Where the curvature stands still

Two derivatives of a motion settle the curvature of every point’s path. That is Euler and Savary’s relation and it is where the classical treatment of this subject usually stops being elementary.

The next question is the one a designer actually has. A curvature is a number attached to an instant; what matters in practice is how long it lasts. If a coupler point’s path is momentarily on a circle of radius $\rho$, does it stay near that circle for a degree of crank, or for forty? The answer is decided by how fast $\kappa$ is changing, which is a third derivative, and the points where it is not changing at all form a curve.

The condition #

Path curvature is $\kappa = N/D^{3/2}$ with $N = \operatorname{Im}(\overline{z'}z'')$ and $D = |z'|^2$. Differentiating and clearing the denominator, $\kappa' = 0$ becomes

$$S(w) \;=\; \operatorname{Im}(\overline{z'}z''')\,|z'|^2 \;-\; 3\,\operatorname{Im}(\overline{z'}z'')\,\operatorname{Re}(\overline{z'}z'') \;=\; 0,$$

using $N' = \operatorname{Im}(\overline{z'}z''')$, which holds because $\operatorname{Im}(\overline{z''}z'') = 0$.

The three quantities $z'$, $z''$ and $z'''$ are each affine in the point’s offset $w$, since

$$z''' = A''' + (i\varphi''' - 3\varphi'\varphi'' - i\varphi'^3)\,w.$$

So $S$ is a product of things that are affine in $w$ and their conjugates, arranged so that the highest terms are of degree four. It looks like a quartic. Its zero set ought to be a quartic curve.

The quartic terms cancel #

They do not. The two halves of $S$ each have a $|w|^4$ term and the two terms are equal, so they cancel identically.

Taking the leading coefficients in turn: $\operatorname{Im}(\overline{z'}z''')$ has $|w|^2$ coefficient $3\varphi'^2\varphi''$, and $|z'|^2$ has $|w|^2$ coefficient $\varphi'^2$, so the first half contributes $3\varphi'^4\varphi''\,|w|^4$. In the second half, $\operatorname{Im}(\overline{z'}z'')$ has $|w|^2$ coefficient $\varphi'^3$ — the coefficient that made the inflection locus a circle — and $\operatorname{Re}(\overline{z'}z'')$ has $\varphi'\varphi''$, so with the factor of three the second half contributes $3\varphi'^4\varphi''\,|w|^4$ as well.

Equal, and subtracted. The locus is a cubic, and that is why the classical name for it is the cubic of stationary curvature rather than the quartic of stationary curvature.

That cancellation is exactly the kind of fact the site does not take on trust. A general bivariate quartic — fifteen coefficients — is fitted to $S$ on a scaled grid, and the norm of the five degree-four coefficients is compared against the norm of the four degree-three ones. Over five crank positions the worst ratio is $1.2\times10^{-14}$, against a degree-three norm of order one. The quartic part is not small; it is absent.

A double point at the pole #

The cubic passes through the pole, for the same reason the inflection circle does: at the pole $z' = 0$, so every term of $S$ vanishes. But it passes through it twice.

The evidence is in the polar form. Along a ray from the pole in direction $\psi$, $S$ is a cubic in the distance $r$:

$$S(P + re^{i\psi}) = s_0 + s_1 r + s_2 r^2 + s_3 r^3,$$

and both $s_0$ and $s_1$ come out at rounding — worst $4.4\times10^{-14}$ of the largest coefficient, over every ray at five positions. So $S = r^2(s_2 + s_3 r)$ and each ray meets the curve at exactly one point other than the pole, at $r = -s_2/s_3$.

A double point is a self-crossing, and the two branches there have definite tangents. What they are is worth knowing: one of them is the pole tangent and the other is the pole normal. So the cubic crosses itself at the pole with its two branches along the two directions the whole field is organised by — which is a statement one could hope to prove and which the site takes as measured, since the polar form gives it directly.

The practical consequence is that the curve is trivial to trace. No marching squares, no contouring, no root-finder: one evaluation of four coefficients per ray, and the ray’s crossing follows by a division. That is what makes the figures cheap enough to be draggable.

Two points on it are always hardware #

The cubic passes through both moving pins at every position of the mechanism, and the reason is the plainest available.

A pin travels on an exact circle. A circle has constant curvature. Constant is certainly stationary, so $\kappa' = 0$ at the pin, and the pin is on the cubic. It is also on every higher locus of the same family — the curvature’s second derivative vanishes too, and its third, because they are all zero.

