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.

Controlling the 3 speed quadrupole motor

In response to my last post, I have to give you a little more detail. 

Lets assume that coils A, B and C are installed on the outer stator and U, V and W are installed on the inner stator.

Here you will find down 3 possible systems. Your comments are welcome.

Scheme 1, option 1

In scheme 1, option 1, 3 contactors are use in series with the each coils and each coils are connected in parallel. Coil A is in parallel with it's sister coil U, coil B is in parallel with it's sister coil V, and coil C is in parallel with it's sister coil W. Switches are use to control the the contactors in with, the contacors will allow current to the coils, either the inner, the outer or both stators coils. In any condition, the should allways be at least on contactors energize.

Scheme 1, option 2

In scheme 1, option 2, the difference with the shceme 1, option 1, 3 contactors are connected in parallel with each coils and each coils are connected in series. Therefore, Coil A is in series with it's sister coil U, coil B is in series with it's sister coil V, and coil C is in series with it's sister coil W. Switches are use to control the contactors. In turns, the contacts of in parallel of the coils will be shorted to reroute the current, either the inner, the outer or both stators coils. In any cases, should both contactors, C1 and C2 be shorted at the same time.

The last one, two inverters are use. So what is the best system?
 

Scheme 2.

For systems 1, I do not know what will be the reaction of the inverter to rapid change. The answer certainly come from the manufacturer of the inverter.

Maybe the easiest would be 2 inverters.

A three speed motor

Bonjour tout le monde!

some news: the final simulations with FEMM are almost completed. I now have a pretty good idea on how the motor will perform and will very soon be able to order my laminated steel ... 

How time passes and thanks

It has been over two years since I started working on the Quadrupole electric motor. I had a few people exchanging knowledge with me and also, giving me good advice. So, I would like to take this opportunity to thank each of them.

The non-obvious behaviour

During my first simulations way back then, I noticed that BackEMF in the inner and outer coils of the stators were not equal. This meant that the airgaps of the two stator were also "unballance". So I tried to eliminate it and led me nowhere. I abandoned that path because it was time consuming.

So my adventure continued without paying attention do this behaviour of the Quadrupole. In this situation, I was more interested to have the Back-EMF waveform to mimic the voltage sinusoidal wave source.

To recall; the main objective with the Quadrupole motor is to achieve high torque, high speed, and not use as much as possible torque weakening technique to achieve speed.

I discovered that by disconnecting for instant the outer coils from the simulation it gave more speed but lower torque, so a pseudo-field weakening technique without the controller/inverter being involve, simple logic. I have never pay any attention in my early simulations since I did not quiet understood the torque and counter electromotive force constants. It is not obvious for non-practitioners in this field of research, but the Quadrupole motor can act as a three speed electric transmission!

How is this possible? Just because the Quadrupole motor is radially design! The radius and the circumference length of each airgap are always unequal. The magnetic flux densities of the air gaps will therefore uneven. Comparing this to an axial flux motor having two stators, the axial flux in the airgaps of both stator are always equal. So, what was not obvious became clear and hit me!

Three values for the torque and back-EMF motor constants in one motor

Another interesting thing about the Quadrupole motor is that its torque and BEMF constant will vary according to wire phase configurations.

If the outer and the inner stator coils are configured to be connected in series, the torque constant of the latest design, is 0.375 Nm/A and its BEMF constant is 0.375 V/rad∙s^-1.

If only the outer stator coils are used, the torque constant will now be 0.205 Nm/A and its BEMF constant is 0.205 V/rad∙s^-1.

If only the inner stator coils are in use, the torque constant will now be 0.145 Nm/A and its BEMF constant is 0.145 V/rad∙s^-1.

So the higher value of the torque constant, the higher will the torque be. On the other hand, the lower the BEMF values will be, faster the rotating speed be.

Configuring the new winding schemes

By modifying and reconfiguring the wire connections of the inner and outer winding, it is possible to have a three speed electric transmission-motor that could give a sporty-like drive as if the vehicle was equip with manual 3 speed mechanical transmission.

At the date of the publication of this post, the results of the latest Quadrupole motor are described below.

1) At first speed for high torque at 37.5 Nm and low speed of 3700rpm: the coils of the inner and outer coils are connected in series. This means that the windings of phases 1, 2 and 3 with respect to coils A, B and C of the outer stator are connected to their sister the windings of phases U, V, and W of the inner stator respectively.

2) At second speed for medium toque of 20.5 Nm and speed of 6700rpm, only outer stators windings A, B and C are active.

3) At third speed for low torque of 14.5 Nm and high speed of 9500 rpm, only inner stator windings U, V and W are active.

The above figures are achieve with a 144 Vdc and 100A peak (71 Arms).

Now, how will all this be controlled?

