Hi Mitch,

I like your low Tec gravity motor. Something that just came to me was that you could use 2 big bellows on the shore that would use the weight of the falling arms (2 pistons) to force air in the pistons that are needing air to make it rise up in the water. You could have a lot of leverage on the end rod to fill up the piston with air with very little movement of the arms and have your piston be like a collapsible chamber (like a hand organ) that would at TDC be fully extended and full of air with the bottom being able to float so it would want to come to the water surface and it would force the air back into the depleted bellow that blows it up as the weight of the piston arm fall to BDC in its travels. This would make it so you do not need any motors to make it work.


Ron in Redding
----- Original Message ----- From: "Mitch" <[EMAIL PROTECTED]>
To: "Interact" <interact@listserv.capital-master.com>
Sent: Friday, November 03, 2006 5:01 PM
Subject: [Keelynet] free energy


    Monday, October 30, 2006

Hello! Here is one machine that I have been thinking about building. On paper, this machine appears to operate over unity. I am also going to supply what I think of as 'proving numbers', verifiable by anyone. I have looked at many blueprints and descriptions of machines, but they never come with any numerical verifications. I have learned that it is worthless to postulate anything without some kind of 'proof'. I have never written a paper like this before, so please bear with me.

   I will start with some 'facts'.
One cubic foot of water weighs 62.5 lbs. Conversely, one cubic foot of air has a buoyancy ( negative gravity) of 62.5 lbs., due to the fact of displacement. These figures are used routinely to build ships and subs. For every foot of water depth, the pressure increases by .445 p.s.i., so that at 33 feet of depth, the pressure is increased to 14.5 p.s.i. over the ambient pressure. One atmosphere at the surface of the water, two atmospheres at 33 feet. A bubble of air will decrease its volume by 6% for every foot of depth due to the pressure exerted on it by the water. Pressure is omni-directional, so to calculate the pressure exerted on a given cylinder, it would be expressed as follows; the cylinder is 6 inches in diameter, with a pressure of 10 p.s.i., so in order to move the piston you will need 6" x 6" x .7854 = 28.77 sq. in. x 10 p.s.i. = 283 lbs. of force, because only the 'face' of the piston is what has to move. A bubble of air rises at approximately one foot per second.* (unverified, the info is scarce) However, It is known that it takes an amount of time for a buoyant object to rise when under water, because the water must 'get out of the way', to get under the buoyant object. A lever is a simple machine, a force magnifier, and the work it does can be expressed as follows; there is a 11 foot lever, with a fulcrum at 10 feet. 100 lbs. of force is exerted on the long end, moving the lever 10 feet. The object to be moved is one foot beyond the fulcrum, so the lever can do 100 x 9 = 900 lbs. lift for one foot, or 900 foot-pounds. This is known as a 'first class' lever. A 'third class' lever has the load between the fulcrum and the lifting force. Horse power is figured to be 550 foot pounds torque on a shaft per second, or 33,000 per minute.
  These are known facts that are used everyday.

I am proposing a machine that utilizes these facts to produce power in excess of that needed to operate. This machine will be fairly large, as it is a gravity machine, and those tend to be quite large if expected to do any useful work.

