No there is a brief period where both gears are driving simultaneously. The torque transmitted by either gear is never static during this period, the lower gear has reducing torque and the higher gear has ingreasing torque.
If you understood Engineguy's analogy in post #178 and imagine the bumpers on the pushing trucks to be thick rubber, you will see what I mean/ Whether we are talking rubber or hardened steel, the result is the same - only the deflections and the duration of the handover period are different.
Alternatively look at Cheapy's post #180, diagram #2 with both pawls in conact simultaneously. Perhaps imagine them with rubber faces. I hope you get the idea.
Maybe it helps, to picture two small springs of equal stiffness in Cheapy´s illustrative sketch, to account for elasticity in the material and in the drive train in general.
In the first picture, the spring in front of the "blue square/dog", would be fully compressed, pushing the light blue trapez, transmitting the torque.
Now, when the "red square/dog" enters into the gap between the dogs, it´s spring is fully extended, but as it is faster (higher gear) it closes onto the brown trapez (dogring) in front of it.
In the moment, the spring of the "red square/dog" makes contact with the brown trapez (dogring) it´s spring will start to compress, and it will start to transmit torque.
For the same amount the spring infront of the "red square/dog" compresses, the spring in front of the "blue square/ dog" will relax.
As long as both springs are in contact, both will share the total torque tranfered.
There will be a point when both springs have the same length, this is the point where both dogs/gears a driving/transmitting equal torque.
After this point, the spring in front of the "red square/dog" will compress more (spring gets shorter) and the one in front of the "blue square/dog" relaxes futhers, get´s longer.
This process continues until, the spring in front of the "blue square/dog" is fully relaxeed/extented and the spring in front of the "red square/dog" is fully compressed.
If this point is reached all the torque to drive the car/bike, whatever is supplied by the "red square/dog" (higher gear) and the "blue square/dog" (lower gear) is fully unloaded.
Force at both ends of an spring can be defined as F = s x K ( displacement of the spring (s) multiplied by spring stiffness K) e.g.: 5000 N = 5 mm X 1000 N/mm
the example given is a case of two springs in parallel.
At the moment when both springs sharing the load it would be: 5000 N = (2.5 mm x 1000N/mm) + (2.5 mm x 1000N/mm)
Maybe this analogy makes it easier for some to understand the concept.
Edited by TC3000, 29 February 2012 - 04:27.