# Variable Gear System [closed]

I have a variable speed gear system with 4 primary parts. I need to find the relationship between input rotational speed (wi) to output rotational speed (wo).

Fig. 1 shows three of the primary parts - input wheel (radius ri) , satellite wheel (radius rs) and connecting arm with a slot. The connecting arm can rotate "relative" to the input wheel about its axis A. The satellite wheel is on other end of connecting arm mounted on a ratchet bearing that allows spinning only in counter-clockwise (CCW) at O. When "wi" is in clockwise (CW) direction satellite wheel spins about axis $O$ in CCW direction at free spinning speed $\omega_s$ without rotating about $A$. When $\omega_i$ is in CCW direction satellite wheel is locked by its ratchet bearing at $O$ and hence two wheels stay at the same teeth contact location and whole system rotates at $\omega_i$ in CCW as a rigid body about $A$, $\omega_c=\omega_i$ (see Fig. 2).

Fig. 3 shows the 4th part of the system – output arm (radius ro). Output arm axis (B) is first placed "in front of" (or behind) the axis “A” and its pin is slidably connected to the slot in the connecting arm as shown in Fig 4. When axes A and B are aligned the system either moves as explained in Fig. 1 or Fig. 2 based on the input wheel "wi" direction of rotation. Now, we shift the axis B by a distance “x” (Fig. 5) and the input wheel ROTATES at a constant speed (wi) in CCW direction. Then depending on its location (how far from B) the satellite wheel either locked and rotates about A or it rotates at aspeed "wc" about A while spinning (ws) about O at varying speeds. i.e. wc >= wi and ws >= 0. With this the output arm also rotates at variable speed (wo). Can someone please show me the relationship of wi to wo in this system? I appreciate if the key equations and steps are shown.

## closed as off-topic by ACuriousMind♦, John Rennie, Kyle Kanos, JamalS, Qmechanic♦Mar 30 '15 at 0:57

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• This is an incredibly complex "first question"! I suggest that you draw a little reference "dot" on every part, and try to figure out how each part moves relative to each other part if the input wheel moves by a small amount. In other words, draw the system in two states that are just a short distance apart. Geometry should then do the rest. I have difficulty following the operation in fig 4/5 - but you obviously understand it so you are probably the best person to answer your own question... – Floris Mar 28 '15 at 18:33
• I need a clarification: Is B fixed to and rotates with gear A? or the axis A-B must stay always horizontal? – user66432 Mar 28 '15 at 18:37
• No. The axis A is the input wheel axis and to it the connecting arm is mounted with a bearing so that the connecting arm can rotate independently about axis A. The axis B is the output arm axis which is parallel to the axis A but completely independent. The Output arm is only connected to the rest of the system by its pin to the slot in the connecting arm. Thanks. – mo jayas Mar 28 '15 at 18:59
• Hi Bruce, The axis A is the input wheel axis and to it the connecting arm is mounted with a bearing so that the connecting arm can rotate independently about axis A. The axis B is the output arm axis which is parallel to the axis A but completely independent. The Output arm is only connected to the rest of the system by its pin to the slot in the connecting arm. Output arm rotates about axis B while transnationally held at a distance "x" from axis A. – mo jayas Mar 28 '15 at 19:10
• Hi Bruce, What I am looking for is an equation for wo in terms of wi and other geometric parameters such as x, radii (ri, rs, ro) and time t. In your equation wc is also an unknown quantity. – mo jayas Mar 28 '15 at 22:25

I do not believe the pin has any influence in the motion of the gears. At least not if the pin is light enough. The reason is that in such a case internal forces will be zero. Let us look at the two extreme cases:

1) Minimum $x$: $x=0$, then the pin (and the arm) will simply rotate parallel to the bearing at $\omega_c$

2) Maximum $x$: $x=r_0$, then the pin will be always at A and the arm will remain horizontal.

For any other intermediate cases you can calculate $\omega_0$ using the equation of the straight line and the fact that the bearing always rotates at $\omega_c$

The position of the pin can be calculated as the intersection of the two curves: the straight line (the bearing): $y=\tan (\omega_c t) x$ and the off-circle $y=\sqrt(r_0^2-(x-x_0)^2)$ (for $y>=0$) and $y=-\sqrt(r_0^2-(x-x_0)^2)$ (for $y<0$). Where I call $x_0$ to what you called $x$.

• Hi Bruce, x=0 condition is correct. I am sorry that part I have not made clear. x < ro. every 360 degree turn of output arm pin goes around "A" in an elliptical path and satellite wheel comes closer to when the pin is on the "right side" of B and satellite wheel is far away when the pin is left side of "A". The distance from axis "B" to the satellite axis "O" is in the range (ri + rs - x) and (ri+ rs + x). – mo jayas Mar 28 '15 at 20:04
• Hi Bruce, x=0 condition is correct. I am sorry, 2) condition I have not made clear. Always x < ro and every 360 degree turn of output arm, the pin goes around "A" in an elliptical path and the satellite wheel comes closer to "B" when the pin is on the "right side" of B and the satellite wheel is farthest away when the pin is left side of "A". The distance from axis "B" to the satellite axis "O" is in the range (ri + rs - x) and (ri+ rs + x). So, the output arm is forced to rotate by the pin. – mo jayas Mar 28 '15 at 20:11
• Hi Mo, it is fine, I just want to know if you agree to the fact that the motion of the satellite gear is not influenced by the presence of the pin, or if you believe it will be affected. – user66432 Mar 28 '15 at 20:14
• In addition, the path of the pin around A must be a shifted circle, not an ellipse (well I know a circle is an ellipse, but that is not what I meant). Do you agree? – user66432 Mar 28 '15 at 20:17
• Hi Bruce, I agree. The pin rotates about B in a circle with the output arm of radius ro, and about A in an off-centered circle (Not exactly an ellipse). Thanks. – mo jayas Mar 28 '15 at 20:24