Weight question
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re: Weight question
Marchello...
I took another look at your animation and studied how the weights were moving around in the design. Although I was enthusiastic over the design initially, I think I can see an obvious reason why it would not work.
I noticed that at any instant three of the four weights are rising while one is dropping. Of the three rising weights, two are moving vertically to the top of the wheel while the other is sliding around the curved path to begin its vertical ascent to the top of the wheel to.
The problem I see is that while the single descending weight is farther from the axle of the wheel than the ascending weights, there are three times as many weights ascending as are decending. In other works, the changes in potential energy as the four weights move about is always equal to zero so there can be no acceleration of the wheel.
ken
I took another look at your animation and studied how the weights were moving around in the design. Although I was enthusiastic over the design initially, I think I can see an obvious reason why it would not work.
I noticed that at any instant three of the four weights are rising while one is dropping. Of the three rising weights, two are moving vertically to the top of the wheel while the other is sliding around the curved path to begin its vertical ascent to the top of the wheel to.
The problem I see is that while the single descending weight is farther from the axle of the wheel than the ascending weights, there are three times as many weights ascending as are decending. In other works, the changes in potential energy as the four weights move about is always equal to zero so there can be no acceleration of the wheel.
ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
Re: re: Weight question
It's like saying: because we have now the Fast-Fourier-Transform it will be less likely to get a PM... :sGeorg Künstler wrote:So Bessler was not able to used this transformation.
I meant, the PE en KE are periodically exchanged: [PE<-->KE], this will be balanced. So we need another, I don't know: maybe Inertia of a swinging Pendulum, or you know what. So we get [PE-->KE,IE] [KE-->PE,IE] [IE-->PE].
(Rupfen -Beschreibung)(Rupfen). Bessler named his device 'RUPFERE'.
Grobfädiges, locker eingestelltes Jutegewebe. Einsatz als Wandbespannung, für Dekoration und Polsterei etc., auch als Hessian bezeichnet. Jutegewebe
---
Looks like a reference to an oil stained cloth...?
the future is where you're looking at.
The three ascending weights times 1 quarter of the wheels diameter -versus- 1 descending weight times eehm...ken_behrendt wrote:the single descending weight is farther from the axle of the wheel than the ascending weights, there are three times as many weights ascending as are decending
hmm, have to be at home(/rest in mind) for that... sorry
Question, how do I lever 3 weights with 1 weight ? What are the sizes, and what are the rotation angles?
The animation was my solition, but I could be wrong.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Weight question
Yet another failure.. :-s
Another variation of a doublespeed wheel in WM2D
Another variation of a doublespeed wheel in WM2D
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
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re: Weight question
Hi Me,
you can find such developements, wheel in wheel with different speed, at Prof. Everts homepage
http://www.evert.de
He is the opinion that it works, I am not, so in that point we have different views.
the future has begun
Georg
you can find such developements, wheel in wheel with different speed, at Prof. Everts homepage
http://www.evert.de
He is the opinion that it works, I am not, so in that point we have different views.
the future has begun
Georg
re: Weight question
That's very interesting George, thanx
What's your view?
What's your view?
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Weight question
Hi, please can anyone find and draw COG point on this example picture ? There is eight not equidistant weights...How can I learn to calculate COG on draw like this ? ..thanks
Claudio
Claudio
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re: Weight question
Claudio...
There are two simple mathematical expressions for the center of gravity of a set of n weights.
Assuming that the weights all have the same mass, you can find the coordinates, X,Y, of a set of n weights by using the individual x, y coordinates of each of the weights. Or,
X = ( x1 + x2 + ... + xn ) / n
Y = ( y1 + y2 + ... + yn ) / n
Drawing your design on graph paper can make getting the coordinates of its weights fairly easy to do. Next, apply the above expressions and use a calculator.
Another method which I am using is to simply make a model of my design on a CAD program like WM2D. Then I activate the control that puts a CG marker for the system of weights on the wheel. Next, I pin the wheel so that it can not turn and then Run the simulation to make the CG marker appear. I can then rotate the model wheel a little and see what effect that action has on the position of the CG marker. Of course, this is only a "static" test. For a wheel in motion, the CG marker might behave differently due to the effect of centrifugal forces shifting the models weights around.
ken
There are two simple mathematical expressions for the center of gravity of a set of n weights.
Assuming that the weights all have the same mass, you can find the coordinates, X,Y, of a set of n weights by using the individual x, y coordinates of each of the weights. Or,
X = ( x1 + x2 + ... + xn ) / n
Y = ( y1 + y2 + ... + yn ) / n
Drawing your design on graph paper can make getting the coordinates of its weights fairly easy to do. Next, apply the above expressions and use a calculator.
Another method which I am using is to simply make a model of my design on a CAD program like WM2D. Then I activate the control that puts a CG marker for the system of weights on the wheel. Next, I pin the wheel so that it can not turn and then Run the simulation to make the CG marker appear. I can then rotate the model wheel a little and see what effect that action has on the position of the CG marker. Of course, this is only a "static" test. For a wheel in motion, the CG marker might behave differently due to the effect of centrifugal forces shifting the models weights around.
ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:
Vaver = -2(√2)πdωcosφ
Vaver = -2(√2)πdωcosφ
re: Weight question
The CoG is where the object is inbalance. You can say it is the average of the weight distribution (as many left as right).
CoG.X = SUM(X[1..n])/n ; or AVERAGE( X[n] )
CoG.Y = SUM(Y[1..n])/n ; or AVERAGE( Y[n] )
Simple as that.