   Physical Machine and Components.
There is a water tank that is 12 feet deep, and 15 feet in diameter. The top is open. In the bottom center of the tank is cut a hole, where an 8" diameter pipe is fitted that extends vertically into the tank by 12 inches. In this pipe is a ball valve (checkvalve) that will keep the water from leaking out. Also in the bottom of the tank is a chamber that is 5 feet in diameter, and 12" tall, with a gradually rising conical top ( 1-12 pitch). This chamber has a volume of 20 cubic feet. In the center top of the chamber is a pipe fitted, with a ball valve, to keep any air in the chamber from leaking out. This pipe extends 12 inches out of the top of the chamber. The bottom of the chamber is open to the water, and this chamber is exactly centered over the hole in the tank. The chamber is placed 1 foot off of the bottom of the tank, held there by 3 studs equally spaced, and water can enter and exit this chamber freely. Connected to the ball valve in the bottom of the tank is a pipe that will carry air, and this pipe is 24 feet long, extending away and outside of the tank. This pipe is connected to a single action air pump ( simple cylinder pump). When this pump is activated, it pumps air into the bottom of the tank, through the check valve, filling the chamber with air. This pump is connected to the lever at the 21.7 foot mark, making it third class, but the lever could be extended, and the pump placed beyond the fulcrum, to make this a 'first class' connection. Also in the tank is another chamber, which I shall call the 'piston'. The piston is an exact duplicate of the chamber, with two exceptions. It has an actuator pin that directly lines up with the checkvalve in the chamber, and this piston measures 60 in. in diameter and is 18 inches deep, and can hold 29.5 cubic feet of air. This piston moves up and down in the tank, from the top of the tank to the top of the chamber. The piston has channels that it rides in, to keep it centered. The piston also has a rod connected to it that is 12 feet long, and extends out of the top of the tank, where it is connected to a crankshaft that has a throw of 10 feet, ( 5 foot radius), and is only connected to a flywheel for precise control over the piston. The fly wheel is 6 feet in diameter, weighs 300 lbs., and is geared to travel at 10 times the speed of the piston. The flywheel's sole purpose is to drive the piston down to the chamber, and hold it there while the air in the chamber transfers to the piston. Connected to the rod, just below where it connects to the crankshaft, is attached a lever that is 30 feet long, and is perpendicular to the rod. (90 degrees) At this point, the rod is slotted, with the lever riding in the slot, so that the lever can slide laterally as the piston rises and falls. This lever has a fulcrum located at its 28 foot mark, and another crankshaft at 30 feet. This is where the extra usable power will be taken. The throw on this crankshaft is .357 feet. 28' to fulcrum lever divided by a 10 foot movement of the lever at 28 feet = 0.357. This crankshaft has a power take off pulley. This whole thing is framed together, from the tank to the fulcrum, and the p.t.o.
 ____________________________________
  Theory of operation.
The piston, as it moves down in the water, is driven there by the weight of the lever, and on the way down, the flywheel is helping to drive it. When the piston gets near the chamber, the actuator pin in the bottom of the piston opens the checkvalve in the chamber, releasing the air, which flows into the piston, filling it with air. Now the piston begins to rise, pushing up on the lever, and spinning up the flywheel. The lever actuates the pump to fill the chamber, re-gauging the machine, and also operates the second crankshaft, where usable power is derived. The piston should take about 3 seconds (?) to reach the top, with a one second hold there, so the time of revolution is 8 seconds. When the piston gets to the top of the tank, another actuator pin, which is attached to the tank, opens the checkvalve in the top of the piston, releasing the air out of the top, and allowing the piston to fill with water, when it will again fall to the bottom, repeating the process. The air should readily flow upwards, into and out of the piston, due to the pressure differences, and the buoyancy of the air, and the large valve sizes.

