In
my last post, Going Direct Drive  II, I have shown a few numbers on different
scenarios where each scenario represented 4 different designs of the quadrupole
motor.
However,
I was unclear if each different design of quadrupole motor were powerful enough
to move my Toyota Yaris. Now, since I know the available torques with different
design, it’s now time to calculate the required toque and power according to
the vehicle mass and parameters imposed in the design. From there, it will be
easier to design the motor’s minimum specifications.
Also,
a lot was to deal with the weight and the mass definitions. Reviewing a few of
my notes and searching on the internet for some of my unanswer questions, I finally
found does answer about the weight and mass. As an example, when you take the
weight on a scale, you are not measuring your weight but your body mass!
Weights take into account the gravitational pull and the unit should not be in
pound not kilogram.
Because
there was a lot of confusing around the mass and the weight, mass is express in
kg (lbs) and weight will be express in N or kg/m^{2} (lbs ft) in this
blog.
First,
a few data about the car.
Description

Symbol

Target / requirements

Gross Vehicle Mass (GVM)

m_{v}

1 515kg (3 300 lbs)

Vehicle rolling resistance
(excellent / fair / poor)

c_{r}

0.008 / 0.015 / 0.03

Vehicle average
acceleration

a_{v}

1.85 m/s^{2} (6.069
ft/s^{2})

Vehicle acceleration time

t_{a}

15 s

Vehicle maximum speed (at
1% grade slop)

v_{v}

39 m/s (140 km/h – 87 MPH)

Vehicle frontal area

A_{v}

1.9 m^{2}

Vehicle drag coefficient

c_{d}

0.29

Rolling radius

R

0.302 m (P175/65R14 tires)

Number of driven wheels

n_{v}

2

Three
important forces act on the vehicle and are calculated according to these
formulas.
The
natural force or the actual weight in N
(1) F_{n} = m g sin (ϴ)
The
rolling resistance force of the tires:
(2) Fr = F_{n} C_{r} cos (ϴ)
The
aerodynamic drag force:
(3) Fd = ½ ρ c_{d} A v^{2}
And
the slope:
(4) F = F_{n} + F_{r} + F_{d }
Power requirement on a 1% slopped road and
100km/h (acceleration is null):
First,
let’s start by calculating the power requirement. The equation is below:
(5) P = F v
P is the mechanical power in watt (W) and v is the
speed in m/s. The total force (whitout acceleration) while be 854.7 N. So the
power requirement for a speed of 100km/h (27.8 m/s) will yield at 23 760W.
For a 2 wheeler, each will have to power half of the
total force which is 11 880 W. This is nominal or minimum power to move my
Toyota Yaris.
To covert speed (linear velocity) to rev/s (rotational
velocity), we simply use equation 5 below:
(6) ω_{w} = v / r_{w}
Where v is the linear velocity of the car in m/s and r_{w}
is the radius of the wheel in m.
In a worksheet like excel, I’ve put in does numbers
and plotted the trend that is shown here.
For a speed of 100 km/h, the power requirement is
23 740 W. For each wheel, it will yield to half the power; 11 870 W.
Below, the force required per wheel is plotted below.
At 100km/h, the force yields to 427N.
Below is the torque per wheel trend.
Let’s compare this to the 4 scenarios of my last post.
First, the power capability of the quadrupole is
calculate from this equation 76.
(7) P_{e} = T_{e} G_{r} v/r_{wheel}
Where Pe is the power from the electric motor, Te is
the torque capability of the motor in Nm, Gr is the gear ratio between the
wheel and the motor, v is the vehicle velocity in m/s and r_{w} is the
radius of the car wheels traction in m. The gear ratio is 1:1 (1) since there
is no gear reduction between the motor and the wheels.
Electronic
speed

RPM

rad/s

T_{e}

G_{r}

r_{w}

v

P_{e} = T_{e} G_{r}
V/r_{w}


Scenario
1

1st
espeed

3327

348,4

39,3

1

0,305

16,9

13 691

2nd espeed

5742

601,3

22,8

1

0,305

29,2

13 709


3rd
espeed

7676

803,8

17

1

0,305

39,0

13 664


RPM

rad/s

T_{e}

G_{r}

r_{w}

v

P_{e} = T_{e} G_{r}
V/r_{w}


Scenario
2

1st
espeed

1746

182,8

37,4

1

0,305

8,9

6 838

2nd espeed

3058

320,2

31,4

1

0,305

15,5

10 055


3rd
espeed

3698

387,2

17,7

1

0,305

18,8

6 854


RPM

rad/s

T_{e}

G_{r}

r_{w}

v

P_{e} = T_{e} G_{r}
V/r_{w}


Scenario
3

1st
espeed

950

99,5

34,4

1

0,305

4,8

3 422

2nd espeed

1679

175,8

19,5

1

0,305

8,5

3 428


3rd
espeed

2076

217,4

15,7

1

0,305

10,6

3 413


RPM

rad/s

T_{e}

G_{r}

r_{w}

v

P_{e} = T_{e} G_{r}
V/r_{w}


Scenario
4

1st
espeed

921

96,4

70,9

1

0,305

4,7

6838

2nd espeed

1614

169,0

40,5

1

0,305

8,2

6845


3rd
espeed

1952

204,4

33,5

1

0,305

9,9

6847

So, all current designs do not meet the requirements excepted
for Scenario 4 at first speed but I will never be able to reach top speed of
140 km/h because the load per wheel needs to triple. In scenario 4, the
current is set at 50A load. I will need at least 222Nm of torque to reach the
top speed of 105 km/h. The currents designs cannot handle that much current.
So, one easy thing to do is to triple the depth of the motor; passing from 76mm
to 228mm, almost 9 inch long.
But, the quadrupole was first design to handle speeds
up to 6000RPM and to be connected to a reducing gear like a car transmission and
differential. Also, I have another
project; I intend to build an XR3 3 wheeler vehicle from Robert Q. Riley Enterprises.
According to their specifications, the Scenario 4 will do the job because I
need 13 kW to reach 140 km/h. So with minor modifications such as adding winding
to the outer stator will enhance the MMF.