i get that larger rear sprockets produce more low end climbing/take off power, and large rims will produce more top end speed. but, how does wheel size fit into the torque equation?? does a 26" rim produce more low end climbing power than a 29" setup with identical sprockets?? seems like smaller wheels would be better on steep hills.

Yes, the smaller wheel will have more climbing ability. But by how much, I don't know. There are members here who will know and there might be a chart or formula available. check in the wheels and hubs section

The final drive ratio is affected by the circumference of the drive wheel (2 x pi x radius). A 26" wheel has about 81.7" circumference and a 29" a 91.7" circumference, which is a 12% difference. Of course the other components (sprockets) could be adjusted to make the final drive ratios equal.

so, because of the difference in circumference, a 26" would have approximately 12% more low end pull than a 29" assuming all other components are equal?

That's correct. However gear ratio is not usually the most important factor in sellecting wheel size. It's much easier to change a sprocket later on than it is wheels.

It is about torque. A torque is just a moment about an axis. An engine can deliver a certain highest torque at an rpm. A tensioned chain leaving a sprocket always leaves tangentially, which is 90deg to the radius line from the axle and is the maximum torque angle. The tangential force (tension) at both sprockets is the same, but on the front sprocket you have a smaller distance from center than on the wheel sprocket. So the torque on a wheel is increased by the use of reductions. The wheel exerts the same torque on the asphalt. The force felt by the asphalt is only the force pushing you forward- that kick to the stomach in a fast car. A 26" wheel has a shorter distance from the center which makes the larger part of T=f*d*sin(a) the force pushing you forward (f). So to calculate, let's say you have an engine rated to 5lb-ft torque at 2500rpm. Drive is 11t and 88t #41 sprockets. The factor between them, 8, is the same for pitch radius and tooth count (drive ratio). So wheel torque = (5lb-ft)*(8:1 ratio) = 40lb-ft 26" wheel radius in feet: 1.085ft 29" wheel radius in feet: 1.21ft forward force 26" = (40lb-ft)/1.085ft = 37lb forward force 29" = (40lb-ft)/1.21ft = 33lb the 26" wheel will pull 12% harder than the 29" at 2500 rpm.

Here's a link to the download page of the Gear Ratio Calculator program. http://jimsitton.net/ratiocalc/ A 29" with a 40 tooth acts just about like a 26" with a 36 tooth. The difference can be thought of as 4 teeth. Simple, no? On my stock engined bike, a 29" with a 40, I am geared to go 39 at 7500, but there's no way the engine can hit 7500 geared this way with me on it. The 40 is the tallest gear usable, and it only hits about 30. The low RPM at cruise is very nice. I suspect the 36 is the tallest usable gear on a 26 with a stock engine. I'd like to try a 48 tooth on my bike sometime, just to be able to putt real slow sometimes. I can slow to about 16/18 and have the engine pull smoothly. Below that it's rough and lugging.