Lets look at some numbers.
_____________________________________
Our piston measures 60" dia. x 18", and that is 60 x 60 x 0.7854 x 18 / 1728 (cubic inches, 12 x 12 x 12 = 1728) = 29.45 cubic feet. 29.45 x 62.5 = 1840.8 lbs. upward lift. But we lose 6% volume per foot of depth, and the piston starts 10 feet deep, so 1840.8, 1730.3, 1626.5, 1528.9, 1437.2, 1350.9, 1269.9, 1193.7, 1122, 1054.7, 991.5 at 10 feet, or about 50%. We will take the average of the top and bottom numbers, 1730.3 + 991.5 = 2721.8 / 2 = 1360.9, or about 75 % The piston is sized for the maximum capacity, because the air bubble will increase in size as the piston rises. The lever weighs 300 lbs. , but it is much heavier at the fulcrum end, so the weight that the piston has to overcome is 125 lbs. The lever is constructed out of aluminum tubing, like a 'truss', so its actual weight will be less than this. The fulcrum is at 28 feet, so we will calculate the weight x 26 feet, and the p.t.o. is at 30 ft., 2 feet beyond the fulcrum, so we must subtract 2 from 28, so 26 feet is our calculation point for actual weight deliver to crankshaft. The pump has a travel of 18", and is located at 21.7' on the lever, in the 'third class' on the lever. 26' minus 4.3 x 0.3357 = 1.53' @ 21.7 ft. The crank is 2 feet beyond the fulcrum, and has a throw of 2 x 0.357 = 0.714, and is first class on the lever. We are connecting to a crank shaft, however, and this means that we must use the radius of the crank to calculate h.p. 0.714 / 2 = 0.357 The fly wheel will consume about 15% of the work from the piston, 1360.9 x .85 = 1156.7 - 125 (lever weight) = 1031.7 1031.7 is what is left to pump the air and produce horse power. This figure is variable, depending on how many pistons are connected together. With many pistons running in time, a fly wheel will not be needed, as one piston's rise will force another to the bottom, resulting in more efficiency. The pump is 60" in diameter, and that is 60 x 60 x 0.7854 = 2827.5 square inches times the pressure at 10 feet, 4.45 x 2827.5 = 12582.3 lbs. of pressure to collapse the air pump. We must assume a higher pressure to pump all of the air in 3 seconds, and I will assume a 25% increase- 12582.3 x 1.25 = 15727.9 The pipe and valve are large- 8" diameter, to ease air passage. Once the water pressure is overcome, the air should move rapidly, without a huge pressure increase, due to the large size of the pipe & valve. It should almost be like popping a tire, so I think that this figure is actually too high, because once the air begins to flow, and the initial pressure is over-ridden, the pipe and valve, (or valves), allow such a volume of air through them as to offer very little more resistance.

weight lever point pump pressure lever point 1031.7 x 21.7 = 22387.9 minus 15727.9 = 6660 / 21.7 = 307. This is what is left over at the piston after running the machine has been satisfied.

weight fulcrum point 8 second revolution ( power is only derived on the upstroke) 307 x 26 = 7982 x 3 sec. rise = 23946 x 7.5 r.p.m. = 179595 x 0.357( crankshaft) = 64115.41 / 33000 = 2 horse power left over. (I have rounded, instead of printing every decimal.) When calculating horse power, every second of the minute must be accounted for, including the time that no power is produced, and this is counted as a loss. The only way to get true horsepower and torque ratings, is with the proper test equipment, under actual running conditions. If we tabulate the total power, minus the lever, flywheel, and pump, it is 1360.9 x 26 = 35383.4 x 3 = 106,150.2 x 7.5 = 796,126.5 x 0.357 = 284217.16 / 33000 = 8.6 total h.p. So, 8.6 - 2 = 6.6 hp. used for machine operation, which means that this machine seems to have a C.O.P. of about 130% in its present form. This figure does not, however, take into account for any bearing and gear losses, but these should be negligible. If there were 4 pistons and levers hooked together running in time with each other, like a V- 8, we could eliminate the flywheel and its losses, resulting in a slightly higher C.O.P. Lets see what happens with a faster revolution, say 5 seconds. 368 x 26 = 9568 x 1.75 sec. rise = 16744 x 12 = 200928 x 0.357 = 71731.3 / 33000 = 2.17 h.p. That is not much change, and it will be similar results with a longer revolution than 8 seconds. The pump pressure will change, as well, with different piston speeds, and I think that the slower the piston rises in the water column, the more horsepower created, because of the power over time plus the lower pump pressure at the slower speed, but again, this difference is small. Lets move the fulcrum to 27 ft. and see what happens. 10 ft. / 27 ft = .370, so we have to move the pump to 21.5, and the crank radius has also changed, because it is now 3 feet beyond the fulcrum, so 0.370 x 3 / 2 = 0.555 for crank radius. 1031.7 x 21.5 = 22181.55 - 15727.9 = 6453.6 / 21.5 = 300.16 x 24 feet. = 7204 x 3 = 21612 x 7.5 = 162090 x 0.555 = 89959.95 / 33000 = 2.7 h.p. . The movement of the fulcrum has a direct effect on the efficiency of the machine, to an optimal placement.
 _______________________________
The fundamental reason that this machine appears to work 'on paper', is leverage. It appears that through this leverage, it is possible to pump more air, (more weight), than the force exerted to do the work, and that air has no weight, except when in a liquid, then it negatively weighs exactly the amount it displaces, so we are moving weight that is WEIGHTLESS. We only have to overcome the pressure. This machine also uses one lever directly over another, ( the crankshaft), to magnify the forces. Now, it is obvious that many things about this idea can be changed. Bigger piston, different piston shape resulting in 'streamlining' for various piston speeds, longer lever, different class levers, reversing levers, gears, or pulley's and cables, using compressors instead of a simple pump, (with the p.t.o. powering the compressor), using two changeable pistons on one rod, etc. Also, it should not make much difference as to exactly how fast the piston will rise in the water column, because all that will change is the 'power over time' equation, and as I have shown, the total power remains about the same. If this works the way the numbers say it should, it can be scaled up to any size that is physically possible, with as many pistons and levers as there is room for. It should also work on a small scale, all you must do is balance all of the forces with the materials. It will also have to be hand started, but an automotive type starter motor can be installed with the flywheel for remote starting on a large machine. ( The pipe must be pressurized). Of course, hopefully you only have to start it ONCE, and it should run until some man-made material fails.

