The wheel that forbids nothing

Nine essays of this field are about what a rolling constraint costs. This one is about what happens when it is removed, which is the cleanest way to see what it was doing.

A mecanum wheel is an ordinary wheel with a ring of barrel-shaped rollers around its rim, each free to spin about an axis at forty-five degrees to the wheel’s own axle. An omni wheel is the same idea with the rollers at ninety degrees. Machines built on them go sideways as easily as forwards, turn on the spot, and do not shuffle. They are not exempt from anything: they obey exactly the same equation as every other wheel in this field, with one number changed.

One row for every wheel #

Put a wheel at $(x, y)$ in the vehicle’s frame, steered to $\delta$, with rollers whose axes make an angle $\gamma$ with the wheel’s own axle. The contact point moves at $v + \omega \times p$, and the component of that across the roller has to be whatever the wheel’s own rotation supplies:

$$n \cdot (v + \omega \times p) \;=\; -\,r \sin\gamma \; \omega_{\text{wheel}}, \qquad n = (-\sin(\delta + \gamma),\ \cos(\delta + \gamma)).$$

Every wheel in this field contributes that row and no other. The wheel’s own speed appears only on the right-hand side, multiplied by $\sin\gamma$, and everything follows from that factor.

At $\gamma = 0$ — no rollers — the right-hand side is zero whatever the wheel is doing, so the row is a constraint on the vehicle: the body may not move across the wheel, and the wheel’s speed has dropped out of the statement entirely. That is the constraint the whole field is about.

At $\gamma \neq 0$ the right-hand side is not zero, so the row does not restrict the twist at all — any twist can be accommodated by choosing the wheel’s speed. The row has become a statement about the wheel rather than about the vehicle.

That is the whole difference between a machine that shuffles and one that slides sideways, and it is a factor of $\sin\gamma$ in one row. No new mechanism, no new equation, and no exemption from anything.

The rank, which goes to zero #

Collect the rows that are constraints — the ones with $\gamma = 0$ — and take the rank of the matrix they form. It is the number of components of the body’s twist that are held.

Four fixed wheels all pointing forwards leave a one-dimensional family of twists — straight ahead, and nothing else. The vehicle can go forwards and backwards and that is all it can do, which is what a supermarket trolley with its castors welded would be. A steered car is the same rank arrived at differently: two rows rather than four, because the four axle lines are made to meet and two of the four rows are then redundant.

A platform on wheels with rollers has no constraint rows at all. Its rank is zero, its null space is the whole three-dimensional space of planar twists, and there is nothing in this field’s machinery left to say about it: no forbidden direction, no bracket, no growth vector beyond a single entry, and no manoeuvring. It is holonomic, in the sense the second essay measures, for the simplest possible reason — there is nothing to integrate.

The forty-five degrees #

The mecanum wheel’s rollers are at forty-five degrees on every machine ever built with them, and the reason is worth deriving rather than accepting, because it is a genuinely optimal choice rather than a convention.

With the standard four-wheel layout — rollers at $+45°$ on one diagonal and $-45°$ on the other — the wheel speeds needed for the three unit twists are

$$\begin{aligned} \text{forwards } 1\ \text{m/s}: &\quad (+20,\ +20,\ +20,\ +20) \text{ rad/s} \\ \text{sideways } 1\ \text{m/s}: &\quad (-20,\ +20,\ +20,\ -20) \\ \text{turning } 1\ \text{rad/s}: &\quad (-6.6,\ +6.6,\ -6.6,\ +6.6) \end{aligned}$$

for wheels of 50 mm radius at $\pm 180$ mm and $\pm 150$ mm. The first two rows are the same four numbers with different signs — identical magnitudes, to the last digit of double precision, which is the arithmetic saying that a metre sideways costs exactly what a metre forwards costs.

That is not true at any other roller angle. Sweep $\gamma$ and the ratio of the worst sideways wheel speed to the worst forward one is 1 at forty-five degrees and larger everywhere else, running away as $\gamma$ approaches zero — where the rollers stop helping and the platform stops being able to go sideways at all.

The design consequence is direct: at any other angle the machine’s motors have to be sized for its worst direction, and the machine is slower in that direction than it needs to be in the other. Forty-five degrees is the unique angle at which the machine is isotropic in translation and nothing is wasted.

What is paid for it #

A mecanum platform has four wheels and three twist components, so the four speeds are not free: they satisfy one relation. Command four speeds that do not satisfy it and there is no twist that fits, and the platform pays the difference in scrub — the same currency a locked axle in a turn pays in.

