Calculating the torque required in an electric bicycle

AI Thread Summary
The discussion focuses on calculating the torque required for an electric motor in a pedal-assisted bicycle, particularly when climbing a 35-degree incline. The original poster (OP) estimates resisting torque using the formula Tau = mu * 0.6 * W * Cos(35) * r, arriving at a value of 167.22 Nm. Responses emphasize that while torque is important, the key factor is the power needed, which is determined by the total force acting against the bicycle and the desired speed. Participants clarify that power remains constant through the system, and achieving the desired torque may require adjusting gear ratios. The conversation underscores the importance of understanding both torque and power dynamics in electric bicycle design.
motivehunter
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Hi everyone,

I just completed my bachelor's in mechanical engineering and have taken up a small project which involves a pedal assisted bicycle operating on both torque and cadence values.

Having completed the design and relevant estimations, I have now gotten stuck at a problem which seems to threaten my very understanding of the concept of torque. I am trying to calculate the torque required of the electric motor used by the bicycle and have taken the following design approach:

To determine the maximum possible torque required of the motor, it is assumed that the bicycle is moving up a plane inclined at 35 degrees to the horizontal. As the force which is the weight component acting parallel to the inclined plane (W*Sin(alpha) where alpha is the angle of inclination) acts along the centre of gravity and rear axle of the bicycle, it is assumed to have no torque in resisting the motion of the bicycle. However, the frictional forces resisting the motion of the rear wheel (which being the driving wheel, is being studied) apply a resisting torque by attempting to rotate the wheel in the counter-clockwise direction. The maximum torque required is therefore that torque required to be applied on the rear wheel which will just about counter the effect of this 'resisting torque'. Please look at attached images to better understand this.

In my model, I know that 60% of the bicycle's weight acts on the axle of the driving wheel and I have used this to calculate the resisting torque which is given by Tau = mu * 0.6 * W * Cos(35) * r where 'r' is the radius of the wheel. The resisting torque value comes up to be 167.22 Nm.

Is this approach a valid approach to determining the torque? If not, could you please point out the flaws in this approach so that I may correct my understanding of this?

Thank you in advance!
 

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Everything you did is correct, but what you really want to know is how much power you need, not how much torque. Reaching the wheel torque you desire can be achieved with any motor size; it is just a matter of properly match the power vs rpm output.

But you can have as much torque as you want, what will determine your need is how fast your wheel will turn at that torque. The faster you want to go, the more power you will need. How fast do you want to go up that hill?

This acceleration simulator (the theory is all explained at the bottom of the page) will help you understand what you need to determine, performance wise. It is made for a car, but all the theory apply just as well for a bicycle.
 
Hi!

Thank you for your reply.

I understand that you're saying that any torque value can be achieved by using an appropriate reduction ratio for a given power output? So then the maximum power output will be shifted to a lower speed (rpm) as the torque is increased.

Is my understanding correct?
Thanks!
 
There are 2 things to take into account: how much force you will need and how fast you will want to go. The force is basically the sum of the grade resistance (Wsinα), the drag resistance (½ρCdAv²), the rolling resistance (frW) and the inertia (ma). The product of this sum of forces and the speed of your vehicle is the power you need from your motor/engine/pedaler (not considering inefficiencies).

Since that power will go through the wheel, it will also be equal to the product of the wheel torque and wheel rpm (which you can easily find with the already known force and speed of the vehicle and the radius of the wheel).

If your motor is not producing this amount of power at the same rpm as the wheel, then your will need some sort of reduction ratio to link the two. Since the power is always the same no matter what, as rpm increases, torque will automatically decrease proportionally and vice versa.

Power is the key. The torque/force and rpm/velocity can be adapted for every component of the vehicle as needed.
 
1. Why should I include the inertia force? Isn't the inertia force actually equal to the sum of the grade, drag and rolling resistances?

2. How will the power always be constant? Motorcycles develop their maximum power at certain speeds like 125hp at 7500 rpm etc. Don't power, torque and rpm vary as per the design of the engine/motor irrespective of P = T\omega?
 
motivehunter said:
1. Why should I include the inertia force? Isn't the inertia force actually equal to the sum of the grade, drag and rolling resistances?

if Fv is the force acting at tire-road contact patch propelling the vehicle:

Fv = Wsinα + frW + ½ρCdAv² + ma

Or:

ma = Fv - (Wsinα + frW + ½ρCdAv²)

If Fv = Wsinα + frW + ½ρCdAv², then the acceleration is zero and the vehicle speed is constant.

motivehunter said:
2. How will the power always be constant? Motorcycles develop their maximum power at certain speeds like 125hp at 7500 rpm etc. Don't power, torque and rpm vary as per the design of the engine/motor irrespective of P = T\omega?

I didn't say that engine power is constant. But the power produced by the engine (whatever it maybe at whatever rpm you choose) is the same power passing through the transmission, the same one going through the wheel and has the same value propelling the vehicle.

