The shift that evens the wear

Moving the cutter out shifted a small pinion’s cutting rack away from its centre to stop the rack’s tip eating into the flank it was forming, and then listed what else a shift changes. One item in that list was a reason for shifting a pair even when neither gear is undercut: “to balance the specific sliding between pinion and wheel so that they wear at the same rate.” The essay named it as one of the classical reasons and moved on. It did not say how much shift that takes, or whether it is anything like the shift that cures undercut.

Rolling at one point only measured what is being balanced. Two involute flanks in mesh roll on each other only at the pitch point; everywhere else along the path of contact they slide, and the specific sliding on each flank — the sliding speed divided by that flank’s own rolling speed — is worst on the smaller wheel. This essay puts the two together and finds the shift.

Where a flank slides worst #

The involute makes the sliding easy to write. Along the line of action, a contact a distance $u$ from the pinion’s base tangency is on the pinion’s flank at radius of curvature $u$ and on the wheel’s at $L - u$, where $L$ is the length of the line of action between the two tangencies. Each flank’s material passes the contact at its own angular speed times that radius of curvature, so the pinion’s rolling speed is $\omega_1 u$, the wheel’s is $\omega_2 (L - u)$, and the sliding is their difference. Divided by each flank’s own rolling speed:

$$\zeta_1 = \frac{\omega_1 u - \omega_2(L-u)}{\omega_1 u}, \qquad \zeta_2 = \frac{\omega_2 (L-u) - \omega_1 u}{\omega_2 (L-u)}.$$

Both vanish at the pitch point, where the rolling speeds are equal. The pinion’s grows without limit as $u$ goes to nought — as the contact approaches the pinion’s base circle, where its flank has no radius of curvature and its rolling speed vanishes while the sliding does not. The wheel’s does the same near the wheel’s base circle.

A real pair never reaches either base circle, because contact runs only between the two tip circles. It begins where the wheel’s tip crosses the line of action, which is low on the pinion’s flank, near its root; it ends where the pinion’s tip crosses, low on the wheel’s. So each flank’s worst sliding is at its own root, at one end of the path of contact, and how bad it is depends on how close that end comes to that flank’s base circle.

For a 17-tooth pinion driving a 51-tooth wheel, cut standard, the pinion’s root reads 11.1 and the wheel’s 1.38. The pinion’s flank near its root slides eleven times as far as it rolls, and eight times as hard as anything on the wheel. That is the imbalance a shift is meant to correct.

What a shift of ±x does to the contact path #

Shift the pinion’s cutter outward by $x$ modules and the wheel’s inward by the same. The pinion’s teeth get longer addenda and the wheel’s shorter; the two tooth thicknesses change in opposite directions by the same amount, and — because the shifts sum to nought — the gears still mesh at their standard centre distance with the standard pressure angle. Moving the cutter out found that an unbalanced shift moves the centre; this one is chosen not to, so that only the teeth change.

What moves is where the tip circles cross the line of action. The wheel’s tip circle is smaller, so contact begins later — further from the pinion’s base tangency — and the pinion’s root is spared. The pinion’s tip circle is larger, so contact ends later too, closer to the wheel’s base tangency, and the wheel’s root is loaded harder. The path of contact slides along the line of action towards the wheel as the shift grows.

So the two roots trade. The pinion’s worst sliding falls steeply as its root is lifted away from the base circle where the sliding runs away; the wheel’s rises gently, because its root is on a large wheel whose base circle is far from where contact ends. They cross at a shift of 0.3574, where both roots read 1.961. The pinion’s worst sliding has fallen by a factor of 5.6 and the wheel’s has risen by 42%.

The balance is a condition on the two ends of contact #

The crossing can be found by bisection, and it has been, but it can also be written down, and the written form says what the balance is.

