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Design of Electric Vehicle Drives
 
 
There are two principal considerations in designing an electric vehicle, namely power needed for (a) max. speed on level (b) hill-climbing and acceleration.
At low speeds tyre and frictional forces predominate. They are approximately constant so the power required is proportional to the speed.
Air resistance is proportional to the square of the speed, so power required (Force x Speed) is proportional to the cube of the speed. Thus at high speeds, air resistance predominates.

On the Lynch electric buggy (weight about 400 Kg inc. batteries) we found that air resistance approx. equalled rolling resistance at 28-30 mph - about 1 Kw power each, total power 2 Kw. Total power at 50 mph was over 6 Kw, the greater part being increased air resistance.

In short, a reasonably light vehicle with small, hard tyres and without gearbox, back-axle etc running in lubricant, should travel at about 30 mph on 2 Kw of power. Hill climbing and acceleration are entirely dependent on vehicle weight for a given motor torque.
This torque may be limited by what a motor can safely provide without burning out, or it may be limited by e.g., an electronic controller which limits the amount of current either the motor can draw or the battery can give.

When climbing a gradient the additional power required is given by the work done in unit time in raising the vehicle's weight against the acceleration of gravity. If s is the slope or gradient in % - e.g. a slope of 1 in 6 rises 1 unit for every 6 units up the slope - the gradient is then said to be 16.6% or 0.166. Thus the vertical velocity against gravity will be s x v - where v is the vehicle speed up the slope - in this case 0.166V. The power required to run up the slope is:

P = MgsV (1) where M = mass of vehicle Kg
g = accel. gravity M/sec
s = slope %
V = speed M/sec
P = power in Watts

Electric Vehicle Motors
The power to raise the vehicle against gravity is supplied by the torque at the vehicle wheel.
If d = diameter of wheel in meters,
n = speed of rotation in revs per sec,
T = torque at wheel in Newton Meters,
then P = 2 p n T (2)

Ignoring, for the moment, torque needed to drive the vehicle at constant speed, the above values
of power will be equal then
MgsV = 2 p n T (3) [(1) = (2)]
But V = p n d (4) d = dia of wheel in M.
Therefore T = Mgsd/2 (5) [substituting (4) in (3)]

Note that the above is independent of speed (V), so that for a constant torque device, like a permanent magnet motor, if the torque is too low for a given slope the vehicle will not climb it at any speed. T is the torque at the vehicle wheel. This will be equal to the motor torque, t, multiplied by any reduction gear ratio, R (assuming that the motor is spinning faster than the wheel, which is usually the case).

Let us work out the implications of this in the case of two examples (1) a lightweight 30 mph runabout and (2) an American electric car capable of 100 mph. (See tabbed panel below for more details).

 
  • Lightweight Runabout
  • Electric Racing Car

Example (a) Lightweight Runabout

Weight 250 Kg without batteries. Range required 50 miles at 30 mph. Assuming 2 Kw required at 30 mph, for 1 hour 40 mins total energy is 3.33 Kw hrs.

We know from experience that a good lead-acid battery will give about 25 watt hrs per Kg* at the 1 hr rate, therefore 3.33 Kwhr will require 3333 divided by 25 = 133 Kg of battery. (* allowing 80% depth of discharge).

Suggested battery pack 4 off 12 V monoblocs weighing 33-35 Kg Total Wt: vehicle 250 & batteries 133 + driver 75 Kg Total wt = say 460 Kg.

Assuming 20 inch dia wheels (0.51 M) speed at 30 mph is 506 rpm. If motor speed is 3400 rpm on 48 Volts, gear ratio (R) is 3400 divided by 506 = 6.7 Torque at wheels T = Mgsd/2. Assuming 16% slope T = 465 x 9.81 x 0.16 x 0.51/2 = 186 NM Torque at motor is 1/6.7 of this, i.e., 27.6 NM.

