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!
