The horsehead is a drum

A beam pumping unit is the machine most people picture when they think of an oil field: a beam nodding on a post, a counterweighted crank turning slowly at one end, and at the other a curved steel head, shaped a little like a horse’s, from which a cable hangs down into the well. The cable carries the polished rod, and the polished rod carries a string of rods a kilometre or more long, with a pump on the bottom of it.

Kinematically the unit is one of the oldest machines there is. The crank, the pitman that runs up from it and the back half of the beam form a crank-rocker: a crank that turns and a rocker that swings. A swing and a time ratio designed exactly this kind of machine for a shaper. What the pumping unit adds is the head, and the head is the reason this essay exists, because it changes what the rocker’s swing does to the rod. It is not a lever. It is a drum.

The machine #

The unit drawn here is representative rather than taken from a catalogue, in metres: a crank of 0.95 on the gearbox shaft, a pitman of 3.6, and a walking beam pivoted on the samson post with a back arm of 2.8 to the pitman. The head on the front is an arc of radius 3.4 about the beam’s pivot. The pivot sits so that at mid-swing the back arm is level and the pitman hangs straight down onto the crank’s centre, which is the plainest arrangement and the one a first design would reach for.

Turn the crank and the beam rocks through 39.7°, from 19.8° head-down to 19.9° head-up. The ends of the swing are the two dead positions every crank-rocker has, where the crank and pitman lie in one line: stretched out at the bottom of the stroke, folded back on each other at the top. The rod, meanwhile, goes straight up and down on a line 3.4 m in front of the beam’s pivot, and the cable always leaves the head at the same height.

That last observation is the one worth stopping on. The head moves through forty degrees, and the point where the cable comes off it does not move at all.

Why the rod’s line never moves #

A flexible cable under tension, pulled away from a curved surface it lies on, leaves the surface along a tangent. A drum is a size, a wrap is a shape made that the whole of how a strand meets a body: what matters is where it leaves and how far round it goes. The rod hangs straight down, so the cable leaves the head where the head’s surface is vertical. On a circle about the beam’s pivot, the vertical tangent is always at the same place: level with the pivot, one radius in front of it. Turn the circle and a different part of its rim is at that place, but the place is the same.

So the head does not need to be a whole circle, only enough of one that the part at the vertical tangent is always steel. The rest of the geometry follows. The cable from its fixed end at the top of the head to the leave point lies on the arc, and its length there is the radius times the angle between the two. When the beam turns head-up by an angle Δψ, the fixed end moves round by Δψ and the leave point stays put. So the length of cable lying on the head grows by $A\,\Delta\psi$, and the rod, on the end of a cable of fixed length, rises by exactly that.

The figure does not assume the tangent. At each of thirteen beam angles the cable’s path is found as the shortest one from its fixed end, round the head without passing through it, to a point far down the well. A taut cable takes that path. Every time, it leaves within 10⁻¹⁰ m of the line 3.4 m in front of the pivot, and the length of that shortest path changes between any two angles by 3.4 times the angle between them, to 2 × 10⁻¹⁵ m. The rod’s position is therefore

$$s = A\,(\psi - \psi_{\text{bottom}}),$$

with no approximation: linear in the beam’s angle, on a fixed vertical line.

The comparison makes the point. Put a pin on the end of a straight front arm 3.4 m long and hang the rod from that instead. The pin moves on the same circle, so at the ends of the stroke it is 20.4 cm behind the rod’s line, and the rod’s top would be dragged back and forth through the stuffing box at the wellhead by that much every stroke. Its height would go as $A\sin\psi$ rather than $A\psi$, and it would cover 2.310 m instead of 2.357 m.

The straight-line problem took the long history of trying to guide a point along a straight line with pins alone, and found Watt’s linkage an approximation and Peaucellier’s cell an exact answer with eight bars. The horsehead is an exact answer too, and it gets there without a single extra pin, by not being a linkage. It is a sector of a winding drum, with the drum turning back and forth instead of round.

It is worth being clear about why this is not cheating. Sarrus’s linkage gets an exact straight line by leaving the plane; the horsehead gets one by leaving rigid bodies. A cable can only pull, so the answer holds only while the rod’s weight keeps it taut, and a pumping unit’s rod is always hanging from it. On a machine that had to push as well as pull, the head would be no use at all, which is why nobody puts one on a steam engine’s crosshead.

