Container 'A' consists of tubes with different diameters (4,6 en 8 squares). Container 'B' is a tube with a diameter of 6 squares.
Container 'A' is equal in weight to container 'B'
Both containers are at the bottom connected via a small tube.
Both containers are at the top connected via strings on a pulley.
Mass 'M' is a solid cylindrical object which loosely fits in the smallest diameter of container 'A' (e.g. slightly smaller than '4')
Blue squares represent a fluid.
Green squares represent the part of mass 'M' which is not submerged in the fluid.
Orange squares represent the part of mass 'M' submerged in the fluid.
On the left side the blue squares for 'A' and 'B' are counted and added up on the top resulting in a total nbr. of squares of fluid of 372
For the weight calculation of 'A' and 'B' de portion of 'M' submerged (orange) is added, as per Archimedes, to the volume of fluid.
Counterweights on the pulley cancel the weight differences between 'A' and 'B' (60 and minus 60)
The 3 positions (1A/1B, 2A/2B en 3A/3B) are each at equilibrium, therefore a small additional force applied to the right side of the pulley will transition 1A/1B clockwise via 2A/2B to position 3A/3B.
The end result is that mass 'M' is submerged in the fluid while maintaining the same level of fluid in containers 'A' and 'B'
Attached scenario2 is basically the same as scenario1 from the previous post.
The only difference is the addition of a solid cylindrical mass 'V' which is suspended above and which loosely fits in container 'B'
Containers 'A' and 'B' and the solid cylindrical masses 'M' and 'V' could be constructed from clear acrylic with a specific density of 1.19 in which case the fluid could be a salt water solution with the same density of 1.19
Attached scenario3 starts with the ending position 3A/3B of the previous scenario2
Container 'A' and 'B' remain in the same position (pulley is kept in fixed position)
Mass 'V' with the same density as the fluid is slowly submerged.
The portion of 'V' which is not submerged (green) is kept in balance with a changing counterweight and is therefore weightless/exerts no force (how this can be achieved will be expained later)
Each position 3A/3B, 4A/4B, 5A/5B is at equilibrium therefore a small force at mass 'V' will complete the transition from 3A/3B to 5A/5B
As a result the volume of fluid in container 'A' will move upwards.
Mass 'M' in container 'A' which has the same density as the fluid will float upwards.
Attached scenario4 starts with the ending position 5A/5B from scenario3.
Mass 'V' remains in the same position. Container 'A' moves upwards and container 'B' moves downwards.
Each position is at equilibrium as a result of the counterweights on the pulley. As a result a small force on the left side of the pulley will transition 5A/5B anti clockwise to 6A/6B and 7A/7B
Mass 'M' is 'pulled up' against gravity untill no longer submerged (which is cancelling out the positive gravitational force from scenario2 where mass 'M' was submerged in container 'A' )
Attached scenario5 starts with the ending position 7A/7B from scenario4 as per the previous post.
The pulley is fixed and 7A/7B, 8A/8B and 9A/9B stay in the same position.
Mass 'V' is moving upwards whereby the portion which is not submerged (green) is kept in balance at each stage with a counterweight (how this can be achieved will be explained later) As a result the not submerged (green) part of mass 'V' should be considered as being weightless.
Position 7A/7B, 8A/8B and 9A/9B are each at equilibrium. A small negative force on 'V' will complete the transition.
Mass 'M' is fixed at the top of container 'A'
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.
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.
@ 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...
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...
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.
QuoteHow does D move up and down to compensate?
Container D does not move up or down but stays in a fixed position.
The change in fluid level in D is as a result of C moving upwards (as per the string on the pulley) and the fluid level equalizing through the connecting tube at the bottom of C and D.
QuoteAs soon as you add the second set of pulleys so the containers move up and down it adds another level of complication
This is only for the example and is not how the part of V not submerged in the design will be balanced.
Everything you have drawn is easy enough to reproduce if you have the will to do so.
Here is a link to one of my last buoyancy tests.
https://www.besslerwheel.com/forum/viewtopic.php?p=177387#p177387
I have attached a pic below.
I then up-scaled the principle to a new model
Unfortunately I have no skills and would need to commision for someone to build this.
Besides I believe someone should be able to disprove this design based on the details provided.
