Three chambers and a constant

The rotor touches its housing at three points and nowhere else. Those three contacts cut the space between the two bodies into three chambers, and everything the engine does is a consequence of what happens to their areas as the shaft turns.

Both boundaries of every chamber are known exactly. One is a piece of the rotor’s flank, which came out of the envelope; the other is the arc of housing between two apexes, which has a closed form. So a chamber’s area is a polygon integral over two curves rather than an estimate, and the areas can be measured rather than described.

A chamber, defined precisely #

Before any of that is measurable, “chamber” has to mean something exact, and the definition is worth spelling out because it is where the arithmetic could go wrong without anybody noticing.

Take the rotor at a shaft angle, transformed into the housing’s frame: rotated by a third of the shaft angle, then displaced by the eccentricity. Its boundary is a closed curve with three marked points on it — the apexes, which the generation places exactly. Take two consecutive apexes. The chamber they bound is the region enclosed by the piece of rotor flank running from one to the other, closed by the piece of housing running back.

Everything in that sentence is a curve with a parameter, so the region is a polygon whose vertices are computed and whose area is a shoelace sum. There is no meshing, no triangulation, no numerical integration of anything but the two boundaries, and the only approximation is that both curves are drawn with finitely many points.

The total that does not move #

Take the three areas at seventy-two shaft angles round a full turn. Each one swings enormously — the largest chamber is twenty-five times the smallest — and the sum is $15{,}257.45$ mm² at every one of them, with a spread across the whole set of $5\times10^{-11}$, which is three parts in a thousand million million.

The number it should be is the housing’s own area less the rotor’s: $33{,}263.1 - 18{,}005.5 = 15{,}257.59$ mm². The measured total agrees with it to a part in a hundred thousand, and the small difference is the polygon approximation of two curved boundaries rather than a disagreement.

It would be easy to dismiss that as a tautology. A rigid rotor inside a fixed housing encloses the same total area however it is placed, so of course the three chambers add to a constant.

The reason it is a measurement is that the three chambers only partition that region if the corners are on the wall. The chamber polygons are built by walking the rotor’s flank from one apex to the next and then coming back along the housing between the two apexes’ positions. If an apex were not touching — if the generated rotor were slightly too small, or its corners in the wrong place, or the envelope computed on the wrong branch — those polygons would overlap or leave a sliver between them, and the sum would not be the total. It is the tightest available check that the three seals of a rotary engine are geometrically seals.

What each chamber does #

Follow one chamber. Over a full turn of the rotor — which is three turns of the shaft — its area rises from a minimum of $412$ mm² to a maximum of $10{,}395$ and back. Multiply by the rotor’s width and that is a chamber that inhales, compresses, is pushed on and exhales, once per rotor revolution.

There are three of them, at a hundred and twenty degrees of rotor phase, so the engine completes one such cycle every shaft turn. That is the property the design is bought for: one power stroke per turn of the output shaft, against one per two turns for a four-stroke piston engine, and with no valve gear, no reciprocating mass and no crank throw.

The swept area of a chamber is the swing: $10{,}395 - 412 = 9{,}983$ mm², times the width. On this geometry it is two thirds of the total space between the bodies, which is a remarkably high proportion — most of the volume in the machine is doing something.

The compression ratio the geometry gives #

Divide the largest chamber area by the smallest and the answer is $25.2$.

No production rotary engine has a compression ratio of twenty-five. They run between about nine and ten, which is ordinary, and the difference is not a discrepancy — it is deliberate, and it is the clearest example this field has of a geometric optimum being thrown away on purpose.

The minimum chamber volume is what sits between the rotor’s flank and the housing’s waist at the moment of closest approach, and on the pure geometry it is very small: $412$ mm², four per cent of the maximum. A real rotor has a recess machined into the middle of each flank — a shallow bathtub — and that recess adds volume at every position, which barely changes the maximum and multiplies the minimum. Compression falls to something an engine can be run on without detonation, and the recess also gives the flame somewhere to travel, which the geometric minimum does not.

