Please comment on my Hydro-Atmospheric Assisted Machine

Please comment on my Hydro-Atmospheric Assisted Machine, HAAM.

Patent Applied for

Experts on aerodynamics and hydrodynamics are invented to comment on my HAAM concept as logically as possible especially if it doesn't work please prove it. If it works it will generate a revolution in aviation and energy industry.

HAAM operates on the ground and in air and water. It is a two parallel operating machine, one of them is a usual machine operates with an external input and the other is hydro-atmospheric, operates in response to the other without input. The total output of both machines should be more than the external input of the usual machine. Full details of HAAM is in the link bellow:

http://www.thrilling.me.uk/haam.html

An opportunity to make £20k or build a big business:

http://www.thrilling.me.uk/rwrd.html

You make a number of assumptions in your calculations. You show two separate flows at varying pressures meeting in one output and the two pressures adding to a higher pressure of the two. If you have 80 psi and 40 psi air pressure meeting and traveling down one tube they would equalize to a lower combination of the two not meet and increase.
if that was the case then having two lines of 80 psi meeting would increase to 160 psi, which will not happen. Everything looks for a way to balance out in equalibrium. Anything at a higher pressure will look for the path of least resistance to a lesser pressure.

Roger Brown wrote:You make a number of assumptions in your calculations. You show two separate flows at varying pressures meeting in one output and the two pressures adding to a higher pressure of the two. If you have 80 psi and 40 psi air pressure meeting and traveling down one tube they would equalize to a lower combination of the two not meet and increase.
if that was the case then having two lines of 80 psi meeting would increase to 160 psi, which will not happen. Everything looks for a way to balance out in equalibrium. Anything at a higher pressure will look for the path of least resistance to a lesser pressure.

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Many thanks for taking time to reply.

The idea of HAAM is to get atmospheric or hydraulic input without cost to get more out put.

In the figure above both fans marked by cross 'X' have the same thrust. Both take an input which have a cost. When both operate they create a lower atmospheric pressure at LP point. This lower atmospheric pressure at LP is counteracted by higher natural atmospheric pressure at HP point. This generates an extra output in addition to the outputs generated by both fans.

Sorry the figure doesn't seem loading. It is figure 9 at this url: http://thrilling.me.uk/rc5.jpg

or click here

Kind regards

When you run the fans you are creating a higher pressure on the output side due to the force of the fans and yes it is creating a lowering pressure on the input side as it sucks the air on that side towards it. The issue is that atmospheric pressure is generally 14.7 psi. So unless you use some outside force using some other power source your input will stay the same. So I don't see where you are getting the free increase.

Roger Brown wrote:When you run the fans you are creating a higher pressure on the output side due to the force of the fans and yes it is creating a lowering pressure on the input side as it sucks the air on that side towards it. The issue is that atmospheric pressure is generally 14.7 psi. So unless you use some outside force using some other power source your input will stay the same. So I don't see where you are getting the free increase.

Thanks for your comment.

The free power is the differentiation between lower atmospheric pressure at the front and natural atmospheric pressure at the back. Usually natural atmospheric pressure is 1.00kg/1 centimetrer sq on sea level or 14.7 psi as you sate. If this pressure differentiation doesn't work this concept, HAAM, will not work.

Best regards

That is my point. With the outside pressure being 14.7 you are not going to gain anything since that is the starting value anyway even with the fan turned off. Consider a tire that is inflated to 35psi. if you tried to inflate it more by sticking a air pump to it at 14.7 psi the tire would not inflate at all. In fact the internal pressure of the tire would try to force itself back towards the air tank since it is a lower PSI.

Roger Brown wrote:That is my point. With the outside pressure being 14.7 you are not going to gain anything since that is the starting value anyway even with the fan turned off. Consider a tire that is inflated to 35psi. if you tried to inflate it more by sticking a air pump to it at 14.7 psi the tire would not inflate at all. In fact the internal pressure of the tire would try to force itself back towards the air tank since it is a lower PSI.

Thank you for response.

Theoretically it sould work but this can be proved or disproved only with a protype model. I will do the prototype by myself soon. I have already got all the components. I hope I do it before next year if no one else does it.

In the figure of the link bellow both fans have the same thrust. If both have an area of 100.00 square inch air passage the area between LP and HP points will be 100 square inch too. Total atmospheric pressure on each side becomes 1470.00 pounds. Assume both fans can lower atmospheric pressure at least by 1/10th at LP point, this makes pressure at HP point 147.00 pounds higher.

http://thrilling.me.uk/rc5.jpg

Click Please

Kind regards

Take your figures to an engineer and have them check them out. Using the assumption you are using saying that you would multiple 14.7 times 100 equals 147 doesn't sound right. If that was correct then how do you explain the normal air outside? You don't multiple 14.7 by every square foot of space outside. It is 14.7 at sea level pretty much every where you go.

Roger Brown wrote:Take your figures to an engineer and have them check them out. Using the assumption you are using saying that you would multiple 14.7 times 100 equals 147 doesn't sound right. If that was correct then how do you explain the normal air outside? You don't multiple 14.7 by every square foot of space outside. It is 14.7 at sea level pretty much every where you go.

Many thanks Roger.

I think I will take your advice and go to an engineer. But first I will make the model and and test it. If it worked I would put it on youtube for every one to see.

From my understanding your 14.7 psi means 14.7 pound per square inch. So 100 square inch creates 14.7 x 100 =1470.00 pound per 100 square inch.

Best Regards


















kind regards

A better introduction to HAAM:

http://haam.thrilling.me.uk/