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### Author Topic: Buoyancy calculations – making use of an exception to Archimedes' principle?  (Read 4661 times)

#### Novus

• Newbie
• Posts: 48
##### Buoyancy calculations – making use of an exception to Archimedes' principle?
« on: April 29, 2023, 04:43:59 PM »
Wikipedia: There is an exception to Archimedes' principle known as the bottom (or side) case. This occurs when a side of the object is touching the bottom (or side) of the vessel it is submerged in, and no liquid seeps in along that side.

Picture 1
A (‘brown’) container contains a fluid.
Submerged in de container is a trapezium shaped mass ‘A’
The trapezium consists of two separate parts which fit air tight and can slide laterally. As a result mass ‘A’ can change in volume.
The sides of ‘A’ are kept airtight to the wall of the ‘brown’ container by some kind of guiding rail system without liquid seeping in between (= exception to Archimedes’ principle)
Tube ‘B’ connects ‘A’ with the air outside of the container.
In picture 1 the combined density of the walls of mass ‘A’ and the volume of air trapped inside is buoyancy neutral.
The width of ‘A’ is 1. The width of the ‘brown’ container is more than 1.
The downward buoyance force on top of ‘A’ =  Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
The upward buoyance force on the bottom of ‘A’ =  Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 8 * 1 * 9.8 * 6 = 470.4
Fb1 = Fb2

Picture 2
‘A’ has increased in volume. The combined density of the walls of mass ‘A’ and the volume of the air trapped inside is now positive buoyant.
The downward buoyance force on top of ‘A’ =  Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 4 * 1 * 9.8 * 10 = 392
The upward buoyance force on the bottom of ‘A’ =  Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
Fb2 > Fb1 + positive buoyancy results in ‘A’ moving upwards.
Each interval between picture 1 en 2 will results in an increase in positive upwards buoyancy force.

Picture 3
The ‘guiding rail system’ is disconnected and the sides of ‘A’ are no longer kept airtight to the wall of the brown container.
The two separate parts of trapezium ‘A’ are locked.
Fb2 > Fb1 + positive buoyancy results in ‘A’ moving upwards.

Picture 4
The two separate parts of trapezium ‘A’ are unlocked.
The force on the sides of ‘A’ reduces the volume of the trapezium to the same size as in picture 1.
The combined density of the walls of mass ‘A’ and the volume of air trapped inside is buoyancy neutral.
‘A’ will ‘float’ down to the same position as in picture 1.

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #1 on: April 29, 2023, 05:43:28 PM »
How are you changing the volume of ‘A’?

Without some mechanism (and energy input) the transition between steps 2 & 3 will not occur in the manner as stated.

“unlocking” the width of A, will simply compress ‘A’ to its’ minimum width at any point in the diagram. Water pressure at every depth = > 1 ATM
This forces air out through the tube and ‘A’ will remain at its minimum volume.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #2 on: April 29, 2023, 07:07:27 PM »
The key point, as per the exception to the Archimes principle, is that there are no lateral forces in step 1 and 2 and therefore object A will not compress to its minimum volume even when the 2 parts of the trapezium A are not locked.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #3 on: April 30, 2023, 01:43:50 AM »
In your opinion will object A move up, down or remain stationary based on calculating Fb1, Fb2 and any other factors which you believe are relevant?

#### panyuming

• Jr. Member
• Posts: 82
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #4 on: April 30, 2023, 01:55:59 AM »
Thanks to Novus for reminding Archimedes buoyancy calculation.
Imagine that this graph can rotate itself?
Thank you!

#### Tarsier_79

• Full Member
• Posts: 118
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #5 on: April 30, 2023, 05:02:56 AM »

This guy tried to use the same principle... No pressure on the sides. It didn't work.

The important thing about buoyancy, it is the result of water moving downward with gravity. If Mass moves downwards, it will move. If it can't, it won't. Mass can't move downwards throughout the entire cycle unless you are filling it back up at the top and releasing some from the bottom.

