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Gravity powered devices => Gravity powered devices => Topic started by: Floor on May 31, 2019, 09:03:00 PM
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@ All readers
This device utilizes
1. A first lifting by gravity, of the flotation element of the device within a fluid to an intermediate position (work out).
2. A first 180 degree rotation of the device, while it is in a balanced (in gravity) state.
3. A second lifting by gravity, of the flotation element of the device within a fluid to an upper most position (work out).
4. Removal of the float element without affecting the pressure within the flotation fluid, nor the lowering of the height of that fluid (work in).
There is no work done against fluid pressure.
5. A first fall by gravity, of the flotation element of the device (external to the fluid), to an intermediate position (work out).
6. A second 180 degree rotation of the device, while it is in a balanced (in gravity) state.
7. A second fall by gravity, of the flotation element of the device (external to the fluid), to a lowest position (work out).
8. Insertion of a float element without affecting the pressure within the flotation fluid, nor the raising of the height of that fluid (work in).
There is no work done against fluid pressure.
9. Start / stop actions (there is no continuous motion). All motion sets complete before a next motion set begins.
10.. more mechanical work out, than is input.
floor
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PNG drawings
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Second drawing
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No tariffs on these materials. Not yet !
floor
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I'\'m looking for some one to find an error in this.
floor
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I had previously thought that gravity power had been explored suffuciently and also
that gravity was simple enough to expect there would never be a gravity powered device.
I gave it another shot, and saw a way to loop it O.U..
This presentation is that sucess. Some seemingly impossable obsticles
and the a eurika, and then a WOW
check it out
cheers
floor
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@ At all readers
I cleaned up / added more detail to the flow diagram / drawings and added a minor detail to the
written explanations page.
floor
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QUOTE from a users personal message
"It's the gravitational potential of the small amount of water that rushes in to fill the space left by the float as it leaves the bottom of the tank. So, when you rotate the tank, you actually have to perform some net work to also rotate that additional amount of water through its increased vertical displacement. Though it's a VERY SMALL amount of vertical displacement, it's still not zero, which I think you need it to be for the idea to work." END QUOTE
Thanks very much for your observations / comments.
Your observations would be correct except that.
9. All / each action is a start / complete / stop
action.
There is no rotation untill after the float is stopped
/ latched / held at the center of the float tank.
.......................................................................................
While the float is at the bottom, the tank is unbalanced
While the float is riseing, the tank is unbalanced.
While the float has stopped at center, the tank IS IN BALANCE.
Rotation occurs only while the float is at center height.
Sorry that my diagram doesn't state / detail this.
Thanks again for bringing this oversigt forward.
floor
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More detail
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While the float is at the bottom, the tank is unbalanced
While the float is riseing, the tank is unbalanced.
While the float has stopped at center, the tank IS IN BALANCE.
Rotation occurs only while the float is at center height.
That is partially true. In step 6 to 7 of your illustration the float is outside and the tank is no longer balanced so it will take work to rotate the tank as it has become bottom heavy.
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That is partially true. In step 6 to 7 of your illustration the float is outside and the tank is no longer balanced so it will take work to rotate the tank as it has become bottom heavy.
Incorrect,
The tank is balanced in steps 6 and 7.
floor
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In the drawings. The device is balanced by the end of step 6 but also the device is balanced
throughout step 7 (the rotation).
floor
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@ All readers
There is a PDF file just below the above drawing, it contains all of the above action sets / drawings.
floor
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what is the weight of a float?
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what is the weight of a float?
answers
1. The weight of the device can be any amount, as long it is no more than the structural / support members of the device can handle.
2. The weight would depend upon the scale of the device.
3. Less than an equal volume of the flotation liquid.
floor
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Here, below, are the drawings, a flow diagram, a written description of this free energy
device and process / method. Nothing unconventional is claimed, except for the end result.
There are several PNG files
There is one PDf file which contains all of the drawings etcetera, collected together for convienience.