So the number to quote for the geometry is $25.2$ and the number to quote for the engine is nine, and the gap between them is a pocket somebody cut. It is the same shape of decision as a cam’s base circle, where the geometry says the smallest cam that can be cut and the designer takes a larger one for reasons the geometry does not contain.

What one number does to the numbers #

The whole geometry is set by $K = R/e$, so the chamber arithmetic is a function of it. Measured across the useful range at a fixed generating radius:

Two things are visible and both matter to a designer.

The compression ratio is a steep function of K and the swept area is not. Going from $K = 12$ to $K = 4$ triples the swept area — barely — and divides the geometric compression by nearly six. The proportion that decides how much gas the engine handles and the proportion that decides how hard it squeezes it are the same proportion, and they move at very different rates.

A deep-lobed engine has an engine-like compression ratio without a recess. At $K = 4$ the geometry gives 9.8, which is what a petrol engine wants. That sounds like the obvious choice and is not one: the housing at $K = 4$ has a far tighter waist, its curvature there is severe, and the apex seal has to negotiate it three times per rotor turn. Production engines took the shallow housing and cut the compression back with a pocket, which is a trade of a geometric property against a sealing one.

Why the areas are computable at all #

It is worth being explicit about what makes this measurement available, because the equivalent measurement on a piston engine is trivial and on most mechanisms is impossible.

The chamber’s two boundaries are the two bodies, and both bodies are known as curves rather than as pictures. The housing has an equation. The rotor came out of the meshing equation as a list of contacts, and — this is the part that matters — its corners are known exactly rather than sampled: the generation returns them at the generating radius, at $0°$ and $\pm120°$, as contacts that hold at every shaft angle. So the split between one chamber and the next is made at a point that is known to the last digit, not at the nearest sampled point to a corner.

That distinction is what makes the invariant a sharp check instead of an approximate one. Splitting at a nearby sample would leave slivers of area unassigned, at random, and the sum would wander by an amount that depends on the sampling rather than on the geometry — which is precisely the kind of error that looks like a real physical effect.

The mistake the invariant caught #

The first version of this measurement reported chamber areas of about thirty thousand square millimetres each, totalling $95{,}586$ — against a housing area of $33{,}263$. Three chambers inside a housing, each nearly as large as the housing, adding to three times it.

The fault was in one decision. A chamber is bounded by the arc of housing between two apexes, and there are two such arcs: the short way round and the long way. Taking the wrong one gave a polygon that wrapped the whole engine and enclosed the rotor as well, three times over. The total was constant to eleven decimal places the whole time, because a wrong-but-consistent rule is still consistent — the invariant was satisfied by a set of regions that were not the chambers.

Two things saved it. The total was compared against a number computed independently — the housing’s area less the rotor’s — rather than only against itself. And the areas were compared against the sizes of things visible in the drawing, which is the check nobody writes down and everybody uses.

An invariant that holds is evidence about consistency and not about correctness. It says the computation is doing the same thing every time; whether that thing is the right one takes a second measurement of a different kind, and the second measurement here is the one that has the housing and the rotor in it separately.

Where the volume goes, and what is not claimed #

Three things about this that a reader of engine literature will want, and that this field cannot supply.

Nothing here is a pressure. The chambers change size; what the gas in them does is thermodynamics, and every number above survives with the chambers empty.

The recess is stated, not designed. How large a recess to cut, where to put it, and what it does to the flame front are combustion questions. The geometry can say what the recess costs in compression ratio and what the swept volume is with and without it, and that is all.

Leakage is not geometry either. The three corners are geometrically points of contact, and a real apex seal is a strip in a groove with a spring behind it. What gets past it depends on the seal, the wall’s finish and the pressure difference — none of which is a shape.

What the geometry does say, and it is worth having, is that the sealing problem is three lines. A piston engine seals one chamber with a ring round a circular bore, an operation the whole industry has spent a century perfecting; this engine seals three chambers with three corners sliding along a wall that is not circular and whose curvature varies round the turn. That is a harder problem for a reason the shape makes plain.

The cycle, in shaft angles #

It is worth putting the phasing down in numbers, because the engine’s headline property falls straight out of it and is often stated as though it were a design choice.

