2015/01/06

Going direct torque II



The new results

As mentioned in my last post “Going direct”, I have shown that it was possible to go direct torque by eliminating all unnecessary gear box like the transmission and the differential (transaxle). I was also questioning myself if 5000 or 600RPM was required to achieve 160 km/h. The answer was no since a 1500 RPM will meet the needs.
By running new simulations with FEMM, the torque constant Kt and the Bemf constant Ke (both are equal) where change. I have been able to achieve interesting results. As you will recall, in my last discussion, the quadrupole motor is design to produce a MMF = 500A/t (100A current and 5 turns per coil). Further down, there are four (4) different scenarios. The first will be the original quadrupole motor 5 turns at 100A. The second will be a 10 turns at 50 A. The third, it’s the same as scenario two but doubling the stack of the quadrupole motor from 38mm (1-1/2 inch) to 72 mm (alsmost 3 inches) and reducing the current as low as 25A. Since the numbers of turn have not been change, then a final scenario was to re-establish the current at 50A with a 76 mm length double the torque compared to scenario 1.



Scenario 1
MMF
500 A
Turns per coil
5
Design peak current
100 A
Design DC Bus voltage
As per table – included 5% voltage drop due to the inverter
Stack length
38 mm (1.5 inch)

1st speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,393
100
39,3
1000
41,1
96
2218
Inductance [µH]
0,0019
at 3327 rpm
2000
82,2
144
3327
Resistance [Ω]
0,003783

3000
123,4
250
5776
Impetance [Ω]
0,000009

4000
164,5


Efficiency [%]
99,91

5000
205,6





6000
246,7



2nd speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,228
100
22,8
1000
23,8
96
3828
Inductance [µH]
0,0006
at 5742 rpm
2000
47,6
144
5742
Resistance [Ω]
0,001892

3000
71,5
250
9970
Impetance [Ω]
0,000002

4000
95,3


Efficiency [%]
99,96

5000
119,1





6000
142,9









3rd speed






Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,170
100
17,0
1000
23,8
96
5117
Inductance [µH]
0,0004
at 7676 rpm
2000
47,6
144
7676
Resistance [Ω]
0,001892

3000
71,5
250
13326
Impetance [Ω]
0,000002

4000
95,3


Efficiency [%]
99,96

5000
119,1





6000
142,9










Scenario 2
MMF
500 A
Turns per coil
10
Design peak current
50 A
Design DC Bus voltage
As per table – included 5% voltage drop due to the inverter
Stack length
38 mm (1.5 inch)







1st speed 
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,748
50
37,4
1000
78,4
96
1164
Inductance [µH]
0,0036
at 1746 rpm
2000
156,7
144
1746
Resistance [Ω]
0,003783

3000
235,1
250
3031
Impetance [Ω]
0,000016

4000
313,4


Efficiency [%]
99,96

5000
391,8





6000
470,2









2nd speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,427
50
21,4
1000
44,7
96
2039
Inductance [µH]
0,0010
at 3058 rpm
2000
89,5
144
3058
Resistance [Ω]
0,001892

3000
134,2
250
5309
Impetance [Ω]
0,000005

4000
179,0


Efficiency [%]
99,98

5000
223,7





6000
268,5









3rd speed 
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,353
50
17,7
1000
44,7
96
2466
Inductance [µH]
0,0009
at 3698 rpm
2000
89,5
144
3698
Resistance [Ω]
0,001892

3000
134,2
250
6421
Impetance [Ω]
0,000004

4000
179,0


Efficiency [%]
99,98

5000
223,7





6000
268,5













Scenario 3

MMF
250 A
Turns per coil
10
Design peak current
25 A
Design DC Bus voltage
As per table – included 5% voltage drop due to the inverter
Stack length
72 mm (3 inches)

1st speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
1,375
25
34,4
1000
144,0
96
633
Inductance [µH]
0,0066
at 950 rpm
2000
288,0
144
950
Resistance [Ω]
0,003783

3000
432,0
250
1650
Impetance [Ω]
0,000030

4000
576,0


Efficiency [%]
99,98

5000
720,0





6000
863,9









2nd speed 
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,778
25
19,5
1000
81,5
96
1119
Inductance [µH]
0,0038
at 1679 rpm
2000
163,0
144
1679
Resistance [Ω]
0,003783