Identify when in your system mass moves downwards and when water mass moves upwards. Then you will know when it isn't going to move.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #6 on: April 30, 2023, 11:02:17 AM »

@ Tarsier, nice find. Impressive build. The principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.

See below for the principle on how I understand buoyancy calculations works.

The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.

« Last Edit: April 30, 2023, 02:32:12 PM by Novus »

#### Willy

• Full Member
• Posts: 236
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #7 on: April 30, 2023, 03:10:11 PM »
some observations...

panyuming design would function given that..

1. Seeping of the fluid into in between the fixed wall and the outer
surface of the pontoons can be prevented without creating to much friction.
2. The exception to Archimedes's is a valid exception.

A rubber like surface on the outside surfaces of the pontoons and upon the
inner face of the fixed wall might prevent the fluid's seepage.

The design referenced by Tarsier-79 in the video

is only, in part, the same principle as is the Novus's design.
The design referenced by Tarsier-79 in that video does not
incorporate the bottom or side case exception to Archemedes's principle,
in that the sides are missing from the floats, at all times during its operation.

i.e. during both the submersion and the rising.

#### Tarsier_79

• Full Member
• Posts: 118
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #8 on: May 01, 2023, 02:38:47 AM »
Quote
.... The principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.

See below for the principle on how I understand buoyancy calculations works.

The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.

In the video, his motor runs in the wrong direction. The larger surface area on the tubes should be rising, not falling. There used to be a video of this device test running in water.

Explain to me the "equal loss"? as it relates to your principle.

Your Buoyancy calculations match this I think:
(https://image.slidesharecdn.com/lecture03-archimedes-140114101600-phpapp01/95/lecture-03-archimedes-fluid-dynamics-3-638.jpg?cb=1389694713)

Also, if we use your buoyancy calculations on the device in the video, won't it work?

Quote
The design referenced by Tarsier-79 in that video does not
incorporate the bottom or side case exception to Archemedes's principle,
in that the sides are missing from the floats, at all times during its operation.

i.e. during both the submersion and the rising.

Does that actually make any difference?

#### Willy

• Full Member
• Posts: 236
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #9 on: May 01, 2023, 05:30:54 AM »

Does that actually make any difference?

Do you mean...

1. is the Archimedes principle exception valid ?
or
2. that, assuming the Archimedes principle exception is valid, "does that actually
make any difference ?"

Answer to # 1. I don't know if the Archimedes principle exception is valid or not.

It is an idea I myself pondered in the past, but didn't have enough balls to consider
that I might actually be correct / had not heard of it from any source / had not the means
to test the idea.
It makes sense to me.

Answer to # 2. Yes, it makes a difference,

In the Panyuming design @

https://overunity.com/19459/buoyancy-calculations-making-use-of-an-exception-to-archimedes-principle/msg577132/#msg577132

the aluminum pontoons touching the fixed wall, will be less buoyant than the
pontoons that have all sides exposed to the liquid.  The wheel will have an
imbalance in the forces causing pontoons to rise and therefore, the wheel will
have a tendency to rotate clockwise, if the Archimedes principle exception is valid.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #10 on: May 01, 2023, 10:33:12 AM »

Quote
The principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.

Quote
Explain to me the "equal loss"? as it relates to your principle

The design in the video is essentially a symmetrical circular 'wheel' type design whereby my 'guess', without making any calculations, based on experience, is that this will never work since any 'gain' on one site is offset by an identical 'loss' on the opposite site.

The Panyuming design is clearly different in the sense that the volume of the pontoons do not change. It might have merit but should probably be presented as a seperate topic.

Quote
The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #11 on: May 02, 2023, 11:01:57 AM »
Quote
The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.

Anyone knows the answer to this question?

Let me know if the question is not clear and/or if relevant information is missing.

Thanks.

#### Tarsier_79

• Full Member
• Posts: 118
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #12 on: May 02, 2023, 11:44:39 PM »
Willy, I don't think panyuming's design will work. The pressure removed from the equation is on the circumference, and missing force is 90 degrees to the rotation, so won't make any difference to the rotation.

