Forces applied to a leaning levitating bicycle

[Qwerty42]

Hi!

So I'm building an electromagnetic bicycle where the frame is levitating over the wheels, thus reducing the rolling friction to almost zero by removing the bearings. Magnets have also been placed alongside the wheels to provide side to side stability. Black is the frame, red are the magnets on the wheel and yellow are the magnets attached to the frame to stabilize side to side movement from the wheel. Se the attached image. Since the wheel is held in place purely by magnetic forces, it can move if enough force is applied to it.

My question is: How much force would affect the axle magnet and side-to-side magnets when the bike is turning and thus leaning at let's say 45 degrees? Let's assume that the person riding the bike is weighing 50 kg.

Thanks!

 

[CWatters]

I think bikes lean over so that the net torque on the bike due to the combination of centripetal force and gravity is zero. I think this means the net force on the bike and rider will be directed through the contact point on the ground. So the side to side force is notionally zero.

The load on the axle magnets is the vector sum of the centripetal force and the force due to gravity.

Using Pythagoras that gives

N =√{(mv²/r)² + (mg)²}

where
m is the mass of the rider and bike
v = velocity
r = radius of the turn
g = acceleration due to gravity

Can someone check I have this right as it's been a long day.

 

[CWatters]

Obviously the side to side load won't always be zero in reality but in that simplistic model it is. In the real world there is acceleration as the bike leans over not to mention bumps etc etc.

 

[A.T.]

I'm building an electromagnetic bicycle where the frame is levitating over the wheels, thus reducing the rolling friction to almost zero by removing the bearings.

Rolling resistance comes mostly from the deformation of the tires, not from friction at the hub bearings. But most loses come from air resistance anyway.