Cycling power of a bike with engaging/disengaging flywheel

In summary: Is the work done by the shaft on the flywheel equal to the difference in the kinetic energy of the flywheel and the work done by the legs on the pedals?I'm sorry, I don't understand what you're asking.
  • #1
Shune2001
4
0
Dear Friends
I’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 don’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= ½ * Inertia_Flywheel * Angular_Velocity_Flywheel(2) - ½ * Inertia_Flywheel * Angular_Velocity_Flywheel(1)
Then take the derivative of the work to get the power of the flywheel.

What I’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 divide 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= ½ * Inertia_Flywheel * Angular_Velocity_Flywheel(2) - ½ * 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= ½ * Inertia_Flywheel * Angular_Velocity_Flywheel(4) - ½ * 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 – W_eng ?

I’m really confused and need your help..

Many thanks
Shune
 
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  • #2
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 ?
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.

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 ?
Where is the difference?

then the derivative is the difference between the two values ? or I need to divide 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 ?
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
Thanks for your reply..

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).

Many thanks for your cooperation
 
  • #4
Shune2001 said:
Thanks for your reply..

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 ?
Looks fine.


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

work=1/2 * I * (W2-W1)
There is a square missing.
dWork/dt= 1/2 * I * (Acc2-Acc1)... is that what you mean ?
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
Thank you for your reply and for being patient with me.
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)
http://books.google.co.uk/books?id=...onepage&q=derivative of work flywheel&f=false

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
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?
 

Related to Cycling power of a bike with engaging/disengaging flywheel

What is the purpose of the flywheel in a bike?

The flywheel in a bike serves as a rotating mass that helps to store and release energy as the rider pedals. It helps to provide a smoother and more consistent pedaling experience by reducing fluctuations in power output.

How does engaging/disengaging the flywheel affect the cycling power of a bike?

Engaging the flywheel makes it harder to pedal as more energy is required to overcome the added resistance. This can help to build strength and endurance in the rider. Disengaging the flywheel reduces the resistance and makes it easier to pedal, allowing the rider to go faster or for longer periods of time.

What are the benefits of using a bike with an engaging/disengaging flywheel?

Using a bike with an engaging/disengaging flywheel can provide a more challenging and versatile workout. It allows the rider to vary the resistance and intensity of their ride, which can help to improve overall fitness and performance.

Can the flywheel resistance be adjusted while riding?

Yes, most bikes with engaging/disengaging flywheels have a resistance knob or lever that can be adjusted while riding. This allows the rider to change the resistance level to their desired intensity without having to stop and manually adjust the flywheel.

Are there any potential drawbacks to using a bike with an engaging/disengaging flywheel?

One potential drawback is that it may take some time for the rider to get used to the added resistance when engaging the flywheel. This can make the initial pedaling feel more difficult. Additionally, if the flywheel is not properly maintained, it can affect the overall performance of the bike.

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