Probably, the easiest way to control the speed and torque by is be re-configuring on the fly the wiring connections by using a motor contactor. High power SSR could be use but a minimum of three will be required for each phase. Motor contactor is probably the best way to go for testing. However, how the inverter/controller will react to a fast change is unpredicted and will certainly be a nice challenge,

Or again, two controllers might also be a good option. One controller per stator and each controller will see a single motor. Having them connected in parallel, it will raise the efficiency of the whole scheme at the expense of raising the current, since they will be connected in parallel from the point of view of the battery pack. The controller / inverter will have to be "disconnected" by changing the value of the analog input of the speed demand. For example, we have the controller / inverter A connected to the phases of the outer stator and controller / inverter B connected to the inner stator phases and a button or handle could be used for electrical shift applications. First, the two inverters are used for the first mechanical/electrical speed and the throttle is use to sens the analogue signal to both of the inverters. By manipulating the button or handle, we move to the second speed by sending an analog 0 volt signal to the inverter B at and inverter A will continue to receive its signal from the throttle. For the third speed analog entered the inverter A will subsequently be set to 0 volts and the inverter B now receives its analogue signal from the throttle.

Advantages

The biggest advantage is that directly coupling the motor to the transaxle drive system of the vehicle will ease the coupling to an electric motor. Also avoiding to the manual transmission will also reduce mechanical losses and reduce weight.

Is this great or what?

Why concentrated magnetic field?

Concentrated the magnetic field is not a new idea.In fact, Dr. T.A. Lipo has worked extensively with this principle by developing reluctance motors.The magnetic field was concentrated by installing the stator permanent magnets whose dipoles were repulsing. Mr. Flynn also use this principle to patent the technology of the magnetic path. Moreover, Mr. Flynn explained very well in one his papers the how and why (unfortunately, it has seem to have been removed from flynnresearch.net website).


In the Quadrupole motor/generator (and also other project like the LRK project on the openvolta website), I have been using the PMs (permanent magnets) in the same fashion as M. Flynn and Dr. Lipo however, in the rotor, not into the stator.

If we have 2 PMs having the same dipole facing each other, a repelling force is produce. However, if you put a block between each, it will concentrate the magnetic flux within this block and if the block in not saturated of magnetic flux, then the PMs will stick to the block. If close the loop, each blocks will have a single pole, they will emulate a homo-polare magnet.

The advantage of doing this is adding force. So if you have both magnet with a unit 1 force, adding them face to face with dipole alike, you'll get 2 forces of unit! Right?.. No! you will get 4 forces of units!

A bit of math!  

No, we are not going back to school but if we need to understand how we get 4 times more force, we have to know about one basic equation.

Note that the equation is valid only for cases in which the effect of fringing is negligible and the volume of the air gap is much smaller than that of the magnetized material (less then 5mm).

The mechanical force between two nearby magnetized surfaces can be calculated with the following equation.

F = μ₀ H² A / 2

where:
 F is the force, N (Newton),
A is the area of each surface, in m²,
H is their magnetizing field, in A/m,
μ₀ is the permeability of space, which equals 4π×10⁻⁷ T·m/A

Since we are dealing with magnets, it is easier to use the following equation:  

F = B² A / 2μ₀
where:
B the flux density, in T (Tesla).

An easier way to express this is: F = 400 000 B² A.

Proof of concept:

As show here, the magnet is on the left side inserted in the 2 parts to create the core and the attracted armature on the right side. The airgap is 1 mm.

One part core is 40 mm x 100m x 20mm depth. The total area A = 40mm x 20mm x 2 poles= 1 600 mm². The area has a flux density of approximately 0.4 T.

F = 4 000 000 ­x 0.4²x 0.001 6 = 102.4 N
If we verify with FEMM (finite element method software), the computed value is 108.1 N, which it is normal since the exact flux density is 0.411856 T.

If 2 PMs are put in parallel has shown here, we can see that the field density in the core and the armature has increased. If we use the same equation, one would think that since we have 2 PMs, we just have to multiply by 2 the force where 2 x 115 N = 203N. This not correct.

Since we have 2 PMs which faces the same dipole, the equation should therefore be restated to reflect this physical aspect. Therefore,  F = 400 000 x (2B)² x A. So the true value of this attracting force will now be F =  400 000 x (2 x  0.4)² x 0.001 6 = 409.6 N. So, 409.6 N / 102.4 N gives up to 4 time the initial force!

The magic is here, since we do have 2 PMs, we have two times the flux density (T);  2 x 0.4 = 0.8 T. Then the square root of 0.8 is 0.64 T. So basically what we are doing here if we assume that on magnet as 1 unit of force (1force x 2) x (1force x 2), we can now see that we quadrupled the force if we look only the components "2".

The results with FEMM yields to 431.6 N. Again 431.6 N / 108.1 N = 3.992 N, rounds up to 4 time more force.

The Quadrature

The quadrature is basically an arrangement of magnet that focuses the magnetic field in the centre of the quadrature. This is also know as a quadrupole magnet. This basic idea is use from which it inspired to design the rotor

 

The rotor

The Quadrupole motor / generator utilizes the PMs in parallel the quadrature, like the quadrature magnet technique to concentrate the magnetic flux within segment. When looping the segments with a PMs, the active production of torque is 4 time higher than using magnet only.

Also, putting 2 stators, in inside, the outer, outside, we will radially that full advantage of the rotor, not just in or out of the rotor. This is like having 2 motors / generators in one machine.