Let $a_1$ be how far the pinion’s tip circle reaches along the line of action from the pinion’s own base tangency, and $a_2$ how far the wheel’s reaches from the wheel’s. Contact ends at $u = a_1$ and begins at $u = L - a_2$. At the start, the pinion’s flank has radius of curvature $L - a_2$ and the wheel’s has $a_2$; at the end, the pinion’s has $a_1$ and the wheel’s $L - a_1$. The angular speeds are in the inverse ratio of the tooth counts. Putting those into the two sliding formulas at the two roots and setting their sizes equal gives

$$\frac{a_1/(L - a_1)}{a_2/(L - a_2)} = \left(\frac{z_1}{z_2}\right)^{2}.$$

Two flanks, one law found that the two flanks’ radii of curvature at a contact are tied, summing to the length of the line of action between the base tangencies; here each side of the fraction is the ratio of those two radii at one end of contact — the flank that is near its tip over the flank that is near its root. The condition says those two ratios must differ by the square of the tooth ratio: one factor because the two flanks turn at different speeds, and a second because the sliding on each is measured against that flank’s own rolling. At the bisected balance of every unequal pair in the table the condition holds to $10^{-9}$; with the tooth ratio unsquared it misses by a third or more.

That is also the cleanest way to see why the balance is nought at one to one and grows with the ratio. At one to one the right-hand side is one, and the symmetric path $a_1 = a_2$ satisfies it with no shift. At a ratio of three it is a ninth, so the end of contact has to sit much closer to the wheel’s base tangency, relative to the pinion’s reach, than the start does — and moving the path of contact towards the wheel is exactly what shifting the pinion out does.

The balancing shift is not the undercut cure #

The same figure carries the shift undercutting and the seventeen-tooth rule is about. A 17-tooth pinion at a 20° pressure angle is just on the edge of undercut — the rule’s exact count is 17.1 — and the smallest shift that clears it is 0.006. The balancing shift is 0.352 larger.

Across pinion sizes the two diverge. The undercut shift, $1 - z\sin^2\alpha/2$, falls through nought at seventeen teeth and goes negative: a larger pinion could be cut with its rack moved inward and still not be undercut. The balancing shift stays positive and falls slowly — 0.494 at ten teeth, 0.357 at seventeen, 0.269 at twenty-five and still 0.182 at forty. The gap between them grows with the tooth count. A pinion that needs no shift at all to be cut cleanly still needs about a third of a module to wear evenly with a wheel three times its size.

That is the answer to the question the earlier essay left open, and it is the more useful half of the shift story. Undercut is a small-pinion problem with a sharp threshold; uneven sliding is every pinion’s problem, fading only slowly with size, and the shift that addresses it is not a correction to the undercut shift but a different, larger number.

The table’s first two rows show what happens below seventeen teeth. A 12-tooth pinion on a 36-tooth wheel, cut standard, does not just slide hard at its root: the wheel’s tip reaches the line of action before the pinion’s base tangency, contact would begin inside the pinion’s base circle where it has no involute, and the pair interferes. That is undercut seen from the mesh rather than from the cutter, and the undercut shift of 0.298 is the least that makes the mesh possible. The balancing shift, 0.446, is again larger.

It is nought at one to one and grows with the ratio #

The balancing shift is a property of the pair, not of the pinion.

At one to one it is exactly nought, for any tooth count: the two gears are identical, the path of contact is symmetric about the pitch point, and the two roots already slide equally. The table’s 20–20 row reads 4.095 at both roots unshifted, which is the worst figure in it and cannot be improved by this kind of shift at all — shifting one gear out and the other in only unbalances them. As the wheel grows the balancing shift rises steeply to a ratio of about three and then flattens. For a 17-tooth pinion it is 0.20 at a ratio of 1.5, 0.36 at three and 0.43 at six.

The shape has a plain reason. The imbalance comes from the pinion’s root flank being the smallest radius of curvature in the pair and the wheel’s the largest, and a larger wheel makes the second larger still. But once the wheel is large, its flank near the end of contact is already nearly straight and making it larger changes little, so the shift saturates.

A reduction of three, cut both ways #

A 20-tooth pinion driving a 60-tooth wheel is a pair nobody would shift for undercut: twenty teeth is well clear of the seventeen-tooth rule, and the pinion could be cut with its rack moved inward by 0.17 of a module and still be clean. Cut standard, its root flank slides at 4.47 and the wheel’s at 1.15, so the pinion’s root is doing nearly four times the wheel’s worst work.

Balanced, at a shift of 0.319, both roots slide at 1.57: the pinion’s worst sliding has fallen to about a third of what it was and the wheel’s has risen by a third. The contact ratio falls from 1.671 to 1.605, and the pinion’s tip is 0.56 of a module thick. The centre distance, the ratio and the pressure angle are unchanged, and so are the cutters — the same rack cuts both gears, which is the property one rack and every wheel found makes shifting possible at all. The only thing that changes in the shop is how far the rack is set from each blank.