What if we had a huge one, with say, 10 pistons? Lets assume a 1000 cubic foot piston,( that is only 10' x 10' x 10'), and a 60 foot lever, with the fulcrum at 55 feet. 15 ft / 55 ft = 0.27. The piston can be any shape, as long as it holds the required amount of air, and has the appropriate valving. Crank radius; 0.27 x 5 = 1.35 / 2 = 0.675. 1000 x 62.5 = 62,500 lbs x .75 (pressure diff. average, at depth) = 46875 x 50 feet = 2,343,750 x 4 sec. rise(piston starts deeper) = 9,375,000 x 6 revs = 56,250,000 x 0.675 = 37,968,750 / 33,000 = 1150.5 h.p. x .25 ( to get usable h.p.) = 286.25 h.p. x 10 pistons = 2862.5 h.p. Obviously, I have extrapolated these numbers based on the first machine, but if they worked there, they should work here, too.

This will fit on any city lot, and will run how many homes? Let's use the figure from the 1000 c.f. x 10 piston machine. 2862.5 x 746 = 2,135,425 watts, or 2135 k.w., or 2.1 megawatts. What is a city lot? 120' x 80'? A quarter acre? Lets put these on ten acres- 2135 k.w. x 4 x 10 = 84,500 k.w. 84.5 megawatts of fuel-free power. Better yet, lets just use a few thousand feet of sea, or lake shore. I don't know about where you live, but up here in Alaska, my power company owns hundreds, if not thousands of acres of land. If a guy had a nice pond in his backyard...

I just had a brain storm! I just thought of another machine. Lets build a huge bellows, say 15 feet by 15 feet by 10 feet, with the collapsing and opening being ten feet, and inside this bellows is a scissor action jack, with a 100 horsepower ( or bigger) electric motor that turns the screw of the jack to open and close the bellows. Put this at the end of a 60 foot HOLLOW lever, so the air in the bellows is common with the lever. At 15 ft. deep, the motor turns and opens the bellows, at the top of the water, the motor reverses, and closes the bellows. The overall pressure in the bellows and lever is 7 p.s.i., and since the air is common with the lever, collapsing the bellows will not result in an appreciably higher p.s.i., due to the total volume overall. Or, the end of the lever could be open to the air, since it is above the water level, so the air can enter and exit at will. Any way, lets see if this will work. 15 x 15 x 10 = 2250 cubic feet, times 62.5 = 140625 lbs. x 60 feet = 8,437,500 x 3 second rise = 25,312,500 x 7.5 revs. = 189,843,750 x 0.9 = 170,859,375 / 33,000 = 5177 horsepower!! (I think I will pursue this idea a little further.) Please bear with me while I do so, as I am going to do my thinking right now, in this paper.

