A machine to convert gravity to mechanical energy # 2

[hansvonlieven]

Yes it is.

Study hydraulics and then you might understand what you are trying to do. Any book on elementary physics will do.

You could always try the library if you don't want to spend the money.

Hans von Lieven

[AB Hammer]

This is getting silly.

People do laugh at what they don't understand. I think since you won't take Hans words of experience, and you seem to not understand hydraulics. You need to study the friction that submarines have to deal with to go through water, and then you might understand what Hans is saying.

Hans

I will be building my buoincy device sometime in the future. The designs are done.

[mondrasek]

@ Mondrasek

You said: The internal pressure of your cylinder will never exceed that of the surface atmosphere on the way down and during the expansion process.  That pressure is 14.7psi.  However, at every depth in water the pressure is greater than 14.7 psi, and doubles every ~33 feet.

This is not correct.

Pressure doubles with doubled distance.

At 10 cm depth the pressure is 10 grammes per centimeter squared, at 20 cm depth it is twice that and so forth.

The pressure is equal to the weight of the water column above. As the weight doubles, so does the pressure.

Hans von Lieven

Hans, you are quite right to correct me there.  I did not type what I meant to say well at all.

At 33 ft deep the pressure is 2 atmospheres.  At 2 x 33 = 66 ft it is 3 atmospheres (not doubled to 4 as I misstated).

M.

[brian334]

Imagine a tank that displaces about 1 cb ft of water submersed to a depth of about 10 ft.
The dimensions of the tank are about 12 in x12in x 12 in.
At the top of the submersed tank is a 62 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 62 lb lead weight can fall 11 inches before it hits the bottom of the tank.
Now imagine a piston with a surface area of 1 square inch is attach to the bottom of the lead weight.
The external water pressure at a depth of 10 ft. is about 4.6 lb/ sq in.
Could 62 lb weight overcome the 4.6 lb/sq. in. external water pressure and push the piston with the 1 sq. in. surface area out of the tank? Yes it could. How many pistons with a surface area of 1 sq. in. could 62 lb. Push out of the tank?
62 lbs. Divide by 4.6 lb/sq in = 13.4
So, as the 62 lb weight falls a distance of 11 ins. It  will  push 13.4 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 ins.
What is the volume of the 13.4 pistons?   1 sq in x 11 ins x 13.4 pistons =  147 cubic inches
So, the displacement of the tank would increase by 147 cubic inches.
Water weights .036 lbs/cu. In.
So how much less would the submersed tank weight after the 13.4 pistons were pushed out of it when the 62 lb weight in the tank  falls 11 in ?
147 cu. In. x .036 lbs/cu in = 5.3 lbs.
Now add to that the momentum  of the 64 lb weight falling 10 ft.

[AB Hammer]

brian334

It is time for you to build the design. It is the only way you will understand. If you are trying to interest other people to build it for you, good luck. I don't want to build it, for I see to many problems and don't want to work on something unless it passes a minimum of good possibilities.