Where the jaws put it

A chuck is a hold that closes. Its jaws advance together until they all touch, and where the part ends up is not chosen by anybody — it is decided by the part’s own shape and by how it happened to be put in.

Where the jaws stop #

Let the $j$ jaws advance along directions $u_k$ spaced evenly round a circle, each a flat face perpendicular to its own direction, all at the same distance $d$ from the chuck’s axis. The part’s own centre is at $c$ and its support function about that centre is $h$, so the jaw $k$ touches when

$$c \cdot u_k + h(u_k) \;=\; d .$$

Sum those over all $j$ jaws. Evenly spaced unit vectors satisfy $\sum u_k = 0$, so the $c$ terms vanish and

$$d \;=\; \frac{1}{j}\sum_k h(u_k),$$

which is the jaw distance, immediately. And evenly spaced unit vectors also satisfy $\sum u_k u_k^{\mathsf T} = \tfrac{j}{2} I$, which inverts the remaining system in one line:

$$c \;=\; -\frac{2}{j}\sum_k h(u_k)\,u_k .$$

That is the whole thing. The offset is a weighted sum of the part’s own support function, sampled in the jaw directions.

For a circle $h$ is constant and $\sum h u_k = h \sum u_k = 0$: perfectly centred, which is what a chuck is sold as doing. For a hexagon in three jaws the three sampled values are equal at every orientation, because 120° is one of a hexagon’s symmetries, so again the sum vanishes. For a square in three jaws they are not equal and the offset is not zero.

How large #

Swept through a full period of the section’s own symmetry, the worst offsets in three jaws are:

• triangle, hexagon, nonagon, twelve-sided: below $10^{-15}$ of a circumradius, which is the arithmetic’s floor;
• square: 0.1725 — 17.3 per cent of the circumradius;
• pentagon: 0.1128;
• heptagon: 0.0586;
• octagon: 0.0450.

On a 25 mm bar the square’s worst is 2.16 mm of runout and the pentagon’s 1.41 mm. Those are not small numbers and they are not the chuck’s fault: a chuck with no error whatever produces them, and the quantity is a property of the section.

The trend down the list is the useful part. The offset falls with the number of sides — a section closer to round is centred better — which is what a machinist would guess, and the exceptions are not the ones a machinist would guess: the nonagon, at nine sides, is exactly centred while the octagon, at eight, is out by 4.5 per cent.

The rule that is nearly right #

The workshop explanation of all this is about symmetry, and it is worth stating in its own terms because it is what everybody carries: a chuck centres a section when a turn of $360^\circ/j$ is one of the section’s own symmetries. Three jaws centre what is unchanged by 120°, which is the multiples of three and the circle.

That rule predicts every row of the table above and it is wrong, and the case that shows it is not exotic.

Put a square bar in a six-jaw chuck. Sixty degrees is not a symmetry of a square. The offset is nought at every orientation, to fourteen figures.

What the right condition is #

The formula says so directly, once it is read as what it is.

$c = -\tfrac{2}{j}\sum_k h(u_k) u_k$ is a discrete Fourier sum over $j$ equally spaced directions, and such a sum annihilates every harmonic of $h$ except those whose index is $\pm 1$ modulo $j$. A regular $n$-gon’s support function has period $2\pi/n$, so its harmonics are the multiples of $n$. The multiples of $n$ modulo $j$ form the subgroup generated by $\gcd(n, j)$, and that subgroup contains 1 exactly when the divisor is 1 — and contains $-1$ with it, since a divisor of $j$ that divides $j-1$ divides 1.

$$\text{a } j\text{-jaw chuck centres a regular } n\text{-gon exactly when } \gcd(n, j) > 1 .$$

Check it against the exceptions. A square in six jaws: $\gcd(4,6) = 2 > 1$, centred. A square in five: $\gcd = 1$, out by 0.136. A pentagon in six: $\gcd = 1$, out. A hexagon in four: $\gcd = 2$, centred — and 90° is not a symmetry of a hexagon either.

Swept across every section from three sides to twelve and every chuck from three jaws to eight — sixty cases — the divisor predicts the measured offset every time, and the symmetry rule is wrong on eleven of the sixty.

The two rules agree whenever the jaw count is prime, since then a common divisor greater than one means the count divides the sides. Three is prime. So the wrong rule has never caused any trouble on the chuck everybody uses, and it becomes wrong exactly where nobody looks. That is the shape of most of what this site’s drawn-wrongly field collects: a rule that is right on the case it was formed on and is stated as though it were general.