The rod’s travel is the rocker’s angle #

The consequence for the machine is immediate. The rod’s height is the head’s radius times the beam’s angle, and the beam is the four-bar’s rocker. So everything about the rod’s motion is the four-bar’s rocker motion, multiplied by 3.4 and by nothing else. The stroke is the rocker’s swing times the radius: 39.7° is 0.693 rad, times 3.4 m, is 2.357 m. That number can be had two ways: from the two dead positions in closed form, where the crank and pitman lie in one line, or from the traced height of the rod over a full turn. They agree to 3 × 10⁻⁷ m, which is as close as samples a tenth of a degree of crank apart can come to a turning point.

The quantity a pumping unit’s designer works with most is the torque factor: how far the rod moves per radian of crank. It has two readings. Kinematically it is a velocity ratio, the rod’s speed over the crank’s. Statically, by virtual work, it is the torque the crank must supply per newton of load hanging on the rod. A gearbox is sized by the peak torque it must deliver, so the torque factor’s peaks on each stroke are the numbers that decide how big the gearbox is.

For a pinned lever this would be a mess of sines. For the drum it is the head radius times the four-bar’s angular velocity ratio, the rocker’s rate over the crank’s. That ratio has a classical two-pin form, which follows from the instant centre where the crank’s line and the rocker’s line meet, the construction every point has a centre uses for velocities generally:

$$\frac{\omega_{\text{rocker}}}{\omega_{\text{crank}}} = \frac{R\,\sin\alpha}{C\,\sin\beta},$$

where α is the angle between crank and pitman and β the angle between pitman and back arm. So the torque factor is $A R \sin\alpha / (C \sin\beta)$. The line in the figure is that formula. The dots are the slope of the rod’s traced height, found by differencing positions with no formula for velocity in it at all, and the two agree everywhere to 3 × 10⁻⁹. The angle β is the transmission angle: where the pitman comes close to lying along the back arm, sin β is small and the crank needs a large torque to move the beam. On this unit it never gets close.

The pin would have spoiled this as well. Hung from a pin, the rod’s vertical travel per radian of crank would be the same velocity ratio times $A\cos\psi$ rather than A, so the torque factor would sag by 6% at each end of the stroke, where cos ψ is 0.94, and the neat product would become a product with an extra factor that changes as the beam swings.

The dead positions sit at 87.9° and 266.4° of crank, measured anticlockwise from the direction of the well. For this unit the upstroke takes 178.4° of crank and the torque factor peaks at 1.185 m on the way up and 1.212 m on the way down. The two strokes are nearly the same length and nearly as hard as each other, which is what the plain arrangement gives: the pitman hangs straight down at mid-swing, and the machine is close to symmetric. The time ratio, the upstroke’s share of a turn over the downstroke’s, is 0.983.

Moving the crank #

In a machine this simple, there is exactly one thing a designer can change about the motion without changing the stroke much: where the crank’s centre sits under the beam. A swing and a time ratio found that a crank-rocker’s time ratio comes from the angle between its two dead positions. Move the crank’s centre sideways and the two dead positions stop being opposite each other, so one stroke gets more of the crank’s turn than the other.

A pumping unit has a reason to want that. On the upstroke the rod string lifts the column of fluid in the tubing as well as its own weight. On the downstroke it only has to sink through the fluid. So the load is larger going up, and a stroke that is slower going up lifts that load with less torque at the crank. A shaper wants the same thing for the same reason, a slow stroke where the work is done and a quick one back, and a quick return that cuts evenly found how far a crank-rocker alone can go before it needs a second linkage in front of it. A pumping unit asks for far less. The question is how much the crank’s position can buy, and what it costs.

Moving the crank’s centre towards the well slows the upstroke. At 0.9 m the upstroke takes 185.7° of crank, a time ratio of 1.066, and its peak torque factor falls to 1.154 m, 2.7% below the unit with no offset. The downstroke pays for it: it now has less of the turn to cover the same distance, and its peak rises 9.6%, to 1.327 m. Moving the crank the other way does the opposite, and fast: at −0.9 m the upstroke’s peak torque factor is 1.372 m, 16% worse than with no offset.

