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Gravity powered devices => Gravity powered devices => Topic started by: Low-Q on April 22, 2018, 10:29:27 PM
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Hi there!
I found this video on youtube. I just love the concept - and the guy is somewhat funny :-)
https://www.youtube.com/watch?v=wI7j6YYZ8-I (https://www.youtube.com/watch?v=wI7j6YYZ8-I)
Can anyone see why it cannot work, given no leakage if the machine is filled with water?
There are seemingly no surfaces that can counterforce the area where the wheels are narrowing and shorten the buoyant tubes. Due to the straight uniform tubes, it cannot be sideways forces that prevent the wheels to decrease the buoyant part of the tubes.
The tubes have greater volume and surface area in touch with water at the farthest position than the closes position. So buoyancy would be greater on that side right?
Water pressure is the same on both sides.
This is almost as difficult to explain as how boiled spagetti is possible to suck in your mouth even if there is seemingly no directional forces pushing it in.
Vidar
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Thanks for sharing this brain twisting variable... full of possibilities .
of course it works.....Until ?
I guess it wouldn't matter if it was a smaller leaky test bed .Or would it ?
and yes ...Mercury ?[his mention ]
he does seem a fun guy
respectfully
Chet K
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Thanks for sharing this brain twisting variable... full of possibilities .
of course it works.....Until ?
I guess it wouldn't matter if it was a smaller leaky test bed .Or would it ?
respectfully
Chet K
This has bugged me all night. I guess the 3D printer have some work to do quite soon.
Just make a small 20cm diameter model with ballbearings. Even if there is leakage, I can let the tapwater keep filling the tank. This is just to prove or disprove the concept. Nothing more.
I'll start the production of parts today.
Vidar
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yes
I lost some sleep too with this one
maybe he could take it to a fire dept and ask them to test run their water hoses in the parking lot ,should fill up fast enuff to compensate for all the leaks.
those fire guys luv to play with their hoses ...and test the equipment.
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I thought maybe some sealed "bellows" would do the same job. I assume, since the two wheels are fixed, there is no force that will compress them more at the bottom than the top.
The pressure on their bellow surface should cancel out in both directions. What I assume, is that there is no water pressure that will try to shrink the bellows. And therefor no counterforce anywhere.
However, there is greater volume where the bellows are stretched out, and less volume where they are compressed.
An open "airway" inside the wheels allows air to flow around from bellow to bellow.
Submerging this thing under water will probably cause more buoyancy on one side than the other.
What do you think?
Vidar
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i think it takes energy to expand the bellows at the bottom.all the gain is lost on reset.
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i think it takes energy to expand the bellows at the bottom.all the gain is lost on reset.
That crossed my mind too. On the other hand, that force is perpendiculary to rotation because the pressure outside each bellow is angular to the surface of the discs and can seemingly not force counter rotation.
Just one way to find out.
I print out bellows with some flexible filament I have, and see what happens 🙂
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If one end of the tube were a solid ball fitting into a socket on the wheel and the other wheel had a ball and socket with the tube going through the ball you would only need to seal the one side,, that is the ball and socket and the sliding tube.
Just a thought,,
Yeah. That would be an approvement. I think the biggest concern is the leak around the circumference of the large disc.
If just the device could pump the water leakage back into the "tank". If that device worked, and the leak is small, it could be possible.
However, what bugs me most is that I have no clue why it shouldn't work. Thinking hard, and disturbing my day job too.
I hate puzzles like this ;D
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What about replacing the tubes with flexible air ducts.
The membrane on these are very flexible, but will they shrink in length by applying pressure to the surface, or will they just try to squeeze together without getting shorter?
I want a design that is somewhat rigid in the diameter, but flexible in length.
Putting a plastic tube that fits nicely inside these, and the leakage and squeezing problem is solved for that part. The plastic tubes can be long enough to get through and some more, just like in the video.
