The relationship between power and performance is derived in this Appendix. The propulsion or tractive force to accelerate a vehicle can be calculated from the sum of the tire rolling resistance, the aerodynamic drag, and the inertial force for the vehicle on a level road as follows (Gantt 2011):
|Ttr = Frolling resistance + Faerodynamic drag + Finertia||(1)|
Expanding this equation yields the following:
|Ttr = Crr m gc + ½ ρ Cd Af V2/gc + m dV/dt/gc||(2)1|
The following parameters were used in Equation 2 for a typical 3,500 lb midsize car:
Crr = 0.0060 (tire rolling resistance)
m = 3500 lbm (mass of the vehicle)
r = 0.075 lbm/ft3 (density of air)
Cd = 0.30 (aerodynamic drag coefficient)
Af = 25 ft2 (frontal area of vehicle)
V = 60 mph (88 ft/sec)
t60 = 0 to 60 mph acceleration time
During a wide-open throttle acceleration, the tractive force can be expressed as following at the 60 mph condition:
Ttr lbf = 0.0060 × 3,500 lbm × 32.2 ft/sec2/32.2 lbm-ft/lbf-sec2 + ½ × 0.075 lbm/ft3 × 0.30 × 25 ft2 × (88 ft/sec2)2/32.2 lbm-ft/lbf-sec2 + 3,500 lbm × 88 ft/sec2/(t60 sec × 32.2 lbm-ft/lbf-sec2) (3)
(Note the cancellation of units leaving lbf for each term of Equation 3.)
Power to propel the vehicle is obtained by multiplying the tractive effort force in Equation 3 by velocity at 60 mph (88 ft/sec) and converting the product to horsepower (hp = 550 ft lbf/sec), which yields the following equation for power:
|Hp = 14 + 1530/t60||(4)|
Applying Equation 4 yields the following relationship between 0 to 60 mph time and horsepower:
|T60||% Change in 0 to 60 mph time||Hp||% Change in Hp|
|7.2 seconds||− 10%||227||+ 10.7%|
These results show that approximately a 10 percent decrease in 0 to 60 mph time requires approximately a 10 percent increase in power.
Allen, J. n.d. Concept Review: Unit Systems. Michigan Technological University. http://www.me.mtu.edu/~jstallen/courses/MEEM4200/lectures/energy_intro/Review_unit_systems.pdf. Accessed April 3, 2015.
Gantt, L. 2011. Energy Losses for Propelling and Braking Conditions of an Electric Vehicle. VPI MS Thesis, May.
1 The force exerted by a mass on earth is given by the following equation, which requires the constant, gc, which is equal to 32.2 lbm-ft/lbf-sec2 (Allen n.d.):
F = m × a/gc
Therefore, a mass of 3,500 lbm (lb mass) has a weight of 3,500 lbf (lb force) on earth, where a = 32.2 ft/sec2, as shown by substituting in the above equation:
F = 3,500 lbm × 32.2 ft/sec2/(32.2 lbm-ft/lbf-sec2) = 3,500 lbf
(Note the cancellation of units leaving lbf.)
This relationship is used throughout Equations 2 and 3 of this Appendix.