Oxygon
Although you design is very scientific, it seems likely you need to build a simple test.
The weight of water is not enough to lift and squeeze.
You have an engine with intake and exhaust, but you are not igniting the fuel, you are just dumping the fuel in the cylinders and hoping it will have enough weight at 10* after top dead center to push down on the piston.
Oxygons Earth Day PMM
Moderator: scott
re: Oxygons Earth Day PMM
JB Wheeler
it exists I think I found it.
it exists I think I found it.
re: Oxygons Earth Day PMM
Oxygon, compliments! your idea is very interesting.
I would like to try in practice this model very much but unfortunately it is difficult to build. To your place I would connect every bags of water to the diametrally opposite bag. Without this trick maybe there is too much resistance in the compressing bag because of the pressure of the water of all the connected full bags. If you connect one bag to the diametrally opposed bag the weight of the full bags (left side) could be able to pump the water from the low bag to the high one.... maybe.
An other good solution is to put the pipe of water in the high part of the bag (when the bag is in its high position and empties) so that the weight of water in the bag in filling does not create too much resistance. Doing this way the resistance (the effort) is always only the weight of the water in the pipe and the weight of the bag in compression. This system is very slow but I really think that it works !!! and revealed here the mystery of Bessler ? Optimum work Oxygon! Good luck
Edit: and obviously using good bearings in every bag so as minimize the friction in the position of the fixed curved shape.
Edit2:
hope that also with my very bad English you will understand what here am saying.
Paul
I would like to try in practice this model very much but unfortunately it is difficult to build. To your place I would connect every bags of water to the diametrally opposite bag. Without this trick maybe there is too much resistance in the compressing bag because of the pressure of the water of all the connected full bags. If you connect one bag to the diametrally opposed bag the weight of the full bags (left side) could be able to pump the water from the low bag to the high one.... maybe.
An other good solution is to put the pipe of water in the high part of the bag (when the bag is in its high position and empties) so that the weight of water in the bag in filling does not create too much resistance. Doing this way the resistance (the effort) is always only the weight of the water in the pipe and the weight of the bag in compression. This system is very slow but I really think that it works !!! and revealed here the mystery of Bessler ? Optimum work Oxygon! Good luck
Edit: and obviously using good bearings in every bag so as minimize the friction in the position of the fixed curved shape.
Edit2:
hope that also with my very bad English you will understand what here am saying.
Paul
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Last edited by Paul on Tue Apr 25, 2006 4:49 pm, edited 4 times in total.
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ken_behrendt
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re: Oxygons Earth Day PMM
Oxy...
I do not think that replacing the water filled drum with a "universal hub" will make that much difference aside from making the final device lighter in weight.
The problem occurs during the 22.5° of CCW rotation when the bottom bellows is supposed to contact the entrance of the ramp and be closed. During this time, one bellows full of water will be pumped up to an opening bellows near the top of the water wheel through the universal hub. However, one must ask oneself how much water is simultaneousl descending in the filled bellows on the left side of the wheel.
In order for your device to produce a net CCW torque that would keep it in rotation, you would have to have more than one bellows full of water descending during that critical 22.5° of CCW rotation than is pumped up from the bottom closing bellows to the top filling bellows.
Although there are seven filled bellows descending during the 22.5° of CCW rotation of the left side of the wheel, in actuality they each only descend a distance less than the diameter of the wheel. This means that, for computational purposes, only a single bellows full of water really descends for a distance equal to the diameter of the wheel as a single bellows full of water is pumped up to the top opening bellows through the universal hub.
In other words, the amount of water flowing to the top of your device would exactly equal the amount descending in the filled bellows on the left side. If this is the case, then the system can not produce any net energy to overcome bearing friction and air resistance and thereby keep itself in motion. And, of course, there would be no extra energy to perform external work. If the system is not outputting some net energy, then there will be no net torque present in it.
ken
I do not think that replacing the water filled drum with a "universal hub" will make that much difference aside from making the final device lighter in weight.
The problem occurs during the 22.5° of CCW rotation when the bottom bellows is supposed to contact the entrance of the ramp and be closed. During this time, one bellows full of water will be pumped up to an opening bellows near the top of the water wheel through the universal hub. However, one must ask oneself how much water is simultaneousl descending in the filled bellows on the left side of the wheel.
In order for your device to produce a net CCW torque that would keep it in rotation, you would have to have more than one bellows full of water descending during that critical 22.5° of CCW rotation than is pumped up from the bottom closing bellows to the top filling bellows.