Novus, I don't know the answer. I suspect if there was no seal on the sides it would lift. Creating and removing the seal might be a problem. I suspect if you don't interrupt the seal, it will stay put, but if you let water in it will float.

Your other problem, to get it down to start with you need to make the assembly denser than water. It sinks to the bottom and wedges itself against the sides and somehow we seal it. Then to move back upwards, it needs to be less dense to float up and expand..... I think that is a killer of your design.

#### Novus

• Newbie
• Posts: 48
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #13 on: May 03, 2023, 02:41:16 PM »
Quote
Novus, I don't know the answer. I suspect if there was no seal on the sides it would lift.

Given that the density of the object is less than the density of the fluid the standard Archimedes principle is applicable and the object would rise.

Quote
I suspect if you don't interrupt the seal, it will stay put,

I find this hard to believe since it would need to apply to any given depts, surface areas and density of the object in relation to the density of the fluid.

The following factors relate to the question raised in relation to the example scenario;
1.   There are no lateral forces on the trapezium shaped object when the exception to Archimedes’ principle applies to both sites?
2.   The force on the bottom of ‘A’ = h2*p*g*A which in the example given equates to Fb2 = 6*p*g*8 = 48pg
The force on the top of ‘A’ = h1*p*g*A which in the example given equates to Fb1 = 4*p*g*10 = 40pg
Since Fb2>Fb1 the result would be a net upwards buoyancy force? Since ‘the density of object A is slightly less than the fluid’ the object would start to rise and increase in volume? As per below picture 1 we would exchange a loss in Fb for an increase in volume?
3.   Any other forces/factors which are applicable to the scenario?

Quote
Your other problem, to get it down to start with you need to make the assembly denser than water. It sinks to the bottom and wedges itself against the sides and somehow we seal it. Then to move back upwards, it needs to be less dense to float up and expand..... I think that is a killer of your design.

See picture 2 below based on the initial post which would result in a net gain when the answer to the example would be that the object rises upwards and increases in both volume and buoyancy.
As for the practical implications and the challenges with sealing the sites water tight and lock and unlock the seals I propose to leave this aside for now and focus on why this should fail even in theory.

Information on the internet on the bottom (and side) case exception to Archimedes’ principle which I have been able to find is limited with no example found when the exception applies to both sides of an object:

https://arxiv.org/pdf/1110.5264.pdf

"The existence of exceptions to Archimedes’ law has been observed in some simple experiments in which the force predicted by AP is qualitatively incorrect for a body immersed in a fluid and in contact to the container walls. For instance, when a symmetric solid (e.g., a cylinder) is fully submerged in a liquid with a face touching the bottom of a container, a downward BF is observed, as long as no liquid seeps under the block [2, 13, 14, 15, 17]. Indeed, the experimental evidence that this force increases with depth (see, e.g., Refs. [2, 15, 16, 17]) clearly contrasts to the constant force predicted by AP. These disagreements led some authors to reconsider the completeness or correctness of the AP statement, as well as the definition of BF itself [2, 14, 15, 16, 18], which seems to make the things more confusing yet."

@ Tarsier (and anyone else who wants to join the discussion) – can you maybe have an other look at this since we both agree that a gravity/buoyancy based design will never work.

#### Willy

• Full Member
• Posts: 236
##### Re: Buoyancy calculations – making use of an exception to Archimedes' principle?
« Reply #14 on: May 03, 2023, 11:31:22 PM »
Looks like this is a valid exception to Archimedes principle...

In that
some particular mathematical approaches to solving for buoyancy do not
work in specific circumstances, i.e. the bottom and side exceptions.

Real world experimentation demonstrates that there is
NO VARIATION IN BUOYANCY
in the circumstances prescribed  for a bottom and / or sides exception.

I, (living in the good old U.S. of A and all) easily acquired the near perfect apparatus for
testing the "exception".

Real world... THERE IS NO EXCEPTION THERE AFTER ALL.

PS
Did you get some local help with the math  ?