Then use a much larger air duct to seal space between the discs. I'm not sure how the buoyancy will work if the water is trapped inside, and rotating with the tubes.
Vidar
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Would be good if he could take some measurements around the clock/wheel
so the displacement disparity and variables can be shown on paper
most likely a good Simulator could tell us the raw potential of this scenario
prior to losses
just for inspiration,
I absolutely believe it will show a gain...and if not ... maybe find solutions or??
a real brain twister with many possible considerations.
respectfully
Chet K
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Hello everybody !
as I see the device - 2 disks, located at some angle to each other.
each disk rotates on its axis.
N pipe pairs are fixed between the discs.
in each pair of pipes one slides inside the other through the ring seals (minimum - 2) with minimal friction.
from the other end these tubes are SEALANT, and fixed through the ball joints on their disks.
all pairs of pipes are connected together by flexible tubes for free flow of air - the pressure in all pipes is constant.
This design does not need side walls - it just needs to be immersed in water.
with respect.
p.s. sorry for my english - translation by Google Translate
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I think the calculation for buoyancy is easy. Generally it is the volume in front vs the rear that counts. However, without the ends being inside the water, maybe the tubes aren't buoyant at all???
It should, somehow, be something that obeys conservation of energy in this device. So what is it? A "true" physicist would probably answer that question in a second, by saying that energy must be conserved.
I have twisted my mind, and will try to analyze the device this way (And hopefully get the sleep I need):
The volume that is submerged into water, is the volume that the pipes is displacing. So far so good.
However:
Since the wheels will shape a triangle, seen from above, there is greater surface area in front of the pipes than behind them. Since there is no ends to the pipe that is inside the water, the water pressure will push more on that side with the greater area - against rotation. Because the greater area is pointing towards the longest pipes, and the pressure is greater at the bottom. The correlations makes sense this far.
On the other hand, these pipes, as each of them might be more buoyant in front of them (because the front half of the pipe is lighter than the rear half) will also try to rotate WITH the wheel, in the right direction, and possibly cancel out the difference in the surface force between front and rear - I'm talking about each pipe separately.
As I write this, my printer is making parts for a device that is similar. Actually, I will make two models. One model with the bellows, and one replica of the device in the video. The drawback with these models, is that they are small with just a few grams of buoyancy. The flexible filament I use, do have resistance that will be hard to fight against. If I squeeze the material/bellow, it takes some time for it to recover into initial shape (and the first sample got destroyed due to poor print quality). The other drawback is the leakage in the other model, but that one will have much less resistance. Finding a model that have no leakage and little friction is something I'm working on.
Vidar
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Well
your image in Post #4 seems to show not half but maybe 1/3 disparity between
up side and down side.
if we were using gravity instead of buoyancy the lift/weights [100 lb for example when "large"
potential of the wheel would be huge [fantasy land musings]
Yes open pipes but ...Hhmmm
going to have to ponder your Triangle [small side comments,hasn't sunk in yet]
Please forgive me if I post too much here,had eye surgery early this AM and can only look down for a week or two [during healing]
I appreciate the distraction and am totally smitten with all the whatifs
I promise not to make a pest of myself ,but will be doing lots of thinking
Chet
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Here is the first bellow. Sloooow printing and 0.1mm layerhight did the trick.
Vidar
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No problem Chet. Post as much as you want in this thread. Everything that contribute to solve this mystery is welcome :)
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I was asked to put in a message from #magneat, as he for some reason got a moderator message when he tried to post here.
Just so everybody knows:
I have not put in any moderations to this thread. Every contribution and questions are welcome.
Here is the message from Magnetman:
"Hello everybody !
As I see the device - 2 disks, located at some angle to each other.
Each disk rotates on its axis.
N pipe pairs are fixed between the discs.
In each pair of pipes one slides inside the other through the ring seals (minimum - 2) with minimal friction.
From the other end these tubes are SEALANT, and fixed through the ball joints on their disks.
