Bessler wheel lesson 1

[Besslers Assistant]

The power of Bessler’s Wheel

Bessler wheel gains its’ power from centrifugal forces and here is how it is done.

First the math.

My example is of a pendulum held at 90 degrees form the base, the base being the lowest point or the 270 degree mark. Numbers used is only to make the calculations easier.
I have a pendulum 10 meters long with a mass of 10kg, gravity is 9.8 m/s^2
Potential energy of my pendulum at 90 degrees to the base: P.E. = mgh = 980 J
Velocity at this point is v = square root (2gh) = 14 m/s
The 980 J is the energy I gain from gravity while falling to the rest position
At the base if I let it fall one meter, my P.E. will be = mg * 11 = 1078 J
Velocity at this point is v = square root (2g * 11) = 14.7 m/s
The 1078 J is the energy needed to restore the pendulum to its original height.
The difference in energy is 98 J, so I need 98 J to restore the pendulum to its original height. But what am I gaining at the base. At the base I am moving one meter downward, My energy will be.
P.E. = (mg + CF)*h
CF = (mv^2)/h = (m * (2gh))/h = 2mg
Therefore ; P.E. = (mg + 2mg) *h = 3mg*one = 294 J
Therefore I gain more energy in dropping my mass one meter downward than that of gaining some length of my pendulum. From the above calculations I only need 98 J to restore my pendulum to its original height (294 J – 98 J = 196 J.), that leaves me with 196 J extra. But I still have to restore my pendulum mass to its original 10 meter length and pay Mr. Friction. I believe 196 J can do this and more.

Second the mechanism.

We have a pendulum and a disc on the same axle, see the picture. From the furthest end of the pendulum arm we have the pendulum mass. Above the pendulum mass we have a small geared wheel and above this we have the disc. The small geared wheel is on the pendulum arm, also there is a restoring spring form the pendulum mass to the pendulum arm. There is a trigger plate connecting and holding the pendulum mass to the pendulum arm. There is a hold, lock and transfer mechanism connected to the pendulum arm that sits between the small geared wheel and the disc. There is a geared arm connected to the pendulum mass and the small wheel.



The pendulum is pulled up to the 90 degree position. At that position we have the HLT box holding and locking the disc and the pendulum together. Left to fall and rotating clockwise, both the pendulum and the disc moves together. When they reach the base, the HLT box releases the lock and just holds the pendulum and the disc together; the trigger plate is hit and the pendulum mass falls. The falling mass pulls on the geared arm and turns the small wheel clockwise (which is the same direction as the pendulum). The small wheel interacts with the HLT box and causes the disc to turn counter-clockwise, because of these motions, the pendulum and the disc share the energy of the pendulum mass moving downward. Just after this the HLT box allows the disc to spin freely. When the pendulum reaches its highest point and changes its direction to counter-clockwise, the HLT box holds and locks the disc to the pendulum. Since the disc is moving counter-clockwise the remaining energy of the disc is put into the pendulum. So at this point you have almost all of the energy (minus friction) of the falling pendulum mass being put back into the pendulum’s swing.

Please Comment.
Bye for now.

[eccentrically1]

The way I'm seeing it, if the small wheel turns CW and then interacts with the CW disc, it would be a brake on the disc.

[Tarsier79]

You still need to regain the height you have dropped the pendulum mass.

[Besslers Assistant]

@eccentrically1, the small wheel sits on the pendulum, the pendulum and the disc moves the same, so the small wheel sees the disc as a stationary object.
when the small wheel goes clockwise ( remember the force you are applying 3mg ) the disc will go counter-clockwise to the outside observer.


@Tarsier79, From the example's math, there is an excess of 196 J ( or I can lift my pendulum mass two meters) since I would be at the other 90 degree mark.
(if I freeze this position a sec.)it will take a fraction of this energy to restore the pendulum mass to its 10 meters position (since I would be moving the
mass horizontally).friction could take half this energy ( 98 J) and I believe I would still have some remaining to gain a little more height than that I had originally started with.


does this help.
Please more Comments.

[daxwc]

If this was true wouldn’t a bungee jumper that jumped pendulum style go higher than his starting point as long as his cord had spring? You can’t bungee him down and lock in energy as now you increased pendulum length. If you dropped 11m you have to raise 11m not 10, I do believe.

So you are saying because of your frame of reference you can lock in more energy than a 0ne meter drop.... now how do you harness it to reset it?