3000
244,4
250
2915
Impetance [Ω]
0,000017

4000
325,9


Efficiency [%]
99,98

5000
407,4





6000
488,9









3rd speed 
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,629
25
15,7
1000
81,5
96
1384
Inductance [µH]
0,0030
at 2076 rpm
2000
163,0
144
2076
Resistance [Ω]
0,003783

3000
244,4
250
3605
Impetance [Ω]
0,000014

4000
325,9


Efficiency [%]
99,98

5000
407,4





6000
488,9










Scenario 4

MMF
500 A
Turns per coil
10
Design peak current
50 A
Design DC Bus voltage
As per table – included 5% voltage drop due to the inverter
Stack length
76.2 mm (3 inches)

1st speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
1,497
50
74,8
1000
156,7
96
582
Inductance [µH]
0,0069
at 873 rpm
2000
313,4
144
873
Resistance [Ω]
0,003783

3000
470,2
250
1516
Impetance [Ω]
0,000031

4000
626,9


Efficiency [%]
99,98

5000
783,6





6000
940,3









2nd speed 
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,810
50
42,7
1000
89,5
96
1090
Inductance [µH]
0,0039
at 1519 rpm
2000
179,0
144
1529
Resistance [Ω]
0,003783

3000
268,5
250
2654
Impetance [Ω]
0,000018

4000
3357,9


Efficiency [%]
99,96

5000
447,4





6000
536,9









3rd speed
Ke, Kt
Current [A]
Peak torque [Nm]
RPM
Bemf [V] (Ke)
Vdc
Max. RPM
0,669
50
35,3
1000
74,0
96
1233
Inductance [µH]
0,0032
at 1849 rpm
2000
148,0
144
1849
Resistance [Ω]
0,003783

3000
221,9
250
3210
Impetance [Ω]
0,000015

4000
295,9


Efficiency [%]
99,96

5000
369,9





6000
443,9











Comparison of scenarios


The table below presents only the peak torque at first speed for each scenario

               
Scenario 1
Scenario 2
Scenario 3
Scenario 4
Peak Torque
39,3
37,4
34,4
74,8
Number of turn per coil/pole
5
10
10
10
Amps
100
50
25
50
Stack length
38
38
72
76



But first, a recall. All these scenarios are based on the BackEMF and the Toque constants; Ke and Kt are always equal.

Equation (1)  Kt = Ke = N Bp D l p where:

                Kt is the torque constant in Nm/A

                Ke is the backEMF in V/rad s-1

                N is the number of turns per coil

                Bp is the average peak flux density in the airgap in T

                D is the diameter of the of the rotor in m

                l is the length or if you prefer the depth of the lamination stack in m

                p is the number of pole pairs.

Equation (2): T= Kt i sin (ϴm) where:

                T is the torque in Nm

                i is the peak current in A

                sin (ϴm) is the mechanical angle of the rotor

Equation (3): E = Ke ωm sin (ϴe) where:

                E is the bEMF in V

ωm is the rotational velocity if the rotor in rad/s

sin (ϴe) and electrical angle in rad



Scenario 1 is the base scenario.


By changing the amount of turns and keeping the same MMF, we can easily increase or decrease the current. Since I want to reduce the amount of current to a bare minimum, I have decided to conduct a few more simulations with FEMM simulation software. By increasing the amount of turn, it also adds more BackEMF.

In scenario 2, I reducing the current by two and doubling the number of turns which maintained the same MMF, we can see that we lost only 4.8% of the initial torque. So we can conclude that by changing the count turn and adjusting the current, the loss is minimal and acceptable.

In scenario 3, however, doubling the stack length up to 72mm and reducing the current to 25A did not maintain the torque level of scenario 1. There is a loss of 12.5%. I thought that doubling the stack I could again reduce the current by half, but again, it is close. Since there is a change, this is explained by the fact that the airgaps magnetic flux is not necessarily proportional as per the torque and backEMF constants for reason that I personally cannot explained. Nevertheless, for I, it’s acceptable compromise because of scenario 4.

In scenario 4, I only opted to double the current as per scenario 3 and I also adjusted the stack depth from 72 mm to 76mm. By doing, so we almost double the torque compared to scenario 1; up to 75 Nm at 873 rpm (and 144Vdc of course) . This is great because with 2 quadrupole motors with a stack of 76mm and 50A, we double the available torque and adds up to 140 Nm. My Toyota Yaris torque has a torque of 140 Nm at 4500 RPM. We get a torque of 140Nm from no speed up to 100 kmh/s.  But the greatest advantage is that I get to keep the same battery pack, 100A/h at 144Vdc.

