- #1
Johnk690
- 3
- 0
I am designing a bicycle (theoretically), and i have my max torque on the crank 245Nm
(Ignor this unless you think iv made a mathematical mistake)
(1400N*.175m), which provides a torque on the drive cog, (245Nm/0.085m)= 2882N which is the tension in the chain, multiplied by the raduis of the back cog 0.035= 100Nm of torque at the back cog, dividing this by the radius of my back wheel i get 305N
Is this 305 N my linear force forward? and divided by the total mass 120kg my acceleration 2.54 m/s^2
I know I have no friction forces ie rolling resistance or air resistance friction at bearing etc.
I thought I understood moments of inertia but i didnt know how to include the mass of the person plus bicycle in them, any help would be much appreciated
(Ignor this unless you think iv made a mathematical mistake)
(1400N*.175m), which provides a torque on the drive cog, (245Nm/0.085m)= 2882N which is the tension in the chain, multiplied by the raduis of the back cog 0.035= 100Nm of torque at the back cog, dividing this by the radius of my back wheel i get 305N
Is this 305 N my linear force forward? and divided by the total mass 120kg my acceleration 2.54 m/s^2
I know I have no friction forces ie rolling resistance or air resistance friction at bearing etc.
I thought I understood moments of inertia but i didnt know how to include the mass of the person plus bicycle in them, any help would be much appreciated