Cycling power of a bike with engaging/disengaging flywheel

1. Dec 22, 2013

Shune2001

Dear Friends
Iâ€™m in need to solve a problem in physics, which might be easy for you but it is not for me as my background in physics is not good.

I have a bike (drawn in simulation platform) with a flywheel connected to the crank (or shaft). The flywheel is not always fixed (or engaged) with the crank, it sometimes engages and others disengages with/from the crank via electrical clutch, and ofcourse there is gear between the flywheel and the crank.

I need to calculate:
1- Cycling total power (includes the effect of the flywheel while engagement/disengagement)
2- The power of the flywheel only (when it engages/disengages) to understand the effect of the flywheel on the total cycling power.

Now, due to simulation constraints, I donâ€™t have a sensor to measure the forces of the legs on the pedals, nor I want to derive it mathematically (for complexity); therefore, I used the following method:

First I calculated the torque on the crank through:
Torque = Crank Acceleration * Inertia of the bike and the humanoid (I can get these values from the program)

Then, I calculated the power as:
Power= Torque * Crank Angular Velocity (rad/s)

Is this method correct to calculate the total cycling power including the effect of the flywheel on the crank during engagement and disengagement? Or do I need to add the inertia of the flywheel to the equation of the torque while the flywheel is engaged? Or do I need to calculate separately the power of the flywheel and add it to the equation of power mentioned above ?

Concerning the second request, The power of the flywheel only, I know that the work done by the shaft on the flywheel is equal to the difference in the kinetic energy of the flywheel
Work= Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(2) - Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(1)
Then take the derivative of the work to get the power of the flywheel.

What Iâ€™m confused about is the angular velocity(1), angular velocity(2) and the derivative to get the power:

For example, if the flywheel engages with the crank for 5 seconds (or 5 samples), should I say that Angular_Velocity(1) of the flywheel is the one before or at the time of engagement ? and Angular velocity(2) is the one at time= 5 seconds(when disengaged) ? then the derivative is the difference between the two values ? or I need to devide by 5 to get the derivative ? or should I calculate the work at engagement and the work at disengagement and take the derivative (the difference) between the two values ? like:

W_eng= Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(2) - Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(1)

Where: Angular_Velocity_Flywheel(2) is the velocity after engagement,
Angular_Velocity_Flywheel(1) is the velocity before engagement

W_diseng= Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(4) - Â½ * Inertia_Flywheel * Angular_Velocity_Flywheel(3)

Where: Angular_Velocity_Flywheel(4) is the velocity after disengagement,
Angular_Velocity_Flywheel(3) is the velocity before disengagement

Then
Flywheel power = W_diseng â€“ W_eng ???

Iâ€™m really confused and need your help..

Many thanks
Shune

2. Dec 22, 2013

Staff: Mentor

If the flywheel is connected, its inertia value will have an additional contribution (as you have to account for its rotation) and in general this contribution will have a different angular velocity and acceleration as well. In the same way the wheels have more inertia than their simple mass, but that is not a large effect.

Where is the difference?

Nothing like this, you have to calculate the derivative of the equation and evaluate it for a specific point in time, as the derivative can change.

3. Dec 23, 2013

Shune2001

Concerning the first part, to account for the flywheel's effect during engagement, is it correct if I calculate the power of the flywheel alone as dwork/dt and add it to the cycling power equation as:

Cycling power = Torque * Crank Angular Velocity (rad/s) + dwork/dt

where dwork/dt = 0 when the flywheel is disengaged ? if not, then how to add the effect of the flywheel to the cycling power when it is engaged?

Also, for the second part, you mentioned that I have to derive the equation of work which is:

work=1/2 * I * (W2-W1)
dWork/dt= 1/2 * I * (Acc2-Acc1)... is that what you mean ?

if so, is Acc2 is the acceleration of the flywheel at disengagement, and Acc1 is the acceleration of the flywheel at engagement ? I'm confused about the time at which I should consider the W1, W2 (or Acc1, Acc2).

4. Dec 23, 2013

Staff: Mentor

Looks fine.

There is a square missing.
That's not the derivative of the equation above.

It is easier to do this with the total energy - work is just a change in this energy, so the derivative of the work is the derivative of the energy as well. What is the total energy? What is its derivative in time?

5. Dec 24, 2013

Shune2001

While searching the internet for how to derive the work, I found a similar problem in a book, shown in the link below: (Example 10.11)

They have calculated the power released from the flywheel during 50 sec brake. After calculating the work = 1/2 * I * (W2^2 – W1^2), they calculated the Power as Work/sec = Work/50.
So in my case, with engaging /disengaging flywheel if I say:
W1= initial angular velocity of the flywheel AT engagement with the crank
W2= Final angular velocity of the flywheel AT disengagement moment from the crank
And use the work equation, work = 1/2 * I * (W2^2 – W1^2), to calculate the difference in kinetic energy of the flywheel.
Then
Power of flywheel = Work/ period of engagement, to get the power of the flywheel (similar to the period of the break in the example above)
Then to calculate the total power:
Total cycling Power = Crank Torque * Crank Angular velocity + Power of flywheel
Where, Crank Torque = Inertia of bike and humanoid *Crank Acceleration
And If the flywheel is not engaged with the crank then we consider the Power of flywheel=0 in the total cycling power equation.
Is this a correct solution?
If so, the power of flywheel might be –ve or +ve due to successive engagement/disengagement, does the +ve value mean the flywheel has absorbed energy from the crank, and the –ve means the flywheel discharged some of its energy into the crank?

Many thanks

6. Dec 24, 2013

Shune2001

Hi again,
Concerning the derivative of the work of the flywheel,
Work= ½ * I * ( W2^2 – W1^2)
d/dt work= ½ * I * d/dt (W2^2) –d/dt (W1^2)..........(1)
d/dt (W^2)= d/dt( W.W)= d/dt (W) * W + W * d/dt(W) = acc*W + W*acc = 2 * acc * W
Back to (1)
d/dt (work)=1/2 * I * [ (2 * acc2 * W2)- (2 * acc1 * W1)] = I * [ acc2*W2 – acc1*W1 ]
Is this correct?
Are the acc1 and W1 the values, acceleration and angular velocity, of the flywheel at the time of engagement? and the acc2 and W2 the values of the flywheel at the time of disengagement?