Bus height, hbus = 3.743m
CG height =hCG = 3.743/3 = 1.248m
CG tilt, htilt = 2.60/2 = 1.30m.
Guardrail height, hG = 1.05m
Change in CG height,
h = hG + htilt – hCG = 1.102m
d = 10.52m
The angle that will give the minimum speed is calculated using equation
“θ”=1/2 cos^(-1)[(-h)/√(d^2- h^2 )]*
Velocity of a vehicle at the point of take off is
v=d√(g/(2cos”θ” (dsin”θ” -hcos”θ” )))*
From the data and formula used, calculated angle,
“θ” = 48°. This gives the minimum speed of the bus at the point of take off as V = 38.5 km/hr.
However, if the angle is taken from the site measurement which is “θ” = 34.1°, the calculated speed is V = 41.2 km/hr.
Therefore, the possible speed of the bus at the point of take off is 41.2 km/hr.