Friction Drive Bicycle -- Required Torque

[Henry Chiansky]

Hi,

I am working on a project regarding the friction drive on bicycles. I would like to design an assembly which has to be compact enough, and has to be able to produce torque that will move the bicycle from a stationary position, but as it turns out with my calculations torque required is really high, or I am calculating something wrong?

My values:

Weight – 100kg

Bike wheel size – 26”

Torque result – 647Nm

Could you please help me find the answers (right values), or guidance in this problem as I think I am not taking into account some variables (friction value, weight distribution, etc.) that will change my result?

 

[Baluncore]

Welcome to PF.
Is the 100 kg the weight of the rider standing on a pedal ?
How is the 26" wheel driven ?
We need a sketch of your mechanism so we know what your numbers refer to.

 

[Lnewqban]

Welcome, Henry!
I would calculate the forward force that is needed to achieve the desired acceleration of the total mass.
That should be the tangetial force on the motorized wheel.
The torque output of your motor or engine would be that force times the radius of the roller plus some safety factor to account for slipping and deformation of the tire at the point of contact.

 

[jrmichler]

There are two paths to calculating rear wheel torque on a bicycle:

1) Specify the desired acceleration, then calculate the torque to get that acceleration.
Or
2) Given the torque of your drive motor, gear ratio, the friction wheel diameter, calculate the tangential force on the wheel. From that, calculate the torque at the wheel.

Your numbers do not makes sense because 647 N-m torque on a 0.66 m diameter wheel would be 1960 N of thrust. If the tire/road friction coefficient was at least 2.0, the acceleration would be 19.6 $m/sec^2$. That's impossible because any normal bicycle would do an instant wheelie. It's also impossible because no bicycle tire has that high a friction coefficient, so it would burn rubber while doing a wheelie.

 

[hutchphd]

diameter

If one uses the radius, what has been calculated is the torque that must be supplied to the bicycle wheel to pedal up a vertical wall (were it possible).

 

[jack action]

or I am calculating something wrong?

Maybe if you show us your calculations, we could tell if they are right or wrong.

 

[berkeman]

I am working on a project regarding the friction drive on bicycles.

Do you have the ability to remind whoever is requesting this that friction drive of a bicycle is a very bad idea? Just sayin'...

 

[russ_watters]

...but as it turns out with my calculations torque required is really high, or I am calculating something wrong?

My values:

Weight – 100kg

Bike wheel size – 26”

Torque result – 647Nm

Could you please help me find the answers (right values)...

Maybe if you show us your calculations, we could tell if they are right or wrong.

I'm pretty sure the answer is wrong, even if the calculation isn't(though it probably is too). The answer looks to be about twice the maximum torque a 100 kg rider can apply if they use all of their weight to push down a pedal (which isn't a good/realistic idea...).

 

[jack action]

I'm pretty sure the answer is wrong, even if the calculation isn't(though it probably is too). The answer looks to be about twice the maximum torque a 100 kg rider can apply if they use all of their weight to push down a pedal (which isn't a good/realistic idea...).

I know the answer is wrong since it's what is said in the OP. The question is about the calculations, which we have no idea what they are. Where did this 967 N.m come from?

You can produce any level of torque you want with sufficient leverage (i.e. gearing). Thus the calculations can be right. But the concept might be irrealistic though. He might also have put a plus sign where there should be a minus sign. Who knows? We can guess for a 100 posts or we could see his calculations and probably pinpoint the problem right away.

 

[russ_watters]

The question is about the calculations, which we have no idea what they are. Where did this 967 N.m come from?

You can produce any level of torque you want with sufficient leverage (i.e. gearing).

Good point. I'm only guessing, but it looks like the OP used the wheel diameter as the lever arm, but the radius would be the wrong choice too. But you're right, it could be anything.

 

[Henry Chiansky]

Thank you all for replying, I will post my replies bellow. As you have figured it out, I am lost in a problem and that is why I came to this forum, to seek help from people with knowledge such as yours.

