雙高法

圖文作者
牟善豪
使用軟體
AutoCAD

運用兩次三角高程測量,以間接求得水平距及高程差的方法。

觀測

將經緯儀整置於 $A$ 點,觀測垂直樹立於 $B$ 點之標尺。當經緯儀照準讀數 $\gamma_1$ 時,得垂直角 $\alpha_1$;再照準讀數 $\gamma_2$ 時,另得垂直角 $\alpha_2$。讀數差爲 $\gamma_2-\gamma_1=b$,儀器高 $i$。

仰視:

俯視:

計算

水平距 $D$:

\begin{cases}
\ v_1= D \cdot \tan \alpha_1 & \cdots \cdots \ \enclose{circle}{\kern .06em 1\kern .06em} \\\\
\ v_2= D \cdot \tan \alpha_2 & \cdots \cdots \ \enclose{circle}{\kern .06em 2\kern .06em}
\end{cases}

$$\enclose{circle}{\kern .06em 2\kern .06em}-\enclose{circle}{\kern .06em 1\kern .06em}:$$

\begin{align}
v_2-v_1&=(D \cdot \tan \alpha_2)-(D \cdot \tan \alpha_1) \\\\
\because \ v_2-v_1 &= \gamma_2-\gamma_1=b\\\\
\therefore \ b&=D \ (\tan \alpha_2-\tan \alpha_1) \\\\
D&={b \over {\tan \alpha_2-\tan \alpha_1}}
\end{align}

高程差 $\Delta H_{AB}$:

\begin{gather*}
v_1=D \cdot \tan \alpha_1 \\\\
\Delta H_{AB}=i+v_1-\gamma_1
\end{gather*}

 

例題一(仰視)

於 $A$ 點整置經緯儀,於 $B$ 點整置水準尺,由 $A$ 點經緯儀讀得水準尺下端覘板讀數 $\gamma_1=0.860 \ m$,縱角 $\alpha_1=30°$;上端覘板讀數 $\gamma_2=1.860 \ m$,縱角 $\alpha_2=31°$。已知儀器高 $i=1.500 \ m$,求 $AB$ 點水平距及高程差各爲何?

水平距 $D$:

\begin{cases}
\ v_1= D \cdot \tan \ 30° & \cdots \cdots \ \enclose{circle}{\kern .06em 1\kern .06em} \\\\
\ v_2= D \cdot \tan \ 31° & \cdots \cdots \ \enclose{circle}{\kern .06em 2\kern .06em}
\end{cases}

$$\enclose{circle}{\kern .06em 2\kern .06em}-\enclose{circle}{\kern .06em 1\kern .06em}:$$

\begin{align}
v_2-v_1&=(D \cdot \tan \ 31°)-(D \cdot \tan \ 30°) \\\\
\because \ v_2-v_1 &= \gamma_2-\gamma_1=b\\\\
&=1.860-0.860=1.000\\\\
\therefore \ b&=D \ (\tan \ 31°-\tan \ 30°) \\\\
1.000&=D \cdot 0.02351 \\\\
D&={1.000 \over 0.02351} =42.535 \ m
\end{align}

高程差 $\Delta H_{AB}$:

\begin{align}
v_1&=D \cdot \tan \alpha_1 \\\\
&=42.535 \cdot \tan \ 30°\\\\
&=24.558 \ m \\\\\\
\Delta H_{AB}&=i+v_1-\gamma_1\\\\
&=1.500+24.558-0.860 \\\\
&=25.198 \ m
\end{align}

例題二(俯視)

經緯儀於 $A$ 點觀測 $B$ 點標尺,當儀器中絲照準標尺讀數 $\gamma_1=2.500 \ m$ 時,得縱角 $\alpha_1=-8°$;再照準標尺讀數 $\gamma_2=0.500 \ m$ 時,得縱角 $\alpha_2=-10°$。已知儀器高 $i=1.500 \ m$,$B$ 點高程為 $50 \ m$,求 $AB$ 兩點間水平距、高程差、$A$ 點高程各爲何?

水平距 $D$:

\begin{cases}
\ v_1= D \cdot \tan \ (-8°) & \cdots \cdots \ \enclose{circle}{\kern .06em 1\kern .06em} \\\\
\ v_2= D \cdot \tan \ (-10°) & \cdots \cdots \ \enclose{circle}{\kern .06em 2\kern .06em}
\end{cases}

$$\enclose{circle}{\kern .06em 2\kern .06em}-\enclose{circle}{\kern .06em 1\kern .06em}:$$

\begin{align}
v_2-v_1&=(D \cdot \tan \ (-10°))-(D \cdot \tan \ (-8°)) \\\\
\because \ v_2-v_1 &= \gamma_2-\gamma_1=b\\\\
&=0.500-2.500=-2.000\\\\
\therefore \ b&=D \ (\tan \ (-10°)-\tan \ (-8°))\\\\
-2.000&=D \  (-0.035786) \\\\
D&={-2.000 \over -0.035786} =55.888 \ m
\end{align}

高程差 $\Delta H_{AB}$:

\begin{align}
v_1&=D \cdot \tan \alpha_1 \\\\
&=55.888 \cdot \tan \ (-8°)\\\\
&=-7.855 \ m \\\\\\
\Delta H_{AB}&=i+v_1-\gamma_1\\\\
&=1.500+(-7.855)-2.500 \\\\
&=-8.855 \ m
\end{align}

$A$ 點高程 $H_A$:

\begin{align}
H_A&=\Delta H_{AB}+H_B\\\\
&=-8.855+50.000\\\\
&=41.145 \ m
\end{align}

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