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Date Posted: Thursday, November 27 2008, 19:42:12
Author: Michael
Subject: Re: mercury lens floation
In reply to: Fred 's message, "Re: mercury lens floation" on Thursday, November 27 2008, 18:30:21

Easy, Bob, you’re not the only “defender of Archimedes.” His principle underlies the whole argument. However, my rephrasing him by no means “confirms all” you have said, because much of what you have said is unfortunately wrong (and it gives me no pleasure to have to keep repeating that). What Archimedes said was that a floating body displaces its own weight of the medium it floats in; but he made no attempt to describe how or why this happens. In particular, he didn’t attempt to say that the weight of the displaced liquid made any contribution to flotation. This is just something you have invented, and by golly, you are good at sticking to it. You have asked, if it’s not that weight causing flotation then what is it? Can I suggest you read my last post more carefully, where the explanation is quite clear. Unfortunately, it involves hydrostatic pressure, which you have already dismissed as a contributory factor.

The explanation of why a body floats, or doesn’t float, was left to Pascal in the seventeenth century, who established the laws of hydrostatics. Bob’s interpretations of what goes on are not in accordance with Pascal’s theory.

Pascal was the first to show that the pressure exerted by a static fluid depends only on the depth and density of the fluid and the acceleration due to gravity: P = d.g.h, where d is the density (mass/volume), g is the acceleration due to gravity and h is the depth. Notice what is not included in this equation: not the mass or the total volume of the fluid.

Pascal also showed that at a given depth in a homogeneous static fluid, this pressure is exerted by the fluid equally in all directions. But at the bottom of a floating object there is no downward pressure (because there is no fluid above this surface) and therefore an unbalanced upward force appears: this is buoyancy.

Pressure is defined as force/area. At a given depth and therefore a given pressure, the total force exerted on a small area is less than the total force exerted on a larger one. This is how the hydraulic ram described in Bob’s latest post works: a piston acting in a small tube exerts a certain pressure, F/A say , which when applied to a larger area, 2A say, exerts a larger force (F/A).2A = 2F. It’s all to do with the areas of the pistons in the cylinders and nothing at all to do with the weight of fluid doing the lifting. Bob might like to have a look at the next JCB he sees down a hole somewhere, and then tell me that the mass of the hydraulic fluid in its pipes is the same as the mass of the bucket and a few cubic yards of stones. And of course it needs to be powered from an external source.

Bob’s quite right , if we were trying to lift the lens using a seesaw with equal arms, we’d need 4.5 tons of mercury. (If we used a see-saw with one arm three times the length of the other, 1•5 tons of mercury would do the job.) But we’re talking about floating the lens, not levering it up. The see-saw concept is irrelevant.

The objects of my experiment and Fred’s thought-experiment were simply to show that there is no need for the displaced liquid to remain in the system for flotation to be possible. This is a recurring plank in Bob’s arguments. He is still using it in his last post. He says he is “wondering how this displaced fluid, which does not remain in the bath, actually manages to provide the support required.” Well, this fluid does not provide the support required, and is not called on to do so. The support comes from the hydrostatic forces in the mercury that does remain in the bath; and the amount of this mercury is kept small by the relative sizes of the bath and the float. Thus all his sums about the size of the annulus and its weight are based on a misunderstanding.

In Bob’s experiment with glass and coins, he writes that the coins added represent the float weight. But this isn’t true: the glass with the coins in it represents the float weight. So adding small amounts of water won’t give the right answer; and unless his glass is exactly three-quarters of the volume of the jug his experiment won’t reproduce the conditions at Souter.

I have no difficulties with Bob’s explanation of the stated weights at Souter. The optic weighs 4500kg , and it floats on 1500kg of mercury. Bob asks how this is possible? Crikey, this is where we came in.

I had hoped to get away without doing the real maths behind all this, but I suppose it’s the only way to demonstrate what actually happens.

First we do the sums to find out how big the float must be. Pace Fred’s last post, you can’t get away from Archimedes’ principle, which requires that the floating lens displaces its own weight of mercury. Notice that its weight includes the weight of the float, which might be any thin-walled vessel of negligible weight compared to all the glass and bronze. We have all agreed that 4500kg of mercury occupies 330 litres, so that is the volume of the float; and if we assume that the diameter of the float is 2 metres, we must have the depth of the float = 11cm.

We now look at Pascal’s statement: the pressure P at depth h in a fluid of density d is given by P = h.d.g, where g is the acceleration due to gravity. This does not depend on the amount of fluid present, so long as there’s enough to make it that deep. We then observe that pressure is defined as force per unit area, so that the total force exerted on an area A is given by P.A. Now taking dimensions in cm, we have a 200cm-diameter float, so its area is pi.(100)² sq.cm; and so the total force exerted on this area at depth 11 cm (that is, on the bottom of the float) = 11.13•6.g.pi.(100)² = 4•7.10 exp 6.g. (10 exp 6 means ten-to-the-sixth, because this system won't allow superscript figures apart from those in the ASCII code.)

But the weight of the float (with the lens) is only 4•5.10 exp 6.g. We can cancel the g’s from the two sides of this inequality, and we are left with the interesting fact that the upward force of buoyancy on the float, due entirely to the hydrostatic pressure at its lower surface, is greater than the weight of the float. The maths requires the thing to float, so it damn well floats.

All that’s required is that the mercury is more than 11cm deep, which as I’ve said before we can arrange with as little as 1500kg of mercury by making the bath the right size. So there isn’t any mercury “displaced” by the float; the equivalent amount of mercury was never there, but the space it would have occupied had the bath been full of mercury is taken up by the float.

I’m sorry, Bob, but you’ll either have to take my word for it, or do some serious reading about buoyancy and hydrostatics. Try Googling “buoyancy” for a start.

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