Understanding electric racing
Aeronautic drag and constant speed
'if all other things are equal, the vehicle that is able to maintain a speed closer to it's average speed will need to overcome less aero drag and thus be able to maintain a higher average speed with the same energy than the vehicle who's speed fluctuates more widely'.
The faster a car goes, the more energy required to push through the air. This energy cost is due to aeronautic drag.
A key concept is that aero drag increases by the square of velocity. The drag goes up exponentially as the vehicle goes faster.
This means the theoretically best strategy to go as fast as possible while minimizing aero drag is to go a constant speed. Any aero drag reduction from going slow will be overcome by the higher drag penalty of going faster to achieve the same average speed.
Here is the equation:
FD = 1/2pv2CdA
FD = Resistive Force
Here is an example:
Strategy 1. Average speed 100 MPH consisting of 30 minutes at 100 MPH.
So FD = 100*100*?*Cd*A*1/2. Let's make a constant called Z that equals ?*Cd*A
So FD = 5000Z for 30 minutes
Strategy 2. Average speed 100 MPH consisting of 15 minutes at 75 MPH and 15 minutes at 125 MPH.
So FD = 75*75*?*Cd*A*1/2 and FD = 125*125*?*Cd*A*1/2
So FD = 2,812.5Z for 15 minutes and 7,812.5Z for 15 minutes.
In the second strategy, the average aero drag over the whole 30 minutes is (2,812.5Z + 7,812.5Z)/2 = 5,312.5Z which is 6.25% more drag than the strategy of maintaining a constant speed. With less drag to overcome, the constant speed car will finish ahead of the varied speed car.