Step 1: Understanding the Question:
A dragline repeats a dig, swing, dump cycle over and over, but it is not actually working every second of the hour since there are stoppages and delays. We need its real, practical output per hour after accounting for both the bucket not filling completely and the machine not running the whole hour.
Step 2: Key Formula or Approach:
Instead of scaling the final output by the utilization, scale the working time itself: out of every hour, the dragline is actually productive for only $3600 \times 0.90$ seconds. Multiplying the number of cycles that fit into that productive time by the effective bucket load gives the same answer from a different direction.
Step 3: Detailed Explanation:
The effective bucket load, after the 0.9 fill factor, is $20 \times 0.9 = 18\ m^3$ per cycle.
One cycle takes digging (20 s) plus the full to and fro swing (32 s) plus dumping (10 s), so $t_{cycle} = 62\ s$.
The productive seconds available in an hour, after the 90% utilization, are $3600 \times 0.90 = 3240\ s$.
The number of cycles that fit into this productive time is $3240 / 62 = 52.26$ cycles.
Multiplying by the load per cycle, $Q = 52.26 \times 18 = 940.6\ m^3/h$.
Step 4: Final Answer:
Scaling the working time instead of the output gives the same practical dragline output, about 940.6 cubic metres per hour.