Six-tenths rule on log–log plots — what is the slope? Using the six-tenths factor rule for scaling equipment costs with capacity, if you plot cost vs capacity on a log–log graph, what slope do you expect?

Introduction / Context:
Early-stage cost estimation often relies on capacity scaling laws: Cost_2 = Cost_1 * (Capacity_2 / Capacity_1)^n. The exponent n captures economies of scale. The classic six-tenths rule sets n ≈ 0.6 for many types of process equipment.

Given Data / Assumptions:

• Cost ∝ Capacity^n with n ≈ 0.6.
• Plot is log(Cost) vs log(Capacity).

Concept / Approach:
On a log–log plot, a power law appears as a straight line with slope equal to the exponent. Therefore, using the six-tenths factor rule, the slope equals 0.6. Deviations from 0.6 occur for specific equipment classes and time periods, but 0.6 remains a widely taught heuristic.

Step-by-Step Solution:

Verification / Alternative check:
Any two points (C1, Cap1) and (C2, Cap2) on the line satisfy n = log(C2/C1) / log(Cap2/Cap1). Using data generated with n = 0.6 reproduces the same slope.

Why Other Options Are Wrong:

Common Pitfalls:
Applying n = 0.6 blindly to all equipment; always sanity-check against vendor quotes and technology specifics.

Final Answer:
0.6