**Suggested Citation:**"Appendix L: Relationship between Power and Performance." National Research Council. 2015.

*Cost, Effectiveness, and Deployment of Fuel Economy Technologies for Light-Duty Vehicles*. Washington, DC: The National Academies Press. doi: 10.17226/21744.

Relationship between Power and Performance

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):

T_{tr} = F_{rolling resistance} + F_{aerodynamic drag} + F_{inertia} |
(1) |

Expanding this equation yields the following:

T_{tr} = C_{rr} m g_{c} + ½ ρ C_{d} A_{f} V^{2}/g_{c} + m dV/dt/g_{c} |
(2)^{1} |

The following parameters were used in Equation 2 for a typical 3,500 lb midsize car:

C_{rr} = 0.0060 (tire rolling resistance)

m = 3500 lbm (mass of the vehicle)

r = 0.075 lbm/ft^{3} (density of air)

C_{d} = 0.30 (aerodynamic drag coefficient)

A_{f} = 25 ft^{2} (frontal area of vehicle)

V = 60 mph (88 ft/sec)

t_{60} = 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:

T_{tr} lbf = 0.0060 × 3,500 lbm × 32.2 ft/sec^{2}/32.2 lbm-ft/lbf-sec^{2} + ½ × 0.075 lbm/ft^{3} × 0.30 × 25 ft^{2} × (88 ft/sec^{2}^{)2}/32.2 lbm-ft/lbf-sec^{2} + 3,500 lbm × 88 ft/sec^{2}/(t_{60} sec × 32.2 lbm-ft/lbf-sec^{2}) (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/t_{60} |
(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 |

8 seconds | Base | 205 | Base |

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.

REFERENCES

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.

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^{1} The force exerted by a mass on earth is given by the following equation, which requires the constant, g_{c,} which is equal to 32.2 lbm-ft/lbf-sec^{2} (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/sec^{2}, as shown by substituting in the above equation:

F = 3,500 lbm × 32.2 ft/sec^{2}/(32.2 lbm-ft/lbf-sec^{2}) = 3,500 lbf

(Note the cancellation of units leaving lbf.)

This relationship is used throughout Equations 2 and 3 of this Appendix.