Back to the OP (landuse) questions. The second reference from the other site tries to equate a "gear reduction ratio" to friction drives... but, that approach is meaningless, in regard to friction drives. As the other person mentions, changing the tire size cancels out the 'gear ratio' change.
The other site poster said:
while the roller's surface-speed IS the same as the vehicle's surface-speed, the fact's irrelevant to the math that's used to calculate ratio and speed.
For a friction drive system, Surface Speed is
ALL that matters, and ratio is meaningless. The surface speed is the speed of a point on the circumference (outer surface) of the drive roller. Assuming that there is no slipping between drive roller and tire, or between tire and road:
The surface speed of the spinning roller is EXACTLY the same speed as the surface speed of the spinning tire, NO MATTER THE DIAMETER, because the roller and the tire are in constant contact.
Likewise, the surface speed of the spinning tire is EXACTLY the same as the surface speed of the tire against the asphalt, (or, the road moving past the tire,) because THEY are in contact.
In a system with no slip, the surface speed of the roller is therefore IDENTICAL to the speed of the bike, no matter the tire diameter, and to calculate the maximum theoretical bike speed, you
only need engine RPM and roller diameter.
Even though the RPM of the wheel is much less than the RPM of the drive roller, (By the ratio of Drive roller diameter divided by tire diameter) the linear speed of a point on the surface of BOTH is the same. It HAS TO BE if there's no slipping. They are in contact.
Assuming a 1 inch roller and a 26 inch wheel, the roller RPM is 26 times the wheel RPM. Your roller would HAVE to spin 26 times for the bike to travel 81.64 inches (pi * 26 inch)
If you cut the wheel diameter in half, the roller RPM would now only be 13 times faster than the wheel RPM. But, since each wheel rotation will only take you half as far (40.82 inches,) as with the 26 inch wheel, so your wheel would have to spin twice, and the drive roller would STILL need to spin 26 times (13 * 2 times) to make the bike move the same 81.64 inches as with one rotation of a 26 inch tire.
Remember, with a chain/belt drive, all the power is directed to the drive AXLE, to make it spin at a given RPM. In that case, you NEED to use the wheel diameter to calculate the final speed.
But, with a friction drive, all the power is directed against the outer surface of the wheel, bypassing the axle entirely (From a power transfer point of view.) Since the power is already applied AT the tire surface, the tire diameter is irrelevant.
Now, you COULD go through the RPM calculations at each point, but, it's POINTLESS. Because the RPM of the spinning tire varies in direct relationship to the tire diameter, AND the distance traveled with each rotation of the tire also varies in direct relationship to tire diameter, any change introduced by a tire diameter change is exactly and completely canceled out.