Fundamentalsof Motorsport

Arear wheel drive car of mass 2200kg has a rolling resistance of 150N, a frontal area of 1.8m2, a drag coefficient of 0.80, atransmission loss of 9% and a maximum engine power of 99kW. Assumingthat the traction limit for each tyre is equal to the weight that itcarries (i.e. the tyre traction coefficient =1) and that 60% of thevehicle weight is on the rear wheels during acceleration, Takethe density of air to be 1.2 Kg/m3 and g = 9.8 m/s2

Calculations

Mass=2200kg

RollingResistance=150N

Frontalarea=1.8 m^{2}

DragCoefficient =0.80

Transmissionloss=9%

MaximumEngine power=99kW

Tyretraction coefficient =1

Weighton the rear wheel =60%

Densityof air 1.2kg/m^{3}

g=9.8 m/s^{2}

a)What is the maximum power available at the back wheels? 

*Drivetrain efficiency **Enginepower = (1-0.09)**99 kW**=90.09*__kW= 90,090 Watts__

b)What is its theoretical maximum speed in mph on a level road? 

*Rearwheels power = losses*

*Therefore*

*90.09kW = Aerodynamic power loss + Rolling resistance + power into hills*

*Rollingresistance power = Rolling resistance Force **vehivlevelocity *

*=150 N **V*

*AerodynamicPower loss =Drag **Velocity**= *

*Poweron level ground = 0 *

*90,090W= 150V + (**)*

*90,090Nm/s = 150 V + 0.864V*^{3}* *

*Usingiteration to solve V:*

*40m/s …. 61,296 Nm/s*

*45m/s …. 85,482 Nm/s*

*45.5m/s …88,210.67 Nm/s*

__45.84m/s ….90,099.80 Nm/s__

*45.85m/s ….90,155.78 Nm/s*

*So102.54116mph (45.84 m/s) is the fairly accurate top speed. *

c)What is its theoretical maximum speed in mph up a 1 in 4 hill? 

*90090Nm/s= Rolling resistance + aerodynamic power loss + power into hills*

*Powerinto hills (1 in 8 hill) = mass **g **change in height per unit time*

*Powerinto hills (1 in 8 hill) =2200 **9.8 **(1/4) = 5,390 N*

*Therefore*

*90,090Nm/s= 150V + (**)+ 5,390*

*90,090Nm/s = 150 V + 0.864V*^{3}* + 5,390*

*Usingiteration to solve for V:*

*44m/s …85,588.98 Nm/s*

*44.8m/s …89,796.90 Nm/s*

__44.85m/s …90,064.80 Nm/s__

*45m/s …90,872 Nm/s*

*So100.33 mph (44.85 m/s) is the approximate top speed up 1 in 8 hill.*

d)What is the maximum tractive force that each of the rear tyres couldprovide under acceleration before slipping occurs? (NB this is perrear wheel not per axle!!)

 *Tractiveforce = Coefficient of friction **Downward force *

*Tractionforce = 1**(0.60**2200**9.8) = *__12,936N__

e)What would be the maximum, useable, tractive force at each of therear tyres if full power was applied at 30 mph? (hint: is thislimited by tyre grip or by engine power?) 

*Useabletractive force will be the minimum of (force due to engine power,tire traction force)answer = min(90.09kw / 30mph , 12,936N)answer = min(671.75 N, 5,292 N) = *__671.75N__

f)What would be the wind resistance at 30 mph? 

*Windresistance = *

*=*__155.37N__

g)What would be the acceleration (in mph/s) at 30 mph on the level? 

*Force(F) = Tractive force – Aerodynamic Drag- Rolling resistance force*

*AerodynamicDrag = *

*AerodynamicDrag = 15**5.37N*

*Tractiveforce =671.75 N*

*Rollingresistance force =150 N*

*671.75N – 155.37 N – 150N = 366.38 N*

*=0.3725 MPH/s*

h)What would be the acceleration (in mph/s) at 30 mph up a 1 in 4hill? 

*Force(F) = Tractive force – Aerodynamic Drag- Rolling resistance-Gravity*

*AerodynamicDrag = 155.37**N*

*Tractiveforce =671.75 N*

*Rollingresistance force =150 N*

*Gravitationpull due to the hill = mass **g **(1/4) = 2200**9.8**(1/4)= 5390N*

*671.75N – 155.37 N – 150N -5,390 N= -5,223.62 N*

*=- 2.374**=- 5.311 MPH/s*

*Thenegative sign means it was a deceleration *

REFERNCES

JoshSmith (2013) *Smith’s Engineering.*