This idea works only in a large pool, or lake, because it is only a single long lever. There is still a rod and flywheel, as the piston's top dead center and b.d.c. must be precisely controlled, in concert with the crankshaft at the other end, but the piston is connected directly to the end of the lever, so most of the lever is always under water. I will also calculate to a commercial application. (You will have to calculate for your own needs yourself.)

There is a lever that is 65 feet long. The lever is principally a 4 foot (?, depends on motor diameters) diameter pipe, with trussing along its entire length for maximum strength. At one end is a cylinder that is 12 feet in diameter, and can expand and contract for 6 feet. The cylinder and pipe are commonly sharing air, as the cylinder's piston collapses, the air is forced out through the pipe, and as the cylinder is expanded, the air fills it through the pipe. The air pressure within the pipe and piston will always be the same as the ambient air pressure except for when expanding and collapsing. Within the pipe are mounted six 100 horse power electric motors, equally spaced along the lever's length. Through the center of the all of motors runs a threaded shaft, that is 4 inches (?) in diameter, and has threads at 2 per inch, similar to a jack screw used on a McDonnell Douglas M-D 80 jetliner to control the tailfin. This threaded shaft is connected to the piston in the cylinder. When the motors operate in unison, the screw turns, forcing the piston to either open, or close, depending on the motor direction. All of the wiring for the motors is run through the pipe, and extrude from the end without the piston, where they are connected to the power source. The cooling of the motors is accomplished by the air rushing in and out of the pipe, over the motors. At the open end of the pipe, at 60 feet on the lever ( We count from the piston), is mounted a fulcrum for the lever. At the end of the lever, at 65 feet, is mounted a crankshaft that is geared to an a.c. generator, and this generator powers the motors. The total weight of the lever and cylinder and motors is 20, 000 lbs. But, when under water, the total weight is 16, 500 due to displacement. The fulcrum end is on shore, above the water level. This apparatus is appropriately stabilized to the ground, so it will withstand the forces. Theory: As the cylinder falls to 20 (?) feet deep in the water, most of the lever is also submerged, and the total weight of the machine is reduced accordingly. When the piston is 19 feet deep, and while it is still falling, the motors begin to turn, forcing the piston wide open, to 6 feet capacity. This will take 1.1 (?) seconds to accomplish. The cylinder is 12 feet in diameter, and open 6 feet, so it is now full of 678.5 cubic feet of air. This gives the cylinder a buoyancy of 678.5 x 62.5 = 42,411.6 lbs minus 16,500 = 25911.6. The lever will now begin to rise, and will take 4 (?) seconds to reach the top of the water, where the motors will run in reverse, and close the cylinder, and the cylinders own weight will drive it down to 19' where the cycle will repeat. The six 100 horsepower motors will use 450 kilowatts of electricity to operate, or 600 h.p. Will This work for over unity? 25,911.6 x 55 feet = 1425138 x 4 seconds = 5700552 x 6 r.p.m. = 34203312 x 0.9 crank = 30782980.8 / 33000 = 932.8 h.p. But wait a minute, there is 16,500 lbs. pulling the lever down, as well as the upstroke power. 16,500 x 55 = 907,500 x 4 = 3,630,000 x 6 = 21,780,000 x 0.9 = 19,602,000 / 33000 = 594 H.P. on the downstroke. But wait again. The motors only run for 2.2 seconds per revolution, leaving 5.8 seconds that they are not operating, and not using any power. Maybe they could be run from capacitors, or batteries, and so spread their 600 h.p. usage over time? It appears that the power from the downstroke alone is enough to power the machine. ****See below to understand where I am getting these figures, that is where some of my 'thinking' took place. 932.8 + 594 = 1526.8 total h.p. created with 600 used to operate, a C.O.P. of 255%? It would seem that even if more electric motors were needed to open the cylinder in the time allotted, there is still going to be plenty of power left over. It also seems that no matter what method is used to open the cylinder, there is still plenty of power. What about a shorter r.p.m., say 7.5 seconds, instead of ten? 16,500 x 55 = 907,500 x 3 = 2,722,500 x 8 = 21,780,000 x 0.9 = 19,602,000 / 33,000 = 594. Exactly the same. Lets move the fulcrum a little, to 58 feet, with the p.t.o. still at 65', or 7' beyond the fulcrum. 20 / 58 = 0.344 x 7 = 2.4 ft throw on crank, a radius of 1.2 25,911.6 x 51 = 1321491.6 x 3 = 3964474.8 x 8 = 31715798.4 x 1.2 = 38058958.08 / 33,000 = 1153.3 h.p. (?1?!) and a little more-- to 57 feet, with pto @ 65', 8 feet beyond fulcrum, 20 / 57 = 0.350 x 8 = 2.8 ft. throw, 1.4 r. 25911.6 x 49 = 1269668.4 x 3 = 3809005.2 x 8 = 30472041.6 x 1.4 = 42660858.24 / 33000 = 1292.7 ? Wait a minute. What is happening here? Lets split the lever. 32.5. 0.625 feet movement of lever per foot @ 20 feet. Lever is split , so 20 foot movement on both sides equally. 20 ft. throw, 10 rad. 25911.6 x 3 x 8 x 10 = 6218784 / 33000 = 188.4 Fulcrum at 45' ? - .44 x 20 = 8.8 ft. throw, 4.4 rad.
  25911.6 x 25 x 3 x 8 x 4.4 = 68406624 / 33000 = 2072.9
16,500 x 25 x 3 x 8 x 4.4 = 43560000 / 33000 = 1320 1320 + 2072.9 = 3392.9 H.P. a C.O.P. of 565% (????) Fulcrum at 40 ft. 12.5 throw, 6.25 rad. 25911.6 x 15 x 3 x 8 x 6.25 = 58301100 / 33000 = 1766 .7 16,500 x 15 x 3 x 8 x 6.25 = 37125000 / 33000 = 1125 It would seem that a 45 foot fulcrum is better. Also, as the fulcrum is moved along the length of the lever, the 'at rest' weight of the lever will change somewhat. I could calculate for optimal placement, but I don't see the necessity.

