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Date Posted: Thursday, November 27 2008, 8:49:47
Author: Michael
Subject: Re: mercury lens floation
In reply to: FRed 's message, "Re: mercury lens floation" on Thursday, November 27 2008, 3:41:27

Simplistic explanation, without formulae: For an object to float in any liquid medium, its weight acting vertically downward must be balanced by a force of buoyancy acting vertically upward. Buoyancy is a result of the difference in hydrostatic pressure across the vertical dimension of the float.

Hydrostatic pressure at a given depth in a given medium depends on the depth and the density of the medium, and on nothing else. At a given depth, it acts equally in all directions: so all the horizontal components cancel out, and while the medium remains unbroken the vertical ones do too. But at the bottom of an immersed object, there is no hydrostatic pressure acting downward; so the upward pressure creates an upward force U, called buoyancy.

Similarly, at the top surface of the object there is no hydrostatic pressure acting upward, and so the top surface feels a downward force D. This will always be less than U, because the depth at the top is less than than the depth at the bottom (if normal English means anything!) If the difference of the two forces U – D is greater than the force due to the weight of the object, the object floats. If it’s less, the object sinks. If the top surface of the object is at the surface of the medium, there is no downward force, and the net hydrostatic force is U.

To support any given weight in any given medium, an exactly equivalent weight of the medium must be displaced. Given the density of the medium, the corresponding volume can easily be found. Therefore, given the horizontal surface area of the floating object, the corresponding depth can be found too.

For the object to float, this depth must be equal to the vertical dimension of the float. This why ships are said to “draw” a certain amount: if they don’t have that certain depth of water, they don’t float. The “draw” is the depth of water at which the hydrostatic pressure is great enough to counterbalance the weight of the ship. The “draw” does not change as the surroundings of the ship change: it’s the same in mid-Atlantic as it is in the River Clyde; and this is why a 90,000-ton ship can’t get as far up as Glasgow city: the water’s not deep enough.

The same considerations apply to the float supporting an optic in a mercury bath. It needs the mercury to be deep enough, but beyond that it doesn’t care. If the weight the float has to support remains constant, then the required depth depends on the area of the float; but not on anything else. In particular, it does’t depend on how much mercury is present in the bath, so long as there is enough to surround the float completely and provide a bit of a cushion underneath. This amount will depend on the dimensions of the bath, which is the only constraint the engineer has any control over.


Our particular problem: It’s difficult to provide an explanation not using formulae, because all the dimensions, weights, areas and the rest, are interconnected.

We start with 4500kg of optic and 1500kg of mercury. The optic is known to displace 330 litres of mercury (previously agreed), and the 1500kg occupies 1500/13.6 = 110 litres. The bath must therefore be able to hold 440 litres. A smaller bath would require less mercury, but it can’t be less than 330 litres or the float won’t fit in; a larger bath would require more mercury but there’s a cost constraint, as Bob pointed out.

It can of course be any shape, but for simplicity we’ll suppose it is circular and has vertical sides (that is, it’s a cylinder). Its depth is therefore a function of its radius:

Depth = Volume / Area = Volume / { pi x (radius)² }.

You can pick the diameter (twice the radius) of the bath to be anything you like: I picked 2 meters, which makes the depth 14cm. (You remember I picked a depth of 13cm before, and didn’t get as much as 1500kg of mercury.) Suppose you pick 20cm: then the depth must be 1400cm (and the optic is going to be pretty unstable, very tall and thin, or very much overhanging the float). I like 2 meters and 14cm, because it looks like a good, stable system will result.

Following my “Little Archimedes” experiment, or Fred’s “Boat picked out of the Falkirk Wheel” thought-experiment, we know by now that the displaced fluid does not have to remain in the bath. It’s the acceptance of this idea that is the basis of understanding the problem of the “missing mercury.” Instead of the displaced fluid, the bath holds the float. The rest of the space in the bath is occupied by the 110 litres of mercury, which is enough to surround the float at the sides and underneath, so that it doesn’t touch the sides or the bottom. Nothing mechanical holds the float up, so it must be floating.

You can work out the actual thickness of the annulus of mercury surrounding the float, and of the disc of mercury under the float, because you know the three important volumes, of bath, float and mercury. Not without some formulae, though.

Of course, as I’ve said, if you make the bath any bigger than 440 litres, you won’t have enough mercury to fill up the spaces around the float, and it won’t float. The whole thing depends on there being enough mercury to fill the bath to the top when the float’s in there.

In fact, what it really depends on is the mercury in the bath being deep enough so that the hydrostatic pressure at the bottom of the float is enough to provide the bouyancy required. This will always be the case when there is enough mercury to surround the 330-litre float. You need a bit extra mercury underneath to avoid mechanical friction against the bottom of the bath, but 2cm is fine. Theoretically, two molecules is fine, but let’s not fool about.

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