@MIdone. I have to politely disagree with what you are saying. I could build a "black box" to do exactly what you are asking. At the bottom would be a small platform and another platform at the top. I place a 100g weight on the platform at the top.Then I place a 10g weight on the bottom platform. The 10g weight and its platform drops half an inch. The top platform instantly rise 2 inches, and is prevented from falling by a ratchet.
Does this pass your test?


The bad news is , that hidden inside the black box is the mechanism from a childs toy gun. The lower platform operates the trigger, releasing the spring which raises the 100g weight 2 inches. The even worse news is that I now have to use a force of 250 g over a distance of 2 inches to recompress the spring,replace the energy lost to friction, and reset the machine to its starting point.


So you must analyse a machine over one complete cycle, and return it to its start position.
Do you agree or disagree?

I understand what you are saying neptune.  I'm talking about everything being out in the open and about the total potential energy of the system before and after.  Whatever moves would have to be considered; they would be part of the masses. Like water levels.

But you are right that in order to convince a person who cares less about formulas, that bringing everything back to the first state, plus spitting water up is what would matter to them.

Thanks

Neptune .. MIdone is correct - in your example of the black box you would not have accounted in your energy budget for the PE of the spring & the work done joules to compress it [WD = f x d] - after system losses there would be a net loss of joules.

Parisd .. my 2 cents.

I believe this will probably turn out to be a measurement/interpretation error by Mr Wayne & his team - having said that this is the fun of doing experiments & garnering good data along the journey.

I will give you a clue where that measurement error might be occurring, IF there  is one.

It has to do with Pascal's principles & enclosed hydraulic systems - pressure is transmitted undiminished in an enclosed system & Wayne's system is a mixture of buoyancy & hydraulics, all under pressure - often pressure & force are confused which is unfortunate - so IMO a force is generated & magnified [it has to do with area ratio's] due to Pascal's principles - because it is basically a hydraulic system the force is multiplied in the system which is seen/interpreted as buoyancy potential increase to lift mass - but if, as some might know, when you take that magnified force output & add it to the weight force of the system fluids & compressed air you get a high internal force figure - when this is converted back to pressure [pgh] & then you recalculate the height potential [h] by rearranging the formula to find hypothetical 'h' i.e. Static Pressure = F / A = pgh therefore h = P / pg you may find the calculated 'h' doesn't match the heights of fluid & air in the actual system - this is an anomaly.

In short the Pascal force multiplication is added to the weight force of the system & by some math is converted to a system pressure which deduces a different 'h' than actual.

The upshot is that there appears a potential to lift virtual mass because of the force multiplier effect & the pressurized system.

The trouble is in 'normal world' that for both hydraulics & buoyancy to do work [f x d] then the Effort to Load [work done] ratio is zero sum energy wise i.e. f1 x d2 joules = f2 x d1 joules, less losses.

JMO's at this time - I watch with interest.


........................

See3d .. I don't know what formula's you are working thru but one comes to mind as a possibility.

That might be for a fluid drag formula to take account of viscosity - a simplified one I might build & use would be say ...

Coefficient of Drag 0.36 [change this factor as desired]

Viscosity Drag = Cd . 1/2 . density . velocity squared . area               i.e. Vd = Cd.0.5.1000.v^2.A

Where 1000 is density of water, v is velocity of riser etc, A is area of riser etc.

You would need to include an IF statement in the formula in your sim to account for changing directions & velocities of fluids i.e. viscosity dampening works up & down.

Here is a simple formula I quickly built & would consider using.

Input[Cd]*0.5*1000*if(Output[velocity].y1<0,Output[velocity].y1^2,Output[velocity].y1^2*-1)*0.01

Considerably simplified for a generic sim could be this stripped down one i.e. adding a viscosity dampening effect to buoyancy fairly similar to above in behaviour.

Input[Cd]*if(Output[velocity].y1<0,Output[velocity].y1^2,Output[velocity].y1^2*-1)*100

P.S. viscosity represented as a force is a small system energy loss but probably should be included for sim accuracy & to dampen oscillations.

The density figure has to be adjusted to say 1.225 kg/m^3 for air at 1 atmosphere etc - Vd = 0.36.1/2.p.v^2.A  where p = density

Quote from: see3d on August 30, 2012, 08:25:09 AM
Hi AmoLago,

I have been working hard to make sure I have all the physics details on the updated simulation mathematically correct.  There are a couple of twists that may have escaped many casual observers.  I just have one last formula to iron out before I can release it.  It is easier to know what it should be, than to translate that into a simulation formula.

1.) I just reviewed all my released videos.  I could not see any case where the riser sank while the piston rose.  Perhaps you should advance it frame by frame to make sure. 

2.) The density of the riser material is not a factor, since it is counterbalanced. 

All the videos will have to be redone with the new PDF release.

Hi Se3d,

Thanks for the update, great to hear to the sim's coming along nicely.

With regards to my point 1., maybe it's only an image rendering thing as opposed to what's actually going on, but here's what I see.

At the start of the cycle, the riser is at it's natural floating point i.e. I'm thinking that if the piston didn't move it would just sit there as is. Throughout the whole cycle, the riser is never locked or stopped in any way to restrict movement. Also no weight is added or removed.

At the start, lowest point, I can see that the riser is partially submerged, looks like about 30%. However, by half way through the cycle, highest point, it is now about 60% submerged.

So to my eyes, the riser has sunk about 30% of it's height.

Or am I just confused!?

Amo

Quote from: AmoLago on August 30, 2012, 06:55:24 PM
Hi Se3d,

Thanks for the update, great to hear to the sim's coming along nicely.

With regards to my point 1., maybe it's only an image rendering thing as opposed to what's actually going on, but here's what I see.

At the start of the cycle, the riser is at it's natural floating point i.e. I'm thinking that if the piston didn't move it would just sit there as is. Throughout the whole cycle, the riser is never locked or stopped in any way to restrict movement. Also no weight is added or removed.

At the start, lowest point, I can see that the riser is partially submerged, looks like about 30%. However, by half way through the cycle, highest point, it is now about 60% submerged.

So to my eyes, the riser has sunk about 30% of it's height.

Or am I just confused!?

Amo

Hi AmoLago,

Yes, as the piston goes up, it raises the head around the pod, which increases the output force.  When the riser hits the top stop, it can not rise any further.  However, if the piston continues to rise, the water head around the pod continues to rise and increases the pressure against the top stop.  The actual physical location of the riser should never go down as the piston is going up.