Nevertheless, we still need to compare the top speed at third electric gear!



Scenario 1
Scenario 2
Scenario 3
Scenario 4
Maximum RPM (third egear)
7676
3698
2076
1849
Top speed (km/h) (as per wheel size : 0.610mm)
890 !
428 !
241
225
Current (A)
100
50
25
50
Stack length
38
38
72
72



Now, one last comparison, the available torque!



Scenario 1
Scenario 2
Scenario 3
Scenario 4
Maximum Power (kW)
13.7
6.8
3.4
6.8
Maximum RPM (first egear)
3327
1746
950
873
Current (A)
100
50
25
50
Stack length
38
38
72
76
Torque constant (Nm/A)
0.393
0.748
1.375
1,497
Force (N) (as per wheel size : 0.610mm)
128
123
113
245



So, is low power critical when the power output is as low as presented in in scenarios 2, 3 and 4? Can’t answer this question yet!


Force and torque requirements


Toyota advertises the Yaris of having a curb weight of 1050 kg (2 315 lb). The gross vehicle weight rating (GVWR) is rated at 1515 kg (3 300 lb).

Technically, if it is the weight, then the mass of the vehicle would be 107.14 kg (236 lb) and the gross vehicle mass (GVM) would yield to 155 kg (341 lb) !?! However, it is actually, the mass and the unit is kg. This is the maximum operating mass of the vehicle as specified by the Toyota with the driver, passengers and cargo added to the curb mass.

However, the above numbers are very small!?! If my Yaris weight is in fact the mass, then the gravitational acceleration g will have to be added in the equations.

However, all this is confusing especially in imperial units. The pound, unit lb, is use with weight just as much as mass. Sometime the unit lbf for pound-force, the f is shown and confirms that gravity is in the equation. In this case it is clear that the force g act on a body. To insure that I have the total load force acting upon the car, the gravitational acceleration has to be in the equations.

So, the natural force or weight W = m g where m is the mass in kg (lb) and g is gravitational acceleration and equates to 9.81 m/s2 (31.16 ft/s2).

Weight is the force on the object due to gravity. Weight and mass are scientifically distinct quantities; the terms are often confused with each other in everyday life and I'm certainly confused.

W = 1050 kg × 9.81 m/s2 = 10 301 N (74 507.28 lb ft/s2 or 2315.75 lbf!).

Ok, I’m confused. If 10 301 N = 2315.7 lbf and that 1050 kg also = 2315 lb – Can someone tell me what is going on?!? So lb and lbf are equal? I just mentioned above that lbf took into account the acceleration of the gravity g! So when my doctor’s telsme that I weigh 176 pounds and the scale shows 80 kg, one measurement includes the gravity while the other excludes it?. OK I got it, there’s pound-mass and pound-force or pound-weight. At the doctor’s, he is measuring the mass. However, when I convert my units, I use tool converters available on the web and I noticed that the conversion from pound-force to pound-mass is not available. So, converting Newtons to pond or kilogram remove the gravity in the equation.

With that much force acting on the car, it would take a fair amount of energy!

From now, we will have two tables side by side to be able to compare. One with the natural force W (Fn) for the weight and the other with the advertised weight.

The rolling resistance force factor of a car is on asphalt.

Fr = c W

Where :

Fr = the rolling resistance force (N, lb ft/s2 – short ft-lbs)

c = rolling resistance coefficient – dimensionless (table of available from different sources)

r = radius of the wheel (m, in)

Example 1:
Natural Force W

Example 2:
Advertised weight (which technically, it would already taken into account the gravitational pull
Weight/Force: 10 301 N

Weight/Force: 1050 N

Wheel diameter: 0.610 mm (24 in), r = 0.305 m (12 in)

Wheel diameter: 0.610 mm (24 in), r = 0.305 m (12 in)
Fr = 0.03 ×10 301 N

Fr = 0.03 ×1050 N
Minimum Force to displace is Fr = 309 N

Minimum Force to displace is Fr = 31 N
Minimum Torque at wheels =309 N × 0.302 m = 94.245 Nm

Minimum Torque at wheels =31 N × 0.302 m = 9.455 Nm


Note: since two wheels are involved in moving the car, then the total torques has to be distributed in the two front wheels, so the minimum torque to move the car per front wheel is 47 Nm.

