The mesh with one curvature reversed

An annulus is a gear with its teeth pointing inward. Nothing about the involute changes — the flank is the same curve unwound from the same base circle at the same pressure angle — and yet almost every derived quantity in the field comes out differently, several of them with the opposite sign.

What is reversed, precisely #

One thing is reversed and everything else follows from it. On an external gear the material lies inside the pitch circle and the tooth grows outward, so the tip radius is $r + m$ and the root is $r - 1.25m$. On an annulus the material lies outside and the tooth grows inward: the tip is $r - m$ and the root is $r + 1.25m$.

The base circle is untouched. It is at $r\cos\alpha$ on both, because the base circle is a property of the involute and not of which side the metal is on. So the annulus’s flank is a piece of the same involute, traced over a different range of radii — outward from the tip at $r-m$ to the root at $r+1.25m$ rather than outward from the root to the tip.

Three consequences arrive immediately and they are not independent.

The centre distance is a difference. $a = r_2 - r_1$, not $r_2 + r_1$. For a 24-tooth pinion in a 72-tooth annulus at module 1 that is 24 rather than 48, and the pinion sits inside.

The two shafts turn the same way. An external pair reverses; an internal pair does not, which is the single fact that makes an epicyclic train arithmetic rather than a tangle of sign conventions.

The flanks’ curvatures subtract. The pinion’s flank is convex and the annulus’s is concave, so where an external pair’s relative curvature is $1/\rho_1 + 1/\rho_2$ an internal pair’s is $1/\rho_1 - 1/\rho_2$.

The tooth is the same curve, read the other way #

That last point deserves a sentence of its own, because it is the one that makes the rest of the essay short.

An involute unwinds outward from its base circle, getting flatter as it goes — its radius of curvature is the length of string unwound, so it is zero at the base circle and grows without limit. An external tooth uses the part of the involute just outside the base circle, where it is sharply curved. An annulus’s tooth uses a part further out, because its root is at $r + 1.25m$ rather than $r - 1.25m$, and it uses it from the other side.

So an internal tooth is not a new shape and it is not a mirror image. It is the same curve, further along, with the metal on the other side of it. Everything that is true of an involute is true of it, and every quantity computed from a base circle carries over with the radii reordered.

Where the line of action goes #

The most useful picture is the one that shows the line of action, because every difference above is visible in it at once.

For an external pair the two centres sit on opposite sides of the line of action. Their base tangency points straddle the pitch point, the contact path runs between them, and its full extent is $a\sin\alpha$.

For an internal pair both centres sit on the same side. The two base tangency points are still $a\sin\alpha$ apart, but the pitch point now falls outside the segment between them rather than inside it, because the pitch point is at $r_1$ from the pinion’s centre in the direction away from the annulus’s.

That geometry is what raises the contact ratio, and the mechanism is worth stating carefully because “an internal pair carries more contact” is usually stated rather than explained.

The usable contact path is bounded by the two tip circles. On an external pair the wheel’s tip circle cuts the line of action before the wheel’s own base tangency, shortening the path from that end; the path is $\sqrt{r_{a1}^2-r_{b1}^2} + \sqrt{r_{a2}^2-r_{b2}^2} - a\sin\alpha$. On an internal pair the annulus’s tip circle is inside its pitch circle, so it cuts the line on the far side of the annulus’s own tangency and lengthens the path: the same three terms with the middle one subtracted and the last one added.

What the contact ratio comes out at #

For the 24-in-72 pair the internal contact ratio is 1.931 against the external pair’s 1.707 — a gain of 0.224 on the same two tooth counts, the same module and the same pressure angle.

The gain is not a constant. It grows as the two counts approach each other: at 24 in 40 the internal figure is 2.173 against 1.658, a gain of 0.515, and at 30 in 34 it is 2.492 against 1.667.

The contact is gentler, by a factor with nothing in it #

The curvature statement has an exact form and it is the cleanest result in this essay.

At the pitch point the two flanks’ radii of curvature are $r_1\sin\alpha$ and $r_2\sin\alpha$. So the ratio of an external pair’s relative curvature to an internal pair’s, on the same two counts, is

$$\frac{1/\rho_1 + 1/\rho_2}{1/\rho_1 - 1/\rho_2} = \frac{z_2 + z_1}{z_2 - z_1},$$

with the pressure angle and the module cancelling out completely. It is 2.000 for the 24-in-72 pair, 16.000 for 30 in 34, and 81.000 for 40 in 41.

That runs away as the counts approach, and it is the geometric reason the drives that want a very large reduction in a very small package are built out of internal pairs of nearly equal counts. The contact between a convex flank and a barely-more-convex concave one is the gentlest in the field — and it is gentle for a reason a designer can compute from two integers before anything else is decided.

