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Date Posted: Sunday, November 23 2008, 17:22:21
Author: Michael Spencer
Subject: Re: mercury lens floation
In reply to: Bob Pass 's message, "Re: mercury lens floation" on Tuesday, November 04 2008, 7:22:36

The difficulties here arise from the fact that we tend to consider floating objects in a large body of water, where there is an infinite volume for the displaced water to escape into. The mercury bath in a lighthouse is by comparison a very constrained volume, and we have to consider what happens to the volume of mercury displaced.

Because the density of mercury is 13.6, the amount of mercury displaced by a floating 2500kg (2·5-tonne) optic is 2500/13.6 = 185 litres. So we make a bath with a volume of 250 litres, put the optic in it and fill up with mercury. When we have put 55 litres in, which weighs 55x13.6 = 750kg (three-quarters of a tonne), the volume left over is 185 litres, which we can look on as mercury not there, that is, displaced: which is enough to float our boat. It's these volumes that count.

If the bath fits fairly closely round the part of the optic below the surface, the depth of the bath will be just greater than the depth of the float, so the thing works. If you do the sums based on 240 litres for the bath and 185 litres for the float, you find that you get about 15mm of mercury all round the float and under it. Any more's a waste.

As Sherlock Holmes didn't say, it's the amount of mercury that's not there that is significant, not the amount that is.

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