We don't care the Bessler wheel, the most important is to build a working wheel...

[path_finder]

As explained earlier, I first decided end december to stop my postes on this forum.
There were to much exegesis experts doing reference to the Bessler's papers.
After review of my position I decided to ignore these 'temple keepers' and give to the other members and/or guests additionnal and usefull information.
My opinion is: so long these people will to fence the contents this forum will continue to purr without success.
For shure the Bessler story is interesting, but it's only ONE solution available (Asa Jackson, buzzsaw, Tesla, etc.)
And even if somebody will display someday a working wheel, there is no evidence that this proposal was the real Bessler's way.
I came on this forum just by curiosity, in view to eventually discover how the Bessler wheel could be related with the 'flowerbowl' solution.
I have no information at all that Bessler used the same principe, but after months of reflexion I found some analogies.
My intent is therefore to explain here what I have discovered about the method possibly used by Bessler.
In view to not be criticized and avoid any extra-subject remarks, my explanation will start with a notice:

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Notice: The facts reported hereafter are totally fictious.
Any analogy with any living people or ancient history is pure coincidence.
There is absolutely no connection with the Bessler Wheel nor perpetuum mobile of any kind.
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As for any good story, the revealation uses several steps...
And it is not really funny to get immediately the total solution...

Step #1: Today I will start with the 'botafumeiro'.
http://video.google.com/videoplay?docid ... 0904765402
http://www.youtube.com/watch?v=hBJsaxWiCL8
clue #1: At what position the cord should be pulled?
[/url]

[jim_mich]

You could also add a child pumping a swing. When you pull a swinging pendulum inward at the bottom of its swing it adds energy to the swing. Also the reverse is true.

[path_finder]

Dear Jim_Mich
Could you please precise your analogy.
I don't see any gravity energy exchange in your reference.

[path_finder]

-----------------------------------------------------------------------------------------
Notice: The facts reported hereafter are totally fictitious.
Any analogy with any living people or ancient history is pure coincidence.
There is absolutely no connection with the Bessler Wheel nor perpetuum mobile of any kind.
------------------------------------------------------------------------------------------
Answer to the previous question: at the vertical position of the pulley.
step #2: the 'parametric pendulum'
http://www.sciences.univ-nantes.fr/phys ... meiro.html

Note: click first on the small hand and pull the cord down at the most vertical position, and pull it up when the botafumeiro is at the highest position. It's not easy.
But with some practice the botafumeiro will make a full turn (what is not possible in the Holly Jack Compostel church). At this step you will better understand the basic process.

clue #2: Where is connected the handle of the Merseburg drawing?

[rlortie]

Path_finder,

The 'botafumeiro' has already been discussed on this forum related to the Milkovic machine.

I thank you for posting the above link, it explains the properties very well. It also picks up on the mechanical fallacies presented when a set cranking speed is introduced.

As for the boy on the swing, IMO he is doing the same thing as the 'botafumeiro' only from the other end of the suspension. That is to say that he is shortening the angular movement of the swing arc or moving the center of mass.

Ralph

[path_finder]

Answer to the precedent clue: the handle is connected to the axis of the wheel at the left of the screen.

I'm continuing my explanation,
step #3: Imagine now TWO parametric pendula (exactly equal, not like the Milkovic linkage). When the first is excentred (weight = 0) the second is vertical (weight = maxi). Let's suppose there is a mechanism between the both pendula, a ratchet per exemple, like in the MT51...
clue #3: The Merseburg wheel showed two external pendula. why?

[path_finder]

Answer to the precedent clue: These two pendula where in phase quadrature and connected internally.They are exchanging some energy when their lenght passes through the vertical position.

step #4: The mechanism inside must
1. maintain the move of the both pendula
2. transform the excess of energy in a rotational motion of the whole wheel.
But let's forget the inner mechanism for the moment, we will return back later on this subject.

Just one very simple QUID (select the correct figure, note that a clue for the answer is in my signature (here below of the post):

If 'the son of the carpenter would be able to make a replicat' (said Karl), the template could be:
1. The Khalil-M-Niazi engine seen at http://peswiki.com/index.php/Image:G_Fo ... e_1000.jpg
2. the Lindemann design : visible at http://www.youtube.com/watch?v=0GY0oz0iTiI
3. the Aldo Costa design : visible at http://nseo.com/aldocosta/page3.htm
4. the Bob Kostoff design : seen at http://www.peswiki.com/index.php/Image: ... achine.jpg
5. the Gaspare Malinverno design: http://www.gasparemalinverno.com/
6. the design visible below :

(the answer will be given in my next post, be patient...)

[path_finder]

This topic seems not enough interesting...
Anybody here?
May be, it's because the time difference..
But anyway:

step #5 of my explanation:

The correct answer to the quid is the number 6...
If you got another choice, you should be a very complex person, please consult a psychanalyst...
But there are still two problems with this simple design:

1. There is a contention zone (colored in yellow on the picture).
If you build such as device (like me, sometime an idiot), you will observe a big bazar in the area between 7:00 and 10:00 (if we suppose that the wheel rotates counterwise like the clocks).
The rods do have some difficulties to self organize (remember that mother Nature likes the chaos).