These are improved / clairified renderings, pluss a few new drawings.
floor
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more
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The PDF file is just below
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@ All readers
It should be understood, that in order that the float
will not rise or fall until after each 180 degree rotation is compleated, the
float must be latched or locked into the center position.
floor
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@ All readers
@ All readers
I realize that for many (my self included) it has seemed that a gravity / flotation energy device is too simplistic a concept to waste time examining. That this is a subject, which has been explored for so many centuries and in so many ways, that it is unlikely there would be found, a new approach and certainly not any successful approach.
Please examine the above diagrams and written explanations, and give me some feed back.
In my own reviewing of it, I have found no flaw or error in its design and functioning.
best wishes
floor
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Why no one comments? think it doesn't work?
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Another variation in method
floor
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@Telecom
If the float is only slightly less dense than the fluid, then the work output, would be greatest
when the float falls exterior to the fluid.
If the float is greatly less dense than the fluid, then the output work done during its rise within the fluid will be greater than the work done when the float falls while it is exterior to the fluid.
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@ Telecom
If the float is greatly less dense than the fluid, then the output work done during its rise within
the fluid will be greater than the work done when the float falls while it is exterior to the fluid.
Thanks very much for your last inquiry.
It broadened my focus / bought to my mind ..... that most of the useful work, could be done / taped into when the float
falls while outside of the fluid, rather than while riseing inside the tank / fluid.
If the float is only slightly less dense than the fluid, then the work output, would be greatest
when the float falls exterior to the fluid.
regards
floor
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@ All readers
THIS IS CYCLICAL WORK DONE BY GRAVITY.
really !
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@ All readers
This device utilizes
1. A first lifting by gravity, of the flotation element of the device
within a fluid to an intermediate position .(work out). All motion
is arrested.
2. A first 180 degree rotation of the device, while it is in a
balanced (in gravity) state (work in). There is no work input to
lift the fluid against gravity. All motion is arrested.
3. A second lifting by gravity, of the flotation element of the
device within a fluid to an upper most position (work out).
All motion is arrested.
4. Removal of the float element without affecting the pressure
within the flotation fluid, nor the lowering of the height of that
fluid (work in). All motion is arrested.
There is no work done against fluid pressure.
5. A first fall by gravity, of the flotation element of the
device (external to the fluid), to an intermediate position (work out).
All motion is arrested.
6. A second 180 degree rotation of the device, while it is in a
balanced (in gravity) state (work in). There is no work input to lift
the fluid against gravity. All motion is arrested.
7. A second fall by gravity, of the flotation element of the device
(external to the fluid), to a lowest position (work out). All motion
is arrested.
8. Insertion of a float element without affecting the pressure
within the flotation fluid, nor the raising of the height of that
fluid (work in). All motion is arrested.
There is no work done against fluid pressure.
9. Start / stop actions (there is no perpetual motion).
All actions / motions are completed before a next action / motion begins.
10.. more mechanical work out, than is input.
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@ All readers
I keep revising / making more clear these drawings and explanations.
Here is the latest revision.
floor
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There are more details given as drawings and explanations earlier in this topic.
floor
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In #2 when float raises to the middle, water fills its place.
Water is heavier than float.
In #3 you have to do work to turn 180 degrees.
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Hi Telecom
In the above drawing / diagram "New Floater 4-1b. PNG",
YES, the fluid is heavier than float.
In# 1 the entire device is out of balance. It is top heavy, because
the float is at the bottom of the tank.
Yes, In #2 when the float raises to the middle, water fills its place at the
bottom of the tank.
At the finish of #2, the float is arrested in its motion (latched in place ?).
Once the float has arrived at the center of balance and of rotation, The
entire device is in balance. There is the same amount of fluid above the
center of balance as is below the center of balance.
Yes In #3 you have to do work to turn 180 degrees.
But this work is done to overcome inertia and friction only. There is NO WORK DONE
AS LIFTING OF THE FLUID. This is because, there is an equal amount of downward force (from gravity)
on both sides of the center of rotation, the device is balanced.
Thanks for your thoughts / input
Floor
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This may work, but ideally would be nice to allocate weight to each element,
to make it possible to use some rudimental math.
