The error that is an integral

Every tolerance argument on this site so far has had the same shape. A mechanism’s lengths have ranges; the output is a function of the lengths; differentiate, and the output’s range follows. A four-bar’s coupler point moves by so many microns per micron of link length; a gear ratio has no stack at all, because none of the tolerated quantities appear in it.

In every case the answer is a bound. The mechanism can be out by this much and no more, and running it for longer does not make it worse.

A wheeled mechanism does not work that way, and the reason is that its output is an integral of its input rather than a function of it.

Why the stack does not bound #

A wheeled vehicle knows where it is by integrating what its wheels did. Wheel speeds give a body twist through the wheel matrix; the twist is integrated to give a pose; and the pose is what the machine believes.

Every step of that is a computation on the machine’s assumed geometry — the radii it thinks its wheels have, the track it thinks they are on — and the actual motion happens with the geometry the wheels actually have. So an error in the geometry produces a small error in the twist at every instant, and the pose error is the integral of the twist error over the whole journey.

An integral of a constant grows linearly. An integral of a heading error grows quadratically in the distance, because a heading error tilts the direction of travel and the lateral displacement is the heading error times the distance. So the error budget of a rolling mechanism has terms that grow as $s$ and as $s^2$ where every other field on this site has terms that are constant.

That is not a defect in any wheel. It is the arithmetic consequence of the mechanism’s position being reached rather than solved for, which is the field’s premise turned into an error budget.

The two systematic errors #

For a differential drive with wheel radius $r$ and track $b$, two geometric quantities are assumed and two can be wrong.

Unequal wheel radii. Command both wheels at the same speed and the vehicle does not go straight; it drives an arc of curvature

$$\kappa = \frac{\Delta r}{\bar r\, b}.$$

At $r = 50$ mm, $b = 300$ mm and one per cent of mismatch, that is $\kappa = 0.0333$ per metre — a 30 m radius. Over a four-metre straight run the heading has turned by $4/30 = 0.133$ rad, which is 7.64°, and the lateral drop is $4^2\!/(2\times 30) = 0.267$ m.

Four such runs, one per side of a square, and the heading is out by 30.56° before any turn has been mis-executed.

A wrong track width. Command a turn and the vehicle turns through a different angle than it believes. The error is proportional to the commanded turn, so it accumulates once per corner rather than continuously — a 1% track error gives 1% of every turn, which over four right-angles is 3.6°.

Both are deterministic. Neither is noise, neither averages out, and neither is visible to a machine that only knows what it commanded.

The measurement, and the thing that separates them #

Drive a four-metre square, and drive it again the other way round. The comparison is the whole point.

The heading is the discriminator, and the signature is which one reverses.

A radius mismatch bends the vehicle’s path in the vehicle’s own frame — always to the same side of itself — so driving the square the other way round accumulates the same heading error with the same sign. Its two closure errors point 90° apart and have different magnitudes.

A track error scales every commanded turn, and driving the square the other way reverses every turn, so the heading error reverses exactly. Its two closure errors are mirror images of identical length.

One run cannot tell them apart: a single closure error is one vector and there are two unknowns. Two runs can, and the algebra is a two-by-two solve. That is why the standard bidirectional square test exists and why running it in one direction is worth much less than half of running it in both.

The size of it, against what a machinist would expect #

A one per cent difference in wheel radius is half a millimetre on a 50 mm wheel. On a linkage that would be a coarse tolerance and its effect would be a fraction of a millimetre at the output.

Here it is 1.6 m over 16 m of travel — ten per cent of the distance driven. Three orders of magnitude worse than the same fractional error would produce anywhere else on this site, and the reason is entirely structural: the error enters the heading, the heading multiplies the distance, and the distance is unbounded.

The design consequences follow directly and they are not the obvious ones.

A tighter tolerance buys linearly, not quadratically. Halving the radius mismatch halves the curvature and halves the error at every distance. There is no square root to hide behind, and worst case and quadrature do not help, because these errors do not combine like independent random ones — a systematic bias in one direction is the same bias every time.

