[To understand the material of this document, I assume the reader have a basic knowledge of differentiation.]
From the Maximum Speed section, we noticed that the higher speed we drive our car, the more engine power is wasted to overcome rolling resistance and air drag. It seems that there might be a speed that maximizes the distance you drive with minimal use of gas. It turns out there is such a speed for every car if we make some appropriate assumptions.
The fuel economy of different cars is compared in terms of the amount of fuel required to travel a certain distance. This is usually captured in kilometres per litre of gas. So, to find the fuel saving speed, we should try to maximze the value of kilometres per litre. Kilometres per litre is generally called Fuel Economy.
But it is hard to put amount of gas used into the existing car equations we know, so I opt to assume that the amount of gas used is directly proportional to the energy expended. Let d be the distance traveled and E be the energy expended. Then we are trying to maximize d/E which is equivalent to v/P [Note: E = P*t] where P is the power generated by the engine. This will give us a formula to calculate the Fuel Economy function F(v):
Our goal is to maximize this function with respect to v.
We further assume that we modulate our gas pedal carefully such that our engine barely generate enough power to overcome air resistance and other decelerating forces. If we count only air resistance and tire rolling resistancewe will find that our P equals 0.5cdρAv3 + crrmgv. We soon find out our F(v) maximizes when v=0. Obviously, this doesn't make sense as driving at zero speed doesn't get us anywhere. To fix this, we introduce a constant power loss to charge the car battery. With it, our P now equals 0.5cdρAv3 + crrmgv + Pe. Now we are ready to do the math!
To simplifies our calulations, let a be 0.5cdρA and b be crrmg and c be Pe. Now we have:
| v | ||
| F(v) | = | ----------------- |
| av3 + bv + c |
To find the v that maximize or minimize F(v), we set the first derivative to 0.
| c-2av3 | ||||
| F'(v) | = | --------------------- | = | 0 |
| (av3 + bv + c)2 |
Solve this, we have
To verify v* maximizes F(v), we take the second derivative.
| 6av2(av3 - 2c) - 2b(av3 + c) | ||
| F''(v) | = | ----------------------------------- |
| (av3 + bv + c)3 |
Notice that at v*, 2av3-c = 0. So
| -9acv*2 - 3bc | ||||
| F''(v*) | = | ----------------------- | < | 0 |
| (av*3 + bv + c)3 |
This confirms that v* maximizes F(v). Now we can express our fuel saving speed as
| ( | Pe | ) | 1/3 | ||
| v* | = | ( | ------------ | ) | |
| ( | cdρA | ) |
Now we are ready to calculate the fuel saving speed for our Skyline. Unfortunately, I am not able to find the power of its electrical system. A quick search from the web suggests 400W for the power of a typical car electrical system.
| ( | 400 | ) | 1/3 | ||
| v* | = | ( | ----------------------------- | ) | |
| ( | 0.34×1.29×2.4276 | ) | |||
| = | 7.2155ms-1 |
As a result, the fuel saving speed for our Skyline is 25.98km/h. Apparently, this is too slow for daily driving. Since time is also money, my recommendation would be to drive as fast as you can such that the extra time you saved is roughly equal to the money you can otherwise saved by driving slower. Notice that tThe true fuel saving speed should be even higher if we take into account of other constant power loss that doesn't vary with v.
As an interesting side fact, there was a US law in the 70s that limits the maximum speed on freeway to 55mph (about 88km/h) based on the argument that such a speed limit can save gas. Apparently this law was so unpopular that it was rescinded right after the OPEC crisis.