*****I have a 48" pipe that is 65' long. The fulcrum is at 60 ', so the calculating point is 55'. The piston will start at 20' deep. The piston has a screw in it which is controlled by an electric motor that opens and closes the piston, to create the volume. The piston and pipe are common, with the end of the pipe that is out of the water being open to the air. The piston is 12' in diameter, and has an extension of 6'. 678.5 cubic feet x 62.5 = 42,411.6 lbs. The pipe-lever, piston, and motor have a combined under water weight of 6500 lbs. 42411.6 minus 6500 = 35911.6 x 55 feet = 1,975,138 ft lbs. 20 / 55 = 0.36 x 5 = 1.8 foot throw, and a 0.9 crank radius. 4 second rise x 6 revs.
 1,975,138 x 4 x 6 x 1.8 = 853,259,616 / 33,000 = 2585.6 h.p.
12' diameter is 144" x 144" x 0.7854 = 16286 sq. inches x the pressure at 20 feet 0.445 x 20 = 8.9 x 16286 = 144,945.4 150,000 lbs. 100 hp. motor - 100 x 33000 = 3,300,000 / 60 seconds = 55,000 ft. lbs. per sec. @ 1650 rpm (?) = 27.5 rev. per second. Screw has 10 teeth per inch.
 2.7 inch per second. 0.27 per rev.
We need 3 100 h.p. motors, or one 300 h.p. motor to get more than the 150,000 lbs. of force that is needed to open the cylinder. one 100 h.p. motor weighs 2,000 x 3 = 6,000, plus the 4,500 pipe-lever = 10,500. 42,411.6 minus 10,500 = 31,911.6 x 55 x 4 x 6 x 1.8 = 75,821,961.6 / 33,000 = 2297.6 h.p. minus 300 for motor = 1997.6 Speed. @ 2.7 inches per second, it will take 23 seconds to open the cylinder @ 10 threads per inch. Lets go to 2 threads per inch, on a 4 inch shaft, 27.5 x 2 = 55 inches per second. Also I will double the motors. 600 h.p. x 33,000 = 19800000 / 60 sec. = 330,000 ft.lbs. per second, more than enough to open the cylinder. Now the motors weigh 12,000 + 4500 = 16,500 from 42411.6 = 25911.6 x 4 x 6 x 55 x 0.9 = 30,782,980.8 / 33,000 = 932.8 minus 600 = 332.8 usable, about 1-3rd. extra. This all depends on a 10 second revolution, but the actual revolution time really does not matter, as it will be at least a few seconds. Since this system is a sealed from the water, the displacement figures are absolute, so no figuring for depth ( other than piston pressure), is required.