All pairs of pipes are connected together by flexible tubes for free flow of air - the pressure in all pipes is constant.
This design does not need side walls - it just needs to be immersed in water.
with respect."
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Nice to see you thinking old friend. What filament are you using for the bellow?
Also agree with Magnetman's comment to connect the bellows's to each other on the the other side of the wheel so the whole thing can be submerged.
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Nice to see you thinking old friend. What filament are you using for the bellow?
Also agree with Magnetman's comment to connect the bellows's to each other on the the other side of the wheel so the whole thing can be submerged.
Thanks mate :-)
The bellows are printed in a flexible material that I bought on ebay. Works great for small go-pro camerahousings for drones as a soft "rubber" protection, but it is a little stiff and resistand for bellows. I need to change the shape a little to make it more flexible in length, and stiffer in the diameter. I also have an even more flexible filament. Nearly as flexible as silicone, but that filament is very very slow to print with. As the feeding motor easily will bend the filament before it goes into the heated nozzle, there must be very slow feeding to push that filament out of a 0.4mm nozzle. I can try that, but it takes 3-4 hours to print a small model.
Vidar
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Many concepts flows throug my mind. Here is another approach that might be useful as a "normal" gravitywheel or as buoyancywheel.
The light blue parts can be steel springs or hollow tubes.
The arched shape is, because of the angle of the wheels, forced to point forward/horizontally, where there is least tension.
The green parts is attached to the wheel with bearings.
The orange/red parts is hinged to the green parts so the flexible light blue parts can make an arched shape corresponding to where on the wheel they are.
The yellow arrows indicates the axis for the two wheels.
The idea is to make an effortless distribution of mass around so one side is (not heavier) generating more torque than the other side.
How effordless it is, I can't tell before it is built and tested. Because all we "know" is that mass that goes up and down the same distance cannot generate energy.
I just want to learn - the engineering way :-)
Vidar
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Edit
I see you added an image while I was plunking
comment below not for Image [but does raise a displacement question [same lift potential in image collapsed or open ??]
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The big squeezy hand of pressure is omnipresent
a bellows in the pressure hand wants to collapse with depth.. yes
but will not reopen when you want it to refill with air regardless of ambient vents
the vents must over come the pressure hand to force him to open his grip
they must be under air pressure [linear with depth]
overcoming linear to depth pressure issues are huge energy suckers...
sorry if I am not understanding your path forward [your work around for balloons or bellows]
one thing too[I know it can be engineered around]
Floppy saggy bellows will hoop up with buoyancy .
internal support chassis ?
Delve Spectrum's original design avoids this but ??
here again for newcomers
https://www.youtube.com/watch?v=wI7j6YYZ8-I (https://www.youtube.com/watch?v=wI7j6YYZ8-I)
respectfully
Chet
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I think too that the bellows will collapse a little, therfor I think it is important that the bellows are rigid enough in the diameter to not collapse, but rather let the wheel squeeze it instead (or mainly).
Not shure if that makes any difference, but in order to replicate a stiff pipe/tube, but at the same time make it possible to change its volume, I thought that maybe a bellow would do the same or similar job.
The counterforce is depending on how much the bellows collapse, and what force required to do so.
A 20cm diameter wheel, will cause 20 grams/cm^2 more pressure at the bottom than the top. So the collapse will try to pull the bellow together, and force the bellow in the wrong direction.
The problem with bellows, is that they work in the same way as air motion transducers which is used as expensive tweeters on high end loutspeakers. It is a flat "crinkled" surface that expand and contracts as current flows through inside a magnetic field. So any deformation of a bellow will displace the volume outside and inside of it. That is a problem.
Vidar
Edit
I see you added an image while I was plunking
comment below not for Image [but does raise a displacement question [same lift potential in image collapsed or open ??]