That is not an accident of the four-bar; it is true of any mechanism whose moving plane is carried by pins on fixed centres. It means two of the cubic’s more interesting intersections with other loci are always occupied by hardware, and it is the reason the search for the points with fifth-order circle contact has to separate out the pins before counting anything.

What being on it buys #

The claim so far is that the cubic is the locus of $\kappa' = 0$. That is a statement about a third derivative and is hard to feel. Here is the version with an exponent in it.

Take a coupler point and compute the centre and radius of its path’s osculating circle. Throw the linkage away and replace the point with a crank pivoting at that centre. Drive the mechanism away from the instant and measure the geometric distance from the moving point to the fixed circle. The distance grows like a power of the step, and the power counts how many derivatives agreed:

• an ordinary point: exponent 3.00
• a point of the cubic: exponent 3.94

Nothing in that measurement knows which kind of point it was handed. It traces a path with the solver, measures a distance to a circle, and fits a slope through a decade of step sizes. The two points differ by one in the exponent because the cubic point has one more derivative in agreement, which is exactly what $\kappa' = 0$ means.

That is what the cubic is for. It is the locus of points where a single pivot can stand in for the whole linkage for longest — where a four-bar’s coupler point can be replaced by a crank, or a cam by a circular arc, with the error one order smaller than it would otherwise be.

The check that could have failed #

The assertion behind the cubic has two halves and they fail in opposite directions.

The first half takes a point on the curve — found by the polar form at some ray angle — and checks that the numerator of $\kappa'$ there is at rounding. Measured: $-5.9\times10^{-13}$ against a neighbouring value of order a tenth. That half would pass on any point if the numerator were identically zero everywhere, which is the failure mode of a wrongly derived expression.

So the second half takes a point on the same ray at seven tenths of the distance — a completely ordinary point — and requires the same numerator to be large: $-0.217$, three orders above the first. Without it the check is compatible with the curve being the whole plane.

That pairing is this site’s standing habit and it is not decorative. The conjugate-action check in the gears field was circular in its first version and announced a constant ratio because both flanks had been parameterised by the same variable; it passed with the calculation restating its own definition. A locus check that only ever samples points of the locus is the same shape of mistake.

Why a cubic, and what its shape is doing #

A cubic in the plane is a more varied object than a circle, and it is worth knowing which kind this one is, because the answer explains most of what the figures show.

It is a circular cubic: a cubic that passes through the two circular points at infinity. That is not visible in a real drawing, and it has a consequence that is: a circular cubic meets a circle in six points counted properly, two of which are the circular points, so a circle and a circular cubic have at most four ordinary intersections rather than six. Since the cubic has a double point at the pole and the inflection circle passes through the pole, two of those four are used up there, and at most two ordinary crossings are left. On the site’s four-bar there is exactly one real crossing besides the pole at every position of the crank, which is the count Ball’s point rests on.

The double point at the pole also settles the curve’s global shape. A cubic with a double point is rational — it can be parameterised by a single variable — and that is exactly what the polar form does: each ray direction $\psi$ picks out one point, so $\psi$ is the parameter and the whole curve is swept once as $\psi$ runs over a half turn. The asymptote is where the parameterisation passes through infinity.

None of that is needed to compute anything. It is worth carrying because it says what to expect on the screen: a self-crossing at the pole, one branch running out to an asymptote, and never more than one interesting meeting with the inflection circle.

The third derivative is the expensive one, and it is exact here #

Everything in this essay needs $\varphi'''$, and a third derivative is where a numerical route starts to hurt.

Obtained by differencing a closed form with a seven-point stencil at a well-chosen step, $\varphi'''$ is good to about $3.7\times10^{-7}$ relative — six digits. Obtained by differencing the solver at the same step it is good to $2.3\times10^{-8}$, which is better than expected and is a finding of its own. Obtained from the loop equation it is exact to rounding, because the third-order right-hand side is built from quantities already computed and solved through the same two-by-two matrix as the first.

Six digits would be plenty for drawing the curve. It is not plenty for the measurement the curve is checked by. The quartic cancellation is asserted at $10^{-14}$; a $\varphi'''$ good to six digits would leave a residual quartic part at $10^{-7}$ and the check would have to be loosened by seven orders — at which point it would no longer distinguish a genuine cancellation from a small one, and the essay would have to say “nearly cancels” rather than “cancels”.

That is the general shape of the argument for exactness in this field. The figures would look identical either way. The claims would not.

