VoyForums

[ VoyUser Login optional ] [ Contact Forum Admin ] [ Main index ] [ Post a new message ] [ Search | Check update time | Archives: 1 ]

[ Next Thread | Previous Thread | Next Message | Previous Message ]

Date Posted: Thursday, November 27 2008, 15:23:00
Author: Bob Pass
Subject: Re: mercury lens floation
In reply to: Michael 's message, "Re: mercury lens floation" on Thursday, November 27 2008, 8:49:47

Fred and Michael thanks for all your recent posts. It’s beginning to feel quite lonely as the only defender of Archimedes, but since his theory has persisted for such a long time I feel it is incumbent on me to try a little while longer in this context.

First I believe the explanation for Fred’s example of the boat lift, and Michael’s glass in a bowl of water is that in both cases the volume of water and hence its mass exceeds the mass of the floating body, and had sufficient depth to be greater than the floatation depth of the object even after some water has been displaced. Perhaps on reflection Michael will agree that this was not the most appropriate experiment, though maybe better than many conducted in those circumstances.

Secondly I realise I am not explaining the problem with the stated weights at Souter clearly. My only defence is old age and decrepitude, but I will try again.

Let’s imagine that the 4.5 ton lens at Souter is not floating but instead is on one end of a seesaw, or in the pan of a giant balance scale. As in the case of the lens floating we need to raise it up. We will do this by means of a bucket of mercury on the opposite end of the seesaw/ balance. The question is what weight of mercury will be required. I suggest that no one would believe that 1 ton or 1.5 ton or indeed anything less than 4.5 tons in the bucket would be sufficient. The downward force exerted by the bucket must equal the upward force required to lift the lens or it will not move.

Perhaps we can examine another scenario slightly closer to the floating lens of a lighthouse. For this case imagine two vertical open topped cylinders sitting side by side and connected by a pipe attached to the bottom of each cylinder. Also imagine that the lens is a perfect fit within one cylinder such that no fluid can escape past its sides, but if a fluid is introduced under the lens, via the connecting pipe, the lens will be lifted up the cylinder. This of course requires sufficient pressure in the fluid. In effect we have created a hydraulic version of the seesaw, where the weight in one cylinder i.e. the lens will be lifted by the pressure in the other cylinder. This is where I have to confess that my treatment of hydrostatic pressure in my last post was not really correct. The reason for this is that we will create hydrostatic pressure in the pipe and hence in the lens cylinder, by the introduction of a fluid in the second cylinder. As was the case with the seesaw arrangement the weight of fluid doing the lifting has to be the same as the weight, (the lens), being lifted. As in the case of the seesaw 1 or 1.5 tons will obviously not be sufficient to create the magnitude of fource involved, remember this is just a version of seesaw, or balance, albeit an hydraulic one. Note again that is a downward load i.e. the weight of fluid in the second cylinder which is causing an upwards force on the lens.

This second scenario is almost a system which could work, but packaging would be difficult in a lighthouse, and the piston seal would create friction whereas normal flotation reduces this to a minimum. In any case it is over complicated since the flotation fluid and the lens can be in the same cylinder, i.e. the floatation bath. In this case the piston or float is not a close fit in the cylinder and the fluid providing the force to lift it surrounds it rather than being in a separated connected cylinder. It is however still a form of seesaw or balance and a sufficient weight of fluid has to be supplied to do the lifting, there is no magic lifting force associated with floatation. Since the mass of fluid to do the lifting force with any of the above scenarios has to be equal to the mass to be lifted, sufficient space for this fluid, a total of 330 litres of mercury in Michael’s example must be accommodated in this space hence the reason why a 20mm annulus width which only contains 13.6 litres is not enough.

It has been a long time since I have attempted to use maths to prove anything but here goes. Using Michael’s formulae P=A*h*d. (I Have a bit of trouble with this since I would have said what was calculated was force but I have used it) In his example the area of the annulus is 1244.232 sq. cm, the depth is 11cm, and the density of mercury is 13.6 gr per cubic centimetre,
Hence the pressure applied is 13.6grams*11cm *1244.232sq cm/1000 = 186.1371kg
The mass of the lens to be lifted however is 4500kg so it does not float.
If however floatation bath was 139cm radius then the annular area would be 30530sq cm In this case P=A*h*d, would be 13.6*11*30530/1000 = 5460kg i.e. greater than the lens so it would float. Of course the volume of the annulus would now be 335ltrs, i.e. the volume of fluid displaced, (plus a bit to just lift the float).

If none of the foregoing explanation appeals a quick experiment along the lines of Michaels is to take a glass and some coins in the bottom to stabilise it. Put the glass inside another container, I used a jug, and slowly fill the space in between with water, until the glass just begins to move indicating floatation. Weigh a few more coins (150 grams worked for me) and note their weight. This weight represents the float weight. Place these coins in the glass and note it will settle firmly on the bottom. Next place another container on the scales and add 50 grams of water (or one third the weight of the coins already weighed and added to the glass). This represents the weight of mercury supposedly used at Souter. Now add this to the water already in the second container. If the glass now floats Souter is right and Archimedes is wrong. When I tried it did not begin to float until I added a further 100 grams of water, so the water added equalled the weight of the coins added.

My conclusion is that Archimedes’ reputation is still well founded. For an object to float there has to be sufficient depth of fluid to enable the floatation depth to be achieved, and a sufficient mass of fluid to provide the lifting force or buoyancy

Just to tidy up my comment that 2cm was not enough for flotation, was meant to refer to the side clearance between the float and bath, and hence the volume of mercury, not the clearance between the float and the bottom of the bath. Furthermore my comment on the use of rollers to restrain lateral movement was meant to show that a small lateral clearance between the float and bath was not necessary for functional reasons. Also I appear to have used mass and weight to mean the same thing. This would not please the more academic, but I believe does not affect the meaning of what I have said.

I have written the foregoing before I read Michael’s last posting. However I note his first sentence in his forth paragraph which reads “To support any given weight in any given medium, an exactly equivalent weight of the medium must be displaced.” Would appear to confirm all I have said. Though the first sentence in his eighth paragraph “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.” would appear to contradict this. I am left wondering how this displaced fluid, which does not remain in the bath, actually manages to provide the support said to be required in paragraph four.

As ever I look forward to any further comments from either of you with interest.

[ Next Thread | Previous Thread | Next Message | Previous Message ]

Replies:

[ VoyUser Login ] Not required to post.
Post a public reply to this message | Go post a new public message
 * Notice: Posting problems? [ Click here ]
 * HTML allowed in marked fields.
Message subject (required):

Name (required):

  E-mail address (optional):

* Type your message here:

Choose Message Icon: [ View Emoticons ]

Notice: Copies of your message may remain on this and other systems on internet. Please be respectful.

[ Contact Forum Admin ]

Forum timezone: GMT-8
VF Version: 2.94, ConfDB:
Before posting please read our privacy policy.
VoyForums(tm) is a Free Service from Voyager Info-Systems.
Copyright © 1998-2008 Voyager Info-Systems. All Rights Reserved.