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Aiming for 16.6m/s throughout the race. These calculations of torque requirement are done to maintain this given 16.6m/s speed. 16.6m/s = ~60km/h
Old Data - Torque Requirements
Average torque is 15.7Nm
Average of all the positive torques is 17.2Nm (treating the distance between each 2 points as identical). I asked Emma for this assuming that there were negative torque showing regenerative braking but I don’t think there actually were.
^Note that the above torques are for each of the motors. The average calculation was done as “an unweighted average, where i just took the mean of all the torque required values between navigation points” (Slack message from Emma Wai)
This data is the map from ASC 2018 with waypoints along the route. We will look closer at the hills at the 1.3 and 2.5 markers.
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At 1.5: climbing 1100 over 120km, 13Nm
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How long will these hills take?
120km @ 16m/s = 7500s = 125minutes, so we can assume this to be steady-state.
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So why is the torque on the hills at 1.5 and 2.5 lower than the average torque? - From Emma Wai: I think it's just that there are a lot of short portions where a lot of torque was required, which is raising the average.
Because of this, we should look in to getting better map data or using a weighted average for the torque depending on the distance between points.
New Data - Torque Requirements
Emma Wai (Deactivated) Was able to create a weighted average of the absolute value of all the torque values throughout the race:
Weighted average: 10.99 Nm
Median: 11.03 Nm
And a histogram of all the torque values in the dataset (after removing a few outliers - such as the big hills, notice the graph only goes to 150ish and the max hills were ~200Nm):
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With these weighted average torque values, I am much more confident in the estimates of power dissipation in the motor controllers.
Emergency services testing track hill:
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30m rise over a 210m hill gives a degree of 8.21 degrees
From ASC 2018 Map data
List of the 10 steepest portions between two navigation points and torque requirement per wheel.
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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 = 15.7Nm10.99Nm (using updated data for weighted average)
Power(W) at that torque: = 1510.7Nm 99Nm * 2 * pi * 10.1478 = 1kW
At worst-case bus voltage of 100V, this gives 10A of current.
Vbus = 100V
Io = 10A700.7W
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) = 10.99Nm * pi * 6 / 30 = 6.91A
Here’s a desmos graph with the Torque on the x-axis and the power and phase current plotted: https://www.desmos.com/calculator/jpffkscexo
This second graph is a little more complicated but includes some extra parameters: https://www.desmos.com/calculator/chlmv9grii
Plugging this in to the motor controller power dissipation formula, we get:
P = 9.785
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 = 89.8A, which does not exceed the motor controller’s phase current limit of 100A
And Power:
P = 136.68W
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?