Is example 1 good? Yes. However, there are a lot more other factors that need to be put into the final force equation. Example 2 will not be part of any calculation.

Know, we will now take previous table and compare the force required with the table below.



Scenario 1
Scenario 2
Scenario 3
Scenario 4
Maximum Power (kW)
13.7
6.8
3.4
6.8
Maximum RPM (first egear)
3327
1746
950
921
Current (A)
100
50
25
50
Stack length
38
38
72
76
Torque constant (Nm/A)
0.393
0.748
1.375
1,497
Torque (Nm)
39.3
37.4
34.4
74.8
Force (N) (as per wheel size : 0.610mm)
128 x 2 = 256
123 x 2 = 246
113 x 2 = 226
245 x 2 = 490
Example 1 : Fr = 309 N (227.9 lb-fts)
Failed
Failed
Failed
Passes


Note: since there is no transaxle gear, two motors are required per front wheel.


In the table below, I made an exercise to verify just how much torque was available at the 2 wheels from my Toyota Yaris 2008 taking into the counts the transmission and transaxle ratios. For this, the RPM motor was set to 3000 and the maximum torque at 139 Nm (available only at 4500 RPM because I do not know what is the torque at 3000 RPM since I did not found a chart). First I needed the gear ratios for the transmission and the axle ration as the final gear.



Transmission gear ratio
Motor RPM
1st
2nd
3rd
4th
5th
3000
3,166
1,904
1,392
1,031
0,815
RPM at Transmission
948
1576
2155
2910
3681
Final axle differential gear ratio
4,312
4,312
4,312
4,312
4,312
Final wheel RPM
220
365
500
675
854







Transmission gear ratio
Motor max. torque (Nm)
1st
2nd
3rd
4th
5th
139
3,166
1,904
1,392
1,031
0,815
Torque at Transmission
440,1
264,7
193,5
143,3
113,3
Final axle differential gear ratio
4,312
4,312
4,312
4,312
4,312
Final at front wheels torque
1898
1141
834
618
488
Force (N)
6222
3742
2735
2026
1602
Available output power (kW)
44
44
44
44
44


As for the RPM, the Quadrupole motor is right on it.
Note that the power, it’s only a figure to verify ma equation. Actually, it is less because the maximum torque is at 4500 RPM and not 3000 RPM.

Ok, where it hurts it’s the torque. At first egear of the Quadrupole and taking comparing it to scenario 4, the torque is 70.9 Nm times two yield to 141.8 Nm. It is far from the 488 Nm at fifth gear! And even further more at first gear.

The purpose was to study the action the transmission and transaxle and the way the contribute to elevate the torque and speed. 

Putting directly two motors on the axle is too demanding for two motors at 70.1 Nm each. One way to resolve this is to add more depth so that it will be larger. 

For instant, if I take the Kt constant and apply and double the thickness of the motor stack to 152 mm (6 inch), the motor should almost double the torque at around 140Nm (taking 12% of 159Nm). Having 2 motor under the hood, I will raise the torque up to 280 Nm. This number came fresh from FEMM and not also that the backEMF will also raise.

I just want to point out that you will have 280 Nm from start up to 980 RPM! Just to recall a thermal motor of my Yaris will output it maximum torque at 4500 RPM!
 

Conclusion

I have tried to prove that it is possible to put two motors and remove the transaxle and the transmission all together. The number should be validated by a third party. Nevertheless, my calculation where simply taken from known equations. Without any doubts, I think that it is viable to go take this route. I also demonstrated that adding doubling the depth of the motor at 152 mm (6 inches), it possible to almost double the torque.

Also, on the web there is a lot of research and even motor available mounted directly in a wheel.  I also found this website; you should just peak at it: http://www.electriccarsmotor.com/dc-electric-car-motors.html

Final note

It is not my intention to convert my Yaris. However, I do plane to built a reverse three wheeler with this Quadrupole motor and having a frontwheel drive vehicle. I intend to built the chassis entirely in aluminium. I expect to have a total mass of 600kg. So the motor presented in scenario 1 will do the job just fine!