------------------------------------
The big squeezy hand of pressure is omnipresent
a bellows in the pressure hand wants to collapse with depth.. yes
but will not reopen when you want it to refill with air regardless of ambient vents
the vents must over come the pressure hand to force him to open his grip
they must be under air pressure [linear with depth]
sorry if I am not understanding your path forward [your work around for balloons or bellows]
one thing too[I know it can be engineered around]
Floppy saggy bellows will hoop up with buoyancy .
internal support chassis ?
respectfully
Chet
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Hi there!
a pair of device options based on the picture from the post "Reply # 20 on: April 24, 2018, 01:16:46 PM"
respectfully
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@magneat,
I have thought about the idea you suggest in your previous post. So I am on my way designing the parts for this.
The only "concern" is the stress on the bearings at the wheel. Because these weights wants to fall down, and twist the red arms, and the green parts with them, and twist the bearings out of the wheels. I'll make a light weight model first with only the bearings as weight (Blue parts).
First I want to test the springs. I have lots of stainless steel wire I easily can make the springs I need. Then I can make many of them, and they don't need bearings at the wheel. Just need to glue them in place, since the springs easily can roll around their axis even if they are arched, and keep their arched shape in one direction that increase and decrease with rotation. The tension on the top and bottom half on the wheel will cancel out and leave behind only the vertical displacement.
Vidar
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Something like this @magneat?
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@Vidar
Yes something like that.
but, IMHO, the opposite loads (floats) should be directed as in the picture from the post
"Reply # 23 on: Today at 06:49:18 AM" - either towards each other, or vice versa.
and such pairs should be several - that would pass "dead points".
Your concern about the load on the bearings is in vain - you can always do
correct design calculation.
the main task for the model is to show that the very principle is a worker (self-rotation).
for a "correct" gravitational engine, the power should be much higher than
for "water" - weights can be made from a material with a high density (lead density -
11,34 kg / dm³, density of water - 1 kg / dm³, winning - 11,34 times)
and given the greater resistance to movement of disks in the water than in the air, the gain will be greater.
to quickly build a cheap model (to test the principle), you can take a couple of bicycle wheels in assembly (with tires).
then fasten the knots with bearings to the wheel will be convenient with the help of metal clamps with
screws (used in cars) - 2 yokes per 1 knot.
wheels from bicycles will not suffer - after dismantling the model can again be used for its intended purpose. :D
p.s. I still care about the question of "synchronization of rotation" of the wheels.
respectfully
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I don't want to ruin your day, but I got one question. Why would you ever want to start your OU investigation with a device that has friction? It seems to concept and principle of capturing excess energy from the ambient is somewhat alien to us, so I would suggest planning it as efficient as possible from the start. No moving parts, no friction, no air drag and purest copper to avoid massive copper losses.
Then when you have a idea you refine it to take into account Lenz law and other nasty shit. (Good video on how to remove Lenz https://www.youtube.com/watch?v=w9MiEJ6KI8 )
I was just watching a Joseph Newman documentary and this guy probably has found something, but he is also an angry&bitter old man set in his ways. He says you need mass to produce the OU effect. I think he can harvest extra energy EVEN when he uses massive multi ton rotors. I think he has mistaken the flywheel effect for the source of his OU
So why are you trying to "run 100m record under water"? Just because there is water near you and you own scuba gear?
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@Belfior,
on your own link a black screen
respectfully
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original Contributor here
https://www.youtube.com/watch?v=wI7j6YYZ8-I (https://www.youtube.com/watch?v=wI7j6YYZ8-I)
TinMan doing some tests here
https://www.youtube.com/watch?v=KHxxhzh6kRc (https://www.youtube.com/watch?v=KHxxhzh6kRc)
and potential energy available in Brad's chart below
this one keeps a lot of folks awake at Night.
Belior...
no stone left unturned
respectfully
Chet K
PS
any comments on why this should not work [what are we missing??
