Homework 1:

1- A car has the following data:

- Net engine horsepower 100 kW at 6000 rpm (Pmax, NPmax)

- Maximum torque 150 Nm at 3200 rpm (Tmax, NTmax)

- Maximum speed 185 km/h (Vmax)

- Curb weight 1160 kg (Wc = mc g)

- Gross vehicle weight 1600 kg (Wg = mg g)

- Transmission ratios 3.91/2.17/1.37/1.00 (ig1/ig2/g3/g4)

- Final drive ratio 3.46 (if)

- Wheel base 2666 mm (b = 2.666 m)

- Tire 185/65 R15 87H (w/ar RD” 87/H)

- Weight distribution front/rear 60/40 (Rf = 0.6W, Rr = 0.4W)

- CG height = 0.6 m (h)

Find:

A)   Maximum gradient the car can climb S% or θ

B)   The weight distribution at that gradient R’f, R’r

C)   Maximum acceleration the car can achieve on direct drive at Tmax. If:

-         Car coefficient of air resistance (drag coefficient) is 0.3 Cd

-         Air density 1.12 kg/m2 ρ

-         Car fontal area 2 m2 Af

-         Rolling resistance coefficient 0.015 fr

In case of:

i)                   curb weight, level road, no wind Wc, θ = 0, vair= 0

ii)                 curb weight, level road, against wind of (18 km/h) Wc, θ = 0, vair= 18

iii)               Gross vehicle weight, level road, no wind. Wg, θ = 0, vair= 0

Solution:

A) TE = Tmax ig1 iht / rw

rw = (2 h + D . 25.4)/2 = (2 (w . ar/100) + D . 2.54)/2

= (2 x 185 x 65/100) +15 x 25.4)/2 = (240.4+ 381)/2

= 310.7 mm = 0.31 m

assume ht = 100%

TE = 150 x 3.91 x 3.46 x 1/ 0.3107 = 6531.35 N

RR = fr Wg = 0.015 x 1600 x 9.81 = 235.44 N

SE = TE – RR = W sin θ = 6531.35 – 235.44 = 6295.91 N

sin θ = 6295.91/ (1600 x 9.81) = 0.4  or θ = 23.64o

B) Since Rf / Rr = 60 /40  then x / y = 40 /60

x = 0.40 b , y = 0.60 b

S MB = Rf  b – W cos θ  y + W sin θ  h = 0

R’f  b = W cos θ  y -  W sin θ  h

R’f = W (y/b) cos θ – W (h/b) sin θ

= W ( 0.6 cos 23.64 – (0.6/2.666) sin 23.64)

= W ( 0.5496 – 0.090) = 0.4596 W = 0.4596 (1600 x 9.81) = 7213.88 N

R’r = W – R’f  = W – 0.4596 W

= 0.5404 W = 0.5404 (1600 x 9.81) = 8482.12 N

C)

vcar  at direct drive at Tmax  (Ne = 3200 rpm)

vcar = (2 p rw Ne / if ) (60/1000)  km/h

= 0.377 Ne rw / if   = 0.377 x 3200 x  0.31 / 3.46 = 108 km/h

AR = ½ ρ Cd Af (vcar)2 = ½ (1.12) (0.3) (2) (108/3.6)2

= 302.4 N (no wind) case (i, iii)

AR = ½ ρ Cd Af (vcar + 18)2 = ½ (1.12) (0.3) (2) ((108 + 18)/3.6)2

AR = 411.6 N (head wind 18 km/h) case (ii)

RR = fr Wc = 0.015 (1160) 9.81 = 170.7 N (curb weight) cas (i, ii)

RR = fr Wg = 0.015 (1600) 9.81 = 235.4 N (GVW) case (iii)

TE = Tmax x 1 x if / rw = 150 (3.46) / 0.3107 = 1670.4 N case (i, ii, iii)

TR = AR + RR

SE = TE – TR

ma = SE

a = SE / m

m = 1160 kg case (i, ii)

m = 1600 kg case (iii)

i)

TR = AR + RR = 302.4 + 170.7= 473.1 N

SE = TE – TR = 1670.4 – 473.1 = 1197.3 N

a = SE / m = 1197.3/1160 = 1.032 m/s2

ii)

TR = AR + RR = 411.6 + 170.7 = 582.3 N

SE = TE – TR = 1670.4 – 582.3 = 1088.1 N

a = SE / m = 1088.1/1160 = 0.938 m/s2

iii)

TR = AR + RR = 302.4 + 235.4 = 537.8 N

SE = TE – TR = 1670.4 – 537.8 = 1132.6 N

a = SE / m = 1132.6 / 1600 = 0.708 m/s2