Posted by Øystein Rustad (217.70.229.44) on May 01, 2002 at 01:28:11:
In Reply to: Oystein, calculation you need to see! posted by Davis Landstrom on April 30, 2002 at 15:23:39:
Hello,,,
I very well know that the simple figure don`t speak clearly for it self, is a litle exaggerated to clarly show what I mean.
It don`t describe how the arm can be positioned
so. BUT the basic idea is :
When you take a classic attemtp, from history we know it balances.
Why? Because we know that a single weight can`t lift itself higher than it is liftet on the uppward moving side.
From this we know that as soon as an arm/position is altered in
a positive way, it will overbalance.
Let`s say we by our self liftet the hinged arm in the clasical
Leonardo da Vinci drawings.
When reaching zenith, we flip the arm up and ahead before it can flip itself from the height it was lifted on the uppward going side. The wheel will turn itself, BUT we must apply this energy to make it overbalance. This is simply because now weight*arm on the downward side is always totally bigger. (fact)
Then you say :
"The force I calculated was 19.52N, which means that in order for the wheel to move the 12 'O' clock mass into the 1:30 possition on the diagram the wheel would need to put out 4.88 more Newtons of force than it does."
And that is correct, if we assume that the mass is changing it`s
position only by interfere with an external object.
Ref. MT13 in Besslers Machinen Tractate.
(for those who have it)
Then the wheel would meet a resistance of maybe 19,5 N
(as you calculated) due to interference with an external object!
If we use external objects to move a weight, NO net gain is possible ! When the arm is lifted in my construction NO
external objects is involved, and no resistant force is meeting the wheel, Thus is raise "itself"!
Bessler certainly did know that if an arms motion is altered beyond the positions of an classic attempt, only from forces and work within the wheel itself, it will turn itself, ref. MT 9,10,11,14,15
Quote from MT/Besslerwheel.com :
however, taking various illustrations together and combining them with a discerning mind, it will indeed be possible to look for a movement and, finally to find one in them."
- Johann Bessler, first page of Maschinen Tractate
Here in MT 9,10,11,14 and 15 the arms are moved ahead before it should, also some moving backwards at the bottom, ONLY from the forces from other weights. If the construction principle he used here (tie the weights to eachother) would function, it would turn itself because of constant unbalance.
One way to achieve the motion I describe on my WEB-site (not saying that what I tell here will work) could be to put another
weight on the arm, so the other weight, when going up, pushes the arm backwards at bottom, inwards on the way up, become a counterbalance at zenith. But istead of moving the arm further innwards, a spring is streched, at the top the counterbalance + the spring raise the arm as described.
There you have the motion only from internal forces.
BUT, a BIG BUT, the counterbalance mentioned here will supply negative weight*arm to the wheel. (How could it not?)
So from this we see that we must use principles that can move the arm, without supplying any particular negative unbalace to the wheel. The extra weight/s must need this position for them self to balance through the revolution.
It should be possible?
Øystein
: I have just seen Oystein's web page, and am particulaly interested in the section entitled "Perpetual Motion, HOW?".
: This section shows a classical attempt at perpetual motion using swining weighted leaver arms and has a reason as to why it can not work. Below this there is another drawing of a perpetual motion machine upon which the conservative force field converter is based. This desighn like the classical perpetual motion machine displayed above it has weighted lever arms, however they are arranged in strange possitions, so I must assume that some mechanism moves and keeps them in these possitions during the cycle period.
: Below this perpetual motion machine desighn is what appears to be equa distances between masses and wheel centre of 0.75m, 0.50m and 0.25m respectivly, for the remainder of the masses I assumed distances of 1m)
: As Oystein pointed out there is a moment difference that is acting clockwise, using my figures I calculated it to be 14.64Nm, (I got the figure by subtracting the sum of the clockwise moments from the sum of the anti-clockwise moments)
: The problem arrises when you try to get mass M5 into the possition of mass M6 to perpetuate the motion of the wheel.
: I calculated how much force would be required to move the mass, to do this I re-aranged the moment equation to make force the subject;
: F = moment/distance
: I took the moment difference as the moment value and the difference in distance of masses M5 and M6 relative to the centre point of the wheel. (M6 = 1m from centre, M5 = 0.25m from centre, therefore difference = 0.75m)
: The force I calculated was 19.52N, which means that in order for the wheel to move the 12 'O' clock mass into the 1:30 possition on the diagram the wheel would need to put out 4.88 more Newtons of force than it does.
: I hope that I am wrong, and I hope that Oystein or someone else can show me why and explain the mechanism that would allow the mass transaction.