Need help with gravity/buoyancy design

[Novus]

Attached scenario6 starts with the ending position 9A/9B of the previous scenario5

As per the red arrow mass 'M' has gained potential gravitational energy which should be sufficient for the forces needed to transition from position 1A/1B to 9A/9B (which are all at equilibrium).
We now return to the starting position 1A/1B

Since I do not believe perpetual motion based on gravity/buoyancy is possible there must be a (probably simple) mistake in this design.

From those of you who take the time to analyse this (which is relatively easy to replicate in Excel) I hope to receive constructive feedback. In addition please feel free to ask questions on anything which may not be clear.

For now I will not elaborate on the forces which keep the portion of 'V' which is not submerged in balance since I believe the error must be in the parts of the design I have explained so far.

[Tarsier_79]

Gday.

I perform my buoyancy tests the same way with excel. I have had a quick look at your design.

For a start, water displacement doesn't care if your weights are the same density as water or the density of lead. They will still be lighter by the weight of water they displace You have already said you can balance them, even as each side changes its "weight". The weights are in effect disconnected from the water, except for the submerged portion.

As you displace more water, your water acts as if it is pushing against the displacement.... I am sure you know how it works. eg. 7B will try to push down. When this happens, 7a will rise, dropping the relative water weight, until it looks more like 8a and 8b.

So your counter-weight not only has to compensate for the difference in weight change, but also water changing levels , again changing the weight.

Fact: you can't get anything for nothing in a gravity system.

[Novus]

@ Tarsier_79

"As you displace more water, your water acts as if it is pushing against the displacement.... I am sure you know how it works. eg. 7B will try to push down. When this happens, 7a will rise, dropping the relative water weight, until it looks more like 8a and 8b.

So your counter-weight not only has to compensate for the difference in weight change, but also water changing levels , again changing the weight."

Can you have a look at attached file (not sure how to insert the file as part of the text like you did...);

A counter weight 1C is attached via a string on a pulley with twice the distance as mass 'V'
Therefore only half the weight is required to keep 'V' and 1C at equilibrium.

The orange part of mass 'V' is submerged in 2B.
Since the density of mass 'V' is equal to the density of the fluid the submerged part will float and not exert any up- or downward force.
The part of mass 'V' above the surface is balanced by the weight of 2C

Both scenario's 1 and 2 are at equilibrium.
Any in between stages between the transition from scenario 1 to scenario 2 should be at equilibrium as well.

As a result a small additional weight added to 'V' should be able to complete the transition.

Can you let me know if you believe above statements are incorrect or if you believe this is a different scenario then 7A and 7B.

"Fact: you can't get anything for nothing in a gravity system."

I totally agree.

PS - not sure how to insert your text as quotes with my reply...

[Tarsier_79]

For a quote, highlight the text you want to quote, then press the quote button.

For the image, convert or snip your excel spreadsheet into a JPG, then attach.
Also make sure the image isn't too big...

[Tarsier_79]

How does D move up and down to compensate?

The rest makes sense.

As soon as you add the second set of pulleys so the containers move up and down it adds another level of complication.