Going Direct Drive!

I have been playing with the idea of removing every gear in the drive system in a car by installing directly the motor on the axle of the wheel.

As an example in my calculation, I’ve decided to take my Toyota Yaris. Here the data:

Wheel size (diameter D): 355.6mm (14 inch)

Tire size (diameter D): 610mm (24 inch)

At 100km/h. the RPM of the motor is around 2650 RPM (44 rev/s).

We know that we have gear ratios but let’s put this aside for a minute. We can calculate the speed of a car just by knowing the RPM of the motor and the wheel circumference. In my case, the wheel circumference Cr = πD, 1916.5mm (75.45 inch). The speed v = rev/s • Cr = 84 320 mm/s and yields to 84.32 m/s. Converting this into km/h yields to 303.5 km/h (188.6 MPH)! Ok, this excludes the final gear ratio. The thing is; do I need a high speed motor such as 5000/6000RPM? The answer is no but what is the minimum acceptable?

Let’s start by targeting the maximum speed that I would like to achieve; 160 km/h (100mph). We need to convert all this to m/s and yields to 44.4 m/s. We need to transform the speed equation v = rev/s • Cr into rev/s = v/Cr : 44.4 / 1.916 = 23.2 rev/s = 1390 RPM!

It would explain why most of all the car EV conversion, there’s always a gear ration somewhere in amateur likes the big companies. In the big companies, they have left out the transmission. What I have seen and red in the amateur EV world, often, the transmission and the transaxle are kept. I assume that it’s a way to reduce the cost of the conversion because adding two motors, two inverters, will almost double the conversion. In the big companies like the Mitsubishi Miev, the Nissan Leaf, or the Ford eFocus (and others), all have a transaxle reducing gear. I’m no mechanical engineer but I would assume that having a transaxle ratio of 1:1 would be very large! 

Let's reduce the current to save energy

My objective in this enterprise is to get rid of all unnecessary mechanical parts. 

Motor in wheel are a good choice but I am a bit skeptical about their durability in the long term. I opted to connect two motors to each axle that drives the front or the rear wheels. This is direct drive.

All calculations and simulations made on the Quadrupole motor/generator are based on a 100A and 5 turns. This gives a magnetomotive force (MMF) of 500 At (amp turn). Because it’s a double stator design, three torque constants 3 are available: 0.375, 0.205 and 0.145 Nm/A. Since we could, at 144 Vdc bus, achieve almost 3500 RPM at 0.372 Nm/A and with the calculation above for direct drive, we can reduce the maximum RPM to 1390 RPM. Taking in mind that we need to achieve the maximum RPM with the maximum torque constant. This will reduce even more the current without any compromises.

Let’s start by reformulating everything, first the BEMF constant K_e. The BEFM is the maximum EMF or the generated voltage divided by the maximum angular speed. So for a 144Vdc bus and because of voltage losses, let's arbitrarily assume that a BEMF of 125V peak with a 144Vdc bus.

So;

K_e = 125Vdc / (1390 rpm / 9.55) = 125Vdc / 13 275 rad • s^-1 = 0.859 V/ rad • s^-1

Since the BEMF constant is equal to the torque constant, we can rewrite T_e = T_k = 0.859 Nm / A

The torque constant Kt muliplyed by the currant give the torque as shown in the equation below:

T = K_t • I

Since K_t as change, the value of the current must change. So the currant will be:

I = T / K_t

The maximum continuous torque of the Quadrupole motor is 41 Nm (41 Nm is the last available numbers of the Quadrupole motor/generator).

I = 41 Nm / 0.859 Nm/A = 47.7 A

We need to calculate the number of turn required to maintain the MMF at 500 At as to maintain 41 Nm/A. The torque constant equation is below:

K_t = N • B_p • L • D • p

Where:

• N is number of turns and it is unitless, 
•  Bp is the average flux in the air-gaps in Tesla (T) and equal 0.9718 T 
• L is the depth of the saliency of the each stator pole in metre (m) and is equal to 0.0381 m, 
•  D is the diameter of the air gaps in metre (m). 
•  p is number of pole pairs ( numbers of pole /2) and it is unitless

We have to separately calculate the T_k of the inner and the outer stator and add them together to get the sum of T_k. Since we have two stators, K_t(inner) and K_t(outer) they will very since the D and B_p are not the equal in both stators. The maximum constant torque Kt = K_t(inner) + K_t(outer). However, the number of turns remains the same in both stators.

So, a MMF of 500 At and feeding hit at 100 A of current, this means that I have 5 turns. 

Insolating N from the MMF will be N = MMF / I.

Let's recap. 47.4 A is required for the new toque constant and that MMF = 500 At. Using the above equation, 500 At / 47.7 A = 10.5 turns. We will round it down to 10 since we have taken into account the voltage drop at the motor wires.

Lower amps per torque

So 50 Amps only at 144Vdc for 1390 rpm and 41 Nm, this is very good since the heat loss will be lower and will have less impact for on the magnet. However, since there is more turns, and to accommodate this, we need to reduce the wire size. By adding more turns and reducing the wire size, iIt will increase the resistance of the winding and also decrease the efficiency of the motor.

Yes, only 50 A per motor (2 required for each axle) to achieve an incredible torque at very low power. On less that I'm wrong, it’s not the power that will make the car go forward but the torque.

I will be running further simulation for a 50A but even at a lower current intensity. The cons is that I will need to increase the voltage.