Consider a transmission for a tandem rotor helicopter with two engines and two main rotors such as the one below.

CH-47 Powertrain

Both engines are connected with sprag (one-way) clutches such that an engine's speed can underrun the rest of the system (e.g. engine failure), but normally they move in sync. The two rotors are always synchronized so they don't collide with each other as they spin (in opposite directions).

Let's assume that we know the moments of inertia for each engine and each rotor. There are accessories (not shown in the picture) which are driven by the same synchronizing shaft and there are of course mechanical losses, but we can neglect all those for simplicity. There is, however, a speed change between the engines and the rotors that needs to be taken into account.

In modeling this system, since each engine can underrun the rotors, I need to integrate the local accelerations to get the speed of the rotors and each engine separately. Generally, my question is: what torques and inertias does each "see" for the purposes for the purpose of finding an acceleration from unbalanced torques?

For the case that both engines are operating and driving (i.e. engaged clutches):

  • Would each component (engine 1, engine 2, and the rotors collectively) "see" all four torques and all four inertias (the four being two rotors and two engines)?
  • Or would each engine be unaffected by the torque and inertia of the opposite engine (i.e. due to the one-way clutch) and only "see" itself and the rotors (or maybe some portion thereof)?

For the case when an engine is being shut down and is no longer driving:

  • Would the shut down engine only see its own inertia and small and diminishing torque (and then you'd have to add some losses to actually get it to decelerate)?
  • Would the rotors "see" the torques and inertias of the rotors and the one remaining driving engine?
  • What about the engine that's still driving - the rotors and itself?
  • $\begingroup$ The image link is dead. Could you please repair it? $\endgroup$ – Nick Alexeev Sep 5 '15 at 18:47

In general, for all ridged bodies constrained to rotate at the same speed, their inertias and the torques acting on them can be added together:

$$\sum_i \tau_i=\dot\omega \sum_i I_i$$

Where $\tau_i$ represents the torques acting eternally on the system (like the torque produced by an engine, or the resistance of a propeller by the wind)

You can also model the system as system of equations:

$$\tau_i + \sum_j \tau_{ij} = \dot\omega_i I_i $$

Where $\tau_{ij}$ is an anti symmetric matrix representing the torque at the interface between components $i$ and $j$. The sprag clutches ensure that the torque between the engine and the rest of the system cannot be negative (ideally. However in reality there will be a small drag torque).

To calculate accelerations while both engines are engaged you can use the first equation, then use the second system of equations to determine the torque on the clutches. If the torque is negative for one engine, then that engine would no longer be constrained to rotate at the same speed. That means it can't be included in the first equation so you'd have to calculated it again, but this time exclude that engine. Then use the first equation on just the engine to determine engine acceleration.

If the engine rotational velocity ever exceeds that of the rest of the system then the clutch would reengage so now the velocities are constrained together so they must be treated like a system again (and to bring them back to exactly the same rotational velocity you could take the inertia weighted average of rotational velocity)

Note: if some ridged bodies are constrained to rotate a speed proportional to each other (eg. gears), the same equation can be used, but the inertia, torques, and speeds must be multiplied/ divided by the ratio.

If constraints: $$\omega_i=r_i\omega$$ exist, then: $$\sum_i \tau_i r_i=\dot\omega \sum_i I_i r_i$$

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