Calibration beats manufacture. The curvature is a function of the ratio of the radii, so measuring the ratio once and correcting for it in software removes the whole term without touching the wheels. That is what the bidirectional square test is for: it is a calibration procedure, and its output is two numbers that go into the odometry.

And the calibration is only as good as the surface. The radius that matters is the effective rolling radius, which depends on the tyre’s compression and on the floor. A machine calibrated on smooth concrete and run on carpet has a different pair of radii and its calibration is spent.

Reading the two errors off one picture #

The bidirectional square is a calibration procedure and it is worth setting out as one, because the arithmetic that turns two closure vectors into two corrections is short and is the whole reason for driving the square twice.

Let $\varepsilon_r$ be the radius ratio error and $\varepsilon_b$ the track error. Each contributes a heading error per lap, and the two contributions are

$$\Delta\theta_{\text{cw}} = a\varepsilon_r + b\varepsilon_b, \qquad \Delta\theta_{\text{ccw}} = a\varepsilon_r - b\varepsilon_b,$$

with $a$ and $b$ decided by the square’s side and the vehicle’s track. Adding and subtracting gives each error on its own:

$$\varepsilon_r \propto \Delta\theta_{\text{cw}} + \Delta\theta_{\text{ccw}}, \qquad \varepsilon_b \propto \Delta\theta_{\text{cw}} - \Delta\theta_{\text{ccw}}.$$

The sum isolates the error that does not reverse and the difference isolates the one that does. With the numbers above: the sum is $61.1°$ and the difference is $0$ for a pure radius error, and the sum is $0$ and the difference is $7.1°$ for a pure track error. Two measurements, two unknowns, one two-by-two solve, and no iteration.

What makes it work is that the two errors have opposite parities under reversing the lap. That is a structural fact about which of them lives in the vehicle’s own frame and which lives in the commanded turns, and it would fail if a third systematic error shared a parity with one of them — which is why a third error, if there were one, would need a third experiment rather than a longer version of this one.

What cannot be calibrated away #

Two things, and they are what set the floor.

Slip. The rolling constraint is an assumption, and a wheel that slips has broken it. Slip is not systematic, does not reverse with direction, and does not average out either — it is a one-way loss, since a wheel that slips always over-reports distance. It is also the only term here that depends on what the vehicle is doing rather than on what it is: accelerating, braking and turning tightly all produce more of it than driving straight.

The unmodelled third wheel. Most differential-drive machines have a castor, and a castor that is being dragged sideways while it swivels is scrubbing, which perturbs the vehicle’s heading by an amount that depends on which way the castor happened to be pointing when the manoeuvre started. That is a state the odometry does not carry, and it produces errors that are repeatable within a run and not between runs.

Both are outside this field. Slip is a friction question and castor scrub is a force question, and what this essay contributes to either is the observation that they enter the same integral and are therefore subject to the same accumulation.

What a straight line costs to check #

There is a cheaper experiment than the square and it is worth describing, because it isolates one of the two errors without any algebra at all.

Command a straight line and measure how far the vehicle has moved sideways at the end. The lateral drop is $s^2\kappa/2$ with $\kappa$ the curvature the radius mismatch produced, so

$$\frac{\Delta r}{\bar r} \;=\; \frac{2 b\,y}{s^2}.$$

Four metres of travel and a 267 mm drop, at $b = 0.3$ m, gives one per cent. The measurement needs a chalk line and a tape and it returns the ratio directly, because the track width does not appear in a straight run at all — no turn is commanded, so a wrong track produces no error.

That asymmetry is the same one the square test exploits and it is worth naming as a principle rather than a trick: a systematic error is isolated by an experiment that does not excite the others. A straight run excites the radius ratio alone; a pure spin on the spot excites the track alone; the square excites both and needs two runs to separate them.

The reason the square is the standard test anyway is that a machine that has to be sent out and brought back is testing the thing it actually does, and a closure error is a single number that a floor and a tape measure can produce. The two-run version buys the separation for the price of running it twice.

The general statement #

Strip out the wheels and what is left is a distinction that applies to any mechanism.

A mechanism whose configuration is the solution of an equation has a bounded error budget: perturb the parameters, and the solution moves by a bounded amount determined by the Jacobian, however long the mechanism runs.

