Quote from: LarryC on November 11, 2012, 11:07:58 AM

Yes it is. Good Question.

Instead of Archimedes calculation, let's calculate using the following.

Pod lift = Bottom PSI X Si - Top PSI X SI
Pod lift = (3.72 X 706.858) - (2.42 X 706.858)
918.63 = 2628.31 - 1709.68

I did these calculations on the spreadsheet to use it's internal precision for accuracy, so the numbers would calculate slightly different on a hand calculator.

So the answers are the same.

I agree this is only force. I previously showed the force over distance calculations in my System Rise Calculator and how that starting force can be maintained over the lift distance, when the proper amount of water is added between the Pod and the Pod retainer. It was very complex and the skeptics excuses then were that we can't trust these calculators because it could have mistakes or hidden manipulations.

This is a simple pod and 1 riser with no Visual Basic code and can be understood by all. More to come.

So I'll ask again 'Can anyone show why the logic used to get that result is wrong?'

Regards, Larry

The pod is displacing 919 pounds of "virtual water"; this is the buoyancy force since your pod is weightless. This is the upward force due to buoyancy of the pod alone. This is found in either of two ways: taking the difference in pressure at the top of the water and the bottom of the water (1.3 psi) and multiplying that by the area of the bottom of the pod which is a little over 706 in² to get 919 pounds force upwards, OR taking the weight of the volume of displaced water (36 x 707 x 0.0361 = 918+ lbs force).  The pod is being pushed up by 919 pounds of buoyancy force. You have this value in your spreadsheet. The pod is also being pushed _down_ by a force equal to the pressure of air acting against the top flat surface of the pod. This force, again using your numbers, is 2.42 psi x 706.5 in² = 1710 pounds of air pressure force, acting downward. Isn't it?

Quote from: TinselKoala on November 11, 2012, 11:40:40 AM
The pod is displacing 919 pounds of "virtual water"; this is the buoyancy force since your pod is weightless. This is the upward force due to buoyancy of the pod alone. This is found in either of two ways: taking the difference in pressure at the top of the water and the bottom of the water (1.3 psi) and multiplying that by the area of the bottom of the pod which is a little over 706 in² to get 919 pounds force upwards, OR taking the weight of the volume of displaced water (36 x 707 x 0.0361 = 918+ lbs force).  The pod is being pushed up by 919 pounds of buoyancy force. You have this value in your spreadsheet. The pod is also being pushed _down_ by a force equal to the pressure of air acting against the top flat surface of the pod. This force, again using your numbers, is 2.42 psi x 706.5 in² = 1710 pounds of air pressure force, acting downward. Isn't it?

Wrong, I showed the attached example back on September 23.

Let's see if you can comprehend and relate it correctly to my previous calculations.

Larry

Quote from: LarryC on November 11, 2012, 11:59:54 AM

Wrong, I showed the attached example back on September 23.

Let's see if you can comprehend and relate it correctly to my previous calculations.

Larry

Wrong. I show in my post how you get the same answer using either the differential pressure or the weight of the volume of displaced water, as you show in your example. There is no argument there. Your example does NOT show the force due to pressure acting downward on the top of the displacer, in fact you have it explicitly stated to be zero.

Now, I've worked your numbers in your example given earlier, and I'm in general agreement with the numbers you have but not the way that you are using them. I don't quite understand how you are calculating the "hydraulic lift force" though.
Consider this simplification. You simply have a solid riser of your dimensions, and a tub of water the dimensions of the outer retainer wall. This tub is filled to the level of 36 inches. Now you lower your solid riser into the tub. The outer water level comes up, and overflows, but that doesn't matter; as long as the resultant level in the outer wall chamber is 72 inches, you've displaced your "virtual water" and you have that full buoyancy force available, and it is 0.0361 pounds per cubic inch x 793.926 square inches x 36 inches = 1031.78 pounds of buoyant force. But your riser isn't solid and it has air pressure pushing it up inside. This adds 1920.27 pounds to the upward force experienced by the riser. But the non-solid riser doesn't displace the full 36 inches of water any more, it only displaces 34 inches, as some of the water goes up inside the riser and doesn't contribute to buoyancy -- it's not displaced. This brings the buoyant force down to 974.46 pounds, for a total lift force of about 2895 pounds, in broad agreement with your total lift value of 2839. The difference between what you got for your total, and what I got for my total, is accounted for by rounding errors and the estimate of 34 inches displaced.
Taking your "hydraulic lift force" value as calculated at 2716.65 and comparing it to the total force upwards experienced by the riser that I calculated, you do indeed appear to have a slight advantage of force ... less than ten percent. (2895/2717 = 1.065)
Note that I didn't even have to include any reference to a "pod" in my calculations, except by reference to your psi air pressure values.
But.... and this is the real question.... over what distance does that force advantage act?

