letting go... a little bit ;P

A Bessler, gravity, free-energy free-for-all. Registered users can upload files, conduct polls, and more...

Moderator: scott

Post Reply
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

Fletcher wrote:That's because you have gone over it a thousand times in your head in every detail & it is very familiar to you.
Sorry about that... your right.

Btw, You know the actual "shifting" point is a little more clockwise than shown in (easyaspie2.jpg) picture... but still overbalanced.
"A man with a new idea is a crank until he succeeds."~ M. Twain.
etjoe
Enthusiast
Enthusiast
Posts: 127
Joined: Wed May 26, 2004 8:40 pm
Location: Toronto, ON., Canada

re: letting go... a little bit ;P

Post by etjoe »

My doubts about this are similar to my doubts about Aldo's wheel (the 5 story high wheel in France). IMHO, the wedges that shift the lighter wedges to the outside will not keep the wheel in motion. I also think that when the wheel turns, the inertial forces will move the wedges to the wrong positions. I'm still gonna try it on the sim so I'll let you know if I get anywhere. In my wheel, I use a ratio to get an advantage, I'll try to incorporate some of that wisdom into deciding what the weights should be.
-e
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

etjoe wrote:My doubts about this are similar to my doubts about Aldo's wheel (the 5 story high wheel in France). IMHO, the wedges that shift the lighter wedges to the outside will not keep the wheel in motion. I also think that when the wheel turns, the inertial forces will move the wedges to the wrong positions. I'm still gonna try it on the sim so I'll let you know if I get anywhere. In my wheel, I use a ratio to get an advantage, I'll try to incorporate some of that wisdom into deciding what the weights should be.
-e
I didnt say the outer weights were "lighter" just... "out-er"...

See the weights "weight" is simply similar... the wieght of the inner weight is irrelevant to the perpetuation effect caused by the shifting of the outer weight... the inner weights never alter their distance from axis... (slight variation/negligible)

What you said about the "inertial forces"...

the wheel has no "set speed", the "slower and bigger the better"... since "inertial forces" would interfere...

the design is inherently unstable, just look at the freeze frame of its "outer" weight locations in the "easyaspie.jpg" picture...

the "easyaspie2.jpg" is a bit complicated with soo many "sets" that only serve to illistrate the path of the outer wieght thru the process of shifting...

the "inner" weight doesn't need to be that heavy at all... the "outer" weight shifts laterally... even if the weight shifted at the "level point"(axis-horizontal) it still would be "over balanced"...

I really dont understand the difficulty, My earlier comments aside... its fairly laid out now... why cant anyone see it...?

???
"A man with a new idea is a crank until he succeeds."~ M. Twain.
User avatar
Jonathan
Addict
Addict
Posts: 2453
Joined: Wed Nov 05, 2003 6:29 am
Location: Tucson, Az

re: letting go... a little bit ;P

Post by Jonathan »

Originally I misunderstood the form of the device, but I understood the principle and still think it is like jim's hydraulic devices. The inner weights that do not move radially are at all times making the wheel bottom heavy, and the energy used to shift the outer weights comes from having lifted those inner weights against the bottom heaviness, and letting them fall again as they pass the zenith (and nadir). But the energy to lift them against the bottom heaviness comes from the energy (radially) stored in the outer weights.
Last edited by Jonathan on Fri Jul 16, 2004 12:52 am, edited 1 time in total.
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.
ovyyus
Addict
Addict
Posts: 6545
Joined: Wed Nov 05, 2003 2:41 am

re: letting go... a little bit ;P

Post by ovyyus »

Hi Oxygon,

I've animated your easyaspie diagram. All weights are assumed equal which means transition point is at 45deg rotation (equal weight falling and rising). Coupling of inner and outer weight pairs by whatever efficient means.

Unfortunately, no million bucks - it obviously balances out over a full cycle.
Attachments
EasyAsPieAnimated.gif
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

ovyyus wrote:Hi Oxygon,

I've animated your easyaspie diagram. All weights are assumed equal which means transition point is at 45deg rotation (equal weight falling and rising). Coupling of inner and outer weight pairs by whatever efficient means.

