Energy and Losses
A motor converts electrical energy into mechanical energy. However, in the conversion some of the electrical energy is wasted as heat. Some of this loss is because motors are not perfect, so if heavily loaded, they get hot, some of it is because mechanical systems are not perfect, they have friction and this also causes heating.
The mechanical energy out of the motor is used partly to accelerate the vehicle (it is turned into Kinetic energy) and partly in is used to climb hills (it is turned into Potential energy).
This sounds quite complicated, but if you consider the electrical energy being used in five separate ways things start to get clearer.
- Electrical inefficiency, shows up as heating in controller, wiring and motor.
- Mechanical rolling losses, difficult to calculate!
- Kinetic energy
- Potential energy
KE and PE are actually quite easy to calculate – if you have all units in Metres, Kilogrammes and seconds.
- Windage, difficult to calculate but not important at low speeds.
The other factor of importance to robots is of course torque. We’ll get to that later. But you do first need to understand a bit about what happens to the electrical power you put into the motor.
Electrical energy = volts x amps x time. So a 12v battery giving 10 amps for one minute (60 seconds) will give 12 x 10 x 60, or 7200 Joules and a motor taking 20 amps at 10 volts for 60 seconds will deliver 20 x 10 x 60, or 12000 Joules.
If the motor and controller and gear ratio are chosen correctly, electrical losses are small: motor efficiencies can be between 70 and 95%, controller efficiency much higher – we don’t want them to get hot – in the region of 97-99% range so, in a well designed system, nearly all the power taken from the battery goes to the motor. Remember that, with an efficient system, you can recover useful energy with regenerative braking.
Generally electrical inefficiency shows up as heating. Heating is proportional to the square of the current, so it pays to keep the current down and go for a higher voltage.
Remember motor and battery current are not the same: because our controllers use high frequency chopping, the motor’s inductance sustains and smoothes the current so that it is pure d.c. with very little ripple. However the battery current is chopped on and off, only flowing when the motor is connected to the battery. So at 50% modulation (i.e. at half full speed) battery current will flow 50% of the time, so you will measure a battery current equal to half the motor current.
Mechanical rolling losses
These you will have to measure. Go for an efficient gear train (worm gears tend to be bad). Keep all bearings well lubricated.
Kinetic energy is defined as 1/2 x mass x velocity². Mass should be in kilograms, velocity in metres/second. So a train weighing 250kg travelling at 4.47 metres/sec (which is 10mph) would have an energy of 0.5 x 250 x 4.47 x 4.47, or 2497 Joules
If we require our vehicle to accelerate smoothly to top speed in, say, 60 seconds then current must flow for this full 60 seconds and the electrical energy used in accelerating will equal the kinetic energy gained.
So: volts x amps x time = 12 x amps x 60 = 2497 (the K.E. gained).
Therefore amps = 2497/60/12 = 3.47 amps
Potential energy is Mass x g x height, where g is 9.80 metres/second/second. the acceleration due to gravity. If we have a gradient of 1 in 50, 30 metres long, then the height gained on this incline will be 1/50 x 30 or 0.66 metres. In ascending this incline our vehicle will have gained a potential energy of
250 x 9.8 x .66, or 1617 Joules
At top speed the train will travel at 4.47 metres/second so it will take 30/4.47 seconds to travel the 30 metre incline, i.e. 6.71 seconds. The current must flow for this time so
It doesn’t help at all to go slowly up the incline (unless you have mechanical gear change): if it takes 20 amps of motor current at full speed, then the motor current will still be 20 amps at half speed, because full speed corresponds to 12v on the motor (which we used in our calculation) so half speed will be 6v on the motor. Halving the motor voltage halves the power, so the motor current won’t change. Yes – the battery current will halve, but it will flow for twice the time since the slower machine will take twice as long to climb the hill, so there is no overall benefit. At high motor currents the motors and controller will get hot (wasting power). The power wasted is only down to the motor current: the quicker you get up the incline therefore the shorter the time for which you will be wasting power, so the smaller the overall losses.
Kinetic energy is defined as : ½ x mass x velocity².
Electrical energy = volts x amps x time.
Equating the two and rearranging to get current,
½ x mass x velocity² = volts x amps x time.
Amps = ½ x mass x velocity² / volts / time.
Current = ½ x (vehicle laden weight) x (max vehicle speed)² / battery volts / (time to top speed)
Potential energy is Mass x g x height,
where the vehicle’s mass is measured in Kilograms,
height is in metres and
g is 9.80 metres/second/second, the acceleration due to gravity.
If we have a gradient of T%, then the height gained will be
T/100 x Length (the length of the incline in metres)
the potential energy will be Mass x g x T/100 x Length
Our vehicle will traverse the incline in Length / Speed seconds.
Current must flow for this time so electrical energy will be: Volts x Amps x Length/Speed
Equating electrical to mechanical energy we get
Mass x g x T/100 x Length = Volts x Amps x Length/ Speed
So the motor current must be
Mass x g x T/100 x Speed/Volts
Volts and amps in the calculation must be the motor volts and amps, not the battery volts and amps but, at top speed, motor volts and amps are equal to battery volts and amps and the calculation approximates to
Current =1/10 x (Vehicle laden weight) x (gradient) x (Top vehicle speed) / (Battery voltage)
This is difficult to calculate, but for Greenpower karts and similar vehicles it makes a significant difference.