I'm waiting for the login I've requested for the BW forum to be activated to have a closer look at your buoyancy tests.
Your build of the scaled up model looks really impressive...
I shouldn't have said buoyancy, more displacement.
You can make your structures out of layers of polycarb, don't use acrylic, because it will crack if you are not used to it. Glue them together with super glue. It is clear, so you can trace the shape from an enlarged graph paper model or printout. Clear tubing is cheap from the hardware store, as are fittings designed for irrigation. You can use the same method to build your pulleys. Alternatively if you are more computer savvy, you can get all or some of the parts 3d printed.
If you can simplify your principle to build a simple test jig, that may give you the answer you are looking for.
Thanks for your detailed suggestions, however I've come to accept that I'm really clusmsy when trying to build something.
You are of course right that from trying to build the design (even in a simplified version) I should be able to learn first hand where the mistake lies, however I'm afraid that, based on previous experience, the build would be so poor that the end result would only be wasted time and frustation.
Anyway, without an actual attempt at a build, it should probably not be hard to disprove the design for some of the forum members with an in dept knowledge about physics (in particular about mechanics and buoyancy)
Hi Tarsier,
Thanks in advance for looking at this design again.
To summarize;
Scenario 2 - mass 'M' is submerged in container 'A' while keeping the fluid at the same level. Each interval of the moving pulley is at equilibrium.
Scenario 3 - mass 'V' is submerged in container 'B'. As a result mass 'M' floats together with the rising fluid to the top position in container 'A'. Each interval is at equilibrium. The pulley is kept fixed/locked i.e containers 'A' and 'B' can not move.
Scenario 4 - Mass 'M' is pulled against gravity at the top while the fluid remains at the same level. Each interval of the moving pulley is at equilibrium.
Scenario 5 - mass 'V' moves upwards in container 'B'. Mass 'M' remains in the upper position thus gaining potential gravitational energy. Each interval is at equilibrium. The pulley is kept fixed/locked i.e containers 'A' and 'B' can not move.
Since each (small) interval is expected to be at equilibrium the potential gravitational energy of mass 'M' should be sufficient to overcome inertia en friction to complete the cycle.
I'll try to answer any questions you may have as clearly as possible.
Attached Excel workbook containing all scenarios
Novus, I am still not sure exactly what you are trying to achieve.
I have been looking at scenario 5
On an initial glance, Your calculations look correct.
The calculation you do not have is the apparent weight of the V weight out of the water. 24, 144, 312 consecutively. This is where you have a method to counterbalance the V weight?
So you move the counterbalanced V weight, achieving a different weight left and right in the containers?
The M weight is when we unlock the containers to move them?
QuoteThe calculation you do not have is the apparent weight of the V weight out of the water. 24, 144, 312 consecutively. This is where you have a method to counterbalance the V weight?
That is correct, the portion of V out of the water should be regarded as weightless at each interval.
QuoteSo you move the counterbalanced V weight, achieving a different weight left and right in the containers?
The counterbalanced weight V is moving up and downwards in order for the fluid level in container A to move up and downwards. The weight in containers A and B will change as a consequence of the submerged (floating) part of V.
QuoteThe M weight is when we unlock the containers to move them?
Not really sure what you mean. Mass M is submerged at the lower position in container A as per scenario2 (free of charge, without changing the fluid level), as a next step M floats upwards with the rising fluid level as per scenario3, followed by being extracted against gravity out of the fluid in scenario4 (free of charge, without changing the fluid level), and finally stays at the top position in scenario5 where the fluid level drops and as a result gaining (free) potential gravitational energy.
OK, Im beginning to understand I think.
1. Shows you dropping mass M into the liquid, while maintaining balance with a counterweight, and maintaining fluid level.
2. Is the same, but with added V weight.
3. The containers are locked, and the V weight is lowered, the M weight is raised without effort.
4. Extracting the M weight from the fluid.
5. Raising the V weight, dropping the fluid level below the M weight.
6. Extracting energy from the M weight.
OK. This is not a 5 minute thing to examine. It is quite a complicated setup. Right now I am a little time poor. I am going to need some time to get my head around the entire setup, and confirm everything.
To be continued...