A mechanism whose configuration is the result of an integration has an unbounded one. The error at the end is the integral of the error along the way, and no property of the mechanism bounds it — only the length of the journey does.

Every field on this site until this one is in the first category, which is why every tolerance essay before this one reports a bound. This field is in the second, and it is the same structural difference that makes its mobility count two numbers instead of one: a mechanism with no closure equation has nothing to hold it to a nominal, in either sense.

Why this field’s mechanisms have no nominal #

One more difference from the rest of the site is worth drawing, because it explains why the word tolerance has to be used slightly differently here.

Everywhere else, a mechanism has a nominal: the linkage as drawn, with its stated lengths, and a real one differs from it by a stack. The nominal is a real object — it is the thing the solver solves — and every tolerance argument is a statement about the neighbourhood of it.

A wheeled mechanism’s pose has no nominal. There is no configuration the machine is supposed to be in, because the configuration is wherever the history took it. What has a nominal is the machine’s own geometry — the radii, the track — and the tolerance stack propagates from there into a pose through an integral rather than through a solve.

So the quantity that has a range here is not the output; it is the map from commands to motion, and everything downstream inherits the range and multiplies it by distance. That is why the essay’s numbers are given as errors per metre or per lap rather than as absolute bands, and why a specification for such a machine is written as a drift rate rather than as a positional tolerance.

The heading grows linearly and the position quadratically #

The essay’s headline compares an error against a distance driven — 1.6 m over 16 m, ten per cent — and that ratio is not a constant. It is worth separating the two errors by how they grow, because the separation says what a longer run costs and the ratio does not.

Heading error is linear in distance. A curvature error $\kappa$ turns the vehicle by $\kappa s$ over a run of length $s$, so a straight line commanded for twice as long ends up pointing twice as far wrong. That is the 30.56° after four sides of the square, and it is the quantity that accumulates cleanly.

Position error is quadratic. The vehicle follows an arc, so its lateral departure from the commanded straight line is $(1-\cos\kappa s)/\kappa$, which for small $\kappa s$ is $\kappa s^2/2$. At $\kappa = 0.0333$ per metre and $s = 4$ m that is 0.267 m, which is the drop measured in the straight-line test.

So the ten per cent is a fact about sixteen metres and not about the machine. Double the run and the drop quadruples while the distance only doubles, so the fraction of distance driven grows linearly — twenty per cent at thirty-two metres, forty at sixty-four. A specification quoting a percentage of distance is quoting the value of a growing function, which is the same species of error this site’s wrong field collects, arriving here in a robot’s data sheet.

The quadratic does not grow for ever, and where it stops is worth knowing. The path is a circle of radius $1/\kappa$, so the departure saturates at $2/\kappa$ — 60 m for the one-per-cent machine — and beyond that the vehicle is going round in circles rather than getting further from the line. That is not a consolation: a machine that has completed a lap of a 30 m circle has a bounded position error and a heading error that has passed through every value, and it is lost in the way that matters.

The design reading follows from the exponent rather than from either number. Errors that accumulate in heading are the expensive ones, because position error is the integral of heading error and therefore always one power worse. Anything that resets the heading — a wall to square against, a docking station, a single bearing to a landmark — buys more than anything that corrects position, and it buys it at the lower power.

That is also why the two-run square test is worth its cost. It measures the heading discrepancy, which is the linear quantity and therefore the one that separates cleanly from noise over a modest run, rather than the position closure, which is quadratic and mixes the two errors together at every distance.

What a bounded version would look like #

The way out is not a better wheel; it is a second source of information, and it is worth saying what kind.

Anything that measures the vehicle’s pose against the world rather than against its own wheels turns the unbounded error into a bounded one: a wall it can touch, a mark on the floor it can see, a magnet in the ground. Each of those is an equation the vehicle can solve rather than a rate it has to integrate, and each brings the error budget back into the first category above.

Which of them to use, how often, and how to combine a measurement with an integration is estimation, and it belongs to whoever owns estimation. The kinematic statement this field can make is the one that says why it is necessary: a wheeled machine’s belief about where it is has no fixed point of its own, and something outside it has to supply one.