Welcome to PF.
Is the 100 kg the weight of the rider standing on a pedal ?
How is the 26" wheel driven ?
We need a sketch of your mechanism so we know what your numbers refer to.

This is a weight of the rider + the bicycle, and wheel should be driven by pedaling or by friction drive

There are two paths to calculating rear wheel torque on a bicycle:

1) Specify the desired acceleration, then calculate the torque to get that acceleration.
Or
2) Given the torque of your drive motor, gear ratio, the friction wheel diameter, calculate the tangential force on the wheel. From that, calculate the torque at the wheel.

Your numbers do not makes sense because 647 N-m torque on a 0.66 m diameter wheel would be 1960 N of thrust. If the tire/road friction coefficient was at least 2.0, the acceleration would be 19.6 $m/sec^2$. That's impossible because any normal bicycle would do an instant wheelie. It's also impossible because no bicycle tire has that high a friction coefficient, so it would burn rubber while doing a wheelie.

Ok, I am aware of my bad numbers. I have approached the problem from a simple side, meaning, I have calculated the force that rider is applying by its weight and calculated the torque using the radius (I know I calculated with diameter but it somehow slipped) and got the torque of 647Nm.

Do you have the ability to remind whoever is requesting this that friction drive of a bicycle is a very bad idea? Just sayin'...

Well I would disagree as I have seen ideas even "crazier" than this that came to existence, but nevertheless thank you for your observations.

I know the answer is wrong since it's what is said in the OP. The question is about the calculations, which we have no idea what they are. Where did this 967 N.m come from?

You can produce any level of torque you want with sufficient leverage (i.e. gearing). Thus the calculations can be right. But the concept might be irrealistic though. He might also have put a plus sign where there should be a minus sign. Who knows? We can guess for a 100 posts or we could see his calculations and probably pinpoint the problem right away.

Calculations I made are bad, but that was a starting point I had to take as my knowledge on this problem is limited, that is why asked a question here. I have calculated the torque using the radius as a lever and a force applied by a rider weight to get the torque.

 

[jack action]

Calculations I made are bad, but that was a starting point I had to take as my knowledge on this problem is limited, that is why asked a question here. I have calculated the torque using the radius as a lever and a force applied by a rider weight to get the torque.

So, I stopped being lazy and tried to guess what you did as calculations. Is this what you did:
$$100\ kg * 9.81\ m/s^2 * 26\ in \times 0.0254\ m/in = 647\ N.m$$
If so, it is wrong on two levels. First, you should use the radius and not the diameter as the level arm. Second, even calculated correctly, it is a pretty useless information.

What I think you want to know is what does your power source have to produce to achieve your goal. If this is what you want to know, you should be looking more at the power needed rather than the torque needed. If your goal is to replace the human pedaling, it is rather easy: Humans produce 50 to 150 W, the best may be able to sustain up to 400 W, with peaks reaching 1000 W. (source)

This should help you sizing your power source. Don't forget that your mechanism will probably be less efficient, therefore, more power will be needed. Once this has been decided, you will need to split that power between force and velocity or torque and rpm.

Say you decide to transmit 150 W of (net) power $P$ to the bicycle through the wheel. And say you want your bicycle to go at velocity $v$ of 15 km/h (4.17 m/s). Therefore the force $F$ that propels the bicycle is:
$$F = \frac{P}{v} = \frac{150\ W}{4.17\ m/s} = 36\ N$$
What is the rpm of your 26 in (0.33 m radius) wheel?
$$\omega_w = \frac{v}{r_w} = \frac{4.17\ m/s}{0.33\ m} = 12.6\ rad/s = 120\ rpm$$
For the torque of the wheel $T_w$:
$$T_w = F * r_w = 36\ N * 0.33\ m = 11.9\ N.m$$
You can now check that the wheel power $P_w$ is the same as the power output:
$$P_w = T_w * \omega_w = 11.9\ N.m * 12.6\ rad/s = 150\ W$$
If you have a motor with, say, a 0.050 m radius roller r driving your wheel, you can determine the torque & rpm this way:
$$F_r = F_w$$
$$\frac{T_r}{r_r} = \frac{T_w}{r_w}$$
or:
$$T_r = \frac{r_r}{r_w}T_w = \frac{0.050}{0.33} * 11.9 = 1.8\ N.m$$
And for the rpm:
$$v_r = v_w$$
$$\omega_r * r_r = \omega_w * r_w$$
or:
$$\omega_r = \frac{r_w}{r_r}\omega_w = \frac{0.33}{0.050} * 12.6 = 83.16\ rad/s = 794\ rpm$$
Again , you can verify that the roller power is still 150 W:
$$P_r = T_r * \omega_r = 1.8\ N.m * 83.16\ rad/s = 150\ W$$