I think your point is somewhere in the center of the crossing lines, or at least not very far away from that.
CoG.X = SUM(X[1..n])/n ; or AVERAGE( X[n] )
CoG.Y = SUM(Y[1..n])/n ; or AVERAGE( Y[n] )
Simple as that.
I think your point is somewhere in the center of the crossing lines, or at least not very far away from that.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Weight question
You'd beat me Ken :-)
When you use rods in WM2D then the connection point to the wheel can have an opposite effect of what you'd expected when just looking at the CoG of the connected weight.For a wheel in motion, the CG marker might behave differently due to the effect of centrifugal forces shifting the models weights around.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: Weight question
I need a ratchet that will allow only movement in one direction around a pin joint.
I've been trying out the rotational damper, will this damper act like a ratchet?
Like in a socket wrench.....
I've been trying out the rotational damper, will this damper act like a ratchet?
Like in a socket wrench.....
Robert (The Carpenter's Boy)
There's never time to do it right the first time, but there's always time to do it over again.
There's never time to do it right the first time, but there's always time to do it over again.
re: Weight question
Robert,
A damper can be made to work somewhat like a ratchet. I've tried many methods to make a good reliable ratched with WM2D but all come up short of my expectations.
To use a damper as a ratchet you set the Rot.Damper to "Active when" you want it to dampen the rotation, such as "Body[1].v.r > 0". Then you set how much torque it applies when active. The problem is that if you set the torque too low it "slips" and if you set the torque to high it will rotate the wheel. A weight at horizontal and at vertical require differing amounts of dampening torque. So no matter what you try it never acts like a true ratchet.
If anyone can figure out a way to make a good working WM2D ratchet I'd be very interested.
A damper can be made to work somewhat like a ratchet. I've tried many methods to make a good reliable ratched with WM2D but all come up short of my expectations.
To use a damper as a ratchet you set the Rot.Damper to "Active when" you want it to dampen the rotation, such as "Body[1].v.r > 0". Then you set how much torque it applies when active. The problem is that if you set the torque too low it "slips" and if you set the torque to high it will rotate the wheel. A weight at horizontal and at vertical require differing amounts of dampening torque. So no matter what you try it never acts like a true ratchet.
If anyone can figure out a way to make a good working WM2D ratchet I'd be very interested.
re: Weight question
Ken and Marchello many thanks :-) It is very simple.
Claudio
Claudio
re: Weight question
Hey Jim, I was just playing around with WM2D, and found that the rot. damper will never work, no matter how accurate. I made a little system and told it exactly what damping is necessary (for all positions, since it was a pendulum), and it didn't work. Then I realized, the damping only takes effect if the object is moving. So if your damping is enough to stop it, then it stops for an infinitesimal moment; its speed is zero, so there is no longer any damping, and the outside force (gravity in my case) makes it move. Once its moving again the damping takes place and it stops, etc. So it will always slip no matter what!
I think you can make one work if you apply a torque-motor with Body[1].v.r>=0 and a properly written torque equation. I don't know how to cite a body's angular position in an equation, I've been working off the constraint you posted, and have used -20*19*Body[1].a.r/16, where 19/16 is the value of my system's rotational inertia, and there are 20 frames/second (and negative because it works against the prefered acceleration). It still slips, and uphill at that, but I don't think it would if I used mgrcos(theta) by citing theta.
I think you can make one work if you apply a torque-motor with Body[1].v.r>=0 and a properly written torque equation. I don't know how to cite a body's angular position in an equation, I've been working off the constraint you posted, and have used -20*19*Body[1].a.r/16, where 19/16 is the value of my system's rotational inertia, and there are 20 frames/second (and negative because it works against the prefered acceleration). It still slips, and uphill at that, but I don't think it would if I used mgrcos(theta) by citing theta.
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.
re: Weight question
The most reliable way to make a one-way rachet that I know of (actually Tinhead showed me) is to make a saw toothed drive wheel & have it collide with a spring loaded locking bar. Much like a clock escapement.
The spring pressure allows one way movement but holds tight when opposite torque is applied. It can be finicky & takes perseverance so that it works reliably. I'll try & find a screen shot if I still have one.
One good thing about this rachet method is that it can allow a sort of controlled feed-back pulse into the system a bit like perhaps the stampers on JB's wheels might have done.
Jim-mich already posted the polygon function to create a saw toothed wheel so it could probably be adapted.
EDIT: Here is a feed-back rachet design that is geared to the wheel & is probably easier to make. See attachment.
The spring pressure allows one way movement but holds tight when opposite torque is applied. It can be finicky & takes perseverance so that it works reliably. I'll try & find a screen shot if I still have one.
One good thing about this rachet method is that it can allow a sort of controlled feed-back pulse into the system a bit like perhaps the stampers on JB's wheels might have done.
Jim-mich already posted the polygon function to create a saw toothed wheel so it could probably be adapted.
EDIT: Here is a feed-back rachet design that is geared to the wheel & is probably easier to make. See attachment.
re: Weight question
You can make a reliable ratchet in WM2D by adding a formula to a rigid joint's Active when property.
Something like:
and(Body[1].p.r<0,Body[1].v.r>0)
It's been working fine for me in a number of simulations. Your mileage may vary depending on what your up to.
Then there's always modeling all the geometry of a ratchet. ;-)
-Ed
Something like:
and(Body[1].p.r<0,Body[1].v.r>0)
It's been working fine for me in a number of simulations. Your mileage may vary depending on what your up to.
Then there's always modeling all the geometry of a ratchet. ;-)
-Ed