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Ini adalah tugas mata kuliah Teknologi Motor Bensin tentang Exhaust Brake.
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Ini adalah tugas mata kuliah Teknologi Motor Bensin tentang Exhaust Brake.
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BRAKE CALCULATIONS Under dynamic load conditions Rear axle load = (m.g.x cg – m.a.ycg)/l Front axle load = mg – rear axle load Where m = mass of vehicle g = acc. due to gravity xcg = Distance of CG from front Axle a = deceleration during braking ycg = CG Height l = wheelbase
Brake torque requirement Torque = µ .N.R where R = rolling radius tyre specification 145/70 R-12 rolling radius = 6 x 25.4 + 0.7 x 145 = 254mm For front tires = 496.46 J For rear tires = 400.56 J
Braking Torque developed Line pressure developed Force applied by driver on pedal, Fpedal = 200N Pedal Ratio, PR = 6:1 Force on master cylinder input shaft = F pedal x PR = 1200N Master cylinder piston diameter Dmc f = 0.625 in = 15.875mm Dmc r = 0.75 in = 19.05mm
Pressure developed at front, 2
3
Pf = Fmc / (3.14 x (Dmc f /2) = 6065.744 x 10 N/mm
2
Similarly at rear, 3
2
Pr = 4210.187 x 10 N/mm Brake torque developed
Radius of rotor = 100.5 mm µ pad = .35 Caliper piston dia.= 48mm No. of piston per caliper = 1 No. of calipers = 2 2
Acp = Total caliper piston area = 3.14 x ( D cp /2) x Np x Nc -3
2
= 3.619 X10 m
Fclamp f = Pf x Acp = 21951.9 N Fclamp r = Pr x Acp = 15236.64N Torque developed at front = .35 x 21951.9 x .100 = 768.3 J Torque developed at rear = .35 x 15236.64 x .100 = 533.2 J Torque developed by the brakes is more than the required therefore this results in locking of all wheels.
