A machine to convert gravity to mechanical energy # 2

[brian334]

HOW MUCH WOULD THE TANK EXPAND?
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 64 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 64 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 4.3 lb/ sq in.
Could 64 lb weight overcome the 4.3 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 64 lb. Push out of the tank?
64 lb. divide by 4.3 lb/sq in = 14.9
So, as the 64 lb weight falls a distance of 11 in it will push 14.9 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 in.
What is the volume of the 14.9 pistons? 1 sq in x 11 in x 14.9 pistons = 163 cubic inches
So, the displacement of the tank would increase by 163 cubic inches.
Water weights .036 lb/cu. in.
So how much less would the submersed tank weight after the 14.9 pistons were pushed out of it when the 64 lb weight in the tank falls 11 in ?
163 cu. in. x .036 lbs/cu in = 5.8 lb.
So what we can conclude is that a 64 lb weight falling a distance of 11 in can increase the displacement of a tank submersed in 10 ft of water by 5.8 lb.
The next question is what if the 64 lb weight falls 10 ft instead of 11 in ?
Would the falling 64 lb weight expand the tank by 10 ft x 5.8 lb/ ft = 58 lb. No it would not. The tank expanding 58 lb would be the maximum possible.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.
So how fast would the 64 lb fall? For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water. Drag is based on the amount of liquid moved. When the tanks fall as a continuous column there is almost no liquid being moved. At the top and bottom the tanks are pulled along by the draft of the preceding tanks.
The maximum increase in tank displacement caused by the 64 lb weight falling 10 ft is
58 lb. What if we only got a measly 50 %?   58 lb divided by 2 = 29lb   Than  29 lb - 3 lb wt of the tank = a increase in the displacement of the tank 26 lb/cycle. We could probably do better than 50 % of the maximum.

[pequaide]

When your freefalling mass strikes the bottom of the tube it will share its momentum with a certain mass of water. The water moved will probably have a greater mass than the falling object; this mass ration would depend upon the experimental design.

Lets say your falling mass is one kilogram moving 10 m/sec, that would be 50 joules of energy. As it strikes; the one kilogram shares its momentum with (lets say) 9 kilograms of water.  Now we have 10 kilograms moving 1 m/sec, this is from the Law of Conservation of Momentum which has never been violated. 10 kilograms moving 1 m/sec is only 5 joules of energy. (1/2mv²)  You can not expect 5 joules of energy to do more work than 50 joules.

You are doing the problem backwards. You should start with 10 kilogram moving 1 m/sec and give all the motion to one of those ten kilograms. Then you would have a ten fold energy increase.

[mondrasek]

HOW MUCH WOULD THE TANK EXPAND?
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 64 lb lead weight. Say the lead weight at the top of the tank is 1 inch thick.
That means the 64 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 4.3 lb/ sq in.
Could 64 lb weight overcome the 4.3 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 64 lb. Push out of the tank?
64 lb. divide by 4.3 lb/sq in = 14.9
So, as the 64 lb weight falls a distance of 11 in it will push 14.9 pistons with a surface area of 1 sq. in. out of the tank a distance of 11 in.
What is the volume of the 14.9 pistons? 1 sq in x 11 in x 14.9 pistons = 163 cubic inches
So, the displacement of the tank would increase by 163 cubic inches.
Water weights .036 lb/cu. in.
So how much less would the submersed tank weight after the 14.9 pistons were pushed out of it when the 64 lb weight in the tank falls 11 in ?
163 cu. in. x .036 lbs/cu in = 5.8 lb.
So what we can conclude is that a 64 lb weight falling a distance of 11 in can increase the displacement of a tank submersed in 10 ft of water by 5.8 lb.

However, your tank design requires that the expandable piston(s) be on both ends.  In your drawings on your web page you show the weight nested inside these hollow expandable pistons of ~144 square inches surface area.  In this theoretical model you show that you are only able to expand 14.9 square inches.  I have asked you to show a diagram of a design of a tank where you have the two sets of 14.9 square inch pistons on each end and the weight inside.  If you try you should see that the weight cannot be at one end so as to be able to fall 11 inches.  There are pistons on the top side of your tank that are in the way of your weight being there.  Another consideration is that you have your pulley system that requires your weight to move at a 2:1 ratio to the piston(s).

This theoretical model you keep repeating does not represent a model of the tank you need to run your entire cycle.  You are only analysing a model of a tank that can sink and then rise again.  Just like AllCanadian's falling torpedo, it could go down one time and then rise again.  Nothing magical there.  But your model eliminates the features you require for the system to cycle continuously.

The next question is what if the 64 lb weight falls 10 ft instead of 11 in ?
Would the falling 64 lb weight expand the tank by 10 ft x 5.8 lb/ ft = 58 lb. No it would not. The tank expanding 58 lb would be the maximum possible.
The maximum impact velocity for a object falling 10 ft is about 17 mi/hr.

The maximum impact velocity *for an unrestricted object accelerating due to gravity near the surface of the Earth* is about 17 mi/hr.  Your tanks are restricted by the drag of water.  Your statement about 17 mi/hr impact velocity is not relavant.

So how fast would the 64 lb fall? For 8 ft the tanks fall in a continuous column, when the tanks fall as a continuous column there is almost no drag with the water.

This has been repeatedly explained as not true/possible.

Drag is based on the amount of liquid moved.

This has been explained as not true.

When the tanks fall as a continuous column there is almost no liquid being moved.

It has been repeatedly explained that your tanks cannot fall (or rise) as a continuous column if they are to accelerate. 

M.

[brian334]

The initial velocity of the falling tank is not zero.

[brian334]

The initial velocity of the falling tank is not zero.

The tanks would be feed into the top of the machine at the same rate as they rise.