Every pressure points at the pin

A distributed load’s resultant is its area acting at its centroid, and that substitution works because every element of a load on a flat surface points the same way. Add them up and the sum has an obvious direction.

Pressure on a curved surface does not have that property. Pressure acts normal to whatever it is pressing on, so on a curve every element pushes in a different direction, and there is no area to multiply and no centroid to place it at. The resultant exists, but nothing about the pressure distribution hands it over.

Two free bodies, and no calculus #

Cut a body of water out of the reservoir, bounded by the curved surface, by its own vertical projection, and by horizontal planes at each end. Now write equilibrium of that wedge.

Horizontally, three things act on it: the pressure on the flat vertical face, the pressure on the curved face, and nothing else — because the horizontal planes carry no horizontal force. So

$$F_H = \frac{\gamma H^2}{2}$$

which is the force on the vertical projection of the curved surface, and which is the same number whatever shape the curve is, provided it projects onto the same rectangle. For the gate here it is 176.6 kN/m, acting at $H/3$ above the sill.

Vertically, take a different free body: the column of water standing directly above the curved surface, bounded by the surface below and the free water level above. The only vertical forces on it are its own weight and the pressure on the curved face, so

$$F_V = \gamma \times (\text{volume above the surface}) = 110.5\ \text{kN/m}.$$

Two free bodies, two areas, and the calculus is gone. The integral round the arc agrees to four parts in a million, which is the discretisation and nothing else.

The column that need not contain any water #

The vertical rule is stated carelessly almost everywhere, and the careless version is wrong in the case that matters.

The free body is the column of fluid above the surface, and equilibrium of it says the vertical component equals that column’s weight. If the surface curves so that water genuinely stands on it — the upstream face of a dam sloping back, a gate concave upward — then the column is full of water, the vertical component is downward, and everything is as it sounds.

If the surface curves the other way, the column above it contains no water at all, because the surface itself is the ceiling of the reservoir there. The free body is then the column of water that would stand above it, and the vertical component is that same weight pointing upward. The rule is about a geometric volume, not about a physical body of liquid.

That distinction is why the vertical component on the downstream face of an arch dam is upward and destabilising, why a sluice gate with water underneath it is harder to hold down than one with water beside it, and why the sign of the term is the first thing to check in any of these calculations.

The trunnion, where the whole force vanishes #

Everything so far is true of any curved surface. The next observation is true only of a circular one, and it is why radial gates exist.

Pressure acts normal to the surface. A normal to a circle is a radius. A radius passes through the centre. So every element of pressure on a circular gate passes through the circle’s centre, and if the gate is hinged at that centre, every element has zero moment about the hinge.

The moment of the entire 208 kN/m resultant about the trunnion comes out at $-3.4 \times 10^{-15}$ kNm — the linear algebra’s zero — against 353 kNm for a flat gate of the same height on a hinge at its sill.

What is left for the hoist is what does not pass through the pin:

• The gate’s own weight, which acts at its centre of gravity and is nowhere near the trunnion.
• Friction in the trunnion pin, which is the resultant times $\mu$ times the pin’s radius: $208 \times 0.05 \times 0.25 = 2.6$ kNm/m.
• Seal friction, and whatever silt has collected on the sill.

So a gate holding two hundred kilonewtons per metre of water is lifted by a mechanism sized for its own weight and a couple of kilonewton-metres of friction. 99.3 per cent of the hydrostatic moment a flat gate would deliver never reaches the machinery, and it never reaches it because of where a hole was drilled.

The same argument for something that is not water #

Nothing above used any property of water except that it is a fluid at rest, so the pressure is normal to the surface and grows linearly with depth. Replace it with anything that shares those two properties and every result survives.

Grain in a silo does not: its pressure is not normal to a wall, because it has friction, and the wall carries part of the weight rather than merely being pushed by it. So a curved silo wall’s resultant does not pass through the centre of curvature, and the arithmetic here does not apply to it.

Soil behind a wall does not either, for the same reason and a second one: the pressure depends on how far the wall has moved, so it is not a property of the retained material alone.

Gas pressure in a vessel does, exactly. Which is why the hoop and longitudinal forces in a cylinder come out of the same projection rule, and why a hemispherical head’s thrust on the shell is the pressure times the projected circle and not the pressure times its own curved area.

Where the force does go #

The water has not gone anywhere. It arrives at the trunnion as a force rather than as a moment, and the trunnion has to be founded on something.

That is the trade the arrangement makes, and it is a good one: a force at a fixed point is much easier to deal with than a moment at a machine. The trunnion beam and its anchorage into the pier carry 208 kN per metre of gate at a known point and a known direction, and neither number changes as the gate is raised — the geometry rotates with the gate, so the resultant swings but the pin stays where it is.

There is one more asymmetry worth noticing. The resultant on the gate has a vertical component of 110 kN/m — upward or downward depending on which way the arc curves — and it too passes through the pin. Where the gate is concave toward the water and the pivot is above the sill, as drawn, that component is upward, so the water is helping to lift the gate. Where the arrangement is inverted it is pushing down, and the hoist has to overcome it as well as the gate’s weight.

