2.17 If the car in Example 2.9 had 𝐶𝐷 = 0.45 and 𝐴𝑓 = 25 ft2, what is the difference in minimum theoretical
stopping distances with and without aerodynamic resistance considered (all other factors the same as in
Example 2.9)?
Solution
(a)
1- Calculate coefficient of rolling friction 𝑓𝑟𝑙 using the formula:
𝑓𝑟𝑙 = 0.01 [1 +
V +V
( 1 2 2)
147
5280 ft
1h
(80 mi/h × 1 mi × 3600 sec) + 0
(
)
2
] = 0.01 1 +
= 0.0134
147
[
]
2- Determine the coefficient of aerodynamic resistance
𝐾𝑎 =
𝜌
0.0024
𝐶𝐷 𝐴𝑓 =
× 0.45 × 25 = 0.00135
2
2
3- Calculate theoretical stopping distance with aerodynamic resistance (S) using the formula:
𝛾𝑏 𝑊
𝐾𝑎 𝑉1 2
]
𝑠=
ln [1 +
2g𝐾𝑎
𝜂𝑏 𝜇𝑊 + 𝑓𝑟𝑙 𝑊 − 𝑊 sin(𝜃g )
Here, 𝑊 is total weight of the vehicle, 𝛾𝑏 is baking mass factor, g is gravitational constant, 𝜇 is
coefficient of road adhesion, 𝑓𝑟𝑙 is coefficient of rolling resistance, and 𝜃g is angle of grade, 𝜂𝑏
is braking efficiency.
Substitute 1.04 for 𝛾𝑏 , 80 mi/h for 𝑉1 , 2500 Ib for 𝑊 , 32.2 ft/s2 for g , 80 % for 𝜂𝑏 , 0.70 for
𝜇 ,0.0134 for 𝑓𝑟𝑙 , 5.71° for 𝜃g .
2
5280 ft
1h
0.0135 × (80 ×
×
)
1.04 × 2500
1 mi
3600 sec
𝑠=
ln [1 +
]
2 × 32.2 × 0.0135
0.8 × 0.7 × 2500 + 0.0134 × 2500 − 2500 sin 5.71°
= 2990 ln [1 +
185.856
] = 435.70 ft
1184.76
(b)
4- Calculate theoretical stopping distance by ignoring aerodynamic resistance (S) using the
formula:
2
5280 ft
1h
1.04 [(80 mi/h × 1 mi × 3600 sec) − (0)2 ]
2
2
𝛾𝑏 (𝑉1 − 𝑉2 )
𝑠=
=
= 451.65 ft
2g(𝜂𝑏 𝜇 + 𝑓𝑟𝑙 ± sin 𝜃g ) 2 × (32.2) [( 80 ) (0.70) + (0.0134) − sin(5.71°)]
100
Thus, the theoretical slopping distance without aerodynamic resistance is 451.65 ft