2. The path of the weights is not optimized. For sure the weights between 12:00 and 4:00 are excentred, but the accumulated energy is not sufficient for counter-react the force given by the *!?#-§** weights bad located in the yellow area.

But, but...There are several solutions for solving these two problems.
The answer in my next post. Be patient...and think about a little bit...

[DrWhat]

Pathfinder I am reading with interest!

[path_finder]

HA! DrWhat, many thanks...
If you are the only still reading, I will continue just for you.


Just a crazy point for the fun:
on the earth map of the forum (where are shown all the members and guests) there is now a red dot right in the middle of Africa: THAT'S ME!...
I was there for one month in Yaoundé (Cameroon). I'm pretty sure, because before my departure, Africa was really empty.
I'm really proud to be identified alone in the planet map.
It's a ppitythat (except RSA) the africans are not interested by the gravitic engines.

[path_finder]

step #6 of my explanation: (specially dedicated to DrWhat)

Let's first speak about the optimal path (remember my avatar) for the UNIDIRECTIONAL gravitic engine.
The ideal path should offer two circle active zones and two horizontal exchange zones (as shown in the picture below)
- 1. zone BC : a part of the outer circle, where the weights give a maximal torque
- 2. zone DA : a part of the inner circle, where the torque is minimal but allowing the lift-up of the weights.
- 3. and two exchange paths (AB and CD), absolutely horizontal (the only way do not consume any gravitic energy)
Note: in this case the summary of the gravitic energy is null (if the friction is marginal, the energy won during the descent is just sufficient to lift up the weights during the ascendent path).

[AB Hammer]

path_finder

There are most likely several people reading. It is the way you are writing that makes us pause for the next chapter. I have built similar attempts as your last drawing which shows several problems. But I await what is next as well.


Alan

[path_finder]

step #7 of my explanation:
(Dr What are you still here?)

The theory says that :
1. The work issued from the gravity just depends from the two altitudes (top and bottom)
2. This work is independent of the path chosen between these two levels.
3. The amount of the WORK available is the surface in GREEN in the picture of the previous post. For more complex curves the mathematicians invented the integral calculation (if you have the equation of the curve, just compute the primitives and calculate the delta of surfaces inside the curve). Then from this value of work, depending of the time of circulation of the vector along the curve (basically the rounds per second) you will obtain the power of your wheel.
Unfortunately this a to much theorical (the frictions are not included) and I stop here.

Somebody will argue that even for a translation motion you need a minimum of energy.
But when the weight arrives at the C position, it has a cinetic energy accumulated during the BC travel, and in any case sufficient to go from C to D without to consume any additional energy (if the friction is still marginal). The same at the top between A and B.
THEREFORE THE PATH FROM THE INNER CIRCLE TO THE EXTERNAL CIRCLE (AND RECIPROCALLY) MUST BE ABSOLUTELY AN HORIZONTAL SEGMENT.

This assumption brings some restrictions to the available paths.
On a pure theorical point of view, the path in BLUE will be the most efficient, but it's impracticable for the two good reasons:
- how could you put the half of the weights population on the central axis O, there is no room for everybody
- the lengh of the rod must be variable (R at the point A and R/2 at the point B, and about zero at 3:00 o'clock)
At contrary if we reduce the horizontal parts of the path (like with the RED path) the difference of torque will be very low and not sufficient to overforce the friction.
By the way (I can't demonstrate it here more longer) the best compromise is to use the VIOLET path.
This path will be used in the rest of the discussion hereafter.

[path_finder]

Many thanks to AB_hammer for reading.

step #8 of my explanation:

The VIOLET path is an objective, the reality could be different.
If we keep the proposed design, we observ first that the horizontal segment AB has been transformed into a circular piece (in fact a down translation of the wheel circle). It's not really disconvenient, except that we loose some effective torque.
If the radius of the weights is 'r' we can compute it's optimal value by the following observation:
- We start at 12:00. The rod is vertical and it's weight is just leaving the inner circle.
- the wheel rotates. Between 12:00 and 2:00 the distance between the weight and the axis increases
- at 2:00 the weight reaches the inner rim of the wheel (giving the noise heart by the widnesses)
- but now the weight is sticked to the rim of the wheel and it will still be here until 8:00 ....IF WE DO NOTHING AGAINST...
(this is the reason of the *!?#-§** between 8:00 and 10:00)
SO HERE IS THE MOST IMPORTANT POINT OF THE DESIGN...
Be patient. I will return back on this point hereafter.

But let's suppose for the moment we have solved this question, THE SECOND IMPORTANT POINT IS THE INNER CIRCLE PATH...
Question:
How to oblige the weights to lift up between 6:00 and 12:00 using the inner circle ?
The answer is very simple. See my next post. Be patient...

[path_finder]

step #9 of my explanation

The answer is very simple: cords.
For each weight, you have to attach one end of the cord to the axis of the weight and the other end to the rim of the wheel (at the axis of the precedent rod par exemple).
Like shown in the picture below, where the cords are RED when pulled, and GREEN when loose.
'Elementary, my Dear Watson...'

Now we have solved the question of the return, how can we oblige the weight to follow the YELLOW horizontal path?..
Be patient. This will be in the next post. (I'm tired, I leave you thinking a little bit about...).