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Data we will use for calculations:
A 20.5” diameter tire has a circumference of pi*20.5in * 2.54(cm/in) / 100 (cm/m) = 1.636m
Travelling at 16.6m/s, this tire rotates at 10.1478 rpsThe strategy model is currently using a 26cm radius tire.
For calculating power, we use: (https://www.engineeringtoolbox.com/angular-velocity-acceleration-power-torque-d_1397.html)
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Where P (W), T (Nm), nrps = number of rotations per second
For calculating phase current, we will use these calculations: Motor Controller Output Current
This desmos graph was used to calculate based on the bus voltage: https://www.desmos.com/calculator/kwklxrs8g3
Steady State:
Average torque = 15.7Nm (See note that this may not be 100% accurate due to the distribution of waypoints on our map data)
Power(W) = 15.7Nm * 2 * pi * 10.1478 = 1kWAt worst-case bus voltage of 100V (worst-case according to the desmos graph on Robin’s page linked at the top), this gives 10A of bus current - but this is not useful for the output current calculations.
Vbus = 100V
Robin’s desmos graph on the pages linked above shows that lower bus voltage leads to higher power output, but that is in fact not true. When we decouple the bus voltage from the output current (as it really is), high bus voltages leads to a high power output, so we will use the max Vbus reasonable for our calculations. 145V since the voltage drops off fairly quickly from full charge at 151.2V, and a complete top-balance of the batteries is unlikely.
Vbus = 145V
We’ll use Kv = 6 for the high speed motor coil.
Phase Current (Io) = 15.7Nm * pi * 6 / 30 = 9.86A
Plugging this in to the motor controller power dissipation formula, we get:
P = 11.73W
Transients (On Hills):
Let’s go with a 10 degree, 200m hill, and we climb it slowly, almost as if from a stop. This is chosen since I do not believe that maintaining a 16.6m/s speed up a hill would always be reasonable. There is definitely going to be sometime that the car is only able to travel at walking speed, say when starting and going up a hill.
Choosing an average walking speed of 1.4m/s (https://www.healthline.com/health/exercise-fitness/average-walking-speed#average-speed-by-age), we can find the hill to take 200 (m) / 1.4(m/s) = 142 seconds = 2minutes, 22 seconds.
From above, the torque requirement for a 10 degree hill is around 165Nm per motor per the hills listed above. We’ll also switch to using Kv = 5.2 for the low speed motor coil, as on hilly days, this is what we would require.
This gives a phase current of:
Io = 87.1A, which does not exceed the motor controller’s phase current limit of 100A
And Power:
P = 130.16W
So for transients, I would be comfortable going with a 130W power dissipation for 2 minutes.
What if we have a 36km/h (10m/s) headwind - aero losses will increase, and thus the torque requirement and bus current will increase. Will the torque requirement increase significantly? Should we take the wind into account for this model?