Zeroing in on Bessler's wheel

[johannesbender]

Depends on if your extra mechanisms and weights give constant torque right , keep on trucking.

[Sam Peppiatt]

That's true jb, it's kind of a chicken or egg type situation. Which comes first. For the rollers to do any thing, the wheel has to turn. And the only way it will turn is if they, the rollers, shift the weights back and forth.

Will try it and see what happens-------------------Sam

[Sam Peppiatt]

The work is so difficult and the talk so easy---------------------------Sam

[Sam Peppiatt]

I can see already that it's not going to work. At 9 & 3 it's good but at 12 & 6, it tends to roll the roller backwards. Have to find another way----------------Sam

[Robinhood46]

Oh come on Sam, we were counting on you.
Can you not just pretend it will work, but you've decided not to tell us?
What about, it will definitely work, but we have to wait a bit before you can show us?
There must be some way we can cling onto the hope that you've found it, hope is all some of us have left.
I'm sure you can make up any old nonsense to convince yourself it will work, other members have been doing it for years, i don't see why you can't.
I'm disappointed, i was enjoying the illusion that you might have found the solution.

[johannesbender]

I can see already that it's not going to work. At 9 & 3 it's good but at 12 & 6, it tends to roll the roller backwards. Have to find another way----------------Sam

I guess 12 and 6 is where the weights lift ? It would make sense if it were , yes sideward movement is really easy and lifts arent , however every sideward movement becomes a lift requirement past 6 and again at 12.

[Sam Peppiatt]

Robinhood46,
It reminds me of Casey at the bat; mighty Casey has struck out. No joy in Mudville! Don't worry I'll find a way,(maybe)--Sam

jb, right; when it try's to lift the weights at 6 & 12 it rolls backwards and causes back torque------------Sam

[Tarsier79]

Well discovered Sam. Now you know the problem that exists for any gravity powered weight shift wheel.

[Sam Peppiatt]

Hi Tarsier,
I still think there must be a way to do it. But, I have to admit, I could be wrong--------------Sam

[johannesbender]

I guess one could say the down side of it is upwards ...

If we have 2 wheels , each 1 meter radius , each with a weight of 1kg .

One wheel is a radius change concept that brings the weight in at 6 and out at 12

The other wheel is a non radius change concept , the weight goes around at the same radius .

Both wheels start with the weight out at 12 at the same radius and both swing down to 6.

Both wheels weights will have the same energy at 6 , because they are the same weight traveling at the same speed , they fell the same height under gravity.

At this point , which wheel design needs less energy to get back up to 12 ?

Is it :
A) The wheel with no radius change .
B) The wheel with the radius change .
C) Both needs the same energy .

Why ?

[Sam Peppiatt]

jb,
Seams like they would be different. Not sure. Maybe keeping the weight at the same radius is better.

I want to change gears, and try to do some thing, to make the rollers roll foreword, to turn the wheel.

I found out that, if they do roll backwards, the wheel turns the wrong way. So, maybe I need to find a way to get the rollers to drive the wheel, instead of having the rollers lifting weights------------Sam

ETA
Maybe I should use the toggle to shift the rollers-----------------

[Tarsier79]

At this point , which wheel design needs less energy to get back up to 12 ?

They will be very similar . The PE requires the same to lift in both, but shifting the weights in and out will require a little extra due to CF. Shifting the weights (assuming both are paired) will take a little extra to accelerate one arm faster and one arm slower momentarily. Shifting against CF might require slightly more.

[johannesbender]

I guess one could say the down side of it is upwards ...

If we have 2 wheels , each 1 meter radius , each with a weight of 1kg .
One wheel is a radius change concept that brings the weight in at 6 and out at 12
The other wheel is a non radius change concept , the weight goes around at the same radius .
Both wheels start with the weight out at 12 at the same radius and both swing down to 6.

Both wheels weights will have the same energy at 6 , because they are the same weight traveling at the same speed , they fell the same height under gravity.

At this point , which wheel design needs less energy to get back up to 12 ?

Is it :
A) The wheel with no radius change .
B) The wheel with the radius change .
C) Both needs the same energy .

Why ?

Untitled.png

In the above image , any path taken between A and B , needs the same energy , so work for , the green path = work for purple path = work for blue path .

When its the same height and same weight , then it needs the same energy , so both of them require the same energy , neither has any advantage in terms of energy , which means energy to lift the weights is a different problem.


Wikipedia :

The work done by the gravitational force on an object depends only on its change in height because the gravitational force is conservative. The work done by a conservative force is equal to the negative of change in potential energy during that process.

For example, if a child slides down a frictionless slide, the work done by the gravitational force on the child from the start of the slide to the end is independent of the shape of the slide; it only depends on the vertical displacement of the child.

Taking it directly from another physics web page too, so we may stay correct :

(I think you have to click on the image to animate)

au.gif

Observe in the animation that each path up to the seat top (representing the summit of the mountain) requires the same amount of work. The amount of work done by a force on any object is given by the equation

Work = F * d * cosine(Theta)

where F is the force, d is the displacement and Theta is the angle between the force and the displacement vector.

The least steep incline (30-degree incline angle) will require the least amount of force while the most steep incline will require the greatest amount of force. Yet, force is not the only variable affecting the amount of work done by the car in ascending to a certain elevation. Another variable is the displacement which is caused by this force. A look at the animation above reveals that the least steep incline would correspond to the largest displacement and the most steep incline would correspond to the smallest displacement. The final variable is Theta - the angle between the force and the displacement vector. Theta is 0-degrees in each situation. That is, the force is in the same direction as the displacement and thus makes a 0-degree angle with the displacement vector. So when the force is greatest (steep incline) the displacement is smallest and when the force is smallest (least steep incline) the displacement is largest. Subsequently, each path happens to require the same amount of work to elevate the object from the base to the same summit elevation.

Another perspective from which to analyze this situation is from the perspective of potential and kinetic energy and work. The work done by an external force (in this case, the force applied to the cart) changes the total mechanical energy of the object. In fact, the amount of work done by the applied force is equal to the total mechanical energy change of the object. The mechanical energy of the cart takes on two forms - kinetic energy and potential energy. In this situation, the cart was pulled at a constant speed from a low height to a high height. Since the speed was constant, the kinetic energy of the cart was not changed. Only the potential energy of the cart was changed. In each instance (30-degree, 45-degree, and 60-degree incline), the potential energy change of the cart was the same. The same cart was elevated from the same initial height to the same final height. If the potential energy change of each cart is the same, then the total mechanical energy change is the same for each cart. Finally, it can be reasoned that the work done on the cart must be the same for each path.

Bessler writes something wise too , from which a person can deduce that he knew torque alone is not an answer :

some mobile makers think , If their things just move Out a little further here Than there o! That's how it's going to work;
I experienced this myself With sheer effort many years ago .

While the torque created by the differences in distances ,from the center of rotation to the locations of the weights does create rotation when they fall.

The weights moving in and out does not create extra energy nor does it reduce the energy that is needed to lift the weights back up.

Moving the weight across the same distance from the bottom to the top costs the same energy no matter what path or angle the weight has to travel along to get there.

[johannesbender]

At this point , which wheel design needs less energy to get back up to 12 ?

They will be very similar . The PE requires the same to lift in both, but shifting the weights in and out will require a little extra due to CF. Shifting the weights (assuming both are paired) will take a little extra to accelerate one arm faster and one arm slower momentarily. Shifting against CF might require slightly more.

Tarsier your thinking further than me at this point with CF..

[Sam Peppiatt]

To@,
I'm trying to figure out how to do it, not why it's impossible----------------Sam