Although there are seven filled bellows descending during the 22.5° of CCW rotation of the left side of the wheel, in actuality they each only descend a distance less than the diameter of the wheel. This means that, for computational purposes, only a single bellows full of water really descends for a distance equal to the diameter of the wheel as a single bellows full of water is pumped up to the top opening bellows through the universal hub.
In other words, the amount of water flowing to the top of your device would exactly equal the amount descending in the filled bellows on the left side. If this is the case, then the system can not produce any net energy to overcome bearing friction and air resistance and thereby keep itself in motion. And, of course, there would be no extra energy to perform external work. If the system is not outputting some net energy, then there will be no net torque present in it.
ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
re: Oxygons Earth Day PMM
sorry to say this ken but if you were so analytical and precise also with your drawings you would save quite a lot of time :-)) no offense intended
Edit:
Ken wrote:
Paul
Edit:
Ken wrote:
Ken what you mean in this point ? on the descending side why you consider only the weight of one water bag ? If I am not wrong the torque is given by the difference of weights of the two side of the wheel: full bag on the left side and empty bag on the right side. correct ? therefore we must evaluate if this torque is sufficient to pump all water from the lower bag to the upper bag. if there is sufficient torque to pump water it will work; it would certainly turn very slowly but it would work.Although there are seven filled bellows descending during the 22.5° of CCW rotation of the left side of the wheel, in actuality they each only descend a distance less than the diameter of the wheel. This means that, for computational purposes, only a single bellows full of water really descends for a distance equal to the diameter of the wheel as a single bellows full of water is pumped up to the top opening bellows through the universal hub
Paul
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ken_behrendt
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re: Oxygons Earth Day PMM
Paul asked:
If this is the case, then the descending water will not be able to provide enough torque or energy to pump the water into the filling bellows at the top of the wheel and also provide energy to keep the wheel in rotation and perform other external work.
The design is deceptive because it leads one to think that it is so out of balance that it must turn CCW. That would surely happen if some way could be found to pump the water from the bottom emptying bellows to the top filling bellows without the need for the bottom bellows to contact the entrance of the right side wraparound ramp so that its contents could be forced into the attached "universal hub". That contact, however, dramatically changes the energetics of this device and, basically, renders it unworkable.
I wish that it was possible for me to make a quick WM2D model that would demonstrate this, but such models are very hard to even approximate on WM2D. The closest analogy to Oxy's Earth Day Water Wheel that I could find was a failed model for a device that I designed last year and which, upon first glance, looked like it must work. However, later analysis showed it was useless. The device was called the "Spacer Ball Injector Device" and created a momentary stir on the board. I've attached its image below.
In the device, the blue balls are metal heavy spheres and the yellow balls are light weight plastic "spacer" spheres. The idea was that the semi-circular right side guide containing only the blue metal sphere was obviously heavier than the central ascending column of mixed metal and spacer balls and, therefore, the device's right side descending column of blue metal balls should remain in constant motion.
Unfortunately, the device can not work because, although the central column of mixed balls does weigh about half of what the descending column of blue metal balls weighs, the central column of mixed balls must ascend at about twice the velocity that the descending circular column balls move. This meant that all of the gravitational energy released by the descending metal balls would be used to keep the mixed stream of balls in the central column rising. When one figured in friction, this device would not even be able to produce enough energy to keep itself in motion. It would be a "non-runner".
I think that Oxy's device will fail for much the same reasons. It does not matter whether one is trying to lift a column of mixed metal and spacer balls or microscopic water molecules. The energetics of the device would prevent it from operating as he envisions it would.
ken
I think that in this device, the amount of water flowing up to the top bellows during any 22.5° of CCW rotation will exactly equal the amount of water that descends in the filled left side bellows. The net amount of water that descends is not 7 bellows full of water...it is really only one bellows full. Why? Well, 7 bellows full of water each dropping through 22.5° on the left side of the device is only equal to one bellows full of water dropping a distance equal to the diameter of the wheel which is also the distance that the one bellows full of water rises when it is pumped from the bottom of the wheel to the filling bellows at the top of the wheel....on the descending side why you consider only the weight of one water bag ?
If this is the case, then the descending water will not be able to provide enough torque or energy to pump the water into the filling bellows at the top of the wheel and also provide energy to keep the wheel in rotation and perform other external work.
