Ready for 55 again 4063

Matthew Russotto

How can you tell that the mileage is fairly flat between 60 and 75? Unless you have some very specialized equipment, and you are running the test under very controlled conditions (same road, same direction, same wind conditions, same traffic conditions - none is good), you cannot "know" this. If you believe that on a particular trip where you can drive 75, you get almost the same mileage as on another particular trips where you can only drive 60 - then I can believe that. But if you are saying you get the same mileage under the exact same conditions whether you drive 75 or 60, I am not a believer.

I have tried to look at this analytically. Here are approximate drag power calculations for a 1991 Miata

Cd = coefficient of drag = 0.36 (from AF = Frontal area = 66.1" x 47.3" = 3,127 sq in = 21.7 sq ft (it might be slightly more or less) (2.0 sq meters) (not exactly correct, since the car is not a box, but it will do for this comparison) V = vehicle velocity (1 miles per hour = 0.44704 meters per second) DA = density of air = .002377 slugs-cubic foot (1.225 Kg-m3) at sea level

Drag Power = AF*Cd*(V**3)*DA-2 = (2.0 sq meters)* 0.36 * (V**3) * (1.225 Kg-m3) 2 = 0.44 * (V**3) Kg*m(2)-sec(3) = 0.44(V**3)N*m-s

Speed Drag Power mph m-s nm-s HP 30 13.4 1,059 1.4 40 17.9 2,534 3.4 50 22.4 4,945 6.6 60 26.8 8,468 plus 1 11.3 70 31.3 13,492 18.0 75 33.5 16,541 22.1 80 35.8 20,188 26.9 85 38.0 24,143 32.2 90 40.2 28,656 38.2 100 44.7 39,309 52.4 105 46.9 45,505 60.7 110 49.2 52,320 68 plus 1.8 115 51.4 59,784 79.8

Aerodynamic drag is only part of the power consumption. The other part is rolling resistance. The force to overcome rolling resistance can be represented as:

FR = U*W

Where:

U = coefficient of rolling friction, somewhere in the 0.01 to 0.015 range for most cars W = weight of the car

I'll use 2300 lb. for the weight and 0.01 for the coefficient of rolling friction.

FR = 0.01 * 2300 = 23 lb = 102 N

The power required to overcome rolling resistance is therefore:

Speed Rolling Resistance Power mph m-s nm-s HP 30 13.4 1,371 1.8 40 17.9 1,831 2.4 50 22.4 2,291 3.1 60 26.8 2,741 3.7 70 31.3 3,202 4.3 75 33.5 3,427 4.6 80 35.8 3,663 4.9 85 38.0 3,888 5.2 90 40.2 4,113 5.5 100 44.7 4,573 6.1 105 46.9 4,798 6.4 110 49.2 5,034 6.7 115 51.4 5,259 7.0

So combining the two factors - Rear Speed Wheel Engine mph m-s HP HP 30 13.4 3.2 4.8 40 17.9 5.8 7.8 50 22.4 9.7 12.4 60 26.8 15.0 18.6 70 31.3 22.3 27.1 75 33.5 26.7 32.4 80 35.8 31.8 38.4 85 38.0 37.4 45.0 90 40.2 43.7 53.4 100 44.7 50.8 60.8 105 46.9 67.1 79.9 110 49.2 76.5 91.0 115 51.4 86.8 103.1

I used 85% for the drive line efficiency, and added 1 hp for power to address the accessories (power steering alternator other) when guesstimating the engine horsepower.

My numbers aren't too bad. Mazda claimed 116 HP for the engine and testers got top speeds around 113 mph. I may be too low on the rolling resistance, or too high on the drive line efficiency (I used 85%), or I might not be allowing enough hp for the accessories (I used 1 hp) or maybe the car weighs more than 2300 lb. I can easily jiggle these numbers within a reasonable range to match the reported HP and top speed. For instance if I use a 2500 lb weight, 0.015 for the coefficient of rolling friction, 82% for the drive line efficiency, and buttume the accessories are drawing 5 hp at 115 mph, the rear wheel horsepower at 115 mph is 91.2 Hp and the engine horsepower required is 116. I know my numbers are not perfect, but the trend is reasonable.

buttuming you agree with the trend, if not the exact numbers, it can be seen it takes about 40% more engine power to go 75 versus 60. If you spread the total energy used over a mile, at 60 miles per hour it takes about 614,000 ft-lb of energy to move the car one mile. At 75 miles per hour it will take around 916,000 ft-lb of energy to move 1 mile. Do you really believe that the engine efficiency increases enough as the rpm increases from around 3000 to around 4000 rpm to compensate for all the extra energy required? The torque peak for your Miata engine is around 5500 rpm, so in neither case are you at the most efficient speed. A gallon of gasoline includes somewhere around 90,000,000 ft-lb of energy. Engine efficiencies for converting gasoline into work output are generally in the range of 30%. So you might guess that a gallon of gasoline could do around 27,000,000 ft-lb worth of actual work. At 60 miles of hour this would translate into about 44 miles. At 75 (buttuming the efficiency was still around 30%), this would only be 29 miles. To get the same mileage at 75 as at 60, the efficiency would have to increase from around 30% to 45%. This is not reasonable. Maybe the efficiency could increase by 5%. In this case, a gallon of gas could move the car around 34 miles at 70 mph.

Please don't get hung up on the exact numbers. I am only trying to illustrate trends. The relative numbers are what is important.

Ed