Ask the Best and Brightest: Road Load Power = Av + Bv2 + Cv3?

Robert Farago
by Robert Farago
ask the best and brightest road load power av bv2 cv3

I am not the world’s most mechanically-minded person. Nor am I the world’s worst. But I’m definitely closer to the latter than the former. For example, I can’t make head nor tails of the above formula, even with the following guide [via]:

The letter v represents the velocity of the car, and the letters a, b and c represent three different constants:

The a component comes mostly from the rolling resistance of the tires, and friction in the car’s components, like drag from the brake pads, or friction in the wheel bearings.

The b component also comes from friction in components, and from the rolling resistance in the tires. But it also comes from the power used by the various pumps in the car.

The c component comes mostly from things that affect aerodynamic drag like the frontal area, drag coefficient and density of the air.

Bottom line? “if you double your speed, this equation says that you will increase the power required by much more than double.” Yes, yes, but the question was “What speed should I drive to get maximum fuel efficiency?” Can TTAC’s Best and Brightest provide the required dose of enlightenment, either specifically or generally?

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4 of 28 comments
  • Cdotson Cdotson on Sep 29, 2008

    Well, miked is probably closest but misses it by a bit. The formula is the generic form of the equation that describes the power to maintain a constant speed, we all agree on that. Fuel efficiency does not enter this equation in the least, so let's tackle that later. The equation addresses almost all variables that affect drag (aero, rolling, mechanical friction). There should be an additional scalar constant that is not a product of velocity to be mathematically correct, but I guess it's OK to ignore since it would be small relative to the other products in the equation. Tests are performed on vehicles and the data are used to calculate the value of the constants for any particular vehicle. This is done using a coast-down test where a vehicle is driven to a high speed, shifted into neutral and shut off. The car's velocity over time is recorded and the 3rd-order polynomial curve fit will calculate the constants (excel is really good at this). This is a little different than miked's explanation since you aren't actually measuring power but calculating an instantaneous rate of change of velocity which can be used to determine actual power required to resist that rate of deceleration. Aerodynamic drag is a function of the square of velocity, so aerodynamics will play mostly into the "B" constant. But your initial statement is right; power required to increase speed increases much more rapidly than the square of velocity. I always thought the old Bonneville racers' rule of thumb was that power increased with the cube of speed. Only aerodynamic drag increases with the square of speed. Now maximizing fuel efficiency at highway cruise can be calculated empirically but it will typically require much more information that any individual typically knows about their vehicle. Not only do you have to know the constant values for the above equation but it would be best also if you had an isoefficiency plot. This is a chart where the islands of thermal efficiency are broken down like a topographical map on a plot where the x-axis is RPM and the y-axis is engine load. Since you need to know your engine's load capabilities you also need to know the flywheel torque curve output from a dynamometer. Now that we know the power to maintain any given speed you can calculate the percent torque load required to achieve that power knowing your gear ratios. The gear ratios are the critical conversion between the x-axis of RPM on the isoefficiency plot and the x-axis of speed (velocity) in the road load equation. Overlay the two results on the isoefficiency plot and you can see where, in each gear, your steady cruise load is in the highest area of thermal efficiency. Duplicate that engine RPM in that gear and maintain it and voila, maximum highway fuel efficiency.

  • Landcrusher Landcrusher on Sep 29, 2008

    Wrong question, Robert. Gas is cheap. Unless your funds are really constricted, the number you really want to know is how fast you can drive while remaining safe and not damaging your car. Repairs from overloading the equipment will easily overrun any savings due to mileage that you could build up over a even a single year.

  • on Oct 09, 2008

    Wow. Nobody got it right. Alot of close, but not right. Start at the beginning and consider "Road Load". Load is a force. It's easy to find your road load driving force by a coast-down test (described above but there's no reason to turn off the engine). After coast down, you'll know the driving force (not power, not yet) you need to keep your car moving. The equation is a 2nd (not 3rd) order polynomial that looks like: a + bV + cV^2. a= rolling drag, it's a constant that comes mostly from the tires deforming b= mechanical friction (depends on speed, usually very small compared to the others) c= wind drag, depends on the square of velocity. The question is about POWER. Power = force x velocity. Since we know our driving force (road load), then to find the power, multiply by velocity to get the first equation. aV + bV^2 + cV^3. You can find the a b & c from the EPA website for new cars. Trouble is, this still doesn't factor in efficiency of the motor. Bottom line, the answer comes from experiment or a bunch more calculations if you already know everything about the car. Then, the pumping loss discussion is another good one. OEM's are overcoming the pumping losses by opening the throttle valve and controlling fuel intake (off idle) using the intake valves. That's BMW's valvetronic and Nissan V-VEL and others. If you can precisely control the valve opening, you can leave the throttle plate open, or eventually, eliminate it like diesels or HCCI engines. Drop the hammer and have fun! I drive a 4 cylinder 27 Roadster and get 21mpg. Works for me! Take a look:

  • Michael_Grad_Student Michael_Grad_Student on Dec 12, 2008

    This may be an old thread, but in case anyone is wondering... You can Roughly Calculate Road Load Coefficients A, B, and C!! 1) Find a very flat road. Flat means no slope AND no bumps. 2) Get a partner with a stopwatch and clipboard 3) Wait until there is NO wind and NO traffic 4) Don't do anything illegal, e.g., speeding or blowing stop signs. 5) Use a consistent starting place 6) Drive up to the max speed, hold it steady 7) Same Time: start watch, shift to neutral (or release clutch) so engine doesn't "suck" your speed 8) Let off gas pedal and say "Now" every time the speedometer drops 5 mph (or kph). This may be too fast at very high speeds. 9) Partner jots down time (helps with 'Lap' function) every time you say "Now" (Fill in the corresponding speeds later, or before hand) 10) Go all the way until the vehicle stops. 11) Do this as many times as you can, back and forth -> Collect tons of data! Analyze: You will notice at lower speeds it takes longer and longer to fall 5 mph. This is because the ACCELERATION decreases as speed decreases. Since Accel = Force / Mass, the force is decreasing as speed decreases. By going on a flat road, with no wind, no bumps, no slope, and disconnecting the engine, you'll hopefully just have the one force left acting on the vehicle: Road Load! Convert Speed to meters/second (mps) Calculate acceleration between time steps. A = (V(2)-V(1))/(T(2)-T(1)) Calculate "velocity" between time steps. V = (V(2)+V(1))/2 Estimate the Road Load force by taking A*Mass. F (N) = A (m/s^2) *Mass (kg) Plot F as a function of V. In MS Excel you can just Add Trendline. You can add whatever trendline you want, but the best one to describe your car will be a 2nd order polynomial, then display equation on chart! You'll have to make your best guess on mass. Lookup your vehicle specs, and add passenger mass. You may also want to try a range of masses to see where they are. Notes: 1 mph = 0.447 mps and 1 lbs = 0.454 kg A Midsize Sedan: A = 93.45 (N) B = 3.58 (N)/(m/s) C = 0.858 (N)/(m/s)^2