Manual Transmission calculations – 2 | Tractive forces, accelaration and gradability

Some basic formula for the calucation of various dynamics of vehicle…..for gradability the force component will be resolve in two component one is along the road and one is perpendicular to the road. only along the road component will help in climbing and perpendicular to the road will cause the vehicle to topple. there are various losses in engine , tranmission due to various friction so the result which is gonna come is gonna be in ideal case(i.e no friction etc) so what ever you get after calculation the actual result will be only 60-70 % of that. so calculate accordingly.
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here is sample calculation(  ideal case)

At hill climbing, tractive force available at an inclination of 45°,
F=Wsinθ- μWcosθ ≈2081N
Power =10hp=7460 Watt =F*V
V=3.58m/sec=12.9km/hr
So, our vehicle can climb the hill of inclination 45° with this velocity.
On a horizontal road,
Air resistance (Ra)= Ka*A*(V)²
 Ka= 0.004725 – 0.005 ,depending upon type of vehicle we have considered it to be 0.005
 Area (A) = Projected front area= 1.025 m²
 Velocity = 59.06km/hr
Hence Ra=17.88N
Rolling Resistance (Rr) = K*W
 W=total weight of vehicle= 300kg
 K is a constant and depends upon the nature of surface
 We have considered the road surface to be loose mud whose rolling resistance is 0.0435
Hence Rolling resistance (Rr) = 128.02N
Total resistance= Ra + Rr =145.9N
Traction –
F= (Te* Gear ratio* Transmission efficiency)/wheel radius
 Te=engine torque= 19Nm
 Wheel radius = 0.3175m
Hence F=756.21N (at low speed)
Acceleration:
a=(F-R)/m=2.03m/sec²
So, 0-60 km/hr in 8.21 sec.

 

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