The design is deceptive because it leads one to think that it is so out of balance that it must turn CCW. That would surely happen if some way could be found to pump the water from the bottom emptying bellows to the top filling bellows without the need for the bottom bellows to contact the entrance of the right side wraparound ramp so that its contents could be forced into the attached "universal hub". That contact, however, dramatically changes the energetics of this device and, basically, renders it unworkable.
I wish that it was possible for me to make a quick WM2D model that would demonstrate this, but such models are very hard to even approximate on WM2D. The closest analogy to Oxy's Earth Day Water Wheel that I could find was a failed model for a device that I designed last year and which, upon first glance, looked like it must work. However, later analysis showed it was useless. The device was called the "Spacer Ball Injector Device" and created a momentary stir on the board. I've attached its image below.
In the device, the blue balls are metal heavy spheres and the yellow balls are light weight plastic "spacer" spheres. The idea was that the semi-circular right side guide containing only the blue metal sphere was obviously heavier than the central ascending column of mixed metal and spacer balls and, therefore, the device's right side descending column of blue metal balls should remain in constant motion.
Unfortunately, the device can not work because, although the central column of mixed balls does weigh about half of what the descending column of blue metal balls weighs, the central column of mixed balls must ascend at about twice the velocity that the descending circular column balls move. This meant that all of the gravitational energy released by the descending metal balls would be used to keep the mixed stream of balls in the central column rising. When one figured in friction, this device would not even be able to produce enough energy to keep itself in motion. It would be a "non-runner".
I think that Oxy's device will fail for much the same reasons. It does not matter whether one is trying to lift a column of mixed metal and spacer balls or microscopic water molecules. The energetics of the device would prevent it from operating as he envisions it would.
ken
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- This, too, looked workable at first glance.
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
re: Oxygons Earth Day PMM
...what?Oxygon, you're still using an incorrect drawing.
I quartered off segments...
the bellows are on center...
these bellows are not mechanical assemblies, we have to deal with pressure variables here...
The fluid vessel can be varied in form from hub or conduit, its the diametric nature that was the true origin of this idea...
I will include the evolution from original idea...
which was a "free-floating" vessel inside of a shaft within a wheel or 180 see-saw type design...
When the vessel fell upon itself it displaced its mass upward... from bottom to top...
I believe this is the best idea I have seen...
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"A man with a new idea is a crank until he succeeds."~ M. Twain.
re: Oxygons Earth Day PMM
Some varied designs include...
the first has a middle fluid weigth to help facilitate the exchange...
The second utilizes two fluid weights and each end which also serve as "bellow/reservoirs"...
the third uses an inward protruding cone to further the exchange with greater differences in mass exchange... this also can use a middle fluid weight...
though I dont believe it would be necessary...
all these designs of course can be combined or varied...
these of course are different remember from the "Oxywater wheel" in that they are free-floating vessels...
As in the wheel of chambers... the vessels exchange their masses and certain angles that are determined by the direction of rotation...
In the "OxyVessel" image the (One) vessel between 12 and 3 and 6 and 9 would be in transition...
Also I've added an animation...
the first has a middle fluid weigth to help facilitate the exchange...
The second utilizes two fluid weights and each end which also serve as "bellow/reservoirs"...
the third uses an inward protruding cone to further the exchange with greater differences in mass exchange... this also can use a middle fluid weight...
though I dont believe it would be necessary...
all these designs of course can be combined or varied...
these of course are different remember from the "Oxywater wheel" in that they are free-floating vessels...
As in the wheel of chambers... the vessels exchange their masses and certain angles that are determined by the direction of rotation...
In the "OxyVessel" image the (One) vessel between 12 and 3 and 6 and 9 would be in transition...
Also I've added an animation...
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"A man with a new idea is a crank until he succeeds."~ M. Twain.
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ken_behrendt
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re: Oxygons Earth Day PMM
Oxy...
The problem with the "OxyVessel" design is that as the water is pumped to the top of the wheel by a sinking plunger, the CG of the plunger will sink. This will result in a situation that places the CG of the water pumped to the left side rim left of the axle, but the CG of the plungers will wind up on the right side of the axle. Result...the composite CG of both water and plungers will come to rest below the axle.
ken
The problem with the "OxyVessel" design is that as the water is pumped to the top of the wheel by a sinking plunger, the CG of the plunger will sink. This will result in a situation that places the CG of the water pumped to the left side rim left of the axle, but the CG of the plungers will wind up on the right side of the axle. Result...the composite CG of both water and plungers will come to rest below the axle.
ken
Last edited by ken_behrendt on Wed Apr 26, 2006 4:35 pm, edited 1 time in total.
