How to calculate max car mass and tow capacity from motor specs?

[Grez]

Hi I am trying to do an artisanal EV car and I would like to know how do I choose the right motor. I already have one in mind but I don't know how to calculate the max mass that it can move. (see attached pdf)

Gear Ratio : 2.39
Diff Ratio : 4.88
Wheel Diameter: 26in

I would like to know the formula to calculate the car max weight from motor torque

The car would be RWD

 

[jrmichler]

Is this a school assignment?

Hint: You also need to specify or find the desired performance, the air drag, and the rolling resistance.

 

[Baluncore]

I would like to know the formula to calculate the car max weight from motor torque

There are other approaches to a solution.
If the vehicle is to share roads with other vehicles, then the “power to weight ratio” will need to be similar to those other vehicles. You don't want a reputation of being the “slow coach” that holds up the traffic flow on the highway. At the same time there is no point having extreme performance that cannot be safely exercised.

Look at the power and weight specifications of other vehicles to determine the typical values used.

Are there national standards that must be met?

 

[Grez]

Is this a school assignment?

Hint: You also need to specify or find the desired performance, the air drag, and the rolling resistance.

no this is a personal project

this would be my everyday car

can we use any car air drag to approximate the result even if I know that air resistance would be less that most cars ?

rolling resistance as in friction coefficient?

There are other approaches to a solution.
If the vehicle is to share roads with other vehicles, then the “power to weight ratio” will need to be similar to those other vehicles. You don't want a reputation of being the “slow coach” that holds up the traffic flow on the highway. At the same time there is no point having extreme performance that cannot be safely exercised.

as for max speed I would be ok with 2.39 and 4.88 gear ratio I should be able to go 80kmh @ 7.5krpm and 106kmh @ 10krpm but this gear arrangement is to maximize towing capacity because this would be "low gear" and "high gear" would be 1.11 and 4.88 and would be for highway and everyday use and should be able 115kmh @ 5krpm and 230kmh @ 10krpm this should give me better battery consumption

Are there national standards that must be met?

not that I know I just want to pick the right motor for my projet which is why I wanted to know how to calculate the max weight that an electric motor can move with a specific gear arrangements and if I need to change those gear ratio so that the motor can make my car move

 

[Baluncore]

Any driver will learn that, in order for the traffic to flow smoothly, a car needs a similar acceleration to the other cars on the road. It is clear that acceleration is a function of the vehicle's power to weight ratio. Do you drive a motor vehicle ?

I would like to know the formula to calculate the car max weight from motor torque

Look at the power to weight ratios of typical cars on the road, measured in watts per kilogram. You can then apply that ratio to your vehicle to check your power is reasonably well matched to the weight.

as for max speed I would be ok with 2.39 and 4.88 gear ratio I should be able to go 80kmh @ 7.5krpm and 106kmh @ 10krpm but this gear arrangement is to maximize towing capacity because this would be "low gear" and "high gear" would be 1.11 and 4.88 and would be for highway and everyday use and should be able 115kmh @ 5krpm and 230kmh @ 10krpm this should give me better battery consumption

The speed in different gears is irrelevant when it comes to power, which is the flow of energy. You can select the most appropriate gear ratios, once the power, speed and torque range of the motor is known.

 

[jack action]

Anything producing a maximum acceleration less than 0.3g will feel sluggish. Typical passenger cars can do up to 0.7g easily (full throttle).
$$F_w = \lambda_m ma$$
$$\frac{T_w}{r} = \lambda_m mg\frac{a}{g}$$
$$\frac{\eta\ G_R\ T_m}{r} = \lambda_m W\frac{a}{g}$$
$$T_m = \frac{\lambda_m \frac{a}{g}r}{\eta\ G_R}W$$
In your case (assuming a drivetrain efficiency of $\eta = 0.85$ and a mass factor $\lambda_m = 1.38$):
$$T_m = \frac{(1.38) (0.3) (\frac{13}{12}\text{ ft})}{0.85 \times (2.89 \times 4.88)}W$$
$$T_{m\text{ [lb.ft]}}= (0.0374\text{ ft}) W_\text{[lb]}$$
So if you have 258 lb.ft of motor torque, you will be able to accelerate a vehicle weighing 6900 lb at a 0.3g acceleration (from rest, that is). Repeating the same process for a 0.7g acceleration, you'll find your vehicle must weigh no more than 2960 lb.

and "high gear" would be 1.11 and 4.88 and would be for highway and everyday use and should be able 115kmh @ 5krpm and 230kmh @ 10krpm

The typical amount of power you need for highway (mostly due to aerodynamic drag) is about 25 hp. At 5000 rpm, this means 26 lb.ft (or 35 N.m) of motor torque. According to the efficiency map of your pdf file, this is where you are:

(red dot: 115 km/h; blue dot: 200 km/h; green dot: 230 km/h)​

As you can see, with this motor, your vehicle would be most efficient at cruise speeds of around 200 km/h, which is also the maximum speed that you can use continuously.

The equation used to find those points on the map is found by equating the motor power to the drag power:
$$P_m = P_d$$
$$\eta T_m \omega = \frac{1}{2}\rho C_d A v^2 \times v$$
$$\frac{\eta T_{m\ [N.m]} RPM}{7122} = \frac{\frac{1}{2}\rho C_d A\left(\frac{1}{3.6}\frac{(115\text{ km/h})}{(5000\text{ rpm})} RPM\right)^3}{746}$$
$$T_{m\ [N.m]} = \frac{(7122)\frac{1}{2}\rho C_d A\left(\frac{1}{3.6} \frac{(115\text{ km/h})}{(5000\text{ rpm})} RPM\right)^3}{746\eta RPM}$$
$$T_{m\ [N.m]} = \frac{\rho C_d A\left( \frac{(115\text{ km/h})}{(5000\text{ rpm})}\right)^3 RPM^2}{9.774\eta}$$
$$T_{m\ [N.m]} = \frac{RPM^2}{793057}$$
Where $\rho = 1.23 \text{ kg/m³}$ and I assume the drag factor $C_d A = 0.7 \text{ m²}$

Redoing the previous exercises with a new gear ratio, note that if you could use a higher gear ratio instead of 1.11:1 you would maybe end up in a more favorable zone of your motor where you would gain efficiency at lower speeds while losing nothing at high speeds. You even gain a higher top speed (torque-limited instead of rpm-limited), with only a minor loss in the maximum cruise speed (I imagine you don't need to drive over 185 km/h continuously anyway?). The results are In the following table (some are eyeballed):

1.11:1 /[0.833:1]/ (0.555:1)	Speed	Torque	RPM	Efficiency
Low Cruise Speed	58 km/h [58 km/h] (58 km/h)	8 N.m [11 N.m] (16 N.m)	2500 rpm [1875 rpm] (1250 rpm)	63% [65%] (67%)
High Cruise Speed	115 km/h [115 km/h] (115 km/h)	31 N.m [41 N.m] (62 N.m)	5000 rpm [3750 rpm] (2500 rpm)	77% [80%] (87%)
Max Cruise Speed	200 km/h [200 km/h] (185 km/h)	95 N.m [127 N.m] (185 N.m)	8700 rpm [6525 rpm] (4000 rpm)	93+% [94%] (93-%)
Max Speed	230 km/h [227 km/h] (245 km/h)	126 N.m [210 N.m] (275 N.m)	10000 rpm [7400 rpm] (5400 rpm)	92% [92%] (92%)

 

[Grez]

@jack action thank you for your great answer