Unfortunately, no million bucks - it obviously balances out over a full cycle.
So you animated it... with a gif generator?

that can't be your evidence?...


How does it obviously balance out?...
Jonathan wrote: The inner weights that do not move radially are at all times making the wheel bottom heavy, and the energy used to shift the outer weights comes from having lifted those inner weights against the bottom heaviness, and letting them fall again as they pass the zenith (and nadir). But the energy to lift them against the bottom heaviness comes from the energy (radially) stored in the outer weights.
Bottom heavy? Yes, true... but it also makes it "side heavy"... all it has to do is turn it less than a quarter to cause the other sets to shift... I have done some math regarding this bottom heavyness... and it only comes "close to stall"... and this problem of "near stall" can be further alleviated by a series-connected design...


Not to mention that bottom heavyness reduces with size and "set" number...

the larger diameter and greater set number means that the "inner shift" (bottom heavyness) is reduced with relation to the axis...

"bigger is better" - "more is better"

it would not fail.
"A man with a new idea is a crank until he succeeds."~ M. Twain.
ovyyus
Addict
Addict
Posts: 6545
Joined: Wed Nov 05, 2003 2:41 am

re: letting go... a little bit ;P

Post by ovyyus »

...
Last edited by ovyyus on Fri Jul 16, 2004 2:41 am, edited 1 time in total.
ovyyus
Addict
Addict
Posts: 6545
Joined: Wed Nov 05, 2003 2:41 am

re: letting go... a little bit ;P

Post by ovyyus »

Hi Oxygon,
Great so you animated it... What, with a gif generator? Thanx, but that can't be your evidence? How does it obviously balance out?
Sorry mate, I thought an animation might help make it a little clearer for everyone. Apparently I was wrong.

The gif anim was taken from a cad layout. In this design you could simply measure the horizontal distance of the various weights from the axle line in order to ascertain relative balance across increments of rotation. I don't see the problem?
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

Oxygon wrote:Great so you animated it... What, with a gif generator? Thanx, but that can't be your evidence? How does it obviously balance out?
ovyyus wrote:Sorry mate, I thought an animation might help make it a little clearer for everyone. Apparently I was wrong.
I am sorry for the "impetuousness" of the statement, I did alter it before you responded...

I do thank you for adding an image to the forum...
ovyyus wrote:The gif anim was taken from a cad layout. In this design you could simply measure the horizontal distance of the various weights from the axle line in order to ascertain relative balance across increments of rotation. I don't see the problem?
... Yes I have, but have you?... Perhaps we get different results?

...
"A man with a new idea is a crank until he succeeds."~ M. Twain.
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

re: letting go... a little bit ;P

Post by Oxygon »

Sorry, I am a bit defensive, even though its my easiest idea....

All of my results have determined that it should be possible, at least according to my present understanding/design...

I am looking to build a simplistic model over the weekend...

too settle this once and for all... but in this field you know how promises "like that" go...

We'll see...

I'll be around here during the test...
"A man with a new idea is a crank until he succeeds."~ M. Twain.
User avatar
Jonathan
Addict
Addict
Posts: 2453
Joined: Wed Nov 05, 2003 6:29 am
Location: Tucson, Az

re: letting go... a little bit ;P

Post by Jonathan »

If you're going to the trouble, might I suggest this design? It seems simpler as it has no 3d motion, and you don't need any special joints. The round things are pulleys, the lines between them are strings, the black squares are weights connected to the strings, and the encompassing rectangle is some arbitrarily shaped frame with an axle through the little central circle.
EDIT
It just occurs to me that you don't even need the pulley on the inside of the curve, then the string will make a triangle shape, but it will work the same.
Attachments
ForOxygon.JPG
ForOxygon.JPG (4.35 KiB) Viewed 32568 times
Disclaimer: I reserve the right not to know what I'm talking about and not to mention this possibility in my posts. This disclaimer also applies to sentences I claim are quotes from anybody, including me.
User avatar
jim_mich
Addict
Addict
Posts: 7467
Joined: Sun Dec 07, 2003 12:02 am
Location: Michigan
Contact:

re: letting go... a little bit ;P

Post by jim_mich »

A Discovery... thanks to Oxygon

Two equal weights, one movable in and out along a spoke, the second movable side to side straight line along a cord (a trig. term, NOT a string). The two weights are interconnected to each other so when one move the other moves, each the same distance. Gravity will shift the weights at about 45 degrees of wheel rotation past top and past bottom.

