Hello all, here's "An Exploratory Research Mechanism....



A uniquely balanced experimental mechanical arrangement, the Mechanism's motion is pendulous, but unlike a simple pendulum which has two possible positions of equilibrium (un-stable when up or stable when down), this Pendulum, because of the way it's balanced, actually has four possible positions of equilibrium.... two un-stable positions alligned with the force of gravity (pendulum un-stable when up or down vertically).... and two stable positions perpendicular to the force of gravity (pendulum stable when positioned to either side horizontally).
The gravitational force itself is not switched or turned on and off, the influence that gravity has on the Mechanism is changed by changing the Mechanism's condition.

The reason no mass is explicitly stated anywhere in the analysis is because I didn't see the need. The length of a line represents the magnitude of a force and the arrow itself represents the direction of a force. For example....
The situation graphically depicted in the diagram below won't change as long as any arbitrarily stated magnitude of force for the vector A is uniformly applied as a standard. In other words.... Whether one arbitrarily states for the vector A a magnitude of force equal to two ounces or sixteen pounds the resulting diagramatically shown vector proportions won't change in any way, and the diagram will remain an accurate representation for both scenarios (two ounces or sixteen pounds). So, since any arbitrarily stated magnitude of force for the vector A will result in an identical diagram and identical vector proportions, for the purpose of analysis, there's no need to state any specific magnitude of force for the vector A in the diagram.



It's the same for all of the scale drawings in the analysis. For example....

The situation graphically depicted in the scale diagram below won't change as long as any arbitrarily stated magnitude of force for the vector D is uniformly applied as a standard. In other words.... Whether one arbitrarily states a magnitude of force equal to two ounces or sixteen pounds for the vector D, the resulting diagramatically shown vector proportions in the scale drawing won't change in any way, and the diagram will remain an accurate representation of both scenarios (two ounces or sixteen pounds). Again, since any arbitrarily stated magnitude of force for the vector D will result in an identical diagram and identical vector proportions, for the purpose of analysis, there's no need to state any specific magnitude of force for the vector D in the diagram.



Whenever an arbitrarily stated magnitude of force for the vector D (or any other vector in the diagram) is uniformly applied as a standard, the magnitude of force associated with any of the other vectors in the scale drawings of the analysis can be quickly and easily derived. For example....

If the vector D is made to equal one inch and the arbitrarily stated magnitude of force associated with it is two ounces (one inch equals two ounces), then....

A.... 3/8 inch equals 0.75 ounces
B.... 3/4 inch equals 1.50 ounces
C.... 3/4 inch equals 1.50 ounces
E.... 3/8 inch equals 0.75 ounces
F.... F = C + B.... 0 ounces

If, instead, the vector D is made to equal one inch and the arbitrarily stated magnitude of force associated with it is sixteen pounds (one inch equals sixteen pounds), then....

A.... 3/8 inch equals 6 pounds
B.... 3/4 inch equals 12 pounds
C.... 3/4 inch equals 12 pounds
E.... 3/8 inch equals 6 pounds
F.... F = C + B.... 0 pounds

As you can see, for the purpose of analysis the very same numerically un-adorned diagram serves to describe both of the above scenarios equally well.

In the examples given above, there's no difficulty of description nor is there any appeal to intuition. Of course, one could go ahead and arbitrarily state this or that magnitude of force for the vector D and then carry out a thorough numerical analysis of the diagram, but that wouldn't change the result already shown diagramatically or the relative proportions of any of the various vectors depicted in the scale drawings of the analysis, it would only be a more specific confirmatory restatement of the generalized result already shown.

Using vectors, the diagram (below) illustrates both the direction and magnitude of the various forces arising from the various moving parts of the mechanism individually and shows (FIG. 4) how they ultimately cancel each other out.

FIG. 1 - Schematic representation of the Chassis.

FIG. 2 - The Chassis is fixed in this schematic. The diagram shows the downward force A of the Pendulum and the resulting force B on the Planet Sprocket.

