Question

A tractor has a permissible loaded speed of 200 m/ minute, which can increase by 25% when the load is removed/deposited. Generally, it is operated at 80% of the permissible speed (loaded or unloaded). It works at a location where haul distance is 120 m. Rest allowance per round- trip is taken as 50 seconds on an average. Fixed time per trip, for loading and unloading and turnaround, etc., is 30 seconds. What is its effective cycle time?
1. 157 seconds
2. 161 seconds
3. 173 seconds
4. 182 seconds

Answer 1

Correct Answer - Option 2 : 161 seconds

Concept:

The distance measured along the centerline or most direct practical route between the center of the mass of excavation and the center of mass of the fill as finally placed is known as haul distance.

Rest allowances are scheduled breaks designed to permit an employee to recover from fatigue during the work day.

Mathematically,

Effective cycle time = ((Forward time+ Return time)/Efficiency) + Loading/ unloading time + Rest Allowance

Forward time = Haul distance/forward velocity

Return time = Haul distance/Return velocity

Calculation:

Efficiency = 80% = 0.8

Forward velocity = 200 m/min = 200/60 m/sec

Return velocity = 1.25 x 200 = 250 m/min = 250/60 m/sec

Haul distance = 120 m

Rest allowances = 50 sec

Loading/unloading time = 30 sec

Forward time = Haul distance/forward velocity = 120/ (200/60) = 36 sec

Return time = Haul distance/Return velocity = 120/ (250/60) = 28.8 sec

((Forward time+ Return time)/Efficiency) = (36 + 28.8)/0.8 = 81 sec

Effective cycle time = ((Forward time+ Return time)/Efficiency) + Loading/ unloading time + Rest Allowance = 81 + 30 + 50 = 161 sec