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
re: Oxygons Earth Day PMM
[add.:]... and if the plunger is too light, it will not be able to move with the result the water will stay below.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Oxygons Earth Day PMM
with response to "ME"... about the plunger not being heavy enough to displace...
I believe I covered that with the varying designs above utilizing the vessels shape as a added weight or using a larger resivoir bellow...
I don't believe I understand what you're saying, the CG is left and right?
The CG is left in a uncontrolled version...
the "sinking" of the plunger as you call it can be controlled by various measures... to actuate at vertical point.
To help...
try to imagine the design as frozen water...
static tests (solid state) reveal the design to be overbalanced!
Get it...
If the plunger operates as shown... whalla!
Every attempt I have made to build a "plunger" have failed... The seal keeps breaking with glue and tape, pipe clamps, etc...
Vary the switch angle as much as you want... the imbalance it undeniable...
((I have been told I should elaborate on what I mean by static test))
When I say "static test" I mean a test where the weights or imbalances are tested at certain points of known operation... giving you a degree of imbalance for that frozen position.
For example if you were to take the image below and divide up its frames... those would be "static frames"
The test would be the imbalance of the wheel as in the frozen frame.
I have added a page on my website to further help detail its design and operation...
http://oxygon.googlepages.com/oxyvessel
also images and explainations are given...
I believe I covered that with the varying designs above utilizing the vessels shape as a added weight or using a larger resivoir bellow...
The problem with the "OxyVessel" design is that as the water is pumped to the top of the wheel by a sinking plunger, the CG of the plunger will sink. This will result in a situation that places the CG of the water pumped to the left side rim left of the axle, but the CG of the plungers will wind up on the right side of the axle. Result...the composite CG of both water and plungers will come to rest below the axle.
I don't believe I understand what you're saying, the CG is left and right?
The CG is left in a uncontrolled version...
the "sinking" of the plunger as you call it can be controlled by various measures... to actuate at vertical point.
To help...
try to imagine the design as frozen water...
static tests (solid state) reveal the design to be overbalanced!
Get it...
If the plunger operates as shown... whalla!
Every attempt I have made to build a "plunger" have failed... The seal keeps breaking with glue and tape, pipe clamps, etc...
Vary the switch angle as much as you want... the imbalance it undeniable...
((I have been told I should elaborate on what I mean by static test))
When I say "static test" I mean a test where the weights or imbalances are tested at certain points of known operation... giving you a degree of imbalance for that frozen position.
For example if you were to take the image below and divide up its frames... those would be "static frames"
The test would be the imbalance of the wheel as in the frozen frame.
I have added a page on my website to further help detail its design and operation...
http://oxygon.googlepages.com/oxyvessel
also images and explainations are given...
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"A man with a new idea is a crank until he succeeds."~ M. Twain.
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ken_behrendt
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re: Oxygons Earth Day PMM
Oxy wrote:
Below is attached a quick Paint sketch that shows what the final static equilibrium orientation for your wheel would be.
I do agree with you that trying to actually construct this device is not easy in light of the requirement for a water tight, yet movable, seal between the plungers and the cylinder walls.
ken
I have seen designs similar to the your OxyVessel wheel that used regular sliding weights instead of flowing water and sinking plungers. But, the principle is basically the same and, like those, I do not think your wheel would work.I don't believe I understand what you're saying, the CG is left and right?
Below is attached a quick Paint sketch that shows what the final static equilibrium orientation for your wheel would be.
I do agree with you that trying to actually construct this device is not easy in light of the requirement for a water tight, yet movable, seal between the plungers and the cylinder walls.
ken
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- This is the orientation it will assume IF the plungers are heavier than the water in each cylinder.
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
re: Oxygons Earth Day PMM
Have you gone mad?
I dont mean to sound so harsh but you must be missing the idea here...
the bellows and liquid conents are sealed...
first of all - both bellows cannot contain mass at the same time... and given any previous rotation the weight either rest at one end or the other...
And the wheel as shown in "your" image is entirely top heavy... ???
did you look at the attached website...?
there is twice as much mass at the top, period...
twice as much weight...
You could reduce this idea down to a slight variance in bellow mass difference and still have imbalance...
thats the key! mass is doubled at top...
Any slight rotation favors one side and perpetuates the process...