[Besslers Assistant]

@ daxwc, gravity gives it and gravity takes it away.
A bungee cord is like a spring, if you allow it to slowly absorb energy it will only give you back what it took.but if you give it too much too fast it will only absorb what it can handle and nothing more. My pendulum is 10 meters, but it is my Pendulum mass that is falling one meter. It is falling one meter when it is at the base position and not before. The mechanism above changes with the changing effects of CF. but the extraction point is at the base of the pendulum, because CF is at its' highest there. Remember the higher the CF the faster the pendulum mass will drop, the faster it drops the faster the disc will spin ( because the energy is shared with only the pendulum and the disc). In the picture I use a restoring spring, because it is easier to understand the effect a spring has, but it will cost you 98 J. if there is a different mechanism that uses less energy to do the same thing more energy would go towards swinging the pendulum.


I will give you a heads up, lesson 2 is extraction of energy.

[Ben]

what huh?

[daxwc]

You still increased pendulum length from 9m to 10m then, but you have 1 meter of cf forced locked, yet you have to raise the mass to 10m ... I await the harvest ;)


From the middle of wheel radius, the bob starts 4.5m at noon, but at 6pm it is at 5m… correct? You have to go from where your bob (mass) is not the over all length of the pendulum. It doesn't matter that it hangs out a meter.

[Ben]

who me?

[eccentrically1]

@eccentrically1, the small wheel sits on the pendulum, the pendulum and the disc moves the same, so the small wheel sees the disc as a stationary object.
when the small wheel goes clockwise ( remember the force you are applying 3mg ) the disc will go counter-clockwise to the outside observer.

Maybe when you build it the orientation will be easier to see. The disc may be stationary from the small gear's perspective, but in reality the disc is spinning CW, because in your description the pendulum and disc are locked together. When the box unlocks the various parts and engages the two gears, you'll hear a grinding sound of two gears going in opposite directions. The small gear will lose the battle .

[Besslers Assistant]

@eccentrically1, You are correct in saying the reality of the spinning disc is CW, However the the HLT box does not engage the two gears. The HLT box (Hold, Lock and Transfer box) is there to Transfer the motion of the small geared wheel to the disc, it does not have any power to turn anything. The small geared wheel gets its' power from the pendulum mass dropping, and is much more powerful than the disc's motion. Also when the small geared wheel starts to turn, the disc is stationary (relative to the small geared wheel), so there is no grinding action.

[Tarsier79]

B.A.

I believe Dax has already covered the missing link in your maths. Think about what will happen to your pendulum as you drop the mass, and the rotation of the pendulum slows as the period (and inertia) increases. Your pendulum mass will lose height as you drop it.


Or alternatively, why not build the pendulum part to experiment.

[Besslers Assistant]

@ daxwc and Tarsier79, I hope this explanation works, if not let me know, I’ll try again. Look at the picture below,




I have a 10 meter point and an 11 meter point and the pendulum mass is the small circle. We start at position 1, where the pendulum mass is locked into the 10 meter position on the pendulum arm. When released the pendulum gains energy from gravity and centrifugal forces begin to increase. At position 2 the lock is released and the pendulum mass falls from the 10 meter position to the 11 meter position. The pendulum losses some energy, but because of centrifugal forces the fall of the pendulum mass gains three times the energy the pendulum loosed. This energy is shared with the pendulum and the disc (as per description above). So, according to how massive you make the disc, the pendulum may reach the top at position 3. If the disc is lighter it may spin very fast due to the energy, and you would have to make another device to use the spin of the disc to set the pendulum back to the 10 meter position. In the design above I use a recovery spring, the spring could be set to 60 degrees (half weight of pendulum) or another angle where you would extract less energy from the pendulum/disc energy gain to correct the pendulum mass to the 10 meter position. If you use the recovery spring idea the pendulum mass would correct itself and the remaining energy would be in the disc. But remember the disc will be rotating opposite to the movement of the pendulum. But the pendulum would change direction on the half cycle. So when the pendulum goes down again the disc is stopped by the HLT box (which is on the pendulum arm) and the remaining energy of the disc is put into the pendulum.

Bye for now.

[Tarsier79]

Thanks BA, I'm pretty sure we both got that the first time. My. understanding of Position 2 is that any energy gained by dropping your mass is the same amount lost in rotation. We have pondered over, and worked on this type of mechanism before. One thing I found with this expanding radius mechanism, is the weight must follow a particular path, or you lose much more than you harvest.

[jim_mich]

You seem to indicate that CF will cause the weight to move outward at location 2. This is wrong. The inertial momentum of the moving weight will cause it to continue moving to the left. Gravity will cause it to move downward. The winner depends upon velocity. Many think that CF is an outward force. In a way it is, but in another way it is not. This is because the inertial force is actually in a tangent line. It is in the direction that the weight was moving just before it is released. In this case that direction is sideways to the left. This will cause the weight to move outward simply because the weight tends to move in a straight while the pendulum (or wheel) travels in a curved circular path.