Dear Kartheek,

First work out the rpm of the motor: Power = torque * rotational speed (in Hz).

Then choose a wheel size, a larger wheel will go over rougher terrain with less force, a smaller one will be cheaper.

One wheel rotation will move PI * the diameter, i.e. its circumferential length.

You can work out the required wheel speed by dividing the required top speed into the circumference. (make sure you use the same units, for example, m/s and m)

Then the gear ratio is the two rotational speeds divided together.

There is a problem though. There may not be enough power to take the vehicle to the speed you require.

A better design philosophy, is to start with the torque you need to move the vehicle. Plot this on a graph against forward speed. On the same graph draw another line of power versus speed. (using the equation above). This will give you the maximum speed that a 1250W motor can achieve. (or what power you need for 35kmph).

The required vehicle torque will depend on the roughness of the terrain and the maximum angle of the road you wish to travel.

I suggest you edit your question to improve the English; you may get more people answering. "How much torque..." and " The motor I am using.."

Best Regards

Paul

Dear Kartheek,

First work out the rpm of the motor: Power = torque * rotational speed (in Hz).

Then choose a wheel size, a larger wheel will go over rougher terrain with less force, a smaller one will be cheaper.

One wheel rotation will move PI * the diameter, i.e. its circumferential length.

You can work out the required wheel speed by dividing the required top speed into the circumference. (make sure you use the same units, for example, m/s and m)

Then the gear ratio is the two rotational speeds divided together.

There is a problem though. There may not be enough power to take the vehicle to the speed you require.

A better design philosophy, is to start with the torque you need to move the vehicle. Plot this on a graph against forward speed. On the same graph draw another line of power versus speed. (using the equation above). This will give you the maximum speed that a 1250W motor can achieve. (or what power you need for 35kmph).

The required vehicle torque will depend on the roughness of the terrain and the maximum angle of the road you wish to travel.

I suggest you edit your question to improve the English; you may get more people answering. "How much torque..." and " The motor I am using.."

Best Regards

Paul

In addition to Paul: For calculating the power demand of your vehicle as a function of the driving conditions (speed, acceleration and road inclination), you have to compute the vehicle resistance forces. For detailed information, please see my publication "Basics of longitudinal vehicle dynamics". There you can also find some characteristic parameters and some exemplary diagrams. Besides the longitudinal driving resistances, the motor characteristics significantly influences the gear layout. Due to the fact, that the engine power is a function of the engine speed multiplicated with the engine torque (see Pauls comment), both parameters have to be considered. Finally, losses in the gear and the other drivetrain components may lead to a reduction of torque at the driving wheel. Depending on the motor characteristics, a shiftable gear might be required. Target of the layout is both, sufficient acceleration behavior and maximum speed.

Kind regards,

Mario

Kartheek, just to illustrate what was mentioned inter alia by Mario: Take a look at published efficiency maps for various drivetrains in the scientific resp. engineering literature. These graphs are available for conventional powertrains based on internal combustion engines but also for alternative propulsion based on electric motors. In these maps you may observe the necessary road load or the requirements of a certain test driving cycle like the NEDC. Such an efficiency map is e.g. found in the paper attached below. Best Regards Ulrich

Article Fuel cell vehicles: Status 2007

I agree with all of the above. It is also possible to do the calculation "the other way". With 1250W of power and a target of 36kph (10m/s) the sum of the resistive forces on the vehicle have to be no greater than 1250/10 = 125N. And this calculation assumes 100% efficiency from the drivetrain (which is clearly not possible...). So the sum of the aero drag, the tyre rolling resistances, the mechanical losses must be less than 125N if you want to achieve your speed target, which is quite a challenge, but not impossible. Note that there is nothing you can do with gear ratios that will affect this calculation...

Then, if you assume that the overall gear ratio is a single fixed value, and the motor torque is constant across the speed range, then the specified 125N of tractive force will provide a maximum acceleration rate of 125/220 = 0.57m/s^2 at zero speed (i.e. when the resistive forces are assumed to be zero).

I think that is as much as you can calculate from the figures you provide. But at least it gives you a target for the resistive forces, and a limit on the maximum acceleration capability you can achieve without resorting to a gearbox.

Actually, there is one further calculation you can do. If the motor is capable of 10Nm at its rated 1250W, then it would have a rotational speed of 1250/10 = 125rad/s. Clearly if the car was travelling at 36kph (10m/s) whilst the motor was producing this power/torque/speed then the gearing would need to be 125/10 = 12.5rad/m (The motor would need to rotate 12.5rad for every meter the car moved). A slightly odd unit for gearing, I admit, but this would allow you to specify the wheel size and transmission ratio, which are mutually dependent.