Two routes to the number #

The closed form leans on $\sum u u^{\mathsf T} = \tfrac{j}{2} I$, which is an identity somebody has to get right, so the second route knows nothing about it.

Closing the jaws until they all touch is the same sentence as make $d$ as small as it will go without the part penetrating any jaw, and that is a linear program in the three unknowns $(c_x, c_y, d)$:

$$\text{minimise } d \quad\text{subject to}\quad c \cdot u_k \le d - h(u_k) \text{ for every } k .$$

At the optimum every constraint is tight, which is the statement that all three jaws touch, and that is asserted rather than assumed. The two routes agree to $10^{-9}$ on every section and orientation checked.

That is the same discipline the whole site runs on, and here it earns its place twice over: the identity is the kind of thing that is memorised rather than derived, and an error in it would have moved every offset in this rung by a constant factor and left every one of them looking plausible.

Why the jaws all touch at once #

One assumption deserves checking, because the whole derivation rests on it and it is not obvious: that all $j$ jaws touch.

They do because they advance together. Every jaw is at the same distance $d$ from the axis at every moment — that is what a scroll chuck’s spiral does — so as $d$ falls, the first jaw to reach the part pushes it, which moves $c$, which changes the other jaws’ gaps. The process stops when no jaw can advance further without penetrating, and at that configuration the $j$ constraints are simultaneously tight because there are $j$ of them and $j$ unknowns.

The linear program above proves it rather than assuming it: at its optimum every constraint is active, and the routine asserts that count and refuses an answer where fewer than $j$ jaws touch.

That is worth having because the alternative arrangement — independent jaws, which a four-jaw chuck has — is a completely different problem. There each $d_k$ is set separately, so the part’s position is chosen rather than determined, and the machinist’s job is to solve the inverse problem: given the part where it should be, find the four jaw settings. That has a solution for any position, which is exactly why an independent-jaw chuck can centre anything and takes ten minutes to do it.

A pentagon, and the shape of the error #

The sweep for a section that is not centred is worth reading rather than summarised, because the offset is not a random-looking wobble.

It is periodic with the section’s own period, smooth, and never touches zero — a pentagon in three jaws is out by at least some amount at every orientation, with a minimum as well as a maximum. So there is no lucky angle at which a square or a pentagon sits true in three jaws, which is the thing a machinist would try first.

The reason is the harmonic argument again. The offset is a single Fourier component of $h$ rotated with the section, so as the section turns the offset vector rotates rather than shrinking, and its magnitude varies only through the higher harmonics that also survive. For a square in three jaws the dominant surviving harmonic is the fourth, and the magnitude runs between 0.089 and 0.173 without ever approaching nought.

What this is not #

It is not a hold. Three flat jaws on a round bar have three normals all pointing at the centre; the rank is two, the margin is nought, and the bar can spin — which is the disc’s row in the ledger and holds at any number of jaws. That is the whole reason a chuck is tightened rather than merely closed, and the tightening is friction, which is outside this field.

So a chuck is a locating device analysed here and a holding device by means this site does not compute. The offset above is a locating error and it is exactly what this field can say about a chuck: where the part ends up when the jaws close, before anything is tightened.

And it is not a calibration error. Every number in this rung comes from a chuck with no error at all. A calibration can find and remove a chuck’s own eccentricity; it cannot remove this, because this is not a property of the chuck. Turn the bar in the jaws and the offset changes; the chuck has not moved.

What a machinist does with it #

Three things follow, and the third is the one worth carrying.

Round and hexagonal stock in three jaws is exact, and that is not luck — it is $\gcd(6,3) = 3$. It is the reason hexagon bar is the stock shape it is, and the reason a hex socket is a good drive form.

Square stock in three jaws is not, by up to 17 per cent of the circumradius, and the amount depends on how the bar happened to go in — so it is not repeatable either, which is a worse failure than a bias and is the one a stack-up analysis would miss entirely, since nothing in the parts is out of tolerance. A four-jaw chuck exists for this and independent jaws exist for the general case.

And a section’s own harmonics decide it, which is the same kind of statement a gearbox’s ratios get from its tooth counts — a number stamped on the outside that is a consequence of an integer relation nobody chose for that reason. The rule is not about how many sides a thing has; it is about which harmonics its support function contains and whether the jaw count picks any of them out. A bar with a lobed profile — three lobes, five lobes, the kind of thing centreless grinding produces — has harmonics at its own lobe count, and a three-jaw chuck on a three-lobed bar reads it as perfectly round while a four-jaw chuck does not.

