Illustrated Physics
physics, drawn
Series

Hydrostatics — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Three vessels, one pressure. Three vessels filled to the same depth of 3 m. The pressure on each base is 29.4 kPa — identical, because pressure is set by depth — while the weight of water each holds differs by a factor of 4.7. The base of the flaring vessel carries more force than the water standing over it weighs.

    The pressure that only knows depth

    A litre of water and a swimming pool press equally hard on a floor at the same depth. Pressure in a still fluid is a scalar with no direction of its own, it depends on how far down and on nothing else, and the shape of the container falls out of the arithmetic entirely.

    part 1 · fluids
  2. Force multiplied, distance paid. Two pistons on one body of fluid, of areas in the ratio 16 to 1. A force of 200 N on the small one holds 3.20 kN on the large one, and pushing the small piston 16 cm raises the large one by 10.0 mm. The two products are the same number: nothing is gained except the shape of the bargain.

    Force multiplied, and nothing gained

    A push on a small piston becomes a much larger push on a large one, in the ratio of their areas, with no machinery in between except the liquid. What the liquid will not do is give anything away — the distances shrink by the same factor the forces grow by, and the product is untouched.

    part 2 · fluids
  3. The surface a spin decides. The free surface of a liquid in a dish of radius 0.5 m turning at 10, 20, 40 revolutions a minute. In the rotating frame the surface is a level set of gz − ½ω²r², so it is a paraboloid exactly and not to some approximation, and nothing about the liquid appears in its shape: the same curve is got with mercury, water or oil. The rim stands 14.0 mm, 55.9 mm, 223.6 mm above the centre at those rates. A parabola z = r²/4f has focal length f, so these surfaces are mirrors of focal length g/2ω² — 4.47 m at 10 rpm, 1.12 m at 20 rpm, 0.28 m at 40 rpm. That is checked here on the drawn curves rather than quoted: a vertical ray reflected off the surface at a quarter, a half, three quarters and the whole of the radius crosses the axis at the same height to 0.00%, which is what a mirror with no spherical aberration means. Doubling the spin quarters the focal length, and there is no other adjustment: the dish can only ever look straight up.

    The surface a spin decides

    Spin a dish of liquid and its surface settles into a paraboloid — exactly, with nothing about the liquid in the shape. A parabola of that form has a focal length of g over twice the spin rate squared, so a bucket of mercury turning at twenty revolutions a minute is a telescope mirror figured by a clock instead of by grinding.

    part 3 · fluids
  4. A siphon's pressure, and the 10.09 m it cannot pass. The absolute pressure of the liquid along a siphon, from the upper surface, over the crown, and down to an outlet 1.2 m below the surface, for 4 crown heights. The profile is hydrostatic and depends on nothing but height: the tube's shape, its length and its bore do not appear. At the crown the liquid is below atmospheric pressure by ρg times the lift, and the whole question of how high a siphon can reach is whether that number stays above the liquid's vapour pressure — 2.34 kPa for water at 20 °C, which puts the ceiling at 10.09 m. a 2 m crown sits at 81.7 kPa and holds, a 6 m crown sits at 42.5 kPa and holds, a 9.5 m crown sits at 8.1 kPa and holds, a 11.5 m crown sits at -11.5 kPa and is below the vapour pressure, so it boils. The ceiling is a property of the liquid, not of the mechanism: a degassed liquid that can be pulled into tension has no such limit, and siphons in a vacuum.

    The height a siphon cannot pass

    A siphon will not lift water more than about ten metres, and the usual explanation for the limit is also given as the explanation for the mechanism. It cannot be both. A siphon runs in a vacuum, with degassed water, over a crown no atmosphere could support.

    part 4 · fluids
  5. The wedge that proves it. A wedge of water 4 mm on its vertical side, at a depth of 3 m, with the pressure on each of its three faces as an unknown. The two force balances decide them. Horizontally, the sloping face's push has a component that must exactly cancel the vertical face's, and since the sloping face is longer by exactly the factor its slope reduces the component by, the two pressures are equal — the geometry cancels, at every angle, for every size. Vertically the same cancellation happens except for the wedge's own weight, which needs the bottom face to carry 0.0667 per cent more. That excess falls in proportion to the size of the wedge, so at a point it is nothing and the three pressures are one number. Pressure being the same in every direction is the conclusion of that argument, not an assumption in it.

    The push that has no direction

    That the pressure at a point in a still fluid is the same whichever way the surface faces is not a definition. It is a theorem, and its proof is an argument about how two kinds of force scale with size — which is also the exact statement of when it stops being true.

    part 5 · fluids

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