The first stage does not know what it drives

A quick return that cuts evenly found something a designer would not have predicted. A shaper wants its working stroke slow and its return quick, and a crank-rocker with a 60° swing cannot give a time ratio of 2 on its own without its transmission angle collapsing. Put a drag link in front — a four-bar whose two cranks both turn, one unevenly — and the pair reaches 2. That much was the finding of a drag link ahead of a crank-rocker. The surprise was that the right drag link at the right phase also makes the cut more even than the crank-rocker driven alone at constant speed: the drag link’s slow stretch lands where the crank-rocker’s rocker would otherwise be fastest, and two uneven factors multiply to nearly a steady one.

That essay ended by asking whether this was a fact about crank-rockers or a fact about quick-return trains. It measured a rocker angle, and a shaper’s tool does not ride on a rocker. It rides on a ram, and the simplest mechanism that drives a ram directly while giving a quick return of its own is the offset slider-crank: a crank, a rod and a slide whose line of travel misses the crank’s centre. So the question here is what happens when the same kind of first stage is put ahead of a slide, and the answer turns out to be sharper than “the cancellation happens there too”. The drag link does not need to know what it drives. It needs two numbers.

The slide as a second stage #

The slider-crank’s geometry is short enough to write out. With the crank’s pivot at the origin, crank length $a$, rod length $b$, and the slide the horizontal line a distance $e$ from the pivot, the ram sits at

$$x(\varphi) = a\cos\varphi + \sqrt{b^2 - (e - a\sin\varphi)^2}.$$

The ram stops where crank and rod are in line, exactly as a crank-rocker’s output stops where its crank and coupler are. Extended, that happens at $\varphi_e = \arcsin\big(e/(a+b)\big)$; folded, at $\varphi_f = \pi + \arcsin\big(e/(b-a)\big)$. With no offset the two are exactly half a turn apart and the machine goes out and back in equal times. With an offset they are not, and the difference is the machine’s own quick return:

$$Q = \frac{\pi + \delta}{\pi - \delta}, \qquad \delta = \arcsin\frac{e}{b-a} - \arcsin\frac{e}{a+b}.$$

That closed form is not trusted on sight. On each of the three machines used below, the time ratio read off the ram’s recorded extremes over 7,200 crank positions matches it to $5 \times 10^{-4}$, and the stroke — the difference of the two in-line distances — to the same precision. A version with one sign turned round, taking the folded limit at $\pi - \arcsin$ rather than $\pi + \arcsin$, misses the measured ratio by 0.32, so the agreement is not one that a plausible slip would also produce.

What an offset costs is the rod’s lean. The rod meets the slide at an angle $\beta$ with $\sin\beta = (e - a\sin\varphi)/b$, and the transmission angle — the thing the transmission angle judges a linkage by — is $90° - |\beta|$. The worst of it over a turn is $90° - \arcsin\big((a + |e|)/b\big)$. A longer rod leans less. But a longer rod also needs a larger offset to buy the same quick return, and the offset leans it more, so for every own ratio above one there is a best rod.

How much quick return a single stage gives #

The crank-rocker’s version of this limit was the starting point of the whole run of shaper essays: a swing and a time ratio found that a crank-rocker with a 60° swing keeps 40° only up to an own ratio of 1.207. The slide’s version is found the same way — for each ratio, the rod whose worst angle is greatest, the offset following from the ratio by bisection on the closed form, the rod by golden section on the worst angle.

The slide does better. It keeps 40° up to an own ratio of 1.286, where the crank-rocker stops at 1.207, and at every ratio in between it leans less. The reason is visible in the two curves’ left ends: at a ratio of one the slide’s best rod is infinitely long and its worst angle is 90°, while a crank-rocker with a 60° swing can never do better than 60°, because its rocker has to swing through 60° and its coupler cannot stay square to it throughout. The slide begins with thirty degrees of margin in hand and spends them more slowly.