I couldn't have explained it any better.
The critical part is that the fluid levels remain the same at the bottom and top positions therefore not impacting V.
At the same time the total volume of fluid is the same at each interval and the weight difference changes from + 60 and -/- 60 at the bottom position to -/- 60 and + 60 at the top allowing for the counter balances on the pulley to keep everything at equilibrium.
OK, so at least I am on the right track. I was confused at the start mainly because they are called scenario, which I took as meaning 6 different mechanisms.
Probably since english is not my first language.
Hopefully you or someone else can point out the error in the design so we can put this to rest.
Hi Novus
I did have a little time to look at your design today.
One thing I did notice is that in scenario 3 and 5, your V weight is not counterbalanced as it enters the liquid. Your counterbalance relies on an equal rise in fluid compared to the relative drop in the V weight. I haven't had time to calculate how much of an effect this will have.
Hi Tarsier,
Thanks for looking into this.
QuoteOne thing I did notice is that in scenario 3 and 5, your V weight is not counterbalanced as it enters the liquid.
As I mentioned before, I didn't as yet share the idea on the counterbalance to keep V weightless.
QuoteYour counterbalance relies on an equal rise in fluid compared to the relative drop in the V weight.
The counterbalance doesn't rely on an equal rise in fluid compared to the relative drop in the V weight.
QuoteI haven't had time to calculate how much of an effect this will have.
I believe you will find that when using the method to counterbalance V as per the 'Example' given earlier (with the 2nd pulley at twice the distance from the pivot point) you will find a loss equal to the gain in potential gravitational energy of mass M.
Hi Novus
Quote
As I mentioned before, I didn't as yet share the idea on the counterbalance to keep V weightless.
Quote
I believe you will find that when using the method to counterbalance V as per the 'Example' given earlier (with the 2nd pulley at twice the distance from the pivot point) you will find a loss equal to the gain in potential gravitational energy of mass M.
So you have an additional counterbalance for the V weight?
I have been thinking about the V counterweight. I believe there is a method to counterbalance it with a variable proportional weight.
In #3, the water level raises 12, then 12. The V weight is dropped 8, then 16.
In #5, the water level drops 12, then 12. The V weight is lifted 8 then 16.
The leverage occurs in reverse on the lift and drop. At least the total lift and total drop heights are the same.
A pure mechanical variable solution will be difficult, but probably possible. A simpler (at least for me) method is to have two counterbalance that can occur at different times. We can engage each one when required with a microcontroller. I believe your system already requires one to swap between "scenarios".
Hi Tarsier,
As it turns out my idea for a counterbalance to keep V weightless at each interval doesn't work.
I'll have a look at you proposal but, since we already agreed that a buoyancy/gravity perpetuum mobile does not and can not exist, I somehow doubt this will work.
I guess we agree that, if only we can find a counterbalance for the part of V at each interval which is not submerged, the system should work.
Novus, the solution you proposed initially is unique and interesting. Although I believe a gravity solution is not possible, I try to keep an open mind to an extent, and won't mind finding out if I am wrong.
I believe your method of counterbalance can work in proportion. Both proportions separately.
QuoteI believe your method of counterbalance can work in proportion. Both proportions separately.
I'm working on a solution which looks promising as per your suggestion. If it works I'll be forwarding the details later today/tomorrow.
Hey Novus, were you able come up with a solution for counterbalancing?
I tried sharing a detailed comment but it got held up for moderation. I removed details partly in hopes that my post would go through, but it made no difference. The gist is that I think it's possible to construct a device vaguely similar to yours - having a goal of raising/lowering the fluid level in order to float a buoyant mass upward in a communicating vessels container, locking it in place, then dropping it like shown in 9A -> 1A.
If you haven't come up with a solution for counterbalancing, have you considered using a spring arrangement and determining a spring constant (https://sciencing.com/spring-constant-hookes-law-what-is-it-how-to-calculate-w-units-formula-13720806.html)?
The thing is, I did some math and I don't think there exists a spring constant that would work for your device due to the unusual shape of your "A" container. Also I think that, if using a counterbalancing spring, it'd be easier to raise/lower the B container rather than trying to counterbalance a "V" mass (which acts heavier as it becomes further immersed in the fluid).