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Date Posted: Tuesday, November 25 2008, 17:10:46
Author: Michael Spencer
Subject: Re: mercury lens floation
In reply to: Bob Pass 's message, "Re: mercury lens floation" on Tuesday, November 25 2008, 14:28:20

We’re obviously going to go on arguing here for a long time, until both Bob and I are agreed on the basic physics of buoyancy. Bob’s fundamental mistake is to insist that it’s the weight of the displaced fluid that causes the upward force on the float, whereas it is in fact the difference in hydrostatic pressure between the top and the bottom of the float. This pressure difference causes an upward force called buoyancy. It can’t be the weight of the displaced fluid causing the upward force, because as Bob himself says weight only acts downward.

The pressure at any level in a fluid is dependent only on the depth and the density of the fluid, not on the amount present. That pressure P is given by P = A.h.d, where A is some element of area, h is the depth and d is the density of the fluid. At a given depth, it’s the same whether the column of fluid is an inch wide or a mile wide.

If the pressure difference from top to bottom of the immersed object is less than the weight of the object (which occurs when the density of the object is greater than the density of the working fluid), then it will sink. If it’s equal, the object will float level with the surface. If it’s greater, the object will float with a certain amount of freeboard.

We assume for simplicity that the top of the float is at the surface, so that the upward pressure there is zero. Horizontal pressures on equal elements of area (of either the float or the surrounding fluid) are opposite and equal at any given depth, so that in the first place the float shows no tendency to move from side to side, and in the second place the thickness of the containing wall of fluid is irrelevant. The upward pressure at the bottom, though, is balanced only by the weight of the float. We have to organise the float to have a suitable horizontal surface area, so that its depth is such that the buoyancy acting upward, caused by the pressure difference between top and bottom surfaces, is enough to counterbalance the weight acting downward.

I have shown that the weight of the optic is the same as the weight of 330 litres of mercury, and Bob has agreed, so that fixes the volume of the float. I have also shown that a two-metre float of that size has to be 11cm deep. This sum was of course done backwards, starting with the known horizontal area of the float. It tells us that the hydrostatic pressure difference between mercury at depth X and mercury at depth X+11 is enough to float 4500kg. So long as the bath is deeper than 11cm, the float floats. I have also shown that this can be achieved by fitting the bath fairly closely around the float, and putting about a ton of mercury into the bath. Now these are the figures actually reported by Bob from Souter, so I don’t understand why he thinks I’ve got something wrong.

As to 2cm not being enough to provide flotation: consider Bob in his little boat making carefully up a drying harbour at the end of the day. So long as there’s water under his keel, he floats. When there isn’t, he doesn’t. The boat doesn’t even know how little water there is under its keel. It just keeps going until there’s no water, and then it tips him over the bows.

I daresay that lateral movement of the optic is constrained by rollers on vertical axes, because after all we don’t want the optic leaping about when huge waves make the tower vibrate. But rollers on vertical axes have nothing to do with forces acting vertically, do they?

PS: You’re not going to get any sense out of the National Trust. They are caretakers, not engineers. Try Trinity House, who steered me in the right direction (though it seems Bob wouldn’t agree!)

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