For a simple planetary gear system with a single input shaft of a known torque, the output torques of the other two shaft can be calculated using the equations below (for a steady state system).

Torque ratios of standard epicyclic gearing

If I were to apply torques to two of the shafts how could I go about calculating the torque of the final shaft for a steady state system?

My intuition is that for a steady state system the sum of external torques (the shafts) must be 0. In this case the output torque will be the sum of the two inputs.

Edit: By the above comment I mean Tt + Tc + Ts = 0

Am I missing something?

Planetary gear system

  • $\begingroup$ Not worth a whole answer but you're correct. Some automatic transmissions work on a similar principle. $\endgroup$ – jko Mar 8 at 12:50
  • $\begingroup$ @jko I'm thinking that the sum of the power should be equal ( not of the torques). The torque depending on the diameter would be different? Am I wrong in thinking that? $\endgroup$ – NMech Mar 8 at 13:41
  • $\begingroup$ @NMech power is torque x speed, you usually evaluate the gear train independent of speed for systems like this. You can use either the pitch diameter or number of teeth (N in the above equations) to determine the torque ratio, that is what normalizes the torques relative to radii. $\endgroup$ – jko Mar 8 at 13:57
  • $\begingroup$ That's what I thought. If I understand your reply correctly then the statement from the OP that steady state system the sum of external torques (the shafts) must be 0 is not accurate. $\endgroup$ – NMech Mar 8 at 14:00
  • $\begingroup$ I assumed the intent of OP's statement was the net output torque is the sum of input torques multiplied by their respective tooth ratios. This may require a complete answer after all. $\endgroup$ – jko Mar 8 at 14:29

If by "steady state" you mean the output shaft isn't rotating, you are correct if the applied torques cancel out.

For example, if the sun shaft is your desired output with the ring gear and carrier as the inputs, rotating in the same direction, the net torque will be (Tr x Ns/Nr)-(Tc x Ns/(Nr+Ns)). The output torque is the sum of the input torques multiplied by their respective tooth ratios.

  • $\begingroup$ By 'steady state' I mean that none of the shafts are accelerating. NMech is correct in his comment previously where he interprets my question as 'asking that what would happen on the 3rd shaft, if the other two are excited.' $\endgroup$ – jake love Mar 8 at 15:36
  • $\begingroup$ Oooookay. Yes you are correct. A simple analogy would be 2 gears in mesh, with equal torque rotating in opposite direction. As long as the torques cancel out there won't be any external forces on them causing acceleration/displacement. The same applies for any number of gears. $\endgroup$ – jko Mar 8 at 15:54
  • $\begingroup$ @jake love I am somehow getting lost here. If the sum of external torques equals zero, then the torques must be rotating in different directions, then the output should be zero. What exactly is your question? Also, the sketch lacks the details required to understand the question correctly. $\endgroup$ – r13 Apr 7 at 17:39

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