Fundamentals of Motorsport

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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,&nbspTakethe density of air to be 1.2 Kg/m3 and g = 9.8 m/s2

Calculations

Mass=2200kg

RollingResistance=150N

Frontalarea=1.8 m2

DragCoefficient =0.80

Transmissionloss=9%

MaximumEngine power=99kW

Tyretraction coefficient =1

Weighton the rear wheel =60%

Densityof air 1.2kg/m3

g=9.8 m/s2

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

Drivetrain efficiency Enginepower = (1-0.09)99 kW=90.09kW= 90,090 Watts

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

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 + 0.864V3

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?&nbsp

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 + 0.864V3 + 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!!)

&nbspTractiveforce = Coefficient of friction Downward force

Tractionforce = 1(0.6022009.8) = 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?)&nbsp

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?&nbsp

Windresistance =

=155.37N

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

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

AerodynamicDrag =

AerodynamicDrag = 155.37N

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?&nbsp

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

AerodynamicDrag = 155.37N

Tractiveforce =671.75 N

Rollingresistance force =150 N

Gravitationpull due to the hill = mass g (1/4) = 22009.8(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.

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