PLEASE contribute....here or at TinMan's You tube link above
not worried about friction seals or leaking ATM just issues with the theory ??
???
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I don't want to ruin your day, but I got one question. Why would you ever want to start your OU investigation with a device that has friction? It seems to concept and principle of capturing excess energy from the ambient is somewhat alien to us, so I would suggest planning it as efficient as possible from the start. No moving parts, no friction, no air drag and purest copper to avoid massive copper losses.
Then when you have a idea you refine it to take into account Lenz law and other nasty shit. (Good video on how to remove Lenz https://www.youtube.com/watch?v=w9MiEJ6KI8 (https://www.youtube.com/watch?v=w9MiEJ6KI8) )
I was just watching a Joseph Newman documentary and this guy probably has found something, but he is also an angry&bitter old man set in his ways. He says you need mass to produce the OU effect. I think he can harvest extra energy EVEN when he uses massive multi ton rotors. I think he has mistaken the flywheel effect for the source of his OU
So why are you trying to "run 100m record under water"? Just because there is water near you and you own scuba gear?
If a machine can run from excess energy, or any other energy sources, friction is not a problem. The goal must be to create a machine that can do some useful work. The machine does not know if it is friction or useful work it is doing. So friction is actually important to introduce into the design - more or less.
I still don't believe in excess energy, but the best concept I've seen this far is the video in post #1. This is the machine I want to test very first.
The other concepts that I've drawn are lifting weights up and down the same distance. Even if they look over balanced, they aren't. Because the rear side, where the wheels are closest, and the mass is closest to the center, the wheels circumference must lift that mass the same distance anyways. So in a way, the wheel is in balance. It will be a nice piece of artwork though. I will build it, and place it on my desktop at work :-)
Vidar
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This design https://www.youtube.com/watch?v=wI7j6YYZ8-I (https://www.youtube.com/watch?v=wI7j6YYZ8-I) change buoyancy in front and rear of the wheel.
One counterforce/countertorque is the difference in surface area on the front half side of each pipe and rear half side. Pressure will push more on the greater surface. Say you have 1 bar pressure at the top, and 2 bar pressure at the bottom, and the front side of the pipe is 100cm2 and the rear side is 90cm2, then the difference in force on the highest pipe is 10kg, but 20kg at the bottom. This difference in surface area must correspond to the angle between the wheels, and also corresponds to the difference in submerged volume. Do my conclusion seem right?
Vidar
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This design https://www.youtube.com/watch?v=wI7j6YYZ8-I (https://www.youtube.com/watch?v=wI7j6YYZ8-I) change buoyancy in front and rear of the wheel.
One counterforce/countertorque is the difference in surface area on the front half side of each pipe and rear half side. Pressure will push more on the greater surface. Say you have 1 bar pressure at the top, and 2 bar pressure at the bottom, and the front side of the pipe is 100cm2 and the rear side is 90cm2, then the difference in force on the highest pipe is 10kg, but 20kg at the bottom. This difference in surface area must correspond to the angle between the wheels, and also corresponds to the difference in submerged volume. Do my conclusion seem right?
Vidar
I will make this post here,as i see you are working on it as well Vidar.
The upward force of each pipe is the weight of the displaced fluid.
With fresh water,if your tube displaces 1 liter of water,then the generated upward force is 1KG.
Once again ,the diagram shows my calculated results from my build so far.
Brad
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[/size]I will make this post here,as i see you are working on it as well Vidar.The upward force of each pipe is the weight of the displaced fluid.With fresh water,if your tube displaces 1 liter of water,then the generated upward force is 1KG.Once again ,the diagram shows my calculated results from my build so far.Brad
Actually, the torque will be less than the upward force because the pipe doesn't move vertically. It is 30 degrees between the tubes.
4.1kg is correct
But then (right hand side):
3.06kg
1.75kg
0 at the bottom
Total 8.91kg angular force
Left hand side:
1,07kg
0,905kg
0,66kg
minus 2.64kg angular force.
in total 6,28kg angular force.