If e is the engine, w is the wheel and v is the vehicle (again, not taking into account inefficiencies):

Teωe = Twωw = Fvvv
 
Thanks a lot! Makes more sense now.

Curious: given any motorcycle engine's power curve, can't I plot its torque curve using

P = T\omega ?
 
motivehunter said:
Curious: given any motorcycle engine's power curve, can't I plot its torque curve using

P = T\omega ?

There is usually no other way. With dynamometers, only the torque and rpm are measured. Power is calculated with P = T\omega.
 
Once again, thanks a lot. That helped make so many things clear!
 
  • #10
When a bicycle just starts to move, it's wheels just about begin to roll.

In this case, is it proper to consider rolling friction or sliding friction?
 
  • #11
  • #12
how to calculate the power required by motor if the total weight carried by bike+rider=110kg on a simple horizontal plane and on inclined plane
 
  • #13
Abhisek Karki said:
how to calculate the power required by motor if the total weight carried by bike+rider=110kg on a simple horizontal plane and on inclined plane

For the horizontal plane, use the acceleration simulator. Adjust the power until you reach your desired performance in acceleration and speed.

The simulator also calculates the maximum slope angle (hill climbing). If the vehicle is on an inclined plane of that angle, its acceleration at maximum power will be 0, so its speed will stay constant (if it is at rest, it will not be able to start moving). More info on the hill climbing page.
 
  • #14
4617450373_32ebfaa4d2_m.jpg


OK, all I need some help.

Looking to make a (Workcycles Nijland Classic trike, XL) into an electric bike that will conquer hills with bike fully loaded in Seattle for a business.

Total Weight of bike + Rider = 900Ibs
State Law =
A) motor cannot be more than 1000Watts
B) cannot travel more than 20MPH

I know this is possible but do not have the math skills to figure out. I'm thinking gears may have a lot to do with this scenario to make it possible.
 
  • #16
It's best to start a new thread @bobbyebikeq if you find out the slope of the hill you want to climb we can guide you through calculating how fast 1000w will get you up. Gearing selection comes much later in the design
 
  • #17
1000 Watts won't do much, top speed on the level : 16 - 17 mph
On a 35 ° incline (which is very steep) : 1 mph
 
  • #18
To give some perspective, the steepest street in the world is 19deg. Only elite cyclist can top 1kW and only for a short time
 
  • #19
jack action said:
With dynamometers, only the torque and rpm are measured.
Dynamometers measure the torque and rpm of the dyno apparatus (for example some dynos use a large drum). The torque is determined via a load or by the rate of acceleration of the dyno apparatus. Without knowledge of the vehicles engine or driven tires rpm, a dyno can only calculate the power produced by the driven wheels onto the dyno apparatus. So the short version of this is dynos measure power (as force times speed) during a run, and if additionally given the engine rpms during a run, they can calculate the engines effective torque at the driven wheels versus rpm (of the engine) during a dyno run.
 
  • #20
The steepest street in Seattle is ( apparently ) Northwest 60th Street at 28 % ( 15.64 ° )
(use degrees (°) in your calculations, not %)
You have got about 150 Watts of sustained power through pedal power alone,.
Keep your tyres proper hard ( like 60 psi ), this will minimize the rolling resistance force.
 
  • #21
motivehunter said:
Hi everyone,

I just completed my bachelor's in mechanical engineering and have taken up a small project which involves a pedal assisted bicycle operating on both torque and cadence values.

Having completed the design and relevant estimations, I have now gotten stuck at a problem which seems to threaten my very understanding of the concept of torque. I am trying to calculate the torque required of the electric motor used by the bicycle and have taken the following design approach:

To determine the maximum possible torque required of the motor, it is assumed that the bicycle is moving up a plane inclined at 35 degrees to the horizontal. As the force which is the weight component acting parallel to the inclined plane (W*Sin(alpha) where alpha is the angle of inclination) acts along the centre of gravity and rear axle of the bicycle, it is assumed to have no torque in resisting the motion of the bicycle. However, the frictional forces resisting the motion of the rear wheel (which being the driving wheel, is being studied) apply a resisting torque by attempting to rotate the wheel in the counter-clockwise direction. The maximum torque required is therefore that torque required to be applied on the rear wheel which will just about counter the effect of this 'resisting torque'. Please look at attached images to better understand this.

In my model, I know that 60% of the bicycle's weight acts on the axle of the driving wheel and I have used this to calculate the resisting torque which is given by Tau = mu * 0.6 * W * Cos(35) * r where 'r' is the radius of the wheel. The resisting torque value comes up to be 167.22 Nm.

Is this approach a valid approach to determining the torque? If not, could you please point out the flaws in this approach so that I may correct my understanding of this?

Thank you in advance!

Hi, I am also doing this project, is there any way that you can help with output torque relation to motor, frame analysis?
 
  • #22
waqas Wasim said:
Hi, I am also doing this project, is there any way that you can help with output torque relation to motor, frame analysis?

The OP hasn't been to PF for over two years. Best you start your own thread, include all relevant information and your work so far.
 
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