That last one is the practical version of the whole thing and it inverts the usual advice. A measurement made with three jaws is blind to threefold error, exactly as a two-point micrometer is blind to constant-width lobing. The chuck is not a bad instrument; it is an instrument with a known null space, and knowing which harmonics it cannot see is the whole of using it well.

A support function, which the site already had #

The quantity the whole rung turns on is not new here, and it is worth saying where it came from because that is the reason the derivation is three lines rather than thirty.

$h(\phi)$ is how far a convex shape reaches in the direction $\phi$, measured from a stated centre — the shape’s support function. The strand field built one to compute where a rope runs round a set of bodies, since the taut path’s length is an integral of exactly this quantity, and the meshing field’s envelopes are the same object read as a family of tangent lines.

What makes it the right tool here is that a flat jaw touching a convex part touches it where the support function is attained, and the jaw’s position is the support function’s value. So a chuck is a device for sampling $h$ at $j$ directions, and everything about what it does follows from what those samples contain.

That is worth noticing as a piece of shared vocabulary rather than a coincidence. Three fields on this site — strands, meshing, and this one — reach for the support function, and in each of them the reason is the same: a flat thing touching a convex thing touches it at one place, decided by direction alone.

The condition, exactly #

The gcd rule agrees with the harmonic argument on the cases above and it is not the whole rule, and the exact version is worth writing down because it decides cases the gcd does not — including the six-jaw square that refutes the workshop’s rule.

The offset is a discrete Fourier sum over $j$ equally spaced directions, so it picks out exactly those harmonics of the support function whose order is congruent to $\pm 1$ modulo $j$, and it is blind to all the others. A regular $n$-gon’s support function carries harmonics at the multiples of $n$ and nowhere else. Put those together and the condition is:

A regular $n$-gon is centred by $j$ jaws exactly when no multiple of $n$ is congruent to $\pm 1$ modulo $j$.

Check it against the table. A square in three jaws: $4 \equiv 1$, so a harmonic survives and the bar is off centre — the 17.3 per cent. A hexagon in three jaws: every multiple of six is $0$ modulo three, nothing survives, and the centring is exact. A pentagon in three jaws: $10 \equiv 1$, so it is off. And the case that breaks the workshop rule — a square in six jaws: the multiples of four are $4, 2, 0, 4, 2, 0, \ldots$ modulo six, never $1$ or $5$, so nothing survives and the square is centred exactly.

That last line is the whole refutation in one congruence. The workshop rule asks whether the section’s symmetry is a multiple of the jaw count, which is a divisibility question, and the right question is whether any multiple of the section’s order lands next to a multiple of the jaw count. The two agree when $j$ is prime, because then a multiple of $n$ is either $0$ or runs through every residue, and they part company at $j = 6$, which is the smallest jaw count where they can.

It also says which sections a given chuck is bad at, which is the direction a machinist actually needs. A three-jaw chuck is exact on every $n$ divisible by three and inexact on every other, so it handles round, triangular, hexagonal, nonagonal and twelve-sided stock and nothing else. A four-jaw chuck with jaws advancing together is exact when no multiple of $n$ is $\pm1$ modulo four, which is to say when $n$ is a multiple of four. A six-jaw chuck is exact on multiples of two and of three — round, square, triangular, hexagonal — which is a considerably longer list and is the arithmetic reason six-jaw chucks are the ones fitted where mixed stock is turned.

Where the number goes #

The offset is a pose error, and this field has a rung about what a pose error is worth: a hold makes the set of poses finite and a clearance gives it a size. A chuck is the other case — the pose is a single point, determined exactly, and it is the wrong point.

The two are worth keeping apart. A clearance leaves a set of poses and the part could be anywhere in it, so the error is a range. A chuck leaves one pose and it is displaced, so the error is a bias — reproducible, computable, and removable by turning the bar and taking the mean, which is what a machinist who indicates a square bar in a four-jaw is doing by hand.

And it is a reminder of what the whole field measures. Nothing above involved a force, a stiffness, a coefficient or a material. Three flat faces, a support function and a Fourier argument put a bar’s axis 2.16 mm out on a perfect machine, and the only way to know is to compute it.