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@ Telecom
I will follow your recomendation / request.
thanks
floor
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@ telecom
the energy density is, as should be expected very low, because gravity is a weak force.
assume:
float = 3 meters length by 0.33 meters radius.
approximate 10 feet by 1 foot
cylinder of 0.33 meters diameter or
0.165 meters radius by 3 meters length
= 0.25669285714285694 cubic meters of water.
0.2566 cu meters of water = 256.6 kilograms of weight or
565.7062 pounds of weight.
256.6 kilograms = 2516.3864 newtons.
assume :
float chamber interior is 10.5 meters height, by 10.5 meters width,
by 0.5 meters depth into the page.
this means that the height of the lifting of the float
= 10.5 meters minus 1 float cylinder diameter.
this is = 10.5 m - 0.33 m = 10.17 m
the float is lifted to a total of twice this distance
10.17 m x 2 = 20.34 m total height of the float
...
2516.3864 newtons x 20.34 meters = 5214.899376 joules of energy stored
as weight of the float external to the fluid chamber. OUTPUT
...
assume:
It requires 25 kilograms of force or 245.1663 newtons, over a displacement of
3 meters to INSERT the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
It requires 25 kilograms of force or 245.1663 newtons, over a displacement of
3 meters to REMOVE the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
...
735.4989 newtons x 2 = 1470.9978 joules input work / energy. INPUT
...
assume:
It requires 25 kilograms of force over a distance of 0.33 meters to initiate / accomplish
the 180 degree rotation of the device.
25 kilograms = 245.1663 newtons
245.1663 newtons x 0.33m = 80.904879 joules of work
assume:
It requires 25 kilograms of force over a distance of 0.33 meters to initiate / accomplish
a 180 degree rotation of the device.
25 kilograms = 245.1663 newtons
245.1663 newtons x 0.33m = 80.9048 joules of work
...
there are two 180 degree rotations during each complete cycle.
80.9048 joules x 2 = 161.8097 joules INPUT
...
1470.99 joules input + 80.90 joules input = 1551.9026 joules total input
.............................................................................
5214.89 joules of energy stored
as weight of the lifted float, while external to the fluid chamber.
5214.899376 joules of energy stored - 1551.9026 joules total input
= 3663.0951 excess energy.
assume: a slow operation time to store that energy (not to use it) of 2 minutes.
3663.0951 / 2 = 1831.5475 joules per minute.
1831.5475 joules per minute = 30.5233 watts
conservatively arrived at this is 30.5233 watts total output
or
43,920 watts over a 24 hour period.
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OR
0.25669285714285694 cu meters = 67.78 gallons per two minutes
this is 33.89 gallons lifted to a 20 foot height per minute.
floor
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In step 1, when you insert the float laterally, work needs to be done to expel amount
of water equal the weight of its volume, upward.
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In step 1, when you insert the float laterally, work needs to be done to expel amount
of water equal the weight of its volume, upward.
Incorrect
Because simultaneous to the float's insertion, a volume which is equal to the volume of the float
exits the tank (as the push rod on the opposite side of the tank).
There is no rise in the level of the fluid height nor is there any work done against fluid pressure
as the float is inserted.
floor
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Here are the diagrams and explanations again.
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@telecom
I did the first draft of this / posted it, on the fly. Here below is a revised version.
floor
Revised calculations 1 (only slightly)
the energy density is, as should be expected very low, because gravity is a weak force.
assume:
float = 3 meters length by 0.33 meters radius.
approximate 10 feet by 1 foot
cylinder of 0.33 meters diameter or
0.165 meters radius by 3 meters length
= 0.25669285714285694 cubic meters of water.
0.2566 cu meters of water = 256.6 kilograms of weight or
565.7062 pounds of weight.
256.6 kilograms = 2516.3864 newtons.
assume :
float chamber interior is 10.5 meters height, by 10.5 meters width,
by 0.5 meters depth into the page.
this means that the height of the lifting of the float
= 10.5 meters minus 1 float cylinder diameter.
this is = 10.5 m - 0.33 m = 10.17 m
the float is lifted to a total of twice this distance
10.17 m x 2 = 20.34 m total height of the float rise
...
