Scraper Production and the Optimum Load Time: 24-Bld-A3 Construction Engineering, Undated paper, Q7

Question 7: Scraper Production and the Optimum Load Time (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Field-measured average load volume vs. load time, a fixed haul/return/dump/turn time of 3.0 min (i.e. cycle time = load time + 3.0 min), and a 50 min/hr efficiency factor.

Given: average load volume vs. load time
Load time (min)	0.2	0.4	0.6	0.8	1.0	1.2	1.4
Average load (CY)	11.4	18.7	22.7	24.8	26.1	26.8	27.2

Find. The scraper's production (CY per 50-min hour) at each load time, and the load time that maximizes it.

Approach. A longer load time keeps filling the bowl (more CY per pass) but also lengthens every cycle, so production — volume per unit time, not volume per pass — first rises, peaks, and then falls; compute it at every tabulated load time and locate the peak.

Cycle time and production at each load time. $Cycle = LoadTime + 3.0$ min; $Production = \dfrac{Load\times 50}{Cycle}$ CY/hr. For load time 0.8 min: $Cycle=3.8$ min, $Production=\dfrac{24.8\times50}{3.8}=326.3$ CY/hr. The full table below repeats this for every row.
Locate the peak. Production rises steeply from 178 CY/hr at 0.2 min up to a broad, flat peak of 326.3 CY/hr at 0.8 min and 326.2 CY/hr at 1.0 min — essentially tied — then declines through 1.2 and 1.4 min as the added cycle time outweighs the shrinking marginal gain in load volume. A quadratic curve fitted through the 0.6/0.8/1.0-min points (to interpolate between the tabulated values) peaks at load time $\approx 0.90$ min, production $\approx 328$ CY/hr, confirming the tabulated peak sits between the 0.8 and 1.0 min rows. $$\boxed{\text{Optimum load time} \approx 0.8\text{--}0.9\text{ min, giving a maximum production of about } 326\text{--}328 \text{ CY per 50-min hour}}$$

Load time (min)	Cycle time (min)	Load (CY)	Production (CY/hr)
0.2	3.2	11.4	178.1
0.4	3.4	18.7	275.0
0.6	3.6	22.7	315.3
0.8	3.8	24.8	326.3
1.0	4.0	26.1	326.2
1.2	4.2	26.8	319.0
1.4	4.4	27.2	309.1
Optimum load time ≈ 0.8–0.9 min; peak production ≈ 326–328 CY/hr

Fig. Q7-1 — production vs. load time; the curve peaks in a broad plateau around 0.8–0.9 min (marked), consistent with the tabulated maximum.

Topic: Earthmoving equipment production and the optimum load-time trade-off.

Key relations: $Production = \dfrac{Load(t_{load})\times 60\times E}{t_{load}+t_{fixed}}$, where $E$ is the efficiency factor (expressed here as 50 usable min/hr) and $t_{fixed}$ is the load-independent portion of the cycle (haul, dump, return, turn).

Why this works: Bowl fill volume rises with load time but with diminishing returns (the loader/scraper's cutting edge fills the bowl fastest early in the pass), while cycle time rises linearly with load time — production is their ratio, so it necessarily has an interior maximum rather than increasing monotonically with load time.

Common pitfall: Picking the load time that maximizes volume per pass (here, the longest tabulated time, 1.4 min) instead of volume per unit time — the two optima are different, and the question asks for the latter.

Source: Halpin & Senior, Construction Management, Ch. 6 (equipment production estimating); Hendrickson & Au, Project Management for Construction, Ch. 4.

Question 7 of 7: Scraper Production and the Optimum Load Time

Question 7: Scraper Production and the Optimum Load Time (20 marks)