With 10 watts available in the form of 10 amps at 1 volt, the motor at stall will only take
1.25 amps at 1 volt which is only 1.25 watts input to the motor. (Ohms law) This current
will drop as the motor speeds up reducing the power input to the motor. Let us assume
absolute best case conditions of 1.25 watts input to the motor and motor efficiency of
100%. The best we can achieve is 12.5% of the available panel power being delivered by
the motor shaft.
With 10 watts available in the form of 0.1 amps at 100 volts the motor would like to take
125 amps but the current is limited to 0.1 amp, when we deduct the 0.030 amps needed to
just run the motor we only have 0.07 amps available to drive the load. The next question
is how fast will the motor run? With a 125 volt supply the motor would like to
theoretically run at 148,750 RPM which would destroy the motor. In fact testing has
shown that the motor would use all the 0.10 amps to just run the motor itself at 28,000
RPM. If we assume the motor is running at 15,000 RPM it would produce 0.875 Watts.
This is only 8.75% of the available panel power.
Compare the above with actual test results from a Solarex SX 10 panel configured to give
10.5 volts open circuit and capable of delivering 8 watts. When connected to this panel
the motor to produced 6.25 watts output on its shaft. This is 78% of the available panel
power being delivered by the motor shaft.
From the above it is obvious just how important the choice of solar panel voltage
and current output is.
Gear ratio to match motor power to power to run car: To make best use of the power
available from the motor shaft it is imperative we choose a gear ratio that just fully loads
the motor.
To better explain this look at graph 5 and in particular the power vs RPM graph of motor
direct on panel switched in parallel OCV 10.5 V (blue power graph) Maximum power
occurs at about 6200 RPM. If we aim to run in the power range above 1.1 watts there is
only a 2000 RPM wide band to operate in before power falls below this level.
We must use a gear ratio that reduces the 6200 RPM motor speed to a wheel speed that
pushes the car to the particular speed that just uses the 1.1 watts available.
Great you say, but what is this gear ratio? A good idea of the required ratio can be
obtained by running the “Mathematical Simulation” (See appendix H).
For this though it is essential to have the critical car parameters ie. weight, aerodynamic
drag characteristics and resistances for wheel and guide rollers. Track testing the car is
another and far better option.
Remember when sun level changes so does the required gear ratio.
Extreme care is needed in choosing these components if maximum performance is to
be achieved.
Overall for best car performance the following points must be considered
- Good power to weight ratio
- Low aerodynamic drag
- Accurate construction
- Lots of testing and tuning