, and three more between them, equally spaced. Check that the radii work — the planet’s diameter is the gap between sun and ring — and the picture is right.

It is right, and the gearset may be impossible.

The three conditions #

One: a whole planet must fit. The planet’s pitch diameter is the difference of the ring’s and the sun’s, so

$$z_R = z_S + 2z_P,$$

which needs $z_R - z_S$ to be even. This is the condition every account states, and it is a statement about radii — it is the one a drawing checks, because a drawing is made of radii.

Two: the counts must divide. Put the first planet in anywhere; it meshes the sun and the ring wherever it is dropped. Now go to the next station, $360°/n$ round. The planet there must mesh the sun and the ring, and here is the thing: meshing the sun completely determines its rotation. There is no freedom left over to make it mesh the ring as well. Whether it does is decided in advance, by arithmetic, and the condition is

$$z_S + z_R \equiv 0 \pmod{n}.$$

Three: the planets must clear each other. Adjacent planets sit on a circle of radius $(z_S + z_P)/2$ modules, a chord of $2 \times$ that $\times \sin(\pi/n)$ apart, and each has a tip circle of diameter $z_P + 2$. If the chord is shorter than the tip diameter the planets overlap, and no arithmetic will save them.

The first and third are about lengths. The second is not about any length at all: it is unaffected by the module, by the size of the gearbox, by the material and by every tolerance on the drawing. A perfectly made gearset that fails it fails it exactly as badly as a badly made one.

Why it is arithmetic #

The derivation is worth doing because it explains why the condition looks so unlike the other two.

Sit at the first planet’s station and count teeth going round the sun to the next station. That arc is $1/n$ of the sun’s circumference, so it is $z_S/n$ tooth pitches of the sun. Do the same round the ring: $z_R/n$ pitches. For the planet at the second station to be in the same relationship to both wheels as the first one was, the number of sun pitches and the number of ring pitches passed must both come out whole, or the mismatch has to cancel between them.

Working it through, what has to be whole is the sum: the planet’s required rotation from the sun’s side and from the ring’s side differ by a fraction of a tooth equal to the fractional part of $(z_S + z_R)/n$. Zero, and it goes in; anything else, and it does not.

The remainder is not small when it fails. Its worst possible value is half a tooth, and the six cases above include one that hits exactly that.

Six gearsets #

Every one of them passes the condition that gets stated. Four of the six fail one of the two that do not.

The third row is the instructive one. Take the textbook 24/72 gearset, which works perfectly, and add one tooth to the sun and one to the ring. The planet is unchanged at 24; the radii still close; the ratio changes from 4.000 to 3.920, which is a perfectly reasonable thing to want. And the gearset will now take two planets and not three.

The fifth row is the worst case the condition allows. 30 and 80 sum to 110, which leaves a remainder of 2 out of 4, so alternate stations are out by exactly half a tooth — the planet’s tooth lands exactly on the ring’s tooth. Not near a mesh; as far from one as it is possible to be.

The sixth is the neighbour condition doing its own work. 18 and 90 sum to 108 and divide by 6 perfectly, so the arithmetic is happy; the planets are 36 teeth on an orbit of 27 modules, and six of them at 30° spacing have 27 modules of chord between centres for tip circles 38 modules across. They overlap by 11 modules — a third of a planet — which is not a near miss and would be obvious on a careful drawing and is not obvious on a sketch.

A third arrangement satisfies the same three conditions with a different set of counts, which is worth putting beside the first two: the conditions constrain the catalogue rather than picking a single member of it.

What a drawing can and cannot show #

The reason this is a wrong field essay rather than a design note is the mechanism of the error, which is unusually clean.

A planetary drawing, at any scale a page will hold, draws circles. The sun is a circle, the ring is a circle, the planets are circles at their stations. Every one of the three conditions above is a condition about the teeth, and only one of them — the neighbour condition — has a consequence that shows up in the circles, because overlapping planets are overlapping circles.

The divisibility condition has no consequence in the circles at all. A gearset that fails it is drawn with the correct radii at the correct stations, and a picture of it is indistinguishable from a picture of one that works. Adding teeth to the drawing does not help unless they are drawn at their computed positions — which is what the figures here do and what a general-purpose drawing does not, because a CAD pattern-copies one planet round rather than solving each one’s rotation.