FIG. 3 - The Sun Sprocket is fixed in this schematic. The Chassis and the Planet Sprocket are free to rotate. The diagram shows the downward force D of the planet sprocket. The force C on the Planet Sprocket is the result of the force D after the force E from the oppositely situated Counter Weight (fixed to the chassis) is subtracted, or.... D minus E equals C.

FIG. 4 - The Sun Sprocket is fixed in this schematic. The Planet Sprocket with its attached Pendulum and the Chassis are free to rotate. The equal and opposite forces B and C acting on the Planet Sprocket effectively cancel each other out and equilibrious balance F is the result.



A series of schematic diagrams (below) show how the equal and opposite forces B and C cancel each other out at various points around 360 degrees (the sun sprocket is fixed for this part of the analysis), presented here as an animation....



In order to render the mechanism purturbable the sun sprocket must be free to move. When it's free to move the mechanism's equilbrium (which was stable at all points around 360 degrees when the sun sprocket was fixed) can be purturbed via the chain by a slight change in the position of the sun sprocket by means of the control lever, which is fixed to the same axle as the sun sprocket. This is also the condition in which four distinct positions of equlibrium emerge. I found a video of an older model (balanced the very same way as the current model) that clearly demonstrates the four possible positions of equilibrium that arise when the sun sprocked is freed to rotate (two stable and two un-stable), appearing in the same order as listed below the video. The video also shows how the mechanism can be caused to rotate as easily in one directon as the other....

[size=150]http://www.youtube.com/watch?v=OoF3zUu8G9s

1. Pendulum horizontal to the left, stable equilibrium.... the mechanism can't be caused to rotate by the action of the control lever from this position.

2. Pendulum horizontal to the right, stable equilibrium.... the mechanism can't be caused to rotate by the action of the control lever from this position.

3. Pendulum down vertically, un-stable equilibrium.... the mechanism can be caused to rotate by the action of the control lever from this position.

4. Pendulum up vertically, un-stable equilibrium.... the mechanism can be caused to rotate by the action of the control lever from this position.

This constitutes a perturbable form of balance that can result in immediate onset of rotation (in either direction), presented here as an animation....



A problem then arises as a direct result of the sun sprocket being freed to rotate for the purpose of perturbing the mechanism's equilibrium via the chain. The varying forces arising from changing mass distribution during rotation that was formerly transmitted directly to the stand when the sun sprocket was fixed now come to bear on the control lever instead. The diagram (below) shows the downward force D on the Planet Sprocket. The force H on the Sun Sprocket is the result of the force D, and the force I on the Control Lever is the result of the force H. The Mechanism is not balanced or in equilibrium in this diagram because there is no equal and opposite force to counter the force I.


That's where the calibrated spring comes in....

http://www.youtube.com/watch?v=P_vF3ooVwAU&feature=player_embedded

....it's mounted on the back of the Mechanism (depicted to the right in the diagram below). The lower end X is fixed to the stand the mechanism is mounted on. The upper end Y is connected to the Control Lever. The diagram (below) shows how the equal and opposite forces I and J effectively cancel each other out and equilibrious balance Q is the result, or.... I minus J equals Q. The Mechanism is in a state of compensated equilibrium, the sum of all forces acting on the control lever is zero.



I want to minimize the magnitude of the input force needed to perturb the system.... the calibrated spring variably compensates for and cancels out the varying force coming to bear on the control lever due to changing mass distribution. The sum of the equal and opposite forces I and J coming to bear on the control lever equals zero at all times during rotation as shown (below). This constitutes a compensatory form of balance. It reduces the input force needed to cause immediate onset of rotation to the level of that needed to overcome only inertial and frictional resistance, presented here as an animation....



Analysis is not yet complete, coming up.... Timing, Acceleration, Force, Scale and Possible Applications....

More details.... http://thecolemechanism.blogspot.com/