Hi M. - Nice video 
I know the pressure.....
One thing I didn't see you measure - and I noticed because it's what I've been struggling with today.
Your funnel is well above the head line on the input tube, have you paid a lot of attention to the head values as measured on that tube? - Sorry if I missed it in the earlier posts.

My 1U tests looked REAL good if i simply cycled the same load up and down - teeter-totter style. REALLY looked like same lever arm, same travel distance,,,,, different weights. But lift 5 lbs, remove it with the risers unable to move higher - notice that the input head doesn't change - and try to sink. Now it doesn't look so good.

With the system load staying the same it would lift and sink a total of 18.5 pounds an inch using 3 pounds of water and I could hold the head within an inch to an inch and a half - 13.5" to 15"

Remove 5 pounds at the top and I would still remove 3 pounds of water BUT I would have to drop the head to 10" to sink.

Now at 2U: I added 5 more Lbs ballast, still using a 5Lb brick as the "load" - total weight of POD / riser, ballast, load is 23.5 Lbs. 5.5" POD tube. Lift head is up to 23" ( after tweaking the setup... ) and sink without the brick is 16" head. Fluid transfer is down 1/3 to 2LBs but head differential needed is up.

Even after re-applying the 5Lb lift load the head is at 20".

Looking like the 2U is almost a step backwards taking a 2Lb lift of 3" to lift and remove 5Lb 1"

I wanted to do the build this way - 1U at a time, to be able to test and compare the addition of layers.
Will be real interesting to see what happens when I add the third - maybe tomorrow....

Dale

... OH, and for another future reference on video files.
winff is OK but the real deal is ffmpeg - it is a command line tool but there's a windows build that will convert ANYTHING - Quickly. Fair learning curve but tons of support out there. Another good tool if the cam is DVD friendly is DVDcatalyst. Love it for putting movies on my "phone" and for video conversions.

Quote from: TinselKoala on September 22, 2012, 06:09:27 PM
It would only be a paradox if the bucket of water weighed the same with the battle ship removed _and no water added_. 

That is: you put your battle ship in the weightless bucket full to the brim of water, overflowing the excess,  and weigh it. Your result is W_total.  Now draw a line on the battleship's hull right at the waterline. Remove the battleship and weigh the bucket.... now you see only the weight of the water remaining, W_{total-battleship} = W_water. Right? Now convert this to a volume and subtract that from the total volume of the bucket: that is, the result is the amount of water that the battleship displaced. Right?
Now take your battleship and calculate the volume of the hull below the waterline you marked. Guess what..... the volume of the hull in cubic centimeters is the same as the weight of the displaced "virtual" water in grams, corrected for the exact density of your water.
No paradox at all.

The second part of your statement seems to be at odds with what you said the first time, which is this: That is the part I object to, because that is NOT how buoyancy works and there is no contribution from the water depth psi to the lifting. The head of water pressure raises the air pressure which pushes up against the riser; the depth of the parts extending into the water are not affected by the water "psi due to depth" which acts in all directions equally and so cannot produce movement, but rather by the upward force due to the displaced water being heavier than the volume that is displacing it. This upward force is independent of the "depth psi" but of course decreases as the submerged volume of the riser decreases as it lifts.

TK,  Your method to calculate the lift is valid.  My method to calculate the lift is valid.  They generate the same answer.  I stand by my statements 100%.  First, for the battleship floating in a bucket of water.  Take every square inch of the hull and calculate the PSI at the average water depth for that square inch.  Add all the square inches up and guess what?  The total force lifting the battleship is equal to the weight of the ship.  The weight of the "virtual water" displaced is also the same number. A=1, B=1, A=B.

Let's take a very simple example.  Take a 1 inch cube with a total weight the tiniest bit less than a cubic inch of water.  It will float, but right at the surface.  It will have a virtual water displacement of 1 cubic inch of water weighing 0.036 pounds.  The pressure at the bottom surface one inch deep is 0.036 PSI.  The lift from the pressure differential is 0.036 pounds.  It does not matter if we calculate it with displaced virtual water or pressures at the water depth.  The answer is still the same.    The pressures at depth is the truest mathematical representation.  The buoyancy in virtual water is a geometric equivalent IMHO. 

If you want to argue with this analysis, then please do it with an actual example with real numbers.

Quote from: mondrasek on September 22, 2012, 06:48:14 PM
The video is up on YouTube now.

Mike, Very nice and clear video.  It makes it easier to understand your statements.  One thing to watch out for: When you add water to the funnel, it has one potential energy.  When you drain some water, it has a very much lower potential energy, even though it weighs the same.  You are adding energy into the system with your arm -- every time you take water out from below and add it to the funnel way above.  It is tricky.

@Webby1
Clarification: What I meant by "notice that the head doesn't change" was: I've been concentrating on the head values and using Read_Sunset's method - measure the head on a sight gage. If my load appears to be up against the stop but the head drops when I remove the brick; it ain't there yet....

I've pretty much stopped using that screw jack except for setting up a cycle.

Recent tests have been done this way:
- get the loads set and static level checked, raise that input container slowly until the load hits the stops
- monitor the sight gauge closely and maintain the head levels as low as possible.
- then close the valve below that input cylinder and pull the sight gage tube out of its holder
- slowly slide the tube down the scale and let the water flow into the measuring cup.
- I usually lower the tube about an inch at a time looking for drop and "usable recovery"
- when the whole assembly is back at the starting point I note the difference between highest and lowest heads. I also tend to keep track of how much fluid comes out at each incremental reduction in head. So far, not much recovery after removing the load.
- then I pour the water from the measuring cup back into the input and open the valve to let it lift.

The way I've been trying to track work in and out is by noting fluid transfer weights / volumes and the difference in head pressures required to achieve that transfer.

Is that any clearer?
Dale