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Date Posted: Monday, November 24 2008, 17:15:15
Author: Michael Spencer
Subject: Re: mercury lens floation
In reply to: Fred 's message, "Re: mercury lens floation" on Monday, November 24 2008, 15:22:13

I'm sorry I didn't make myself clear: too elliptical, I suppose! Let me try again, and show the real flaws in Bob's reasoning. Also, I don't know why I took the figures of 2·5 tons of optic and 55 litres of mercury, which is a real lighthouse somewhere else; I should have stuck to the figures from Souter, 4·5 tons of optic and 1·5 tons of mercury.

The other thing I have to say is that I yield to no-one in our right to buy bananas by the pound, but for the present purpose it's easier to think in metric units.

Consider a 4·5 ton (4500kg) optic floating in something. It displaces its own weight of the something, as the old Greek pointed out. In water, then, it displaces 4500 kg of water = 4500 litres by definition. In mercury, however, because the density of mercury is 13.6, it displaces only 4500/13.6 = 330 litres. So the mercury-proof float that supports the optic (part of its overall weight) and the bath it floats in have to be at least as big as that.

We need to put mercury all round the float, to stop it catching on the sides or on the bottom. As Bob points out, mercury is not cheap and we need to minimise the quantity used. So let's say we'll arrange for a minimum of 2cm of mercury between the sides and bottom of the float and of the bath; and we'll also assume that the bath is a simple cylindrical bucket 2 metres across, so that its radius is 100cm; and the float is a simple cylinder contained inside it.

The radius of the float is therefore 98cm, and we'll write d for its depth. Using V = pi.r².d, and expressing the volume in cc, we can write

330000 = pi.(98)².d or d = 11cm.

The volume of the annulus of mercury surrounding the float down to that depth is

pi.(100)².d - pi.(98)².d = pi.d.{(100)²-(98)²} = pi.11.198.2 [because A²-B² = (A+B).(A-B)]

= 13685cc = 13·6 litres.

The volume of the 2cm disc of mercury across the whole of bottom of the bath underneath the float is pi.(100)².2

= 62831 cc = 63 litres.

So the total amount of mercury we need to make a cylinder of mercury 2 metres across, 13cm deep and with a 330-litre hole in it, is 63 + 13.6 = 76.6 litres = 76.6 x 13.6 = 1050 kg = just over a ton. We conclude from this that the tolerances at Souter are slighly more than 2cm all round.

Now the thing that Bob has not grasped in his post is that we have got a ton or so of mercury constrained by the volume of the bath and the volume of the float to accommodate the float completely and be able to prevent it from touching the sides or the bottom. Which is what we observe. The "hole" in the top of the mercury is the same volume as the float needs to be able to float, and the liquid is therefore able to produce enough hydrostatic pressure (is that mercurostatic pressure?) to support the floating optic.

If we had started by filling the 2-metre bath with mercury (using almost six tons of the stuff) and then lowered the optic into it, we should have displaced 4500kg of liquid, which would have splashed all over our shoes, and the optic would have floated on what was left. But we were cleverer than that: we put the right amount of mercury (a ton, about) in so that when the optic was lowered in, the bath filled itself up to the top, leaving mercury all round and under the float. We have fooled the float into thinking it has displaced 4500kg of mercury.

Fred's ideas of working out the density of the materials are all right as far as they go, but you really need to consider the overall density of the floating object (total weight/total volume), allowing for the air within. If you just rolled 4·5 tons of glass and bronze into a tight ball, it would't float in anything!

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