The other pole is on the circle

The other pole introduced the second special point of a moving plane. The first is the instantaneous centre, the point that is not moving; the second is the point that is not accelerating, and over a turn of this field’s four-bar the two are never closer than one and a half coupler lengths. That essay then did something more useful than measure the gap. It sorted the whole field’s apparatus into two kinds. A geometric object depends only on the path a moving plane traces through its positions and is unchanged by driving the input faster, slower, or unevenly: the pole, the centrodes, the circle of points going straight, the cubic of stationary curvature. A kinematic object depends on the schedule as well. It found exactly one of the second kind, the acceleration pole, and concluded that it deserved one essay and nothing built on top of it, since anything built on it would inherit a dependence on the motor.

That classification is right, and it matters because this field opened by saying the mechanism drops out: near an instant, a moving plane is a handful of numbers with no linkage in them. A drive is not a linkage either, and it is fair to ask how many of those numbers it owns.

One sentence in the same essay was not right. Under a figure of the inflection circle, its caption said that both poles are involved in placing the circle, “though neither is its centre and only one of them is on it” — the velocity pole. The caption has since been corrected. The acceleration pole is on it too, always, whatever the drive, and the reason fits on one line. What the drive decides turns out to be much less than “where the acceleration pole is”. It decides one angle.

Why the acceleration pole cannot be anywhere else #

The inflection circle is the set of points of the moving plane whose paths are momentarily straight. A path is straight at an instant when its point’s acceleration has no component across its velocity, and this field writes that condition as the vanishing of one quantity,

$$N = \operatorname{Im}\big(\overline{z'}\,z''\big),$$

where $z'$ and $z''$ are the point’s velocity and acceleration as complex numbers. The circle of points going straight showed why the zero set of $N$ is a circle: as a function of the point’s position it is quadratic, and its square terms come as a single real multiple of $|w|^2$ with no cross term.

The acceleration pole is the point where $z'' = 0$. Put that into $N$ and $N$ is nought. So the acceleration pole satisfies the equation that defines the inflection circle, and it does so under any drive, because nothing in the argument depended on how $z''$ was produced.

That is the whole of the reason. It is a reason of the degenerate kind the caption gave for the velocity pole — a point that is not moving has no path curvature to speak of, and a point that is not accelerating has none either — and it has the same force. The two poles are the two points on the circle that are on it for free.

Where on the circle #

Knowing that the acceleration pole lies on a geometric circle leaves one question: where. The answer comes from writing its position relative to the velocity pole.

Put the velocity pole at the origin. Every point’s velocity is then $z' = i\dot\varphi\,w$, a rotation about the pole at the plane’s angular speed $\dot\varphi$. Every point’s acceleration is the pole point’s own acceleration plus a rotation-and-scaling about it,

$$z''(w) = z''(P) + \big(i\ddot\varphi - \dot\varphi^{2}\big)\,w,$$

and the pole point’s acceleration is $i\dot\varphi\,u$, where $u$ is the velocity with which the pole itself runs along the fixed centrode — the same pole velocity that the two numbers are the curves’ own reads off the centrodes’ curvatures, The first of six things an instantaneous centre is not is the point with no acceleration, and this is the formula that says how far from it that point is. Setting $z''$ to nought and writing $k = \ddot\varphi/\dot\varphi^{2}$ gives

$$w_Q = \frac{i u}{\dot\varphi}\cdot\frac{1 + ik}{1 + k^{2}}.$$

The first factor, $iu/\dot\varphi$, is the inflection circle’s diameter as a vector: its length is the pole’s speed over the angular speed, which is the $\delta$ of the circle’s own essay, and it points along the normal to the centrode. The second factor is exactly what a point on a circle through the origin looks like when it sits at an angle $\chi$ from the diameter with $\tan\chi = k$: $\cos\chi\,e^{i\chi}$. So

$$\tan\chi = \frac{\ddot\varphi}{\dot\varphi^{2}}.$$

The acceleration pole sits on the inflection circle at the angle whose tangent is the plane’s angular acceleration over its angular speed squared, measured at the velocity pole from the circle’s diameter. Everything except that angle is geometry.

Two special cases fall straight out. When the plane is turning at a momentarily constant rate, $\ddot\varphi = 0$, the angle is nought and the acceleration pole is the far end of the diameter: the point classically called the inflection pole. When the angular acceleration dominates the angular speed, $|k| \to \infty$, the angle approaches ±90° and the acceleration pole runs round the circle towards the velocity pole.

The law, measured #

The derivation is short enough that it could be checked by reading it, and it is checked here the way the rest of this field is: by computing the two poles and the circle independently and measuring where they are.

A drive enters the computation through one number. Every motion in this field carries derivatives with respect to its input angle, which is what driving the input at constant speed means. Driving it with a crank acceleration $\kappa$ times its speed squared adds $\kappa$ times the first derivatives to the second, for the plane’s angle and its reference point alike, and the acceleration pole is found from those second derivatives by solving $z'' = 0$ directly. The inflection circle is found from the quadratic $N$ as before, and it does not depend on $\kappa$ at all.

