I think you have an average weight of people, the weight of the bike, the gear ratio, the size of the driven wheel and the length of the lever arm on whatever means of power you intend to use...
Ive been using an old huffy that is in pretty good condition. I've been using around 300 lbs as the weight. In first gear the gear ratio is 1:.93. the drive wheel (smallest chain ring) has a 5 in diameter, and I am going to be a direct drive from motor to the drive shaft. The motor I am using is a 48 volt 750 Watt motor, which produces around 25.35 ft/lbs.
I found a few good formulas online that dealt with "burning rubber". Those help me get a better idea of what it takes to beat the COF of rubber vs pavement which is (.75 N)
so it states:
F= MA
F= (136.08 kg)(9.8 N/kg) = 1333.58 N
F(normal) = 1333.58/2 = 666.8 N
F (friction) = (.75)(666.8N)
F= 500.1
Torque = radius * force
26 in wheel means the radius is .33 Meters
T= .33 * 500.1
T = 165.033 Nm which equals 121.72 Ft/Lbs
So if you take that and factor in the .93 gear ratio I believe that to "burn rubber" you need to have a motor that can produce 113.20 Ft/Lbs
Does that make sence? are my numbers flawed what do you guys think?