None of that is priced in stress here, which needs a modulus and a load and is outside the boundary. What it can say is that the geometric input to any such calculation improves without bound as $z_2 - z_1$ falls, and that it improves for free.

Two counts that nearly match, and what else changes #

The 30-in-34 pair above is worth a closer look, because a tooth-count difference of four is where several of this field’s quantities stop behaving.

Its contact ratio is 2.492 — nearly two and a half tooth pairs carrying at every instant, against the 1.6 an ordinary external pair manages. Its relative curvature at the pitch point is sixteen times gentler than the external pair of the same counts. Its centre distance is 2, which is two modules, so the pinion’s axis sits barely clear of the annulus’s.

And its ratio is 34/30, which is 1.133 — almost nothing. That is the trade nobody states when the gentle contact is being admired: an internal pair of nearly equal counts is a magnificent mesh and a useless reduction on its own. It becomes useful only inside a train where the small difference is the output, which is exactly what a compound epicyclic does with it — a ratio of 2176 to 106 assembled out of meshes whose own ratios are all close to one.

So the gentlest contact in the field and the largest reduction in the field are the same geometry, and the connection is not a coincidence: both come from a difference of two nearly equal integers.

Three floors, and they are one statement #

The counts are not free. An internal pair has three separate thresholds on its tooth numbers, and every one of them is the same sentence — there is no involute inside a base circle — pointed at a different object.

The annulus must have at least 33.16 teeth. Its tip is at $r-m$ and its base circle at $r\cos\alpha$, so the tip falls inside the base circle unless $r(1-\cos\alpha) > m$, which is $z > 2/(1-\cos\alpha)$. Below that count the annulus’s tooth has no flank at all. A 33-tooth annulus at module 1 has a tip radius of 15.500 against a base radius of 15.505, and a 34-tooth one has 16.000 against 15.975.

The pinion must have at least 17.10 teeth. Not because of undercutting — because of interference with the annulus. The contact path must stay on one side of the pinion’s base tangency, since contact on the two sides is contact on the two opposite faces of the tooth. Writing that condition out and squaring it gives

$$z_2\,(2k z_1 - 1) \ge k z_1^2 - 1, \qquad k = \tfrac{1}{4}\sin^2\alpha,$$

and the coefficient of $z_2$ changes sign at $z_1 = 2/\sin^2\alpha$. Below that count the inequality points the wrong way and no annulus of any size will mesh with the pinion: a seventeen-tooth pinion was tried against every ring up to four hundred teeth and meshed with none of them.

And the two counts must satisfy the inequality jointly. Above the pinion floor there is always some annulus that works, and it is large: an 18-tooth pinion needs 161 teeth, a 20-tooth pinion 64, a 24-tooth pinion 40, and a 30-tooth pinion 34.

The middle threshold is 2/sin²α and the first is 2/(1−cos α), and since $\sin^2\alpha = (1-\cos\alpha)(1+\cos\alpha)$ the annulus floor is exactly $(1+\cos\alpha)$ times the pinion floor — 1.9397 times at 20°, so 33.163 against 17.097. Two thresholds derived from completely different constructions turn out to be one identity apart, and the identity is checked rather than admired.

What shift does, and it goes the other way #

Profile shift is the field’s standard cure for an awkward tooth count, and on an internal pair it behaves in a way that follows from the reversal and surprises anyone who has only met it externally.

Shifting an external gear outward moves its tip and root outward together and pushes the operating centre distance up. An annulus’s tooth points inward, so shifting it moves its tip and root the other way — and since the centre distance is a difference of radii rather than a sum, a shift that increases the annulus’s effective size decreases the centre distance where an external pair’s would increase it.

The consequence a designer cares about is that shift is the lever that fixes all three floors above. A pinion below the joint condition can be shifted positively and an annulus shifted with it, and the pair that the unshifted arithmetic refuses becomes buildable. That is why real planetary gearboxes run tooth-count combinations that the plain condition above would reject: they are not exceptions to it, they are not cut at zero shift.

This collection’s internal figures are drawn at zero shift throughout, and the floors quoted are therefore the unshifted floors. Naming that is not a formality: a reader who took 17.10 teeth as an absolute limit on a planet would be wrong about a great many gearboxes, and the limit that is absolute is the annulus’s own tip floor, which no shift can cure because it is about a single gear rather than about a pair.

What the rule of thumb gets backwards #

The advice usually given is that an internal pair needs a large tooth-count difference, and the measurement says the opposite: for a given pinion, small annuli fail and large ones work, so the binding constraint is a floor on the annulus rather than a floor on the difference.