Every instant lies on the diagonal. The check runs over 1,275 instants on five motions — this field’s four-bar and an offset slider-crank through a whole turn in steps of 3°, the elliptic trammel at every 15°, and the linkages of Watt and Chebyshev at their symmetric positions — each under five drives, from a crank decelerating at 25 times its speed squared to one accelerating at 3. The acceleration pole’s distance from the inflection circle is at most $5 \times 10^{-15}$ of the circle’s radius, and its angle matches the law to $10^{-14}$ radians. Instants where the circle is more than ten thousand coupler lengths across are left out, since there the circle is a line to working precision and an angle measured on it means nothing.

The check has a refusal built into it. If the law is read with the plane’s derivatives taken at constant crank speed — ignoring the drive, which is exactly the mistake the earlier essay warned against — it misses by angles up to the whole of the range, 180°. The law is about the angular acceleration the plane actually has, and a drive changes it.

The last figure of this essay lists the result motion by motion; every entry is rounding.

Round one turn #

Driving the input at constant speed is still a drive, and the acceleration pole that the other pole measured was the one it produces. Under that drive the angle $\chi$ is a function of the crank angle, and it can be followed round a turn.

The coupler’s angular acceleration changes sign twice in a turn, at crank angles of 11.99° and 230.71°, found by bisection on the angular acceleration alone. At both, the law says the acceleration pole should be the inflection pole, and measured it is — the two points coincide to $2 \times 10^{-15}$ of the diameter. Between them the angle climbs towards +90° on one side of the turn and falls towards −90° on the other, reaching within a fraction of a degree of both where the coupler’s angular speed passes close to nought and its angular acceleration does not.

Those near-±90° stretches are where the picture needs care, because an angle near 90° means the acceleration pole is near the velocity pole along the circle — and those are the positions where the circle is enormous. At a crank angle of 120°, where the earlier essay drew its circle, the angle is 89.996° and the circle is almost five million units across. The acceleration pole is 292 units from the velocity pole there, about eighty coupler lengths, far off any canvas and still on the circle. That is how the caption came to be written: the figure showed the circle passing through the velocity pole and nothing else, because the other point on it was somewhere in the next county.

The gap between the poles is one number times another #

The other pole measured the distance between the two poles over a turn and reported its minimum: 1.49 coupler lengths. The law says what that distance is. Two points on a circle through the pole, one of them the pole and the other at angle $\chi$ from the diameter, are $\delta\cos\chi$ apart, so

$$|PQ| = \delta \cos\chi = \frac{\delta}{\sqrt{1 + (\ddot\varphi/\dot\varphi^{2})^{2}}}.$$

The measured gap matches the diameter times the cosine at every position of the turn, to $8 \times 10^{-16}$ of the diameter. Its least value is 1.490 coupler lengths at 343°, the earlier essay’s number to its last figure, and the law says where it comes from: at that position the diameter is 7.89 coupler lengths and the angle is −79.1°, so the gap is 7.89 × cos 79.1°.

That also says something the old measurement could not. The poles are closest not where the inflection circle is smallest but where a modest circle meets a steep angle. The circle is smallest at 231°, 2.50 coupler lengths across — and that is within a degree of where the angle is nought, so there the gap is very nearly the whole diameter, 2.496. The minimum gap comes instead from a circle three times larger seen at a steep angle, which is to say where the coupler’s angular acceleration is large compared with the square of its angular speed. A reader asking why the gap never falls below one and a half coupler lengths now has two numbers to look at rather than one measurement.

Two circles through the pole #

The classical construction of the acceleration pole says the same thing from the other side, and drawing it shows which part of it the drive owns.

The inflection circle is where a point’s acceleration has no component across its path. There is a second locus, where the acceleration has no component along the path: the points whose speed is momentarily not changing. It is also the zero set of a quadratic, $\operatorname{Re}(\overline{z'}\,z'')$, whose square terms are again a real multiple of $|w|^2$, so it is also a circle through the pole. It carries the name of Bresse, alongside the inflection circle, and it is usually called the tangential circle.

The two circles meet at the velocity pole and at one other point, and that other point has no acceleration in either direction: it is the acceleration pole. Measured, the tangential circle is a circle — its cross term and the difference of its square terms are below $10^{-14}$ of its leading coefficient — and the acceleration pole lies on both circles to $10^{-14}$.

The difference between the circles is the whole of the drive’s role. The tangential circle’s coefficient of $|w|^2$ is $\dot\varphi\ddot\varphi$, so its size depends on the angular acceleration and it swells, shrinks and swings round the pole as the drive changes; when the angular acceleration is nought it opens out into a straight line through the pole, which is why the acceleration pole is then at the far end of the inflection circle’s diameter. The inflection circle’s coefficient is $\dot\varphi^{3}$ and contains no angular acceleration at all. Bresse’s construction is the drive’s circle cutting the geometry’s circle, and the law $\tan\chi = \ddot\varphi/\dot\varphi^{2}$ is where it cuts.