That is the same failure mode this site has met in six drawings of an escapement and in a sketched four-bar whose coupler changes length by 25%. In every case the drawing is a picture of the parts and the mechanism is a set of relations between them, and the relations that do not show in the parts’ outlines are the ones that get lost.

The condition in the other direction #

The divisibility condition is usually met by adjusting the tooth counts, and there is a second reading of it that is more useful when the counts are fixed: it says how many planets a given sun and ring will take.

For a sun of 24 in a ring of 72, $z_S + z_R = 96$, whose divisors above 1 are 2, 3, 4, 6, 8, 12 and so on. The neighbour condition then removes the large ones — six planets of 24 teeth on an orbit of 24 modules have 24 modules of chord between centres for tip circles 26 across, so they overlap — and what is left is 2, 3 or 4. Five is impossible arithmetically and six geometrically, and the two refusals have completely different characters.

For a sun of 25 in a ring of 73, the sum is 98, whose divisors are 2, 7, 14 and 49. Seven planets of 24 teeth will not fit, so this gearset takes exactly two, and that is the whole of its options.

The library returns that list — counts — for every design, and it is the answer a designer actually wants: not is this legal, but what are my choices. Reading it turns the condition from an obstacle into a menu.

How much of the catalogue it removes #

Sweep sun and ring over the ranges a designer would consider — sun 16 to 60, ring 60 to 120 — and count.

1,822 pairs have an even difference, so a whole planet fits and a drawing can be made. Then:

Two thirds of the drawable gearsets will not take three equally spaced planets. Four planets is the most accommodating count, because $z_S + z_R$ being a multiple of 4 is a common accident once the difference is already even; five is the least, because 5 divides almost nothing here.

The neighbour condition only starts to bite at five and six planets, which is what one would expect — more planets, less room — and it removes 103 designs at five and 282 at six. So at three and four planets the arithmetic condition is doing all of the work, and it is the one nobody draws.

Five planets is the case where the divisibility bites hardest, and it is worth seeing beside the arithmetic that predicts it rather than after it.

What the check has to be able to lose #

The site’s habit is that a condition has to be shown to be independent rather than merely stated. Two conditions that always agreed would be one condition written twice, and a survey where the second never rejected anything the first had passed would be evidence of exactly that.

So the gate asserts that the count passing both is strictly smaller than the count passing the coaxial condition alone — 608 against 1,822 for three planets — and it asserts the two named cases in both directions: that 25/73 is cuttable and will not take three planets, and that 24/72 will. If the divisibility test were accidentally implemented as a restatement of the radii condition, the first assertion would fail. If it were implemented as something that rejects everything, the last would.

What makes this a wrong field essay #

The site’s wrong field holds claims that are drawn confidently and are false, and the entries in it have a common structure worth restating here, because this one fits it exactly.

In each case there is a check the drawing performs and a check it does not. A sketched four-bar’s drawing checks that the links reach; it does not check that the coupler is the same length in both positions, and the sketch that needed a 25% change is the result. An escapement drawing checks that the pallets look like pallets; it does not check that a tooth can reach the second pallet, and six drawings could not alternate. A planetary drawing checks that the radii close; it does not check the divisibility, because the divisibility has no consequence in the radii at all.

What is common is that the check the drawing performs is a check about lengths, and the check it omits is about something else — a closure, a reachability, an arithmetic. Drawings are made of lengths. That is what they are good at and it is the whole of what they are good at.

The site’s answer, in every one of those cases, is the same: compute the thing the drawing cannot check, and let the figure be a consequence of the computation rather than an input to it. The planet stations in the figures here are not placed at $360°/n$ and drawn with teeth pattern-copied; each planet’s rotation is solved from its mesh with the sun, and whether it also meets the ring is then something the picture reports rather than something it assumes.

The same condition, one level up #

The divisibility condition has a relative in this field that is worth naming, because the two together are the whole of what integer arithmetic does to a gearset.

Which tooth meets which is decided by a greatest common divisor: a planet tooth meets $z_{\text{sun}}/\gcd(z_P, z_S)$ of the sun’s teeth and no others. Both conditions are about divisibility of tooth counts; both are invisible in every drawing; both are unaffected by module, size, material and tolerance.