Where the curve runs to infinity #

A cubic in the plane has asymptotes, and this one has one. The polar radius $-s_2/s_3$ runs to infinity at the four ray directions where $s_3$ passes through zero, and the drawn curve breaks there rather than joining up.

That is honest and it is also a trap, and it caught the site once. The natural way to find where the cubic crosses another curve is to compare the two polar radii as functions of $\psi$ and look for sign changes of the difference. But at an asymptote the cubic’s radius jumps from $+\infty$ to $-\infty$, so the difference changes sign without the curves meeting at all.

The first version of the search for Ball’s point returned two answers at every crank position because of exactly that. One of them had a curvature of $10^{-16}$ and a curvature rate of $10^{-14}$ — a genuine point of both curves. The other had a curvature of $10^{-16}$ and a curvature rate of twelve. Both were on the inflection circle, which is what a zero curvature means; only one was on the cubic. The fix took three attempts and ended with a search that walks along the inflection circle rather than across it, parameterised by the circle’s own arc; after it the count is one at 358 of 360 positions and zero at the two where the coupler is instantaneously translating.

The finite version, which is the same curve #

There is a completely different route to this curve and it comes from the synthesis field.

Burmester’s construction for four prescribed positions of a moving plane asks which points of that plane have their four images concyclic — because a point whose four images lie on one circle can be a crank pin, with the circle’s centre as its fixed pivot. The answer is a curve, the circle-point curve, and it is a cubic.

Bring the four positions together and it has to become this curve. A circle through four coalescing positions is a circle of four-point contact, and four-point contact with a circle is exactly $\kappa' = 0$. So the finite construction and the infinitesimal one are the same object approached from opposite ends, and an essay of this field measures the convergence — the gap between the two curves falls as the square of the spread, with a fitted exponent of $2.02$.

That is the strongest available argument that this field is not a separate subject bolted onto the site. The machinery that draws the finite curve was written for the synthesis phase and has been in use since; the machinery that draws the infinitesimal one was written for this field; and the comparison runs the first one rather than a re-implementation of it.

The motion where the whole thing degenerates #

One motion in the field’s ledger breaks the pattern, and it is the trammel: two sliders and a rod, with the rod’s angle as the input.

Its angular rate is exactly $1$ and every higher derivative of its angle is exactly $0$. Put that into $S$ and the third-derivative coefficient collapses, the expression loses most of its structure, and the condition $\kappa'' = 0$ that the next locus needs is satisfied identically along the cubic rather than at isolated points. A search for those points returned 346 of them before the machinery was taught to notice.

Reporting 346 would have been worse than reporting nothing, because four is the most a planar motion can have and a reader has no way to tell a large number from a broken one. The count is now returned as undefined with the reason attached, which is the honest answer: on a motion whose angular rate never changes, the question the count answers does not have a finite answer.

It is a good reminder that a general expression evaluated on a special case returns something, and that the something is not always a number worth printing. The site has met the same shape before — Kutzbach’s formula returns a count for Bennett’s linkage and the count is wrong — and the response is the same: detect the special case, say so, and do not let the general machinery produce a figure about it.

What the cubic does not settle #

Being on the cubic makes a point’s osculating circle fit for one order longer. It does not make it fit for a long time, and the distinction matters.

The exponent is 4 rather than 3, so the error is smaller by a factor of the step — at a step of a tenth of a radian, ten times smaller. That is a real improvement and it is a local one. Over a whole stroke of thirty degrees the difference between a point of the cubic and an ordinary point can be swamped by the coefficient in front of the power, which nothing in this essay controls.

The same caution applies with more force to Ball’s point, where the local improvement is dramatic and the improvement over a whole stroke is, for one of the two classical linkages, almost nothing. A criterion about an instant and a criterion about a range are different criteria, and the field’s habit of measuring both is the only thing that keeps the difference visible.

The quartic terms cancelling is the sort of thing that is easy to state and worth being careful about, because it is where a derivation and a measurement say different things. A cancellation is an algebraic fact: the fourth-degree terms are equal and opposite, exactly, for every motion, and no amount of measuring establishes that. What the measurement at $10^{-14}$ establishes is that the implementation has the cancellation in it — that the terms being computed really are the ones that cancel, and that nothing in the coding has broken the identity. Those are two different claims and only the second is checkable here. That distinction is worth carrying wherever a site computes a classical result. A measurement confirms an implementation and an argument confirms a theorem, and quoting a residual as evidence for the mathematics is offering the wrong kind of support — though it is exactly the right kind for the thing that was actually at risk.