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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” 20.5” diameter tire has a circumference of pi*20in 20.5in * 2.54(cm/in) / 100 (cm/m) = 1.596m636m
Travelling at 16.6m/s, this tire rotates at 104.01rpm10.1478 rps
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
Steady State:
Average torque = 15.7Nm
Power(W) = torque(Nm) * speed(rpm) = 15.7Nm * 624 rpm = ~9800W, which is close to the 10kW peak for our motors. If this is what we use in steady state, then we have a problem. Something is going on here, and I probably just made a simple mistake.* 2 * pi * 10.1478 = 1kW
At worst-case bus voltage of 100V, this gives 10A of current.
Vbus = 100V
Io = 10A
What if we have a 36km/h (10m/s) headwind - aero losses will increase, and thus the torque requirement and bus current will increase?