Solving for the twist is therefore a least-squares fit rather than an inversion, and the residual is not decoration: it is the rate at which the four commanded speeds are asking the platform to come apart. On consistent speeds it is at the noise floor; on inconsistent ones it is metres of sliding per metre of travel.

The conditioning is the other price. The three singular values of the standard mecanum layout are $1.414$, $1.414$ and $0.467$, so the ratio between the easiest and the hardest twist component is 3.03. The two translations are equally cheap — that is the isotropy — and rotation is three times dearer, because the lever arm from the platform’s centre to a wheel is shorter than a metre.

A three-wheel omni layout on a 200 mm circle has its own conditioning and it is a different number, and choosing between the two layouts is choosing which twist to make cheap. Neither is better; they are different distributions of the same total.

Three wheels or four #

Three omni wheels on a circle are the other standard layout, and comparing the two is comparing two ways of spending the same total.

A three-wheel omni platform has exactly three wheels for three twist components, so its matrix is square: the speeds determine the twist and the twist determines the speeds, with no redundancy and no residual. Its singular values, for wheels of 50 mm radius on a 200 mm circle, are $1.225$, $1.225$ and $0.346$ — isotropic in translation, like the mecanum, and a conditioning of 3.54 against the four-wheel layout’s 3.03.

So the four-wheel platform is slightly better conditioned and carries a redundancy the three-wheel one does not. What the redundancy buys is not accuracy — a fourth measurement of a three-component quantity does not sharpen it much when all four are equally noisy — but detection: with three wheels every set of speeds is consistent with some twist, and with four the residual is a live signal that something is wrong. A slipping wheel, a seized roller, a wrong radius: all of them show as a residual that has no business being there, and on three wheels all of them are invisible.

That is the same trade the redundant constraints of an overconstrained loop present, arrived at from the other side. A surplus row is either a nuisance or an instrument, and which it is depends on whether anything is watching it.

The castor, which is neither #

Between a wheel that constrains and a wheel that does not sits the one every trolley has, and it does not fit either row.

A castor is an ordinary wheel — $\gamma = 0$, a hard constraint across itself — mounted on a swivel that is free. Its constraint is real at every instant and its direction is a coordinate of the mechanism rather than of the vehicle, so the vehicle is not constrained by it at all: whatever twist the driven wheels produce, the castor swivels to accommodate it.

The price is that the swivelling takes time and distance. A castor lines itself up over a distance of a few times its trail, which is the towed axle’s settling length applied to a very short rod, and during that distance the castor is dragging sideways and resisting. Anybody who has pushed a supermarket trolley sideways has felt the interval: the trolley does not go sideways, it goes wherever the castors were pointing and then gradually goes sideways.

So a castor converts a hard constraint into a transient. It is the cheapest available approximation to a wheel that forbids nothing, it costs nothing but a bearing, and what it gives up is the promptness that a mecanum wheel has. That is why a machine that needs to move sideways on command uses rollers and a machine that only needs to be pushed uses castors.

What sin γ costs before it reaches zero #

The row draws a clean binary distinction — at $\gamma = 0$ it constrains the vehicle, at any other angle it does not — and reading only that leaves the impression that a roller angle of one degree buys what forty-five buys. It buys the same freedom at a wildly different price, and the price is the same factor over again.

The right-hand side of the row is $-r\sin\gamma$, so the wheel speed a given twist demands is the left-hand side divided by it:

$$\dot\phi = \frac{n \cdot (v + \omega \times p)}{-r\sin\gamma}.$$

Sideways motion is accommodated by spinning the wheel, and the spin required goes as $1/\sin\gamma$. At forty-five degrees the factor is $\sqrt{2}$. At ten degrees it is 5.76, at five degrees 11.5, at one degree 57.3. A platform with rollers at one degree is kinematically omnidirectional and would need motors turning fifty-seven times faster to go sideways than to go forwards.

So the transition at $\gamma = 0$ is discontinuous in kind and continuous in cost, and the two are easy to conflate. Nothing gradual happens to the rank as the roller angle is reduced: the matrix has no constraint rows at any $\gamma \neq 0$ and acquires one the instant $\gamma$ is exactly zero. What happens gradually is that the speeds required to use the freedom run away — and a freedom that requires unbounded speed is a freedom in the algebraic sense and in no other.

