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Date Posted: Tuesday, November 25 2008, 14:28:20
Author: Bob Pass
Subject: Re: mercury lens floation
In reply to: Michael Spencer 's message, "Re: mercury lens floation" on Monday, November 24 2008, 17:15:15

Pleased to see these new posts from Fred and Michael but afraid I think that the answer still eludes us. Fred’s post at 13:04 is in my opinion quite correct. His Later post is correct only as far as “So a ratio of 13,600 divided by 5,200 /m3 = 2.6.”. What this figure really means is that, in the example, the volume of mercury required for flotation is 1/2.6 (0.3846) of the volume of the lens. So the mercury required in this case for the theoretical I m3 lens would be 384.6ltrs. If this Volume is multiplied by the SG of 13.6 the weight of mercury would of course be 5200kg. This is probably a circular argument, but nothing there to indicate that the old Greek was wrong.

Michael has provided further details based on his original assertion that floatation is not dependent on the volume and hence mass of the floatation media His statement that “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” is correct. Where in my opinion he goes wrong is to assume that the displaced mercury has no part in the flotation process, and so need not be accommodated in the bath, whereas actually it is the reason for it. His assumption, that a clearance of 2 cm at a depth of floatation of 11 cm will result in floatation, unfortunately is just that, and has no factual basis. As I said in my last post, if it is not the weight of displaced fluid acting downward, which produces the buoyancy or upward force to achieve floatation, what is it. Hydrostatic pressure does not causes items to float (hydrostatic pressure acts in all directions, so that an item which sinks below the surface does not stop sinking when the hydrostatic pressure reaches a certain value, but continues until it reaches the bottom). Floatation is caused by the displaced fluid, under its own weight, trying to get back to where it was displaced from, for it to do that it must still be present in the container in which it was displaced. This means that in Michael’s example the volume of the annulus of mercury must be 330 litres. Not 13.6 litres, requiring a much greater clearance.
My understanding, from the link provided by Fred earlier, to the Lighthouses of Australia web site is that lateral movement of the lens is constrained by a series of rollers with vertical axis’s, and this seems more likely that reliance on close fits, between the float and the bath, though I suppose the geometry could be such that a close fit was achieved, at some point in the floatation depth Unfortunately the guide at Souter, when asked, could not say exactly how the lateral constraint was achieved, merely replying that it was done by precision engineering.

Incidentally my (still ongoing) contacts with the National Trust with regard to this matter have not resulted, even after all this time, in an explanation.

I had to think long and hard about Michael’s contribution this time, so worthy of debate. Please keep the contributions coming.

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