Add the torque from the 75mm diameter tube itself due to greater volume on right hand side than left hand side.
Then substract horizontal force, going to the left due to the greater surface area on the right hand side of the tubes.
How do this add up at the end?
Don't forget to calculate torque in NewtonMeter at 1 meter radius for every part you calculate.[/size]
I would guess, you end up in zero, but I haven't got time to calculate this yet.
Vidar
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Actually, the torque will be less than the upward force because the pipe doesn't move vertically. It is 30 degrees between the tubes.
4.1kg is correct
But then (right hand side):
3.06kg
1.75kg
0 at the bottom
Total 8.91kg angular force
Left hand side:
1,07kg
0,905kg
0,66kg
minus 2.64kg angular force.
in total 6,28kg angular force.
Add the torque from the 75mm diameter tube itself due to greater volume on right hand side than left hand side.
Then substract horizontal force, going to the left due to the greater surface area on the right hand side of the tubes.
How do this add up at the end?
Don't forget to calculate torque in NewtonMeter at 1 meter radius for every part you calculate.[/size]
I would guess, you end up in zero, but I haven't got time to calculate this yet.
Vidar
The torque on the shaft from each tube will actually be greater than the upward force,due to the leverage distance between the vertical point of each tube in relation to the center of the shaft.
Torque = force x distance
If we take for example the top tube on the active side,with a lifting force of 4.1KGs,at a distance of 400mm from center of tube to center of shaft,the torque at the shaft is 16.08 newtons,or 11.86 ft-lb.
Hope that helps.
Brad
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Calculating the area of the counter acting cylinder part is not that difficult actually because the area is a triangle if you unwrap it. All you need is the circumference of he tube and the angle the wheels are positioned at. However something tells me this sum will tend to go to zero as well ;) .
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It will not work due to water pressure
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a better path forward would be
how can we make it work ....despite water pressure issues ??
http://news.mit.edu/2013/hydrophobic-and-hydrophilic-explained-0716 (http://news.mit.edu/2013/hydrophobic-and-hydrophilic-explained-0716)
and there are OTHER attributes of water not really shared by gravity .
here some test Data
https://arxiv.org/ftp/arxiv/papers/1304/1304.1485.pdf (https://arxiv.org/ftp/arxiv/papers/1304/1304.1485.pdf)
Foil shapes ....which involve boundary layers .. laminar flows Etc etc
maybe the tubes have a useful [less drag]foil in the Middle..on the downside.
and then there is "ION charge" which can be stored in the fluid and put to work ?
attraction /repulsion ....
a very deep rabbit hole where many things need a good looking at...
IMO
respectfully submitted
Chet K
and a Big PS on foils
Brad reminded ...
the foils which drove the google test unit strait into the wind at 2 times windspeed
against the wind
prior to that....you were nutty to even consider such ??
things like FRICTION were given as reasons for much expected failure......
now its old news and well understood ::)
lets be nutty....as Hoppy says
its what we do in the asylum...
until there is a discovery!!!
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It will not work due to water pressure
Yes, you're right. That's what we figured out. Not that I didn't knew in advance that this device could not work, but now we've figured out the technical reason. That horizontal pressure gradient was not obvious at first untill we analyzed the shape on the pipes inside and between the wheels.
And it doesn't help making a notch in the wheel to make pipes with equal surface on front and rear, because that notch would have an area facing forward, and will be that extra surface that prevents the decive from running.
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I was thinking: How is it possible to make those front and rear surfaces equal? What about deviding these pipes into portions of solid material, like hockey pucks, with some space between them and angle parallell to the wheels, so the total surface for each puck is equal on front and rear. If each wheel are 15° angled, and angle each portion so the front and rear surface is the same?
But then we struggle with the sideways pressure to force those pucks in and out of the water...
Vidar