Of course, there's another way to look at it too. LarryC/Travis/Pod lift = 2839. TinselLift, using a simple riser alone, no pod or extra ringwalls, and bubbling air into it to achieve 2.42 psi pressure inside and same outside water displacement = 2895 pounds, for an advantage of TinselLift of 2895/2839  = 1.02.

Quote from: TinselKoala on November 11, 2012, 12:46:10 PM
Wrong. I show in my post how you get the same answer using either the differential pressure or the weight of the volume of displaced water, as you show in your example. There is no argument there. Your example does NOT show the force due to pressure acting downward on the top of the displacer, in fact you have it explicitly stated to be zero.

Now, I've worked your numbers in your example given earlier, and I'm in general agreement with the numbers you have but not the way that you are using them. I don't quite understand how you are calculating the "hydraulic lift force" though.
Consider this simplification. You simply have a solid riser of your dimensions, and a tub of water the dimensions of the outer retainer wall. This tub is filled to the level of 36 inches. Now you lower your solid riser into the tub. The outer water level comes up, and overflows, but that doesn't matter; as long as the resultant level in the outer wall chamber is 72 inches, you've displaced your "virtual water" and you have that full buoyancy force available, and it is 0.0361 pounds per cubic inch x 793.926 square inches x 36 inches = 1031.78 pounds of buoyant force. But your riser isn't solid and it has air pressure pushing it up inside. This adds 1920.27 pounds to the upward force experienced by the riser. But the non-solid riser doesn't displace the full 36 inches of water any more, it only displaces 34 inches, as some of the water goes up inside the riser and doesn't contribute to buoyancy -- it's not displaced. This brings the buoyant force down to 974.46 pounds, for a total lift force of about 2895 pounds, in broad agreement with your total lift value of 2839. The difference between what you got for your total, and what I got for my total, is accounted for by rounding errors and the estimate of 34 inches displaced.
Taking your "hydraulic lift force" value as calculated at 2716.65 and comparing it to the total force upwards experienced by the riser that I calculated, you do indeed appear to have a slight advantage of force ... less than ten percent. (2895/2717 = 1.065)
Note that I didn't even have to include any reference to a "pod" in my calculations, except by reference to your psi air pressure values.
But.... and this is the real question.... over what distance does that force advantage act?

Of course, there's another way to look at it too. LarryC/Travis/Pod lift = 2839. TinselLift, using a simple riser alone, no pod or extra ringwalls, and bubbling air into it to achieve 2.42 psi pressure inside and same outside water displacement = 2895 pounds, for an advantage of TinselLift of 2895/2839  = 1.02.

Love the way you spin the facts, to make it seem like you were correct all along.

But no issue, I'm content to receive this statement 'Now, I've worked your numbers in your example given earlier, and I'm in general agreement with the numbers you have but not the way that you are using them. '

FYI, The way I'm calculating and using the numbers is the same as the Team of engineers at HER, but i'm sure, you still think you know better.

Correct, the percentage is less than 10 as this is only a one riser.

More simple calculators to come.

Regards, Larry

As usual, Larry, you fail to address the MAIN POINT: I obtained the greatest of all three lift numbers without resorting to consideration of "pods" or multiple ringwalls. Is there something wrong with MY math?

All that I required was a riser open on the bottom and a means of bubbling air into it, and standard buoyancy of a single moving part.

And you have failed to tell me how you calculated your lesser, "hydraulic lift" value that you compare with.

But yet, you accuse ME of .... well, whatever it is you are attacking me for now.

I think you've got a problem. Why is MY result even greater than yours?