If yo give the two possible locations for mass and each end of the "plunger" a value of distance which relates to its "power"... (sorry dont know tech. jargon)... lets say a "value" of feet from axis... then the two possible values are 9 and 10... for example.
If one side has a value of 10 and the opposing side a value of 9 and 10 then the 9-10 side has a nearly doubled value and is heavier...
Simple leverage!
I dont mean to sound so harsh but you must be missing the idea here...
the bellows and liquid conents are sealed...
first of all - both bellows cannot contain mass at the same time... and given any previous rotation the weight either rest at one end or the other...
And the wheel as shown in "your" image is entirely top heavy... ???
did you look at the attached website...?
there is twice as much mass at the top, period...
twice as much weight...
You could reduce this idea down to a slight variance in bellow mass difference and still have imbalance...
thats the key! mass is doubled at top...
Any slight rotation favors one side and perpetuates the process...
If yo give the two possible locations for mass and each end of the "plunger" a value of distance which relates to its "power"... (sorry dont know tech. jargon)... lets say a "value" of feet from axis... then the two possible values are 9 and 10... for example.
If one side has a value of 10 and the opposing side a value of 9 and 10 then the 9-10 side has a nearly doubled value and is heavier...
Simple leverage!
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"A man with a new idea is a crank until he succeeds."~ M. Twain.
re: Oxygons Earth Day PMM
Oxygon,
I'm going to try to give you a very simplified example which I'm hoping you will understand.
Think of a vertical square pipe/tube that is 1 inch by 1 inch by 28 inches long. Assume the walls are so thin that the tube has no weight. Now fill the tube with water. The water will be 28 inches high by 1 inch square. Water weighs 0.0361 pounds per cu.in. so the water in the tube will weigh 1.01 pounds and will produce 1.01 PSI at the bottom.
Now assume that you make the middle of this pipe/tube is smaller so it looks more like your vessels with connecting pipe. Remember that we assume the vessels and pipe have no weight. What happens? The weight of the water will be less than the 1.01 pounds while the water pressure at the bottom will still be 1.01 PSI. In order to squeeze the bottom vessel we must add some weight. So now we make the pipe and vessels heavy enough to squeeze the bottom vessel.
If the combined weight of the vessels, pipes and water is less than 1.01 pounds then the vessel will not be squeezed. And if the combined weight is more than 1.01 pounds then the CoG will be below center after the vessel is squeezed and the wheel will not turn as you anticipate.
It seems that no matter how you construct or configure pipes and vessels the results are always the same.
I hope this makes sense? I could go into a lot more detail but I'm trying to keep it simple.
I'm going to try to give you a very simplified example which I'm hoping you will understand.
Think of a vertical square pipe/tube that is 1 inch by 1 inch by 28 inches long. Assume the walls are so thin that the tube has no weight. Now fill the tube with water. The water will be 28 inches high by 1 inch square. Water weighs 0.0361 pounds per cu.in. so the water in the tube will weigh 1.01 pounds and will produce 1.01 PSI at the bottom.
Now assume that you make the middle of this pipe/tube is smaller so it looks more like your vessels with connecting pipe. Remember that we assume the vessels and pipe have no weight. What happens? The weight of the water will be less than the 1.01 pounds while the water pressure at the bottom will still be 1.01 PSI. In order to squeeze the bottom vessel we must add some weight. So now we make the pipe and vessels heavy enough to squeeze the bottom vessel.
If the combined weight of the vessels, pipes and water is less than 1.01 pounds then the vessel will not be squeezed. And if the combined weight is more than 1.01 pounds then the CoG will be below center after the vessel is squeezed and the wheel will not turn as you anticipate.
It seems that no matter how you construct or configure pipes and vessels the results are always the same.
I hope this makes sense? I could go into a lot more detail but I'm trying to keep it simple.
re: Oxygons Earth Day PMM
The Vessel bellows act as a weight...
Or also their is the option of a middle resivoir...
either way according to your logic this would happen...?
Is this what your suggesting...?
Or also their is the option of a middle resivoir...
either way according to your logic this would happen...?
Is this what your suggesting...?
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"A man with a new idea is a crank until he succeeds."~ M. Twain.
re: Oxygons Earth Day PMM
I think Jim is saying the same thing I said before.
Water can be squeezed in any form, but even water needs a force to be lifted because of its weight.
Water can be squeezed in any form, but even water needs a force to be lifted because of its weight.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---