The size the arrangement is worth #

The saving is not a small optimisation, and it is worth putting a number on it in the terms a gate is actually bought in.

A vertical lift gate 6 m deep and 10 m wide holds 176.6 kN/m of horizontal thrust — 1,766 kN in total — and has to be lifted vertically against its own weight while that thrust presses it against its guides. The friction on those guides at $\mu = 0.3$ is 530 kN, which is several times the gate’s weight, so the hoist is sized by the water rather than by the gate. The gate has to be heavy enough to close under its own weight against that friction, so it gets heavier, so the hoist gets larger again.

A radial gate of the same opening replaces guide friction with pin friction, and the pin friction is $\mu$ times the resultant times the pin radius over the gate radius — the last factor is 0.25/8 = 1/32. Same coefficient, same force, one thirty-second of the arm.

That single ratio is most of the case for radial gates on large spillways, and it explains why they get relatively cheaper as they get larger: the resultant grows with $H^2$ and the pin radius grows very slowly, so the friction moment falls further behind the force with every metre of head.

Which free body produced each number #

Three, and each one was chosen so that something inconvenient fell off it.

For the horizontal component, a wedge of water bounded by the gate and its own vertical projection — chosen because the horizontal faces of that wedge carry no horizontal force, so only two terms appear.

For the vertical component, a column of water bounded by the gate below and the free surface above — chosen because the vertical faces of that body carry no vertical force, so again only two terms appear.

For the moment, the gate itself with the pressure on it — chosen with the moment centre at the circle’s centre, so that every pressure has zero arm and the sum is exactly zero rather than nearly zero.

Each of the three is the same technique: draw the body and pick the point so that the quantity being solved for is the only unknown in one equation. It is the technique behind the method of sections, behind taking moments about a support to get a reaction, and behind every hand calculation in this collection.

The check the rule makes available #

There is a practical use for the two projection rules that has nothing to do with gates, and it is the reason they are worth remembering rather than looking up.

They give an independent check on any pressure integral anybody computes. A finite element analysis of a curved dam face, a spreadsheet summing pressures round an arc, a hand calculation with the trigonometry done badly — all of them can be checked in one line each, because the horizontal component must equal γH²/2 on the projection and the vertical must equal the weight of a volume that can be measured off a drawing.

The check here caught nothing, because the integral and the rules agree to four parts in a million. That agreement is the point: an assertion that has never rejected anything proves nothing, and the value of these two rules is that they could reject an integral, from outside it, using no part of the same arithmetic.

Where the model stops #

The fluid is at rest. Water flowing under a partly-raised gate is not hydrostatic: the pressure on the lower part of the gate falls below the static value, there is a downpull force from the accelerating flow, and gates have been lost to it. The static calculation is right for a closed gate and progressively wrong as it opens.

The gate is a surface with no thickness. A real radial gate is a skin plate on curved horizontal girders on radial arms, and the arms are struts in compression whose buckling is the reason the arrangement has an upper size. The pressure on the skin passes through the pin; the compression in the arms does not care.

The pivot is exactly at the centre of curvature. Machining tolerance and settlement of the pier move it, and every millimetre of offset multiplies the whole 208 kN/m by that arm: a 10 mm eccentricity is 2.1 kNm/m, comparable with the pin friction. The zero is exact in the geometry and approximate on site.

There is no sediment, no ice and no debris. Silt behind a gate is a second pressure distribution with a different profile that does not point at the pin, because it is not a fluid and its pressure is not normal to the surface. Ice adds a horizontal thrust with a similar problem.

The water level is at the pivot. For the geometry drawn the two happen to coincide, which makes the arc a quarter circle and the numbers tidy. Nothing depends on it: the projection rule is the projection rule for any arc, and the trunnion result holds wherever the pivot is provided it is at the centre of curvature.

And the drawing shows a gate that is not moving. Every quantity is a statement about equilibrium, which is a statement about a rotation that does not happen — and the whole point of a gate is that it does. What the machinery actually has to supply during operation is a dynamic question with the flow in it, and nothing on this page addresses it.

The ladder from here #

Later rungs on this anchor: the pressure prism on a doubly curved surface, where both components need a genuine surface integral and the trunnion trick generalises only to a sphere. Downpull and the hydrodynamic forces on a partly-open gate, which is where gates are actually lost. The arch dam, where the same curvature argument runs in plan rather than in section and the reservoir’s thrust is delivered to two abutments rather than to a foundation. Buoyancy read as this rule applied to a closed surface — integrate the vertical component all the way round a submerged body and the answer is the weight of the fluid displaced, which is Archimedes derived from a free body rather than from a bath. And the same geometry with the sign reversed: a pressure vessel’s dished head, where the pressure inside a curved surface produces a thrust the cylinder has to take, and where the horizontal-projection rule gives the answer in one line.