Re: Costa Wheel update


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Posted by Smithy (203.38.223.82) on August 26, 2002 at 20:43:30:

In Reply to: Costa Wheel update posted by Scott Ellis on August 26, 2002 at 17:39:21:

G'day Scott,
Lot's of interesting info to be had, you're right, the translation is rather strange but I get the gist of what he is saying. My mass weights are each 100grams and travel only 3.5mm from inner to outer circumference. As far as the mathmatics goes I would be quite lost but all we can do is suck it and see.
Is it possible for me to put some photo's on this site so everybody can see what I'm trying to do, maybe I can get some constuctive criticism, the wheel itself is close to 55cm in dia and the disk is perspex (3mm), I have small aircraft quality ball and roller bearings for the axle and levers etc to help with lowering the friction as much as possible.
However the main friction points are at the "trip" points at the bottom which lift the weights!! Must get back to work...
Smithy.

: Hi Smithy, thanks for your post. Here is the latest update from Rudolf Wehrung. It is translated from French by the Google translation service, which is good, but certainly leaves something to be desired. I am including the original French at the bottom as well.
: -Scott

: ----- Update from Rudolf Wehrung- translated by Google ---

: As you saw on the images, the wheel is roughcast spades. Each spade is an arm of lever which actuates 1 mechanism. There are 2 butted to actuate these levers: - one in bottom of the wheel - one in top of the wheel.

: Description of a mechanism - for a mechanism of bottom - I describe it vertically: with the top/a spring, which pushes on in the medium/a mass of 2,240 kg in bottom/the arm of lever which gears down the force by 10. The goal of the arm of lever is to push the mass of 34 mm upwards. The "graviton" is a small part which comes to retain the mass with this height. Therefore, since there is the spring, 2,240 kg should be pushed Twice, therefore 4,480 kg. And since the arm of lever divides the force by 10, the mass to be pushed is finally 480 grams.

: So that the wheel can turn, the force which must BE PRODUCED must be sufficient to move these 480 grams. It is thus necessary to create a sufficient inertia on all with dimensions of the wheel to exceed these 480 grams. To launch the wheel, it is necessary to make him make a half turn, with the hand, by actuating each lever of each mechanism. Then, half of all the masses of the wheel, blocked by let us revolve, will be more close to the center of the wheel. More nearly 34 mm.

: When the spades arrive in top and touch the second obstinate one, that releases revolve them, and that makes it possible each spring to upwards push back the mass. Thus to create an offset of all the masses of the other with dimensions of the wheel. The wheel is a priori in imbalance. But that is not enough at all. If one remains about it there. It does not move. A calculation (at base of the current formulas of Newton) makes it possible to see that the inertia of a mechanism is 12 grams.

: To raise the 480 grams of bottom, it is necessary 480/12=36,76 mechanisms (If THEY FELL VERTICALLY). But it should then be calculated where is the kinetic point of energy on a revolving wheel. If a mass falls along an arc of circle, its maximum energy is in the medium, with 90°. Thus its average is with 45°. BUT SINCE ON THIS WHEEL, IT There A TWO MASSES WHICH ARE ASSEMBLED At the same time: THAT OF the BOTTOM AND THAT the TOP, it is still necessary to divide by two. The average is with 22,5°.

: To remain in equivalence and simply to balance the wheel, i.e. to provide enough energy just so that it can raise the 480 grams of bottom, one needs 36,76 mechanisms all the 22,5° wheel. I.e. 588 mechanisms in all. It still does not move. So that it moves, it is necessary to put mechanisms in to exceed the 480 grams more. Each mechanism moreover will give 12 grams of additional force, and equivalence is broken. The wheel has enough force to make the effort push the 480 grams of bottom AND be unbalanced. The wheel is thus in permanent imbalance.

: There is certainly a friction in the axis. It is enough to suffic masse à pousser
: est finalement de 480 grammes.

: Pour que la roue puisse tourner, la force qui doit être PRODUITE doit être
: suffisante pour bouger ces 480 grammes.
: Il faut donc créer une inertie suffisante sur tout un coté de la roue pour
: dépasser ces 480 grammes.
: --------------------------------------------------------------------------------
: Pour lancer la roue, il faut lui faire faire un demi tour, à la main,
: en actionnant chaque levier de chaque mécanisme.
: Alors, la moitié de toutes les masses de la roue, bloquées par les gravitons,
: seront plus près du centre de la roue.
: Plus près de 34 mm.

: Quand les piques arrivent en haut et touchent la deuxième butée, ça libère
: les gravitons, et ça permet à chaque ressort de repousser la masse vers le haut.
: Donc de créer un déport de toutes les masses de l'autre coté de la roue.
: La roue est à priori en déséquilibre.
: Mais ça ne suffit pas du tout.
: Si on en reste là. Elle ne bouge pas.

: Un calcul (à base des formules courantes de Newton) permet de voir
: que l'inertie d'un mécanisme est de 12 grammes.

: Pour soulever les 480 grammes du bas, il faut 480/12=36,76 mécanismes
: (S'ILS TOMBAIENT VERTICALEMENT).
: Mais il faut ensuite calculer où se trouve le point d'énergie cinétique
: sur une roue tournante.

: Si une masse tombe le long d'un arc de cercle, son énergie maximale est au milieu, à 90°.
: Donc sa moyenne est à 45°.

: MAIS PUISQUE SUR CETTE ROUE, IL Y A DEUX MASSES QUI SONT MONTÉES EN MÊME TEMPS:
: CELLE DU BAS ET CELLE DU HAUT, il faut encore diviser par deux.
: La moyenne est à 22,5°.

: Pour rester dans l'équivalence et simplement équilibrer la roue,
: c'est à dire fournir juste assez d'énergie pour qu'elle puisse soulever
: les 480 grammes du bas, il faut 36,76 mécanismes tous les 22,5° de la roue.
: C'est à dire 588 mécanismes en tout.
: Elle ne bouge toujours pas.
: --------------------------------------------------------------------------------
: Pour qu'elle bouge, il faut mettre des mécanismes en plus pour dépasser les 480 grammes.
: Chaque mécanisme en plus donnera 12 grammes de force supplémentaire,
: et l'équivalence est rompue.
: La roue a assez de force pour faire l'effort de pousser les 480 grammes du bas
: ET d'être déséquilibrée.

: La roue est donc en déséquilibre permanent.

: Il y a certes un frottement dans l'axe.
: Il suffit d'ajouter suffisamment de mécanismes pour le vaincre.
: --------------------------------------------------------------------------------
: voilà, j'ai dit une bonne partie. Il y a encore des subtilités mécaniques,
: comme la nécessité que chaque pièce de type "bielle" (les parties mobiles du levier)
: soient neutres : c'est à dire avec un contre-poids qui lui donne un centre de
: gravité totalement équilibré.

:
: : G'day all,
: : Just a little update on my attempt to construct a mini Costa Wheel, I've managed to manufacture 2 mechanisms so far, 180 degrees apart and slowly
: : turn the whole unit by hand, with the mechanisms actuated by small levers attached to my work bench, this thing actually turns very "slowly", the only hang-up
: : I can see is the friction required to actuate the mechanisms at the bottom to lift the small mass approximately 3.5mm, maybe a slightly longer lever or more mechanisms will help??
: : Two more mechanisms will be finished soon and I will post the results of my possibly fruitless endeavours in the near future.
: : Smithy.




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