Neither curve comes near 2. A slide asked for an own ratio of 2 leans far past any usable angle. So a ram-driving shaper needs a first stage as much as a rocker-driving one does, and the three slides used here are matched to the three crank-rockers of the earlier essays by their own time ratio: 1, 1.1 and 1.2. The two quick ones take their best rods — 3.717 with an offset of 0.922, and 2.719 with an offset of 0.847, on a crank of 1 — and keep worst angles of 58.9° and 47.2°. The centred one needs a rod, since its best rod is infinite, and it takes the rod of the machine at 1.2, 2.719, so that the two differ only in their offset. It keeps 68.4°.

The same search, on the ram #

The first stage is chosen from the same grid as before: every drag link with a ground of 1 and its other three bars between 1.25 and 5 in quarter steps that turns fully and keeps a transmission angle of 40°, which is 821 drag links. Each is put ahead of each slide at every phase angle in steps of 3° — 98,520 designs per slide — and the ram’s speed is differenced through the whole input turn. The working stroke is the slower of the two strokes, and evenness is read exactly as it was on the rocker: the ratio of the fastest ram speed to the slowest while the ram covers the central 80% of its stroke.

The pattern the rocker showed is there on every row. Driven directly at constant speed the slides give evenness of 2.06, 1.89 and 2.02 — less even than the crank-rockers of the same own ratios, which gave 1.58, 1.80 and 1.91. The drag link with the highest ratio is the same one on all three slides, crank 2.5, coupler 2.25 and output crank 1.5, with an output speed ratio of 5.93, and it is the same drag link the crank-rockers chose. It reaches ratios of 2.30, 2.51 and 2.71 and drives the ram at 5.29, 2.79 and 2.60: on every slide the least even stroke in the table.

The most even designs that reach 2 are gentler drag links at different phases, and they give 1.309, 1.198 and 1.188. On the two quick slides that is about as even as the best crank-rocker designs, 1.21 on both. On the centred slide it is far better than the centred crank-rocker’s 1.54, which is the first sign that something about that pair is different.

Each design keeps both stages at 40° or better, and since a transmission angle has no size that holds for a shaper of any scale. The drag link that evens the quickest slide has a worst angle of 48.2°, and the slide under it 47.2°, so none of this evenness is bought with transmission angle.

Why the ram’s stroke evens out #

The rocker essay took its cancellation apart as a product, and the product carries over with nothing changed but the second factor. The ram’s speed per unit of input speed is the slide’s own ratio — how far the ram moves per radian of its crank, at whatever crank angle the drag link has reached — times how fast the drag link turns that crank.

$$\frac{dx}{d\theta} = \frac{dx}{d\varphi}\bigg|_{\varphi=\Phi(\theta)+\alpha} \cdot \frac{d\Phi}{d\theta}.$$

On the quickest slide’s most even pair, over the central 80% of the stroke, the slide’s own ratio runs from 0.495 to 1.002 — a factor of 2.02, which is why the slide alone cuts at 2.02. The drag link’s output speed runs from 0.533 to 1.066, a factor of 2.00, and it is low where the first factor is high. The product runs from 0.528 to 0.627. Two factors each worth two multiply to 1.19.

The two factors are computed independently of the ram’s speed — one by differencing the slide’s position against its own crank angle, the other by differencing the drag link’s output against its input — and their product matches the speed differenced on the whole machine to $4 \times 10^{-7}$. Read the slide’s own ratio two tenths of a radian away from the crank angle actually reached and the product misses by more than 5%, so the agreement is about where the two stages meet, not about the product being insensitive.

Up to here, this essay could have been the rocker essay with a word replaced. The finding is in why the replacement works.

A rocker and a ram of the same own ratio are nearly one output #

The cancellation needs the drag link’s slow stretch to line up with the output’s fast stretch. So what the drag link sees of the machine behind it is only the shape of that machine’s working stroke: how the stroke is covered against time when the crank turns steadily. If a rocker and a ram cover their strokes alike, a drag link cannot tell them apart.

At an own ratio of 1.2 the two curves lie on each other. The largest gap between them, anywhere in the stroke, is 0.013 of the stroke; at 1.1 it is 0.015. A crank-rocker with a 60° swing and an offset slider-crank are different machines, with different links, different limit geometry and different transmission-angle limits, and over their working strokes they are the same function of time to about one per cent.

