Mathematical Analysis of an Ideal ZED

[TinselKoala]

How many teeth are on the two gears, Webby? Did you forget that gears are circular levers?

Or that the moment arm (horizontal component) of your weights changes as the straight levers change their angles with the horizontal?

[MarkE]

Right, see this is where we are NOT on the same page.

I view the water as a weight falling down, the CoM will fall down 1\2 the height of the column, that fall will be able to raise the other column CoM up that same height.

That is equalization.  Calculate the energy that you have in each column when you start.  Calculate the energy that you have at that point.  It is less.  That is a mathematical fact.

The problem is that allowing the water to flow down and push the other up wastes 1\2 of that potential, that is not a necessary condition.

Come up with whatever mechanism you like then and describe how your mechanism is able to escape the condition.  Hint: Manipulating potential energy will not solve the problem.

If I use all the potential from the top cup of water out of that column to lift the bottom cup of water up one cup height and let the top cup fall all the way down to do that, that is the wasted stuff.


Just in case you missed this picture.

Those are 2 identical weights on 2 different length lever arms. These arms are balanced with each other at the horizontal position.

Pick one weight up

IE apply external work

and the other goes down, let go and they go back to this same position, that is within the constraints this testbed works with.

One weight falls and picks the other weight up instead of allowing gravity to just accelerate the falling weight.

That is called a see-saw.

I have other methods that allow me to use a varying rate of change between to items, so they can start out at say a 1:3 ratio go to a 1:1 ratio and then a 3:1 ratio with no gear changes, kind of like a CVT.

But like all CVTs the range of ratios is limited to finite values.  You cannot get infinite versus zero or vice versa.

[MarkE]

What is it you are saying?  I am reading that you are trying to say that 1\2 the energy is lost when 2 weights are on a teeter-totter, and can therefore not be rotated or anything but dropped.

I actually made one that was termed an IVT, because it could be turned down to zero,, I argued that that was not correct because at some point it would just stop moving the output, but not as a function of 1:0 rather the tolerances within the system would not allow any motion transfer.

Do you understand what a state is?  It is a specific condition.  In order to empty one column you must traverse a state where it is half empty.  In order to fill the other column from empty you must traverse a state where it is only half full.  We know that if the half empty and full states coincide that one half the internal energy has been lost.  So, all you have to do to avoid the 50% energy loss is transfer all of the incompressible fluid from one column to the other without having coincident 50% full states.   Given that the fluid is incompressible you have to put it someplace.  If as you claim you do not "directly" transfer it from one column to the other, then you have to go through an intermediate store.  You will then have to insure that you don't run up against the same problem as with the two columns by themselves. 

The equation that describes internal energy stored in the first column as it empties is:

E₁ = 0.5*(P_{1_START} - K*V_{1_REMOVED})*(V_{1_START}-V_{1_REMOVED})

So, if you are going to try and hold onto as much energy as possible, you will need to deposit the removed fluid into something that exhibits an energy gain that closely complements the column loss equation.  Then whatever mechanism you come up with will have to turn around and deliver the fluid however buffered to the second column that has an internal stored energy equation of:

E₂ = 0.5*(P_{2_START} + K*V_{2_ADDED})*(V_{2_START}+V_{2_ADDED})

This intermediate store will have to lose energy with a function that closely complements the energy gain in the second column.  Don't forget to include the energy expended by whatever "assist" mechanism that you will need to finish filling the second column.

Good luck.

[MarkE]

Thank you for the equations and explanation.

Simple thought experiment for you to consider.

If I take 1 cup out of the raised side and use it to take 1 cup from the lowered side and raise it up to the same height that the first cup was removed from, and this continued for each cup and level,, what is the loss.

Mostly the loss is to your sanity.  How do you remove 1 cup from an empty vessel?  You have a cup at the top of the filled column.  Remove it.  Specify where it ends up:  If it goes into the empty column, it falls to the bottom.  Suppose you had for example some sort of raised platform with a receiving container where that container falls as water enters from the first column.  Then initially the bottom of that receptacle could be just under the top level of the filled column.  Guess where the bottom of your platform is and the water levels are when you have removed 50% of the water from the first column. 

Since in this thought experiment there is no change in CoM height from start to finish, the left side went down and the right side went up the same amount there is no loss.

If you think so then draw it and the truth will be revealed.  Of course the big laugh here is that we have been going around in circles for months with you claiming that you built something that doesn't suffer the N*(X/N)² problem.  Yet, you never manage to actual supply diagrams of what you claim to have built, nor how you supposedly measured the energy in and out to compute the still way under unity 75% efficiency you claim.

In the view you are presenting the highest potential from the raised side is moving the same distance as the lowest resistance from the lowered side, thus the loss due to the disparity of force and its usage.

Is English a second language for you?  The equations succinctly and completely describe the problem.  Absent externally applied work:  Fluids flow from higher elevations to lower elevations.  Fluid flow from a higher elevation to a lower elevation results in a loss of potential energy.

I understand and accept these conditions when these conditions are used.

I also understand and accept that a pendulum converts the force of gravity into an internal potential that is returned.

A frictionless pendulum is able to convert PE from stored on one side of the swing to the other by virtue of the intermediate conversion to KE.  Without that conversion to KE the pendulum would simply find its lowest PE state and stay there.

I also understand that an external resistance can be applied to oppose the force of gravity.

Why do you insist on misuse of terms?  Resistance is an electrical term.  If you mean an opposing force, then say 'opposing force' or more simply: force.

I also understand that in any conversion process work is being done.

More specifically:  the Second Law of Energy condemns all real processes to being lossy.  However, we do not need to rely upon the Second Law of Energy to expose Wayne Travis' / HER's / Zydro's false claims for the bull shit that they are.

I also understand that any packet of f*ds can be changed into any other packet of f*ds providing that they equal the same value.

Sweetener packets are found in restaurants.  Data packets are found in telecommunication networks.  Energy is the integral of F*ds.  Energy / matter may be changed in form.  It may neither be created nor destroyed.  When it is in the form of waste heat that reaches equilibrium temperature with the environment it cannot be converted to another useful form.

[Gabriele]

Hi to all.Webby1 have you filmed the experiment? Can we see what you have produced? Thanks