 

[Baluncore]

Fundamentally, it is the diameter of the drive roller that is important in transforming the motor torque and RPM.

When in friction drive, the force applied by the tyre to the ground will be equal to the force applied by the friction drive roller to the same tyre. The size of the rear wheel is only important when driven by the pedals through the chain.

 

[Henry Chiansky]

So, I stopped being lazy and tried to guess what you did as calculations. Is this what you did:
$$100\ kg * 9.81\ m/s^2 * 26\ in \times 0.0254\ m/in = 647\ N.m$$
If so, it is wrong on two levels. First, you should use the radius and not the diameter as the level arm. Second, even calculated correctly, it is a pretty useless information.

What I think you want to know is what does your power source have to produce to achieve your goal. If this is what you want to know, you should be looking more at the power needed rather than the torque needed. If your goal is to replace the human pedaling, it is rather easy: Humans produce 50 to 150 W, the best may be able to sustain up to 400 W, with peaks reaching 1000 W. (source)

This should help you sizing your power source. Don't forget that your mechanism will probably be less efficient, therefore, more power will be needed. Once this has been decided, you will need to split that power between force and velocity or torque and rpm.

Say you decide to transmit 150 W of (net) power $P$ to the bicycle through the wheel. And say you want your bicycle to go at velocity $v$ of 15 km/h (4.17 m/s). Therefore the force $F$ that propels the bicycle is:
$$F = \frac{P}{v} = \frac{150\ W}{4.17\ m/s} = 36\ N$$
What is the rpm of your 26 in (0.33 m radius) wheel?
$$\omega_w = \frac{v}{r_w} = \frac{4.17\ m/s}{0.33\ m} = 12.6\ rad/s = 120\ rpm$$
For the torque of the wheel $T_w$:
$$T_w = F * r_w = 36\ N * 0.33\ m = 11.9\ N.m$$
You can now check that the wheel power $P_w$ is the same as the power output:
$$P_w = T_w * \omega_w = 11.9\ N.m * 12.6\ rad/s = 150\ W$$
If you have a motor with, say, a 0.050 m radius roller r driving your wheel, you can determine the torque & rpm this way:
$$F_r = F_w$$
$$\frac{T_r}{r_r} = \frac{T_w}{r_w}$$
or:
$$T_r = \frac{r_r}{r_w}T_w = \frac{0.050}{0.33} * 11.9 = 1.8\ N.m$$
And for the rpm:
$$v_r = v_w$$
$$\omega_r * r_r = \omega_w * r_w$$
or:
$$\omega_r = \frac{r_w}{r_r}\omega_w = \frac{0.33}{0.050} * 12.6 = 83.16\ rad/s = 794\ rpm$$
Again , you can verify that the roller power is still 150 W:
$$P_r = T_r * \omega_r = 1.8\ N.m * 83.16\ rad/s = 150\ W$$

Thank you very much for this clarification.
You are right, both with calculations and with wheel diameter mistake.
The starting point for my calculations was to find out what is the torque on the wheel that needs to be beaten in order to get the bicycle going, and that is why I immediately removed the roller out of the equation, obviously I made a mistake in the first step.