That is less mysterious than it sounds. Each output’s stroke runs between two stops, and the own time ratio fixes how much crank rotation the working stroke is given: $360° \times Q/(1+Q)$, or 196.4° at a ratio of 1.2, whichever machine it is. Inside that, both machines cover their strokes roughly as a half-cosine of the crank angle, with a quick-return skew whose size is also set by $Q$. What is left after those two things is a small residual, and at these proportions the residual is about a hundredth of the stroke.

At an own ratio of one the picture breaks. The gap is 0.099, seven times larger. The centred crank-rocker with a 60° swing is nearly symmetric — the member the earlier essays used has a coupler and a ground each about sixty times its crank, which is close to a slide with an infinitely long rod — and covers its stroke almost as a pure half-cosine. The centred slide has a rod only 2.72 times its crank, and the slider-crank shows what a short rod does: it adds a second harmonic to the motion, and within one stroke the ram crawls through the first part and hurries through the second. Two machines with no quick return between them can still have strokes of different shapes.

Six drag links on six outputs #

If the shape of the stroke is what a first stage responds to, then a drag link chosen for one output should serve any output of the same shape. That is a prediction the grid can test directly.

The table reads in pairs. The drag link found for the crank-rocker of 1.2 — crank 5, coupler 4.5, output crank 2.25 — put ahead of the slide of 1.2 and rephased gives 1.198. The slide’s own best drag link gives 1.188. Going the other way, the slide’s drag link on the crank-rocker gives 1.231 against the crank-rocker’s own 1.203. At 1.1 the swaps give 1.215 against 1.198 and 1.257 against 1.210. At a ratio of one, 1.320 against 1.309 on the slide, and 1.535 against 1.536 on the crank-rocker — where the slide’s drag link is, within the grid’s resolution, as good as the rocker’s own. The worst loss from moving a drag link between a rocker and a ram of the same own ratio is 3.8%.

The loss from moving a drag link between own ratios is another thing entirely. The drag link chosen for an own ratio of one, put ahead of the slide of 1.2, gives 1.436 — 21% worse than that slide’s own best. The drag links chosen for 1.1 and 1.2 cannot reach a combined ratio of 2 at all on the outputs of one, which is what the dashes are: a gentler drag link leaves the whole quick return to be supplied by the output, and an output with none of its own cannot supply it.

So the division is the one the drag-link essay found for the ratio, extended to evenness. The second stage’s own quick return decides how much the first stage has to supply and therefore which drag link is right. The kind of second stage — pin or slide, rocker or ram — very nearly does not matter.

The phase moves by where the stroke starts #

A drag link moved between machines has to be rephased, and the phase it wants is a second prediction. If the drag link only sees the stroke’s shape, and the two shapes agree, then the right phase should differ between the two machines by exactly the angle between their stroke starts: the crank angle, in each machine’s own frame, at which its working stroke begins. For the crank-rocker that is the crank angle of its extended limit position; for the slide it is $\varphi_e$.

At 1.1 the stroke starts are 20.42° apart and the best phase moves by 20°; at 1.2 they are 25.25° apart and it moves by 26°. The search is in whole degrees, so both are agreement to within the step. A drag link designed for one machine can be moved to another by a single subtraction of two limit-position angles, each a closed form.

At a ratio of one the prediction misses. The stroke starts are 1.63° apart — both centred machines begin their strokes almost at the dead-centre line — and the best phase moves by 14°. That is the skew again. The centred slide’s fastest ram speed comes late in its stroke, so the drag link’s slow stretch has to be moved later to meet it, and the stroke start, which says nothing about where within the stroke the speed peaks, cannot say by how much.

The skewed stroke helps #

The centred slide’s exception has a second consequence, and it runs in the slide’s favour. Its most even pair reaches 1.309, where the centred crank-rocker’s reaches 1.536. A machine that is less even on its own — 2.06 against 1.58 — ends up more even once a drag link is put in front.