What about a huge compressor to operate the machine. Compressors that operate at 180 p.s.i. usually do so at about a 3 to 1 c.f.m. to horse power ratio. So, if I had a 600 h.p. compressor, it should produce 1800 c.f.m. at 180 psi. 1800 / 60 = 30 cubic feet per second, at 180 p.s.i. If we had a pipe that was 1.5 feet in diameter, and 65 feet long, it's capacity is 18" x 18" 0.8754 = 254.46 x 780" = 198486.288 / 1728 = 114.8 cubic feet x 62.5 = 7179 lbs. buoyancy. The pipe and trussing together weigh 10,000, so we have a net weight of 10,000 - 7179 = 2821 lbs. On the end of the lever we will put a chamber instead of a piston, one that will hold 210 cubic feet of air. At the chamber, there will be several large air valves that will allow air to fill the chamber from the pipe when the lever is at bottom dead center. In the operating respect, this machine will be similar to the first one I described, with the compressor running from the crank, but the pipe is totally sealed to hold air pressure, and the compressor's output is connected to the fulcrum end of the lever. this machine will have a depth of 20', with a revolution of 8 seconds. The pressure at 20' is 8.9 p.s.i. So the air bubble will be about 40% as large as it will be at 1', so 40% of 210 is 84 + 210 = 294 / 2 = 147 x 62.5 = 9187.5 lbs. average. 9187.5 - 2821 = 6366.5 net lift x 25 = 159162.5 x 3 = 477487.5 x 8 = 3819900 x 4.4 = 16807560 / 33000 = 509 on the upstroke, which is not enough. Will this compressor open a piston, instead? 144" x 144" x 0.7854 = 16286 x 8.9 = 144946 lbs. to open, 180 x 16286 = 2,931,480 minus 150,000 = 2,781,480, or easily enough pressure on the piston to open rapidly at depth. As I have already shown the figures for this lever above, I will not recalculate here. It does occur to me that a 12' piston moving 6' in one second will create one hell of a recoil. Maybe I need to make two opposing pistons, each half the size, so there will be no recoil against the lever. But, how do I now get the piston to dump all of the air and close at top dead center? It will be full of air at 180 p.s.i. Maybe there will be a very long air hydraulic cylinder that extends back into the pipe, and around the backside periphery of the main cylinder are valves that will open, and the pressure in the spring cylinder will force the piston closed, as the air is allowed to vent through these valves into the water. Of course, this will increase the opening pressure at the piston, but the pressure was so high to start with, another 200,000 lbs. shouldn't make any difference. In an opposing two piston model, the spring cylinder will have to be connected mechanically to a concentric, as will the pistons. This should work beautifully, with the compressor, or the electric motor machine. I rather like the electric motor idea, though.

Well, I think it is about time to end this exercise in theory and go build something. I hope you have enjoyed wading through this with me, and I hope that I have energized your imagination, as this is principally why I wrote this in the first place. Regretfully, I do not have any drawings as yet, but I will produce some soon. I am not entirely sure of my mathematics, either, ( I am sure that mistakes exist here), and the 'powers that be' say that this cannot possibly work. While I am not entirely pleased to splash one of my ideas all over the net, I think it is about high time that people can have energy that does not require paying a monthly bill, and is green. Since this is now 'out there', ANYONE can build his own. I am not particularly worried about 'patent stealers', because this idea is only one of many, better ideas I have, and that I will hold close, for now. PLEASE contact me with any questions or comments, and criticisms, as that is what will keep me honest.
Mitch.   [EMAIL PROTECTED]







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