But one more question, if I may ask. how should I start the calculations if I decided not to use the data you provided ( data regarding the human provided power) but instead I used the weight of the rider and bicycle itself? I am asking this as, as I can see, not every human can produce the same power, nor it has the same weight. I don't think I could just calculate the power out of the first equation you wrote, and proceed with the same steps you did?

 

[hutchphd]

The torque tells you how rapidly you can get the bicycle going: e.g. to get to 10 m/s in 5 s requires an acceleration of $\frac {10m/s} {5s}=2m/s^2$ . If the mass is 100kg this takes 200N of force supplied to the ground by the wheel.
But the power required maintain this rate of acceleration increases as one speeds up.

 

[jack action]

how should I start the calculations if I decided not to use the data you provided ( data regarding the human provided power) but instead I used the weight of the rider and bicycle itself? I am asking this as, as I can see, not every human can produce the same power, nor it has the same weight. I don't think I could just calculate the power out of the first equation you wrote, and proceed with the same steps you did

You need to calculate the power needed, which will depend on the velocity and mainly on three forces:

• Aerodynamic drag, $F_D$
• Rolling resistance, $F_R$
• Grade resistance, $F_G$

The power needed would then be:
$$P = (F_D + F_R + F_G) * v$$

The drag force is:
$$F_D = \frac{1}{2}\rho C_D Av^2$$
Where:

• $\rho$ is the air density (= 1.23 kg/m³)
• $C_D$ is the drag coefficient
• $A$ is the frontal area (m²)
• $v$ is the bicycle velocity (m/s)

The rolling resistance is:
$$F_R = C_R mg$$
Where:

• $C_R$ is the rolling resistance coefficient
• $m$ is the bicycle + rider mass (kg)
• $g$ is 9.81 m/s²

The grade resistance is:
$$F_G = mg\sin \theta$$
Where $\theta$ is the angle of the hill you are climbing. And $\tan \theta$ = %slope.

You can find typical values for bicycles by a simple google search. Here are some I found:

Note how they only care about the power required.

 

[russ_watters]

But one more question, if I may ask. how should I start the calculations if I decided not to use the data you provided ( data regarding the human provided power) but instead I used the weight of the rider and bicycle itself? I am asking this as, as I can see, not every human can produce the same power, nor it has the same weight. I don't think I could just calculate the power out of the first equation you wrote, and proceed with the same steps you did?

You will need to pick your design requirements, and hopefully you will pick reasonable ones. As something of a cyclist, I can tell you the limiting factor will almost certainly be rider weight, speed and grade (not flat-ground speed). Pick values for them.

 

[Henry Chiansky]

You need to calculate the power needed, which will depend on the velocity and mainly on three forces:

• Aerodynamic drag, $F_D$
• Rolling resistance, $F_R$
• Grade resistance, $F_G$

The power needed would then be:
$$P = (F_D + F_R + F_G) * v$$

The drag force is:
$$F_D = \frac{1}{2}\rho C_D Av^2$$
Where:

• $\rho$ is the air density (= 1.23 kg/m³)
• $C_D$ is the drag coefficient
• $A$ is the frontal area (m²)
• $v$ is the bicycle velocity (m/s)

The rolling resistance is:
$$F_R = C_R mg$$
Where:

• $C_R$ is the rolling resistance coefficient
• $m$ is the bicycle + rider mass (kg)
• $g$ is 9.81 m/s²

The grade resistance is:
$$F_G = mg\sin \theta$$
Where $\theta$ is the angle of the hill you are climbing. And $\tan \theta$ = %slope.

You can find typical values for bicycles by a simple google search. Here are some I found:

View attachment 276859
View attachment 276860​

Note how they only care about the power required.

Hi guys,
Thank you very much for helping me to find the answers for my problem.

I have created a small spreadsheet so anyone can use it if ever comes in a need.

Also, i have done additional calculations in order to get the info on how much force is needed to achieve friction between the wheel and the friction drive as per formulas bellow.

Additional, friction drive minimum width should be calculated, but I left that for futureFriction force needed to push on to wheel

Normal force that is needed to push on to wheel

From there we can calculate friction force

Traction force created