Torque required to power an Electric cycle

[AbyJoseph]

I am having a real tough time determining the torque required to power an electric motor used to drive a cycle. I have calculated the Torque as:

Torque = μ X mg X r(radius of the wheel)


Now I am confused as what is μ. I know it's the coefficient of friction but, not sure whether I have to consider the coefficient of static or rolling friction.

I have taken a considerable amount of time to found the Torque formula and now this is one small step that I am having a hard time deal with it as many sources seems to take different stand on the values.

Any help in this regard would be appreciated.

Regards,
Aby Joseph

 

[berkeman]

I am having a real tough time determining the torque required to power an electric motor used to drive a cycle. I have calculated the Torque as:

Torque = μ X mg X r(radius of the wheel)


Now I am confused as what is μ. I know it's the coefficient of friction but, not sure whether I have to consider the coefficient of static or rolling friction.

I have taken a considerable amount of time to found the Torque formula and now this is one small step that I am having a hard time deal with it as many sources seems to take different stand on the values.

Any help in this regard would be appreciated.

Regards,
Aby Joseph

Where did you get that equation?

A better approach is to use F=ma to tell you what force it takes to accelerate the mass of the bike and passenger. Once you have that, you can calculate the torque based on the wheel radius. The coefficient of friction does not come into the calculation, assuming that the drive wheel is not slipping on the pavement (usually a good assumption for a bicycle).

 

[AbyJoseph]

Thanks a lot for the prompt reply...

I got the equation from https://www.physicsforums.com/showthread.php?t=528133

It tells:

The COF of rubber vs pavement which is (.75 N)

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​

I assumed that the "a" they are referring to is the acceleration due to gravity(g). Correct me if I am wrong.

They have calculated the force (mg?), normal force(divided by two due two two wheels), and the friction force. So doesn't the friction force formula equals to:

F(friction) = μ X m X g
Torque = F(friction) X Radius ?

The discussion too has stated μ is coefficient of friction but hasn't mentioned static or rolling.

Looking forward for a reply.

 

[berkeman]

Thanks a lot for the prompt reply...

I got the equation from https://www.physicsforums.com/showthread.php?t=528133

It tells:

The COF of rubber vs pavement which is (.75 N)

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​

I assumed that the "a" they are referring to is the acceleration due to gravity(g). Correct me if I am wrong.

They have calculated the force (mg?), normal force(divided by two due two two wheels), and the friction force. So doesn't the friction force formula equals to:

F(friction) = μ X m X g
Torque = F(friction) X Radius ?

The discussion too has stated μ is coefficient of friction but hasn't mentioned static or rolling.

Looking forward for a reply.

Blag! That's a confusing thread.

No, the acceleration has nothing to do with gravity, and friction doesn't come into play. The force is the horizontal force on the ground that produces an acceleration horizontally to move the bike.

The torque is how the force is coupled to the ground. The horizontal force and radius of the wheel are used to calculate the rear-wheel torque. That torque is produced by your electric motor, through some gear ration (which changes the motor torque versus the rear-wheel torque).

 

[AbyJoseph]

I completely understood your method of calculating Force(from Mass and Acceleration) and Torque(from Force and Radius)

Please be patient with me but http://www.researchgate.net/post/How_can_we_calculate_the_required_torque_to_move_a_massive_object_by_means_of_gear_assembly2 too advocaties the method of mass, gravity and friction forces.

Would have felt reassured to know why they have considered them or is it me who is missing any point?

 

[berkeman]

No worries. Any mention of coefficient of friction in the context of your question is irrelevant. The bike wheel rolls without slipping, so it's just like 2 gears meshing.

Do keep in mind though that there will be a "rolling resistance" retarding force, which comes from deformation of the bike's tires and wind resistance. You will need to add in extra torque to combat those retarding forces. When you are riding at a constant speed (after accelerating to that speed), you only need enough torque to combat those retarding forces.

 

[AbyJoseph]

Thanks a lot Sir..! I have been on these friction, coefficients and all for quite a lot of time and now finally figured out... Once again... thanks a lot for helping me out.

Aby Joseph

 

[dean barry]

Power required

Work with the power required from your motor to suit a top speed.

You can start by estimating your required level ground top speed (v) ( example : 4 m/s (about 9 mph ) )

So :
v = 4.0 m/s
Drag co-efficient (Cd) = 0.24 ( about the same as a freefaller )
Rolling resistance co-efficient (Crr) = 0.03 ( correctly inflated rubber tyres on ashphalt )
Total mass (M) = 80 kg ( example )
Local gravity rate (g) = use 9.81 (m/s)/s

Drag force due to air drag = v ² * Cd = 3.84 Newtons
Rolling resistance force = M * g * Crr = 23.544 Newtons

Total drag force (f) = total 2 above = 27.384 Newtons

Power requirement at drive wheel = f * v = 109.536 Watts
Build in 10 % power losses in the drive train, so, final power requirement from your motor = 121.71 Watts

Excel sheet if you need it -- PM me.

 

[AbyJoseph]

Sir,
I do appreciate you for taking time to solve my problem.

I would be glad if you could please take a minute to consider the following,

I have a few constraints here for motor,
Maximum Voltage = 48v
Maximum Power = 400W

I have designed by tricycle and it has mass of 400kg (including passenger weight) and wheel radius of 0.32m. As per ur method, I need a power of 906W which is far more than the constraint of 400W.

So I decided to calculate the Torque with the Power and Voltage as the constraint.

Please do provide some light on how to select the motor ratings.Joseph

 

[koolraj09]

For motor sizing, Torque will be calculated using F=Ma. The acceleration 'a' is defined by you. You can assume a suitable top speed and the time you require to reach that speed. This gives you your approximate acceleration.
Now if you are building a 2 wheeled bicycle with one powered wheel, the above calculated force 'F' is to be applied at that wheel. Now Required Torque = Fxradius of your wheel. This gives you torque. This force F, which is the accelerating force also gives you the Peak Power when you multiply this force by velocity.
Now you may perhaps want to run at the top speed (cruise speed) for some time. At the point you just have to overcome gearing losses, internal losses and Rolling resistance.
The main component is rolling resistance, which you can calculate using the co-efficient of rolling resistance and the weight of your bicycle. This is the main force you have to overcome to move at constant speed.
Now Rolling Resistance * Cruise Velocity = Nominal Power Required. So You have the Nominal power required, the peak power required during acceleration and the rpm of the wheel for cruise. You can suitably select a motor from these criteria. The peak power will give you the peak current you need to supply to motor (assuming you are keeping the voltage constant) and the nominal power gives you the nominal current of the motor.
The second part of the question: Why frictional force is to be considered in your calculations?
Now if you observe in first calculation of accelerating force, F, there is a limit to the acceleration you can have. That is governed by the frictional force between the ground and your wheel. If you exceed this force i.e. if F>fric, then your wheels will slip. Practically the wheels will slip instantaneously at the start.

 

[dean barry]

Due to the high all in weight (400 kg) , tyre rolling resistance will be the major factor in drag terms, make certain the tyres are pumped to as high a pressure as is practical and check them for pressure loss regularly ( weekly minimum).