ok rocket scientist, need help calulating force on pinion and driveshaft??

This is how this stuff is designed.

If you are putting 300 ft lbs to the flywheel, multiply that by the gear ratio. In fourth or 1:1, you will have 300 ft lbs trying to rotate the pinion gear. The force applied by the pinion teeth is the torque divided by the radius of the pinion. If you have a 3.73 or so, with a 3in dia pinion, that's 1.5 in/12 in is .125 ft. 300 ft lb /.125 ft is 2400 lbs. How to figure that out over the teeth of the gear would take some more figuring.

Now, with say, a 3:1 first gear, its 3 times that, or 7200 lbs.

I am not a rocket scientist, so feel free to add to this.

In the case of the torque on the dyno I think you are probably confused becuase the dyno converts it but if you were going to take that route you would have to work backwards.

Torque is simply broken down into Force * a Perpendicular Distance. From that you can go pretty much anywhere, gear ratios will just multiply the torque and there will be losses. Could be an interesting project but I dont have the time to draw it all out for it to be accurate but it shouldnt be that bad.