The law has no size in it #

The angle law is dimensionless on both sides: an angle on the left, and on the right an angular acceleration divided by the square of an angular speed, both of which are measured per unit of time squared, so their ratio is a pure number. A curvature is a size with a minus sign found that scaling a moving plane scales its inflection circle and divides every curvature by the scale factor. The acceleration pole follows the circle and keeps its angle, so a machine built twice as large and driven with the same angular schedule has its acceleration pole at the same place on a circle twice the size. A drive, in other words, chooses a shape-free quantity, and every length in the answer belongs to the geometry.

The same holds for time. Running the whole schedule twice as fast multiplies $\dot\varphi$ by two and $\ddot\varphi$ by four, which leaves $k$ and the angle alone. Only a change in the shape of the schedule — a crank that speeds up more in one part of its turn than another — moves the acceleration pole. A motor that merely runs faster does not.

A motor can put the acceleration pole anywhere on the circle, and nowhere else #

Read backwards, the law is a design statement. At any position, the crank acceleration that places the acceleration pole at a chosen angle $\chi$ on the circle is

$$\kappa = \frac{\tan\chi\,\varphi_\theta^{2} - \varphi_{\theta\theta}}{\varphi_\theta},$$

in units of the crank’s speed squared, where $\varphi_\theta$ and $\varphi_{\theta\theta}$ are the coupler’s first two derivatives with respect to the crank angle. The first figure of this essay is drawn that way: each stop of its slider asks for an angle, the crank acceleration is computed from the line above, the acceleration pole is then found from scratch, and it lands on the angle asked for to $10^{-9}$ degrees.

At the pose drawn, placing the acceleration pole at the inflection pole takes a crank acceleration of 0.160 times its speed squared, since at constant crank speed the coupler is decelerating there. Placing it 75° round one way takes −0.893, the other way 1.214. No drive at all can put it off the circle.

That is a sharper version of the earlier essay’s classification. The acceleration pole is still the one object in this field that depends on the schedule. But the dependence is confined to a single degree of freedom on a locus the geometry fixes, and the other objects of the field — the pole, the circle, the pole tangent — carry everything about it except that angle.

The trammel, where the angle is always nought #

The elliptic trammel is the motion in this field whose coupler turns at exactly constant speed when its rod angle is the input, so its natural drive has $\ddot\varphi = 0$ at every position. The law puts its acceleration pole at the inflection pole, always. The other pole found that on the trammel the two poles are exactly one coupler length apart and described the acceleration pole as a special case; the law says why. The trammel’s inflection circle is its moving centrode, the circle with the rod as its diameter, and the acceleration pole sits at the far end of that diameter from the velocity pole — one rod length away, at every position.

Every point has a centre of curvature on the trammel as on any motion, and the inflection circle is where that centre runs off to infinity; the drive has no say in any of it. Driven any other way, the trammel’s acceleration pole leaves the inflection pole and walks round the same circle, like every other motion’s. The ledger’s trammel row includes four drives with crank accelerations and they obey the same law to the same rounding. Nothing about the trammel was special except its drive.

What this does not settle #

A second-order statement stops at second order. The law places the point of zero acceleration. The next derivative has its own special points — the points whose acceleration is momentarily not changing — and where those lie relative to the cubic of stationary curvature, the third-order counterpart of the inflection circle, is a question this essay’s one-line argument does not reach, because the cubic is not the zero set of anything that vanishes when the rate of change of acceleration does.

Measured on five motions. The derivation is general and the measurements are its check, not its proof. Motions with a pole at infinity — a plane momentarily translating — have no inflection circle, and there the acceleration pole exists while the law has nothing to place it on; those instants were left out of the check rather than tested.

Kinematics, not dynamics. The drive here is prescribed. A real motor’s acceleration is decided by its torque, the mechanism’s inertia and the load, and what fixes $\ddot\varphi$ in a running machine is a question for a different field.

Still open: the point whose acceleration is not changing #

The argument here was that a point where $z''$ vanishes cannot help satisfying any condition built from $z''$. The next derivative gives a third special point: the one where $z'''$ vanishes, sometimes called the pole of the third order. Its position depends on the schedule through two numbers, $\ddot\varphi/\dot\varphi^{2}$ and $\dddot\varphi/\dot\varphi^{3}$, rather than one.

The distinct argument would be whether it too is confined to a geometric locus — and if so, which. The obvious candidate is the cubic of stationary curvature, the locus of points whose path curvature is momentarily not changing, since its defining quantity is built from $z'$, $z''$ and $z'''$. But vanishing $z'''$ does not make that quantity vanish, because it keeps a term in $z''$. So either the third-order pole lives on some other curve the geometry fixes, with the two schedule numbers choosing a point on it, or it has two genuine degrees of freedom and is not confined at all. The measurement would be the same as here: the four-bar through a turn under a grid of both schedule numbers, the third-order pole found from $z''' = 0$, and the question of whether the points it reaches at one position fill a region of the plane or lie on a curve.