Rating curve equation, zero-discharge stage 161.0 m

Use the power form $Q = C(G-a)^b$ with $a = 161.0$ m. In logarithms: $\log Q = \log C + b\log(G-a)$, a straight line $y = c + b\,x$.

With $n = 9$ and $\sum y^2 = 62.2104$:

$$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{9(7.4184)-(2.2187)(23.2365)}{9(1.8709)-(2.2187)^2}=1.277$$ $$\log C=\frac{\sum y-b\sum x}{n}=2.2671\;\Rightarrow\;C=185.0$$

Rating equation: $Q = 185.0\,(G-161.0)^{1.277}$ (m$^3$/s, $G$ in m)

The correlation coefficient is $r = 0.986$, so the fit is good.

Check: at $G=163.4$ m, $Q = 185.0\times2.4^{1.277} = 566$ m$^3$/s (observed 650).

Answer: $Q = 185(G-161.0)^{1.28}$.