And they pull in opposite directions. The assembly condition wants $z_S + z_R$ to have small divisors, since it must be a multiple of the planet count. The hunting condition wants the sun, planet and ring pairwise coprime. A design satisfying both is being asked for three numbers that share a factor collectively — through their sum — and none pairwise, which is possible and is not automatic.

The site’s own default gearset shows the tension. Sun 24, ring 72: the sum is 96, which divides by 2, 3, 4, 6 and 8, so it assembles beautifully with almost any planet count. And its planet is 24 teeth against a 24-tooth sun, so a planet tooth meets exactly one sun tooth for the life of the drive. Excellent on one condition, worst possible on the other, from the same round numbers.

The Fibonacci gearset of the equal-step essay manages both — 55 and 89 sum to 144, which divides by 3 and 4, and 17, 55 and 89 are pairwise coprime — which is a third dividend from numbers chosen for a completely unrelated reason.

Where the planets may go, and how nearly equal unequal can be #

The escape through unequal spacing is worth making precise, because “the condition is about equal spacing” leaves it sounding like an evasion and it is a proper statement about where planets may sit.

A planet can only be dropped in where its teeth line up with the sun’s and the ring’s at once. Going once round, there are exactly $z_S + z_R$ such stations, evenly spaced. That is the whole geometry of the situation: the planets are not free to sit anywhere on the carrier, they must sit on that lattice, and the divisibility condition is simply the question of whether $n$ equally spaced planets can be chosen from a lattice of $z_S + z_R$ points — which they can exactly when $n$ divides it.

Turned round, the escape is that any $n$ of the stations will do. The gearset assembles with the planets at any subset of the lattice, equally spaced or not, and the only thing lost is the symmetry. That is a much larger freedom than the divisibility condition suggests, and it is why the condition is a design convenience rather than a law of assembly.

And the departure from equal spacing is bounded, which is what makes the escape practical rather than merely available. The stations are $360°/(z_S + z_R)$ apart, so the best unequal arrangement puts every planet within half a station of where equal spacing would have put it. On a sun of 24 in a ring of 72 there are 96 stations, 3.75° apart, and the worst planet is at most 1.875° from its ideal position. On a finer gearset it is less still.

That number is small enough to change the shape of the design decision. Rejecting a good ratio because $z_S + z_R$ is not divisible by three is trading a ratio for under two degrees of planet position, and a designer who knows the bound will usually take the ratio. Rejecting it without knowing the bound looks like obeying a hard constraint. The two are the same arithmetic read with and without the lattice in view, and only the second reading treats the condition as absolute.

The residual cost is real and it is not kinematic. Planets out of balance put a rotating imbalance on the carrier, and the load no longer divides evenly between them, so the bearing that carries the most is carrying more than a third of the torque. Both are questions about forces and masses, which is where this field stops. What the geometry supplies is the offset in degrees, which is the input those questions need.

The rule that follows #

Two lines, and they are worth having explicitly because the arithmetic is easy to get right and impossible to notice.

A planetary with $n$ equally spaced planets requires $z_S + z_R \equiv 0 \pmod n$. Choose the ratio first, then check; if it fails, the nearest tooth counts that work are usually one or two teeth away and change the ratio by a per cent or so.

Unequal spacing is the escape and it is a real one. Nothing requires the planets to be equally spaced; the condition above is the price of that symmetry. Spacing them at whatever angles do satisfy the mesh gives back the whole catalogue, at the cost of an unbalanced gearset — which is a dynamic objection rather than a kinematic one, and is therefore not this site’s to weigh.

One consequence of the lattice picture is worth recording because it settles a question the divisibility condition invites. Given a gearset that fails for three planets, is there any number of planets it will take equally spaced? Always, and the answer is read straight off the divisors of $z_S + z_R$ — which is precisely the list the library already returns. So the honest form of a refusal is never this gearset cannot have equally spaced planets; it is this gearset takes two, or four, or six, but not three, with the whole list available for the price of factorising one integer. A designer told the first has been told something false, and a designer told the second has been handed the design space.