Gauge-discharge relationship and discharge at 10.5 m

(i) Rating equation

Use $Q = C(G-a)^b$ with $a = 7.5$ m. Taking logs: $\log Q = \log C + b\log(G-a)$, a straight line $y = \log C + b\,x$ with $x=\log(G-a)$, $y=\log Q$.

With $n=14$: $\sum x=-1.2617$, $\sum y=32.3255$, $\sum x^2=3.2388$, $\sum xy=1.6364$, $\sum y^2=81.3787$.

$$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=1.4558$$ $$\log C=\frac{\sum y-b\sum x}{n}=2.4402\;\Rightarrow\; C=275.5$$

Rating equation: $Q = 275.5\,(G-7.5)^{1.456}$ (m$^3$/s)

Coefficient of correlation

$$r=\frac{n\sum xy-\sum x\sum y}{\sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}=0.9913$$

The high value shows a very good fit to the log-log straight line.

(ii) Discharge at G = 10.5 m

$$Q = 275.5\times(10.5-7.5)^{1.456} = 275.5\times3^{1.456} = 1364\ \text{m}^3/\text{s}$$

Answer: $Q=275.5(G-7.5)^{1.456}$, $r=0.991$; $Q(10.5\ \text{m})\approx 1364$ m$^3$/s (the measured value is 1500 m$^3$/s, so the fit is about 9 % low at this stage).