Forces Acting on a Vehicle

AIM : To calculate the forces acting on a vehicle based on following questions.

What should be the maximum speed of the motor used in an electric scooter capable to run at 90 kmph, if the fixed gear ratio is 7 and tire size is 90/100 R10 53J?

Solution –

Here, the tire size is given as 90/100 R10 53J,

To proceed with the calculations we need to determine the tyre diameter from the given tyre size.

To determine tyre/wheel diameter :

The First Number 90 – The number 90 represents the section width of the tyre in millimetres. This is the measurement of tyre at its thickest point.

The Second Number 100 – The number 100 represents the sidewall aspect ratio. This will help in finding the height of the sidewall, from rim to tread. This is also called section height.

Section Height = 100 percent of 90 mm

= (90 x 100)/100

= 90 mm

Section Height = 90 mm

The Third Number 10 – The number 10 represents the diameter, in inches, of the wheel on which the tire is mounted.

Rim Diameter = 10 inch = 254 mm

`therefore` Tyre Diameter = 254 + 90 + 90 = 434 mm

`therefore` Tyre Radius (r) = 434/2 = 217 mm = 217 x 10^-3 m

Linear Velocity is given as;

v = 90 kmph = (90 x 5)/18 = 25 m/s

Angular Velocity (`omega`) is given as;

Angular Velocity = (Linear Velocity) / (Wheel Radius)

`ω= v/r`

`ω= 25/(217×10^(-3) )=115.2` rad/s

Speed of the motor (N);

N = `(ω×60)/(2×π)`= 1100 rpm

Speed of the motor with fixed gear ratio 7;

N = 7 x 1100

`therefore` N = 7700 rpm

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Similarly for tire code is 90/90 R18 51P;

Section Height = 90 percent of 90 mm

= (90 x 90)/100

= 81 mm

Section Height = 81 mm

Rim Diameter = 18 inch = 457.2 mm

`therefore` Tyre Diameter = 457.2 + 81 + 81 = 619.2 mm

`therefore` Tyre Radius (r) = 619.2/2 = 309.6 mm = 309.6 x 10^-3 m

Linear Velocity is given as;

v = 90 kmph = (90 x 5)/18 = 25 m/s

Angular Velocity (`omega`) is given as;

Angular Velocity = (Linear Velocity) / (Wheel Radius)

`ω= v/r`

`ω= 25/(309.6×10^(-3) )=80.75` rad/s

Speed of the motor (N);

N = `(ω×60)/(2×π)`= 771 rpm

Speed of the motor with fixed gear ratio 7;

N = 7 x 771 = 5397 rpm

`therefore` N = 5400 rpm (approx)

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2. Prepare a simple excel calculator to identify vehicle propulsion power based on given inputs and outputs.

Inputs: Kerb Weight (kg), Payload (kg), Coefficient of rolling resistance, Air density (kg/m3), Width (m), Height (m), Drag coefficient, Speed

Output: Total power in kW

What are the limitations of this calculation?

Solution –

Formulas used :

`F_(rr)=μ_(rr) mg`

`F_(ad)=1/2 ρAv^2 C_d`

`F_(hc)=mg sin⁡φ`

Where,

F_rr = Rolling Resistance Force;

`mu_(rr)`= Coefficient of Rolling Resistance;

m = Total mass of the vehicle;

`rho` = Acceleration due to gravity;

F­_ad = Aerodynamic Drag Force;

r = Density of Air;

A = Frontal Area of vehicle;

v = Velocity of vehicle;

C_d = Coefficient of drag;

F_hc = Hill Climbing Force;

`varphi` = Hill slope angle in radian.

Also power is given as;

Power = Force x Velocity

`P=F ×v` …… (kW)

EXCEL SHEET :

Results :

The ‘Air Drag + Rolling Resistance Power’ comes out to be 17 kW whereas the total power considering hill climb comes out to be 79.4 kW.

Limitations :

• Coefficient of Rolling Resistance value keeps on changing based on tyre inflation, tyre pressure and velocity.
• Power is calculated on a particular speed value whereas the speed keeps on changing continuously.
• The value of rolling resistance changes based on the road conditions such as concrete road, unpaved road, soft soil, etc.
• Value of air density changes with respect to temperature.

Assuming that tire pressure reduction by 15% results in increase of rolling resistance effect by two times, how much will be the difference in total traction power keeping all other parameters same?

Solution –

If reduction in tire pressure by 15% increases rolling resistance effect by two times, then the value of coefficient of rolling resistance now becomes 0.05.

EXCEL SHEET :

As seen in the excel sheet, the value of Rolling Resistance force gets doubled.

For Rolling Power,

F_rr = 2*`mu`_rrmg = 2*0.025*1500*9.81

= 735.75 N

P_rr = F_rr x V = 735.75 x 100 = 73375 W = 73.5 kW

So if the coefficient of rolling resistance is doubled the required power changes from 36.78 kW to 73.5 kW which is almost 100% increment.

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4. See the video of Range rover sport dragon challenge. How much is the gradeability? What was the average speed? Which are the forces acting on car?

Soultion :

a. The slope angle was 45 degrees, so the gradeability is 100% or 45^o.

b. The average speed during 99 turns was around 88 kmph and during 999 steps was around 40 kmph.

c. The forces acting on the vehicle during the challenge were centripetal force (while cornering), hill climbing force, aerodynamic drag force, rolling resistance force, linear and angular acceleration force.

Link to google drive :

https://drive.google.com/open?id=1GvVg-X1__SWaAz-7dgeVWu-pBqYT56QG