I mean to say that the torque on the housing (which is the same as the ring gear) is the net torque between input and output.
You have to treat the gear drive like a "black box" and temporarily disregard what's going on inside. In this case there are really only three things acting on the box: the input torque, the output torque and the reaction from the base that the case is mounted to. So, if the motor input is 100 lb-ft (say, clockwise) and the gear train has a total reduction of 5:1, then the load at the output shaft is reacting with 500 lb-ft of torque (counter-clockwise). If you do a free-body diagram, you'll see that this results in a net torque of 500 - 100 = 400 lb-ft CCW (neglecting inefficiencies). This has to be borne by the base so that the gearbox doesn't simply spin in the air.
This is without regard to how many gears, or by what means the reduction is performed; only the overall ratio is important (again, neglecting efficiency, which is generally very high). Since a ring gear remains stationary with the housing in a planetary arrangement, it sees the same torque.
And remember that it all depends on the load as well. Just because a motor is capable of producing 100 lb-ft of torque doesn't mean that's what it will see at a given moment in a given application. If the load is idling or disconnected, you might see only 10 lb-ft at the output, so only 2 lb-ft at the input motor and 8 lb-ft on the ring gear, even though your motor may be quite large.
Don
Kansas City