The rotor turns once for every three turns of the shaft. A chamber’s area goes through one maximum and one minimum per rotor revolution, so one chamber completes its whole cycle — inhale, compress, be pushed, exhale — in three shaft turns. The three chambers are a third of a rotor revolution apart, which is one shaft turn.

So: three chambers, each on a three-shaft-turn cycle, offset by one shaft turn each. One of them reaches its power phase every shaft turn. A four-stroke piston engine fires once every two turns per cylinder, so a single-rotor engine of this kind fires as often as a two-cylinder four-stroke, from one moving part with no valves.

The same arithmetic explains the eccentric shaft’s gearing. The rotor has to turn at exactly a third of the shaft rate, and that relationship is not produced by the contact — it is produced by a fixed internal gear on the housing and a ring gear on the rotor, in the ratio 3 : 2, which is a ratio a gear train can be checked against rather than something the envelope enforces. The shapes hold contact; the gears hold the phase.

The same measurement, elsewhere #

Chambers between two shapes in continuous contact are not peculiar to this engine, and the same computation answers the same question for several machines this site has drawn or named.

A gear pump is two meshing wheels in a close-fitting case: the space between successive teeth and the case is a chamber, it is carried round from inlet to outlet, and its area is the tooth space. The displacement per turn is the number of teeth times that area, and the calculation is a polygon integral over a generated flank in exactly the way this one is.

A lobe pump or blower — two conjugate two-lobed rotors turning in opposite directions — divides its case into chambers the same way, with the sealing done by the contact between the rotors and by their clearance to the case.

A cycloidal pump — an inner rotor with $N$ lobes inside an outer with $N+1$ — has $N+1$ chambers whose areas rise and fall as the eccentric turns, and the same invariant holds: they add to the area between the two rotors.

In every case the geometry supplies a displacement per turn and stops there. What each machine does with that displacement — pressure, flow, slip, efficiency — needs a fluid, and this site has never had one.

The recess is larger than the chamber it opens into #

The geometry gives 25.2 and real engines run at nine or ten, and the essay attributes the difference to a recess cut in the rotor’s flank. It is worth doing the arithmetic on that recess, because the answer is not a detail.

Add a recess of area $r$ and it is present at every shaft angle, so it adds to both ends of the range: the ratio becomes $(10{,}395 + r)/(412 + r)$. Set that equal to nine and solve — $10{,}395 + r = 3{,}708 + 9r$, so $8r = 6{,}687$ and $r = 836$ mm².

That is twice the geometric minimum chamber. The pocket cut into the rotor’s face holds more than the space it opens into at the moment of closest approach, and the mechanism’s compression ratio is set, in the main, by a feature that is not part of its kinematics at all.

Which reverses the natural reading of the whole table. The compression ratio a designer gets is not the geometric one adjusted a little; it is a number chosen by the recess, with the geometry supplying a ceiling. And the ceiling is generous: 25.2 at $K = 3$ is far above anything an engine wants, so the recess is always subtracting from the geometry rather than adding to it.

Read that way, the $K$ table says something a designer would act on. A deep-lobed rotor at $K = 4$ has a geometric ratio that is engine-like without any recess, so the pocket could be small or absent — and the volume it would have occupied is available as swept area instead. A shallow one at $K = 2$ needs a large recess and gives it back in dead volume that never gets swept.

So the choice of $K$ is not only about the rotor’s shape and its sealing. It decides how much of the chamber has to be thrown away at top dead centre to reach a workable ratio, and the amount is computable from two areas and one division. That is a genuinely kinematic contribution to a decision usually made on other grounds — and it is one the geometry can make before anything about combustion is settled.

The invariant as a habit #

The general form of what happened here is worth stating, because it is the cheapest kind of check a computation can carry.

Find a quantity the mechanism cannot change, and measure it while everything else moves. The total area between two rigid bodies in contact is one. The sum of the two flank curvature radii at a gear contact is another. The distance between two joints of a rigid link is the one every solve on this site is checked against.

None of them is interesting as a fact. All of them are load-bearing as checks, because the way to violate them is to have made a mistake somewhere else entirely — and a check that can only be failed by an error nobody was looking for is worth more than one that tests what was already in mind.