If a Lynch motor is used this torque is given at about 220 Amperes, which is just greater than the continuously rated current - permissible for moderate hill climbing & while accelerating.

Example (2) Electric Racing Car, 100 mph

Assuming vehicle weight of 400 Kg without batteries, and low drag coefficient such that 50 Kw are required at 100 mph (67 BHP), then about 21% of this should be required at 60 mph, i.e., 11 Kw. To run for one hour would require 440 Kg of battery @ 60 mph. Therefore total vehicle weight = 400 + 400 + 75 (with one occupant) i.e., 915 Kg.

Assuming 26" dia wheels (0.66 M) speed at 100 mph requires 1302 rpm. Motor speed at different voltages is as follows (assuming 4 motors fitted):

Volts 48 60 66 72
RPM 3400 4250 4675 5100
Power Kw/ motor contin 8.64 10.56 11.48 12.24
Power BPH/motor contin 11.6 14.15 15.4 16.4
Drive Ratio (100 mph...) 2.6:1 3.26:1 3.58:1 3.92:1
Torque req at wheels 492 492 492 492
Torque at motors 189 151 137 126
Torque avail at motors (NM) 260 max 260 260 260
  100 cont 100 100 100

Acceleration of .166 g would give time of 0-60 mph in 16.7 secs. Full current on 60 V gives 0.33 g in 0-60 mph in 8.4 sec. Some overloading of the motors can be permitted in racing. Bigger safety margins should be allowed in normal running. Note that the high top speed gives a low drive ratio, and thus relatively poor acceleration and hill climbing. Greatly improved performance will be obtained by fitting a gearbox or CVT (Continuously Variable Transmission).

NOTES:
1 Continuous 200 A.
2 Calculated for 16% gradient = .16 g acceleration
3 4 motors - absolute max @ 475 A. 62 NM
4 Max current would reduce to 440 Amps.Lynch Motor Co. Ltd., PO Box 919,

To calculate Gear Ratio required by an Electric Vehicle:

Torque = Mgsd/2 (See formula 5)

Plot this out for different values of Slope from 1 in 4 to 1 in 10. Then calculate how fast the vehicle can run up each slope to keep within the power available on 60 v Lynch motor = 10.5KW

Assuming M = 1100 Kg
g = 9.81 M/sec/sec
d = 0.5 M (diameter of wheel)

Then Mg d/2 = 2698
ie Torque = 2698 x Slope (expressed as fraction ie. 25% = 0.25)

Table is then as follows:

Gradient     Torque (NM) Overall Gear Ratio (1) Max speed on gradient M/sec MPH Double Motor Gear ratio MPH
1:4 25% 0.25 674 27:1 4 9 13.5:1 18
1:5 20% 0.2 539 21:1 5 11 10.5:1 22
1.6 16.7% 0.167 447 18:1 6 13 9:1 26
1.10 10% 0.1 269 11:1 10 22 5.5:1 44

NOTES: (1) Ratio required to keep Torque at motor below 25NM (2) Speed below which power is less than 10.5 KW.

We have several Ford Fiestas running for 2 or 3 years with single motors. The motor can be run at higher currents for short periods - eg 2 x continuous torque for 5 secs, eg for climbing ramps and curves. This should not be relied on for continuous running. Nevertheless the double motor gives much more margin on hills and accelerating from rest.

Our double-motor Fiesta, with gearbox, gives over 50 m range with lead-acid batteries and over 40m driven hard in quite hilly country. We do not think the expense of Ni-Cd is necessary. Our customer, John Pochin, of Barkby, Leicester, UK, has been running a Lynch electric Fiesta for three years and is pleased to share his experience. Phone: +44 (0)116 264 0863 or Fax: +44
(0)116 269 5646

NOTE: The double motor uses two Lynch armatures linked together mechanically. The armatures can be connected electrically in parallel or in series, allowing max applied EMF and current of 60 Volts 400 Amp, or 120 Volts 200 Amp respectively. Max. continuous power output in both cases is 21KW.

 
 

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