Quad Cal.xlsx

Selection of Caliper

Market survey and performance was tested before the selection of caliperthe required bore diameter was 25.4mm. Based on this the caliper of Karizma RTR was selected for the front and Pulsar 220 atrear. Calculations for rotor diameter 1) Torque required to stop the vehicle Tb – this torque depends upon dynamic weight transfer and the wheel dimensions. It plays a fair role in calculating the rotor diameter. The following expressions gives the braking torque.

T= Where, Dw =Diameter of tyre Wt = Dynamic axial weight When a vehicle is travelling with a certain speed is suddenly stop. The front suspension goes into jounce where as the rear spring experiences rebound the reason behind this is that a portion of rear axle weight get transferred to the front axle, this is know as dynamic weight transfer. The Dynamic weight transfer is given by the expression.

Wt = Where, Wt= Dynamic weight transfer. h = height of CG from ground M = Mass of vehicle in Kg. b = Wheel base The caliper pressure generated is 800 PSI (According to karizma ZMR manufacturing catalogue) Pa =5515200 N/m2 h = 600mm= 0.6m m = 220 Kg g = 9.81 b = wheel base => 1200mm

220

Wt = = 110 Kg

1128.15

Tb= 110/2*0.558/2*9.81 Tb =150.5344 N-m 1616.14

Torque Generated by caliper on the rotor is given by

T= Iterations were done to find the rotor diameter using this formula. Sample calculations are below Where, Pa=Pressure generated in caliper The caliper pressure generated is 800 PSI according to Karizma ZMR manufacturing catalog. Therefore, Pa = 5000000 N/m2 Pad Friction (f) =0.4 d= Inner diameter of rotor contact m=

0.14

D= Outer diameter of rotor contact m=

0.18

T= T= 2814.867 N-m

The above relationship assumes that pad covers 360 degrees of disc on one side. However the pad covers only a fraction of disc. The pad fraction is the pad contact angle divided by 360 degrees. Pad Fraction = 0.0833333333

Actual T= T*pad fraction = 2814.867 * T = 234.5722 N-m Factor of safety = = FOS = 1.558 Hence the design is safe. 1. Kinetic energy of vehicle , K.E = = =13580.24 Jouls Where, m = mass of vehicle V= speed of the vehicle (m/s) 2. Stopping distance of vehicle a) The maximum friction force, F= m g = 0.7*220*9.81 =1510.74 N Where, H= friction coefficient m = mass of the vehicle g = acceleration due to gravity b) Deceleration of the vehicle (a) = Where, F= maximum friction force M= mass of the vehicle a= 6.86 m/s2

6.86

c) Time taken to stop the vehicle t= = =1.6195 sec d) Distance covered by vehicle in 5 sec in stopping distance = V*5 = 11.11 *5 = 55.55

e) Total stopping distance = V x (reaction time) + 11.11*2.5*(11.11^2/2*0.7*9.81)

1) Tangential braking force (B.F) t =

= 369.429 N 2) Tangential force on each wheel, Ft = = 92.357 N 3) Breaking torque on wheel, Tw = Ft R = 25.76 N-m Where, R= radius of tire 4) Effective rotor radius, Re = = 77.5mm 5) Clamping force, C = = 2312.288 N 19.23 6) Heat Flux In braking system, the mechanical energy is transformed into a calorific energy.

Heat generated or disc brake = Kinetic energy Hg = K.E Where, Hg = md x Cp x md = mass of the disc Cp = Specific heat = temperature difference Hg = 0.4 x 460 x = = 73.8056 Hg-27160.46

110

150.53445

2813.44

234.453333

1.55747295

13577.531

1510.74

1.61953353

55.55

249.623313

T

1

Tb=

2

Wt =

3

T=

4

Pad Fraction =

5

Actual T= T*pad fraction

6 7

Factor of safety =

1.9211799062

8

Kinetic energy of vehicle , K.E =

14194.6915

9

The maximum friction force, F= μΧmXg

1579.41

10

Deceleration of the vehicle (a) =

6.867

11

Time taken to stop the vehicle t =

1.6195335277

12

Distance covered by vehicle in 5 sec in stopping distance = V*5 55.55

13

Total stopping distance = V x (reaction time) +

199.698650211

1) Tangential braking force (B.F) t =

71.0805580558

2) Tangential force on each wheel, Ft =

17.770139514

3) Breaking torque on wheel, Tw = Ft X R

28722.8978632

3105.31351 0.0833333333 258.7761254

4) Effective rotor radius, Re =

0.02

5) Clamping force, C =

57727.0527348

6) Heat Flux Hg = 0.4 x 460 x

14194.6915 77.1450625

BREAKING CALCULATION Tb

wt

1616.3574766

Dw

h (m)

m(kg)

g

b(m)

Pa

f

23inch

0.6

230

9.81

1.2

5515200

0.4

0.6

230

9.81

1.2

0.5842 1128.15

D

d

θ

0.18

0.14

30

F

a (m/sec^2)

v

u

6.86

11.11

0.7