Since a shift is measured in modules, the result is scale-free in the way the module is a size, the ratio is a shape described: the balancing shift for a 20–60 pair is 0.319 modules whether the module is half a millimetre or ten. And what the shift does to the teeth themselves is to thicken the pinion’s and thin the wheel’s by $2x\tan\alpha$ modules at the pitch circle, leaving their sum — and so the backlash the pair runs with — exactly as it was.

What the balance costs #

A shift is not free, and two costs are worth reading off the table.

Contact ratio. Moving the path of contact along the line of action while keeping the centre distance also shortens it slightly, because the pinion’s longer addendum gains less than the wheel’s shorter one loses. For the 17–51 pair the contact ratio falls from 1.636 to 1.558; for the 12–36 pair, from 1.556 to 1.450. None of the balanced pairs comes near the 1.2 that practice treats as a floor, let alone the 1 below which a mesh periodically stops driving, but the loss is 4–7% and it is real.

Tip thickness. The pinion’s teeth, lengthened, get pointier. At the balance a 17-tooth pinion’s tip is half a module thick and a 12-tooth one’s a third; the 10-tooth pinion at its balance of 0.494 is down to a fifth of a module, which is thinner than the quarter of a module most practice asks for. So for the smallest pinions the balancing shift runs into a pointed tooth, and the designer has to stop short of full balance.

Neither cost bears on the finding. It says what shift balances the sliding and what it costs, not whether to use it.

The sliding from the lengths of the two involutes #

The sliding formula above takes each flank’s rolling speed as its angular speed times its radius of curvature at the contact, which is a property of the involute. It is checked here by something that does not use that property.

Step the pinion by a ten-thousandth of a radian at a given contact. The contact moves along the line of action by the pinion’s base radius times that angle, and along each tooth flank by some arc. Measure each arc by summing the length of the sampled involute between the two roll angles, with the step centred on the contact; the specific sliding is the difference of the two arcs over the pinion’s. Over the whole contact path, for the pair both unshifted and at its balance, the arcs give the formula’s value to $3 \times 10^{-11}$ of it. Exchange the two base radii in the arc calculation — the natural slip, since the wheel’s arc is measured from its own base tangency — and the result misses by more than the value itself.

That also settles a point the formula hides. The sliding is a property of where on each flank the contact is, measured by arc along the flank, and a shift changes it only by moving the contact path; the involutes themselves are the same curves at every shift, because a shift moves the cutter and not the base circle. Why a tooth is an involute is why that holds: the base circle is fixed by the tooth count and pressure angle alone.

What this does not settle #

Specific sliding is not wear. It is a ratio of speeds, and a flank’s wear depends on the contact pressure, the lubricant film and the materials as well. Balancing specific sliding is the classical kinematic proxy for balancing wear, and it is the proxy that has been measured.

Only balanced shifts. Every pair here has $x_2 = -x_1$, which keeps the centre distance. A designer free to choose the centre distance can shift both gears outward, which changes the operating pressure angle as well and gives a second knob; the balance with that freedom was not searched.

One pressure angle and one addendum. All the pairs are cut at 20° with the standard one-module addendum. The angle the standard left free found that the pressure angle trades pinion size against contact ratio, and it plausibly changes the balancing shift as well.

Still open: a second knob for the root #

A balanced pair here still slides at almost 2 at each root — the balance equalises the two, but it does not make either small, and a pair of equal large sliding numbers is still a pair that wears.

The distinct argument would be the other way of lifting a root away from its base circle: a shorter addendum on the other gear, or a longer working depth, which is what “short-addendum” and “long-addendum” systems do. With the centre distance, both shifts and both addenda free, the question is how low the worst of the two roots can be brought while keeping a contact ratio of 1.2 and a tip of at least a quarter of a module on each gear. Two things would come out of it: the least worst-root sliding a given pair can have at all, and whether it is reached with the two roots balanced, as this essay assumes a designer wants, or with one of them deliberately worse. The measurement would be a search over the four parameters for the pairs in this essay’s table, with every constraint checked on the drawn tooth rather than on the formulas.