The trade is not symmetric. Slowing the upstroke buys a little on the stroke that carries the load and costs more on the one that does not. Whether that is a good bargain depends on how different the two loads are and on how the counterweights are set, which is the part of a pumping unit’s design this essay does not model. What the kinematics says is only what is available: a few per cent of peak torque on the upstroke, bought with a larger share of it on the downstroke, by moving the crank’s centre.

The stroke itself hardly moves: 2.357 m with no offset and 2.394 m at 0.9 m towards the well. The head radius and the beam do not care where the crank is. All the crank’s position changes is the timing, and the timing is a four-bar’s.

Which way it turns #

A crank-rocker has two senses of rotation, and for a symmetric one they are the same machine. For an offset one they are not.

The rod’s height as a function of the crank’s angle is one fixed curve, determined by the geometry. Turning the crank the other way reads the same curve backwards, which means the bottom and top of the stroke happen at the same crank angles and the stroke is identical — but the arc of crank that was the upstroke is now the downstroke. The unit with its crank 0.9 m towards the well has an upstroke of 185.7° turning anticlockwise and of 174.3° turning clockwise. The durations exchange exactly, and so do the two strokes’ peak torque factors: 1.154 m on the way up in one sense, 1.327 m in the other.

So an offset unit turned the wrong way is not a neutral unit. It is the same unit with its advantage put on the wrong stroke. Its upstroke is the faster one, and the stroke that carries the fluid now has the larger torque factor. That is the kinematic reason a unit built with an offset is specified to turn one way, while a symmetric one runs either way with almost no difference: 178.4° against 181.6° here.

Some things do not change with the sense at all. The acceleration at the very bottom of the stroke, where the rod stops going down and starts lifting, is the second derivative of that one curve at one point, and a second derivative does not care which way the curve is read. At eight strokes a minute it is 0.110 g for the unit with no offset and 0.118 g for the offset one in either sense. Slowing the upstroke did not soften its start. The larger torque factor and the acceleration peak are in different places on the stroke, and moving the crank lowered one and raised the other.

The six units side by side #

The table puts the whole argument in six rows. The stroke column changes only with the crank’s position. The bottom-acceleration column also changes only with the position. Everything else exchanges between the two rows of each pair, because the turning direction reads one curve two ways.

Two rows share a torque factor on the upstroke — 1.154 m — although they are different machines. One has its crank 0.9 m towards the well and turns anticlockwise. The other has it 0.9 m away and turns clockwise, and has the slower upstroke of the two, 193.5°, and a longer stroke. Neither the offset nor the sense is the design variable on its own. What matters is which stroke the crank’s position slows in the sense the unit actually turns, and every entry in the table follows from the rocker’s angle as a function of the crank’s, multiplied by the head’s radius.

What the kinematics leaves out #

The rod string is a spring. Everything here is the motion of the polished rod at the surface. A steel rod string a kilometre long stretches by tens of centimetres under the fluid load, so the pump at the bottom does not follow the polished rod’s motion, and its stroke is shorter by the stretch. The torque factor is exact about the surface; the load it multiplies is a dynamic quantity measured at the well, not computed from the geometry.

Counterweights. A real unit carries weights on its crank, set to balance roughly the average of the upstroke and downstroke loads, so the gearbox sees the difference between the load’s torque and the counterweight’s rather than the load’s alone. The counterweight’s torque is its own sine of the crank angle, and where its peak falls relative to the torque factor’s peak is a design choice this essay does not make.

The cable’s thickness. The cable’s centreline lies a half-thickness outside the head’s steel, so the effective radius is the head’s radius plus that. It changes the number, not the geometry: the line is still fixed and the lift still the radius times the angle.

Still open: a head that is not round #

The drum that is not round showed that a strand feels only the perpendicular distance from the axis to the tangent it leaves along, so a non-circular drum can make any rate the designer asks for, within a limit. A pumping unit’s head could be shaped the same way, to flatten the torque factor on the upstroke instead of accepting the four-bar’s velocity ratio.

The distinct argument there is the price. A circular head keeps the rod’s line fixed because the vertical tangent is always one radius from the pivot. A shaped head’s vertical tangent is a different distance from the pivot at each angle, so the cable would leave from a point that moves sideways, and the rod would be pulled off its line. The measurement would be the head profile that flattens the upstroke’s torque factor for this unit, and the sideways travel of the rod that profile costs — against the 20.4 cm of the pin it was invented to avoid.