2516.3864 newtons x 20.34 meters = 5214.899376 joules of energy stored
as weight of the float external to the fluid chamber. OUTPUT
...
assume:
It requires 25 kilograms of force or 245.1663 newtons, over a displacement of
3 meters to INSERT the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
It requires 25 kilograms of force or 245.1663 newtons, over a displacement of
3 meters to REMOVE the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
...
735.4989 newtons x 2 = 1470.9978 joules input work / energy. INPUT
...
assume:
It requires 25 kilograms of force over a distance of 0.33 meters to initiate / accomplish
a 180 degree rotation of the device.
25 kilograms = 245.1663 newtons
245.1663 newtons x 0.33m = 80.9048 joules of work
...
there are two 180 degree rotations during each complete cycle.
80.9048 joules x 2 = 161.8097 joules INPUT
...
1470.99 joules input + 80.90 joules input = 1551.9026 joules total input
.............................................................................
5214.89 joules of energy stored
as weight of the lifted float, while external to the fluid chamber.
5214.899376 joules of energy stored - 1551.9026 joules total input
= 3663.0951 joules of excess energy.
assume: a slow operation time to store that energy (not to use it) of 2 minutes
3663.0951 joules per every 2 minutes = 1831.5475 joules per minute.
1831.5475 joules per minute = 30.5233 joules per second or in other words 30.5233 watts
(1831.5475 joules / 60 seconds = 30.5233 joules per second)
conservatively arrived at, this is 30.5233 continuous watts gained as output power
or
30.5233 continous watts if used for 1 hour = 32 watt hours
and this is
732.5592 watt hours if continously used for 24 hours
and this is
5.128 kilowatt hours if continously used for one week
and this is
20.51 kilowatt hours if continously used for one month
and this is
246.14 kilowatt hours if continously used for one year
probably actually about 1.5 times this amount of energy.
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Incorrect
Because simultaneous to the float's insertion, a volume which is equal to the volume of the float
exits the tank (as the push rod on the opposite side of the tank).
There is no rise in the level of the fluid height nor is there any work done against fluid pressure
as the float is inserted.
floor
This is a game changer.
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@ telecom
There was an error in my math.
But the outcome is not very significantly effected.
I will post a corrected version of those calcs later.
floor
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Revised calcs.
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revised calculations 3 (there were bad calculations in versions 1 and 2)
the energy density is, as should be expected very low, because gravity is
a weak force.
..........................................................................
assume:
float = 3 meters length by 0.33 meters radius.
approximate 10 feet by 1 foot
cylinder of 0.33 meters diameter or
0.165 meters radius by 3 meters length
= 0.25669285714285694 cubic meters of water.
0.2566 cu meters of water = 256.6 kilograms of weight or
565.7062 pounds of weight.
256.6 kilograms = 2516.3864 newtons.
..........................................................................
assume :
float chamber interior is 10.5 meters height, by 10.5 meters width,
by 0.5 meters depth into the page.
this means that the height of the lifting of the float
= 10.5 meters minus 1 float cylinder diameter.
this is = 10.5 m - 0.33 m = 10.17 m
the float is lifted to a total of twice this distance
10.17 m x 2 = 20.34 m total height of the float rise
............................................................................
2516.3864 newtons x 20.34 meters = 5214.899376 joules of energy stored
as weight of the float external to the fluid chamber. OUTPUT
............................................................................
assume:
it requires 25 kilograms of force or 245.1663 newtons, over a displacement
of 3 meters to INSERT the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
it requires 25 kilograms of force or 245.1663 newtons, over a displacement
of 3 meters to REMOVE the float.
245.1663 newtons x 3 meters = 735.4989 joules of work
735.4989 newtons x 2 = 1470.9978 joules input work / energy. INPUT
...........................................................................
assume:
it requires 25 kilograms of force over a distance of 0.33 meters to initiate
/ accomplish a 180 degree rotation of the device.
25 kilograms = 245.1663 newtons
245.1663 newtons x 0.33m = 80.9048 joules of work
there are two 180 degree rotations during each complete cycle.
