Phi-index from runoff volume 92,800 m3

Runoff depth

$$R = \frac{V}{A} = \frac{92{,}800\ \text{m}^3}{0.8\times10^{6}\ \text{m}^2} = 0.116\ \text{m} = 11.6\ \text{cm}$$

Rainfall

$P = 0.8+1.8+3.0+4.6+3.6+3.2+2.0+1.0 = 20.0$ cm over 8 h (1-h steps). Total loss $= 20.0 - 11.6 = 8.4$ cm.

Trial for $\phi$

Assume the 0.8 cm and 1.0 cm hours are below $\phi$ (no excess). The other six hours contribute (1.8 + 3.0 + 4.6 + 3.6 + 3.2 + 2.0) = 18.2 cm.

$$18.2 - 6\phi = 11.6 \;\Rightarrow\; \phi = 1.10\ \text{cm/h}$$

Check: $1.10 > 0.8$ and $1.10 > 1.0$; all other hours exceed 1.10 cm. The assumption holds.

The excess sums to 11.6 cm, which agrees with the runoff.

Answer: $\phi$-index = 1.10 cm/h.