Steering Calculations 434o3o

STEERING:    Overview: This report is all about the steering system being used in our student formula (SUPRA) with a necessary explanation of the selection criteria.

   INTRODUCTION: The steering system is designed based on the Ackerman Geometry since the average achievable speed on track described in the rulebook is 80 km/h which is moderate speed. The rack and pinion mechanism is preferred as it is easy in design. Due to the ease in manufacturing, lower weight and simple geometry a rack and pinion steering geometry with ratio of 5: 1 was selected. Moreover a higher percentage Ackerman in rack and pinion arrangement was achieved. For selection of components and configuration following criteria were considered:       

Low cost Light weight Simple design High serviceability Positive Responsiveness to driver input Easy to troubleshoot.

   CALCULATIONS: WHEELBASE

65 inch

TRACKWIDTH

50 inch

CG FROM GROUND

15inch

INNER ANGLE (δi)

35o

KING PIN INCLINATION( λ)

3o

CASTOR (ν)

3o

CAMBER

-1.5o

WEIGHT WITH DRIVER

270KG

WHEEL RADIUS STEERING WHEEL RADIUS



10 inch 4.5 inch

Ackerman expressioncotδo−cotδi =w/l Where,

δi = the steer angle of the inner wheel δo = the steer angle of the outer wheel O = turning center. Cot δo−cot 35o =50/65 δo = 24.47o



AVERAGE STEER ANGLE-

δmean

=

(δi + δo)/2

δmean = 29.73o



Turning radiusT.R= (T.W/2) + (W.B/sinδ

=156.05 inches =3.963m

mean

)

  Weight distributionRatio =40:60

Fro

b

nt

c Rear Total weight of the vehicle= 270 kg Weight in front (40 % of total) = 108 kg Weight in rear (60 % of total) = 162 kg

 Distance of center of gravityDistance from front=“b” Distance from rear=“c” For “c”-

Wf = (W * C)/ l c = 26 inch=0.6604m Where, Wf = Weight in front in static condition W = total weight of vehicle with driver C = distance from rear to C.G.

l = wheel base. For”b”-

Wr = (W * b)/ l b = 39 inch=0.991m Where, Wr = Weight in rear in static condition b = distance from front to C.G

 Caster-

Caster angle normally ranges from 0 o to 70,but in our steering geometry caster is assumed + 3o because positive caster creates lot of align torque that straightens the steering wheel when going forward and improves straight line stability of vehicle. Caster angle has a positive effect on steer-camber. With positive caster angle, the outside wheel will camber in a negative direction and the inner wheel in a positive direction, causing both wheels to lean into the turn. Too much caster causes weight transfer that can result in over steering effect.

 King pin inclinationK.P.I angle is different in different vehicles, we have taken it 3° as too much of kingpin inclination causes a lot of rising of the front axle while steering. Keeping scrub radius small will make the car easier to steer. In our upright/knuckle the spindle length is taken positive so that the car will be raised up as the wheels are turned and this results in an increase of the steering moment at the steering wheel.

 Mechanical

trail(MT)-

MT = tan ν * r =tan 3o * 10 = 0.524077 inch=0.0133m Where, r = Tire radius ν = caster angle

Scrub radius(d)-

A

(d) = tan λ * r + AC = tan 3o * 10 + 3.75 =4.27407 inch=0.108m Where, λ= king pin inclination angle AC = king pin offset distance=tyre width ÷ 2 Tyre width=7.5 inches r = tire radius

   Load transfer during brakingW f= w/l( Where, Wf= Load transfer G=gravity Wheel base (l) =65 inches Height of center of gravity (h) = 15inches Deceleration (d) = 2g Gradient angle (θ) = 15o Total weight (w) = 270 kg Distance of C.G. from back(C) = 26 inches Wf = 36.46kg Resultant weight on front = 108 + 36.46 =144.46kg Weightleft = Weightright = 144.46/2 =72.23 kg

   FORCES-

 Lateral forceThe lateral force Fy, acting at the center of tire produces a moment through the longitudinal offset resulting from caster angle as shown in figure below. The net moment produced is:

ML = (Fyl + Fyr) * r * tan ν Where, ν = caster angle, r = tire radius, Fyl , Fyr= lateral forces at left and right wheels.

 Lateral forceF y = CS * FV * α Fv=72.23*9.81=708.5N Fy= 113.36N Ml=3.02Nm Where, Cornering stiffness (CS )= 0.16/1o Load transferred during breaking on left/right wheel (FV) = 70 N Slip angle (α) = 1o

 Lateral momentFzl, Fzr= load transferred during breaking on left/right wheel, λ = K.P.I angle, δ = avg. steer angle,

d = tire radius. Mv=Moment about vertical axis Mv=(Fzl+Fzr)*d*sin λ *sin(δmean) =(708.5+708.5)*0.108*sin3*sin 29.735=3.972Nm

  Self-aligning torqueThe aligning torque, Mz,acts vertically and may be resolved into a component acting parallel to steering axis, since moments may be translated without change in magnitude, the equation for the net moment is: MAT = (MZL+MZR)*cos

MAT = (3.972+3.972)*cos(√18) =7.92N-m

c

 Total steering torqueMAT + Mv + ML = 7.92+3.972+3.02 = 14.912 N-m

  Pinion radiusIn our rack and pinion assembly geometry we are fixing steering ratio 6:1 Then, Input steering moment (I)2 * π * radius of steering wheel Output steering moment (o)2 * π * radius of pinion Then,

Steering ratio = I / o 6/1 = 2 * π * radius of π * radius of pinion =3.83cm

 Steering wheel force

steering

wheel / 2 *