That is also the more useful way to read the forty-five degrees. It is the isotropic angle, and it is as far from the singular angle as the geometry allows: $\sin\gamma$ grows to ninety degrees, and at ninety the rollers’ axes lie along the rim where they cannot carry the wheel forward at all. The angle is squeezed from both ends, and the argument from conditioning and the argument from isotropy pick out the same number — which is the sort of agreement that makes a convention worth trusting rather than merely following.

The residual as an instrument #

Four wheels for three twist components leave one relation among the speeds, and the residual was described above as the rate at which the commanded speeds are asking the platform to come apart. It is worth naming what else that makes it.

A three-wheel omni platform’s matrix is square, so any three speeds are consistent with exactly one twist. Feed it speeds produced by a slipping wheel, by a wheel whose radius is two per cent under nominal, or by a roller jammed with swarf, and it returns a twist — a wrong one, with nothing whatever to indicate that anything is wrong. The dead reckoning drifts, and the drift has no signature to be found.

The four-wheel platform cannot be misled so quietly. Its four measured speeds must satisfy one linear relation, and a fault that changes any single wheel’s contribution violates it. The residual becomes a fault detector with no extra hardware in it — not a diagnosis, since one relation cannot say which of four wheels is at fault, but a detection, which is the half that is otherwise entirely missing.

That is the same trade the fourth wheel offers a steered vehicle, where four constraint rows for two independent conditions leave a residual that vanishes exactly when the wheels agree. It is the trade an overconstrained loop makes with its surplus joints. Redundancy is paid for with a least-squares solve and paid back as the ability to notice, and the choice between three wheels and four is not only about conditioning: it is a choice about whether the machine can tell when it has gone wrong.

The asymmetry is worth stating plainly, because it runs opposite to the usual intuition about part counts. The simpler machine — three wheels, a square matrix, an exact inversion — is the one with no self-check, and the extra wheel that looks like an extravagance is what buys the arithmetic something to be inconsistent about. A mechanism with nothing left over has nothing to compare.

The costs that are not in this field #

Everything above is kinematics and the reasons nobody builds cars on mecanum wheels are not.

A roller is a small wheel. It carries the whole load at the moment it is in contact, on a barrel a couple of centimetres across, and it hands over to the next one as the wheel turns. That gives a mecanum wheel a ride quality nobody would accept in a vehicle and a rolling resistance several times an ordinary tyre’s.

Sideways is expensive in effort even when it is cheap in geometry. Going sideways runs every roller against its own axis, which is a sliding contact rather than a rolling one, and the efficiency is correspondingly poor. The kinematics says the wheel speeds are the same; nothing here says the power is.

Debris. A roller with a gap beside it collects everything on the floor. Mecanum platforms live on warehouse floors and film-studio floors and not on roads.

All three are the reason the mechanism is common in one place — factory floors, camera dollies, wheelchairs, and the robots built for competitions — and absent everywhere else. The kinematic advantage is real and complete, and it is bought with three things that are outside this field entirely.

The row, once more, as the whole argument #

It is worth restating the result in one place because it is unusually compact for a claim of this size.

$$n \cdot (v + \omega \times p) \;=\; -\,r \sin\gamma \; \omega_{\text{wheel}}$$

A car, a shopping trolley, a mecanum platform, a swerve drive, a castor and a locomotive all obey that row at every wheel. The mechanism’s whole character — whether it must shuffle, whether it must be steered, whether it has an instantaneous centre, whether its tracks can be read afterwards — is decided by which of the row’s two sides is zero, and that is decided by $\gamma$. Nine essays of consequences, and one factor.

What removing the constraint removes #

It is worth being explicit about how much of this field disappears when the rows do.

There is no forbidden direction, so no bracket, so no manoeuvre and no exponent. There is no growth vector worth writing: a single entry, 3, which is the same statement as this mechanism has a configuration space of the dimension its count says. There is no off-tracking, because nothing is towed. There is no shortest-path classification, because the reachable set from any placement in any time is a neighbourhood rather than a curve, and the shortest path between two placements is the straight line between them with the rotation done along the way.

And there is no identification from tracks, which is the one worth noticing. A mecanum platform’s wheels leave marks that say nothing about which way it was going, because a wheel’s mark is a scuff in an arbitrary direction rather than a rolling track. The readability of a towed wheel’s track came entirely from the constraint, and removing the constraint removes the evidence.

That last one is the clearest statement of what a constraint is worth. A constrained mechanism is harder to command and easier to read; an unconstrained one is the reverse. Nothing on a mecanum platform’s floor records what it did.