The product explains that too. The drag link’s own output speed, like any drag link’s, is not symmetric within the part of its turn that drives the working stroke: it has a slow stretch and a fast one, arranged by its own geometry. A symmetric output wants a correction that is symmetric about the stroke’s middle, and no drag link supplies exactly that; the best one can do is centre the slow stretch and live with its lopsidedness. A skewed output wants a lopsided correction, and a drag link’s natural lopsidedness can be phased to supply it. The short rod’s second harmonic is doing some of the work a quick return would otherwise have to do.

It would be wrong to read that as a reason to shorten every rod. The centred slide’s worst transmission angle, 68.4°, leaves room; a shorter rod leans more, and the skew that helps the evenness grows with the lean that costs the transmission angle. How far that trade goes is a question with a rod length in it, and the three machines here do not answer it.

What a designer can take from it #

Three practical statements come out of the measurements, each narrower than it might first appear.

A first stage is chosen by the second stage’s own time ratio. Two outputs with the same own ratio and a quick return of their own share their most even drag link to within a few per cent, whatever their mechanism. Outputs with different own ratios do not, and the gap is a fifth of the evenness.

The phase between the stages is carried over by the stroke starts. Between two outputs whose strokes have the same shape, the right phase differs by the difference of their stroke-start angles, which are closed forms for both the crank-rocker and the slide. A table of drag links indexed by own ratio would serve every quick-return output of either kind at ratios where the shapes agree.

The rule fails where the shapes differ, and the shapes can be measured first. The largest gap between two strokes’ shapes is a number that can be computed before any drag link is searched, and where it is ten times larger than one per cent, as it is for the centred pair here, both the transfer and the phase rule should be re-derived rather than trusted.

That last is also the honest limit of the finding. It was shown on six outputs: three crank-rockers with one swing and three slides with two rod choices. The claim that the stroke’s shape is all a drag link sees is supported by those six and by the product decomposition, which is exact. The claim that any two outputs with matching shapes share a drag link is an extrapolation from them.

What the six outputs do not cover #

The ram is the output here, and on a real shaper it is. An offset slider-crank drives its ram directly, so the evenness measured is the ram’s. The Whitworth quick-return, one of the slider-crank chain’s four inversions, drives its ram through a rocking slotted link and a further rod, and was not tried; nor were the six-bar chains that buy a quick return in a single loop. The crank-rockers of the earlier essays still stand one link short of a ram; the finding here, that their strokes and the slides’ agree in shape at equal own ratios, says that such a link would change their evenness only through its own lever-arm variation.

The grid is the same coarse grid. Drag links in quarter steps and phases in steps of 3° locate the best designs well enough to compare them and not well enough to rank designs that differ by less than about 1%. The swaps that lose 1% or less sit inside that resolution; the one that loses 3.8% does not.

One rod per quick slide. Each quick slide is the one whose rod leans least for its ratio. Other rods give other skews, and whether their strokes still match a crank-rocker’s to 1.5% has not been measured.

No dynamics. The ram’s mass, the cutting load and the drive’s compliance all act on a speed profile that is here purely kinematic.

Still open: the rod as a second knob #

The centred slide showed that a skewed stroke can help a drag link rather than fight it: its most even train reached 1.31 where the symmetric crank-rocker’s reached 1.54. The skew comes from the rod, through the second harmonic that a short rod adds, and its price is the rod’s lean.

The distinct argument there would be the trade itself. For a centred slide with a drag link ahead of it at a combined ratio of 2, sweep the rod from very long to as short as a transmission angle of 40° allows, and measure the most even train at each. Two things would come out of it that this essay cannot produce: whether the evenness keeps improving as the rod shortens until the transmission angle stops it, or turns back first because the second harmonic overshoots the correction a drag link can phase against; and whether a centred slide with the best rod beats the offset slides here, which would mean that on a ram-driven shaper the offset — the textbook way of buying a quick return in one stage — is the wrong place to spend the first stage’s help. The measurement would reuse this essay’s grid, with the rod as a fourth dimension and the evenness’s dependence on it compared against the rod’s own second-harmonic amplitude, $a/b$.