Robin did some heat calculations earlier on the motor controllers Motor Controller Heat Dissipation with some estimates for the max steady state power, and incorporating an aluminum plate into the aerobody of the chassis. To avoid the manufacturing of an aluminum plate inside a carbon fiber layup, we are doing some more calculations with less assumptions (or at least more accurate ones) and using a heatsink with a fan and forced air.
How much heat will we generate in the motor controllers - we don’t want to do a crazy over designed cooling solution because we know we can do better.
The motor datasheet specifies the heat dissipation in terms of current and voltage. The current will be higher when we need more torque from the motors, so we look for hills in the route map (ASC 2018 data) and see how much power we will need to dissipate. We will also look at the case of accelerating from stop to full speed, then maintaining that full speed (which we will likely do at some point during testing).
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
Average torque is 15.7Nm
Average of all the positive torques is 17.2Nm (treating the distance between each 2 points as identical).
^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.
At 1.5: climbing 1100 over 120km, 13Nm
At 2.5: climbing 800 over 160km, 9Nm
How long will these hills take?
120km @ 16m/s = 7500s = 125minutes, so we can assume this to be steady-state.
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.
Emergency services testing track hill:
Using a combination of this site for elevation and google maps for distance, assuming the distance google gives is the hypotenuse of the triangle:
https://www.mapcoordinates.net/en
https://www.google.com/maps/dir/43.4365484,-80.5795813/43.434044,-80.576786/@43.4357277,-80.5800187,966m/data=!3m1!1e3!4m2!4m1!3e1!5m1!1e4
369m to 384m over 350m gives a 2.5 degree hill.
Heartland Motorsports Park (FSGP 2020 / 2021?)
Looks pretty flat.
COTA Turn 1 (FSGP 2019)
Roughly a 30 metre elevation gain over 210m, assuming the distance google gives is the hypotenuse of the triangle.
https://www.google.com/maps/dir/Circuit+of+the+Americas,+9201+Circuit+of+the+Americas+Blvd,+Austin,+TX+78617,+United+States//@30.131248,-97.6388914,664m/data=!3m1!1e3!4m9!4m8!1m5!1m1!1s0x8644b03ad152eaf9:0x8ae827dd1ff5e0ed!2m2!1d-97.6358511!2d30.1345808!1m0!3e1!5m1!1e4
30m rise over a 210m hill gives a degree of 8.21 degrees
From ASC 2018
List of the 10 steepest portions between two navigation points and torque requirement per wheel.
Angle 13.125 and torque requirement 213.133 over 59.923 metres
Angle 11.526 and torque requirement 188.117 over 28.771 metres
Angle 11.373 and torque requirement 185.715 over 85.577 metres
Angle 10.542 and torque requirement 172.631 over 88.187 metres
Angle 10.435 and torque requirement 170.955 over 29.996 metres
Angle 10.045 and torque requirement 164.805 over 48.502 metres
Angle 9.878 and torque requirement 162.166 over 89.985 metres
Angle 9.744 and torque requirement 160.053 over 57.792 metres
Angle 9.483 and torque requirement 155.932 over 57.288 metres
Angle 9.47 and torque requirement 155.732 over 42.049 metres
*Note that the data we have is incomplete - some stretches of the map have no data points for over 20km, but this gives us a good estimate.
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 rps
For calculating power, we use: (https://www.engineeringtoolbox.com/angular-velocity-acceleration-power-torque-d_1397.html)
Where P (W), T (Nm), nrps = number of rotations per second
Steady State:
Average torque = 15.7Nm
Power(W) = 15.7Nm * 2 * pi * 10.1478 = 1kW
At 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
What if we have a 36km/h (10m/s) headwind - aero losses will increase, and thus the torque requirement and bus current will increase?