80.9048 joules x 2 = 161.8097 joules INPUT
...........................................................................
1470.99 joules input + 161.8097 joules input = 1632.7997 joules
(previously in error)
...........................................................................
5214.89 joules of energy stored as weight of the lifted float, while
external to the fluid chamber.
5214.899376 joules of energy stored - 1632.7997 joules total input
= 3582.0996 joules ENERGY YIELD
.........................................................................
assume:
a slow operation time to store that energy (not to use it) of 2 minutes
3582.0996 joules per every 2 minutes = 1791.0498 joules per minute.
1791.0498 joules per minute = 29.85083 joules per second or in other
words 29.85083 watts
(1791.0498 joules / 60 seconds = 29.85083 joules per second)
conservatively arrived at, this is 29.85083 continuous watts gained as
output power
or
29.85083 continuous watts if used for 1 hour = 29.85083 watt hours
and this is
716.41992 watt hours if continuously used for 24 hours
and this is
5.0149 kilowatt hours if continuously used for one week
and this is
20.0597 kilowatt hours if continuously used for one month
and this is
240.717 kilowatt hours if continuously used for one year
probably actually about 3 times this amount of energy.
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@ All readers
The designs and methods I have presented in this topic and which are novel,
are given into the public domain.
floor
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Gravity does all of the work of the lifting in this design.
Then the lifted opject falls.
floor
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@ All readers
There is second version of a sucessful gravity based design (presented byCadman)
@ https://overunity.com/18243/cadmans-hydrostatic-displacement-engine/
floor
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This particular device
Gravity does most of the work.
Simple design.
but an
expensive build.
One versions lifts 565.7062 pounds of weight 10 feet into the air.
A variation lifts 565.7062 pounds of weight 20 feet into the air.
Any amount of weight and any height can be achieved within the limits of
what the supporting structure can bear.
Its what we do here at the Over Unity.com forum.
thanks for you time
floor
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In step 6, you cannot turn the system back, because the the bottom will be heavier than the top. The bottom has extra water between the rods, the top hasn't, it's now just air. You have to look the weight at the same levels, but not just inside the container. The device won't work, it's not balanced on the centerpoint in step 6.
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Thank you Codwell.
Yes not balanced in step 6
It fails at step 6
Back to the drawing board.
floor
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In step 6, you cannot turn the system back, because the the bottom will be heavier than the top. The bottom has extra water between the rods, the top hasn't, it's now just air. You have to look the weight at the same levels, but not just inside the container. The device won't work, it's not balanced on the centerpoint in step 6.
Thank you codwell
floor
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In step 6, you cannot turn the system back, because the the bottom will be heavier than the top. The bottom has extra water between the rods, the top hasn't, it's now just air. You have to look the weight at the same levels, but not just inside the container. The device won't work, it's not balanced on the centerpoint in step 6.
Thank you codwell
floor
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@All readers
I think it is sometimes a responceability, to keep a topic up,
when it was a fail, and so, for one last time I am posting
in this topic.
click here
https://overunity.com/18234/new-version-of-flotation-device/msg537516/#msg537516
This is just the way we do it here @ the over unity.com forum.
best wishes
floor
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I'\'m looking for some one to find an error in this.
floor
Regarding this figure at https://overunity.com/18234/new-version-of-flotation-device/dlattach/attach/173024/ (https://overunity.com/18234/new-version-of-flotation-device/dlattach/attach/173024/)
From operation 6 to 7 when the container rotates 180 degrees, you also lift the fluid inside just as much as the thickness of the floater. Because as the floater rises, the fluid flows down, making the container bottom-heavy.
Even if it doesen't take energy to displace nothing as the floter is pushed back in at the bottom, the work required to complete the operation is when the device rotate 180° from step 6 to 7.
If no work is being done during this rotation, you must descend the whole device by the same distance as the floaters thickness for every 6-7 rotation.
Edit: I see now that the problem has already being addressed...
Vidar
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Thanks Vidar !
peer review rocks !
And so here is the New floater fix.
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@ all readers
This embodyment flops as did the last version, and for prety much the same reason.
fllor