The confusion has a source and it is a different failure. A pair whose counts are close cannot be assembled radially — the pinion cannot be dropped straight into the annulus, because the tips foul on the way in — and it may foul again as the teeth pass at a distance from the line of action. Those are real and they do want a difference. They are not the involute-interference condition, they happen at different counts, and only the third of them is computed here.

So the honest summary is that there are at least two distinct failures called interference in an internal pair, that they push the design in opposite directions, and that the one measured here is the one the line of action decides.

What was measured, and how it could have failed #

Four quantities, and each has a check attached that a wrong sign would trip.

The centre distance is required to equal $r_2 - r_1$ to the last bit, which is the cheapest possible test of whether a sum has been written where a difference belongs.

The contact ratio is required to exceed the external pair’s at every wheel size drawn — not at one, but across a hundred counts, with the smallest gain reported. A single count where it failed would mean the tip reaches had been added where they should subtract, and that error is invisible on any one pair because it produces a perfectly plausible number.

The contact path’s endpoints are checked by computing the radius at each of them on both gears: the near end must land exactly on the pinion’s tip circle and the far end exactly on the annulus’s. The first version of this had them landing at 34.64 and 11.30 on a pair whose tip radii are 35.00 and 13.00 — plausible numbers, wrong geometry — and the check is what said so.

The thresholds are searched rather than solved, which is the point: the closed form is what is being checked. The walk over ring counts is compared against $(k z_1^2 - 1)/(2k z_1 - 1)$ at every pinion count, and the two agree to the integer at each.

And the negative half, which is where a requirement earns its keep: a pinion one tooth below $2/\sin^2\alpha$ must be refused by every annulus up to four hundred teeth, and one above it must find one. A threshold with only its positive side tested is a threshold nothing has ever failed.

Why the planetary trains in the fields before this one work at all #

The gears field has been computing epicyclic ratios since its second essay, and every one of those trains has an annulus in it — a sun, some planets, and a ring — with the ring’s mesh treated as an ordinary mesh with a sign on it.

That treatment is right about the ratio, because a ratio is a count and counting does not care which way the teeth point. What it cannot see is the three floors above. The site’s standard planetary is sun 24, ring 72, planets 24, and the planet-ring mesh is a 24-in-72 internal pair: the annulus clears its own floor by more than a factor of two, the planet clears the pinion floor by seven teeth, and the joint condition wants at least forty teeth on the ring against the seventy-two it has. It is comfortable on all three, and nothing in the ratio arithmetic would have said so.

That is the general shape of the relationship between the two halves of this field. The counting arguments are exact and blind; the geometry arguments are where the buildability lives. A train that satisfies the first and fails the second is a correct calculation of a machine that cannot be made, which is a defect met before from the synthesis side.

There is a second counting condition on a planetary that this essay’s geometry cannot see either, and it is worth naming beside the floors so the two are not confused. The planets have to be equally spaced and assemblable at the same time, which requires $(z_{\text{sun}} + z_{\text{ring}})$ to divide by the number of planets — a pure arithmetic condition with no involute in it whatsoever, and the one that decides whether three planets or four. Sun 24 and ring 72 sum to 96, which divides by three, four, six and eight; a ring of 73 would divide by nothing useful and the same gearbox would carry one planet.

Three independent conditions, then, on the same handful of integers: the ratio, which is what the train is for; the spacing divisibility, which decides how many planets; and the three involute floors, which decide whether any of it can be cut. A tooth count that satisfies two of them and fails the third is the ordinary case rather than the unlucky one, and it is why planetary tooth counts look arbitrary and are not.

Still open: the mesh that is not in one plane #

Everything above assumes the two flanks meet along a line of action lying in the plane of rotation, which is true of a straight-toothed pair and of nothing else.

Give the teeth a helix and the contact stops being a point travelling along a segment and becomes a line travelling across a rectangle. The transverse geometry — everything in this essay — survives unchanged in the transverse plane, and a second quantity appears beside it that has no tooth count in it at all. The next essay is about that second quantity, about the fact that it is bought with face width and slant alone, and about a helical pair’s contact never falling below one for reasons that have nothing to do with the profile.

The internal case has a helical version too, and it is the one not drawn here: an internal helical pair is what most planetary gearboxes actually contain. What would have to be worked out for it is whether the three floors survive the substitution into transverse quantities — the transverse pressure angle is larger than the normal one, so $2/\sin^2\alpha_t$ is smaller and both floors fall, which suggests an internal helical pair tolerates counts a straight one refuses. That is a prediction rather than a measurement, and it is the shape of thing worth writing down so that the essay after this one has something to check rather than something to invent.