The discovery... the wheel will be very slightly out of balance! The below calculation comes very close to equaling the OB (Out of Balance torque) found using a computer program that sums the torques at each 1/10 degree full circle.

r = radius from center of wheel to center of small weight.
d = movement distance of a weight
a = angle that the small weight is on either side of the spoke.
OB = Out of Balance torque per one unit of mass

a = ATN( d / 2 / r )
OB = r * ( TAN(a) - SIN(a) ) * 2 * 2

A bigger angle produces a bigger OB so it is best to have the inner weight close to the center. The distance of the outer weight from the axle makes no difference.

Sample numbers...
r = 10
d = 1
a = 2.8624
OB = .0025

Another sample...
r = 4
d = 1
a = 7.125
OB = .0154

I have NOT looked at inertia, centifugal forces, etc. Maybe they cancel out the effect? And I hope I've not made some type of mistake!

Image
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

Jonathan wrote:If you're going to the trouble, might I suggest this design? It seems simpler as it has no 3d motion, and you don't need any special joints. The round things are pulleys, the lines between them are strings, the black squares are weights connected to the strings, and the encompassing rectangle is some arbitrarily shaped frame with an axle through the little central circle.
EDIT
It just occurs to me that you don't even need the pulley on the inside of the curve, then the string will make a triangle shape, but it will work the same.
Will call this "Plan B"...

:)
"A man with a new idea is a crank until he succeeds."~ M. Twain.
User avatar
Oxygon
Aficionado
Aficionado
Posts: 751
Joined: Tue Mar 16, 2004 5:01 am
Location: North of Somewhere
Contact:

Re: re: letting go... a little bit ;P

Post by Oxygon »

jim_mich wrote:A Discovery... thanks to Oxygon

Two equal weights, one movable in and out along a spoke, the second movable side to side straight line along a cord (a trig. term, NOT a string). The two weights are interconnected to each other so when one move the other moves, each the same distance. Gravity will shift the weights at about 45 degrees of wheel rotation past top and past bottom.

The discovery... the wheel will be very slightly out of balance! The below calculation comes very close to equaling the OB (Out of Balance torque) found using a computer program that sums the torques at each 1/10 degree full circle.

A bigger angle produces a bigger OB so it is best to have the inner weight close to the center. The distance of the outer weight from the axle makes no difference.

I have NOT looked at inertia, centifugal forces, etc. Maybe they cancel out the effect? And I hope I've not made some type of mistake!
No mistake,... Slow and large is better... As I am sure these forces would interfere...

... I have been rather tight lipped about other "variations" of this method, It is better to have the inner weights closer to the axle and also it is possible to conjoin the inner wieghts of "12 and 6" and "3 and 9" into two exclusive units, since apart they operate at the same time - in the same fashion, there are many methods of designing this so that the motion of each dual-single inner weight does not interfere with the other at the intersection of the axis... it is possible to create a wheel unit with a "set" on each side of a wheel or... perhaps an offset pair on the lids of a sealed drum...

As far as I know, my ideas are all mathmatically sound.

here is an illistration...
Attachments
easyaspie3.jpg
"A man with a new idea is a crank until he succeeds."~ M. Twain.
User avatar
jim_mich
Addict
Addict
Posts: 7467
Joined: Sun Dec 07, 2003 12:02 am
Location: Michigan
Contact:

re: letting go... a little bit ;P

Post by jim_mich »

Rechecking my numbers I found a small error in calculating angle 'a' but I still get it overbalancing.

a = ATN( d / 2 / SQR((d / 2)^2 + r^2) )
OB = .000622 when r = 10
OB = .003773 when r = 4

The computer program gave me
OB = .000546 when r = 10
OB = .003447 when r = 4

Image
Post Reply