## Monday, November 13, 2017

### Distance to the specific height of object beyond the Horizon

In this tutorial I will give some short explanation about how the object visibility related to the distance from the horizon. For the object need to be visible beyond the horizon line, then its height must be increased.  Look at the picture below:

So based to the picture we get the distance to horizon line:
\begin{eqnarray} R^2 + d^2 & = & (R + h)^2 \nonumber \\ \Longrightarrow & & R^2 + d^2 = R^2 + 2 R h + h^2 \nonumber \\ \Longrightarrow & & d^2 = 2 R h + h ^2 \nonumber \\ \Longrightarrow & & d = \sqrt{h \left( 2 R + h \right) } \simeq \sqrt{ 2 R h} \end{eqnarray} Then from this we get the distance to the specific height of object beyond the horizon line. So the object must have this minimum height to stay visible to the oberver $A$. Then we get: \begin{eqnarray} R^2 + d_1^2 & = & (R + h_1)^2 \nonumber \\ \Longrightarrow & & R^2 + d_1^2 = R^2 + h_1^2 + 2 R h_1 \nonumber \\ \Longrightarrow & & h_1^2 + 2 R h_1 - d_1^2 = 0 \end{eqnarray} then we get (using quadratic formula): \begin{eqnarray} h_1 & = & \frac{ - 2 R \pm \sqrt{4 R^2 + 4 \cdot d_1^2 } }{2 } \nonumber \Longrightarrow & & h_1 = \sqrt{R^2 + d_1^2} - R \end{eqnarray} If we standing in the beach with the height $12 \textrm{ feet}$ or $3.6 \textrm{ m}$ then the distance to the horizon line is (according to wikipedia): $$d = \sqrt{2 R h} = \sqrt{2 \cdot 6371 \textrm{ km} \cdot \frac{3.6}{1000} \textrm{ km} } = 6.7 \textrm{ km}$$ If the object at the distance $17.67 \textrm{ miles}$ or $28.43 \textrm{ km}$ from observer $A$, then it will have a distance $28.43\textrm{ km} - 6.7 \textrm{ km} = 21.73 \textrm{ km }$ from the horizon. Then it must have a minimum height $$h_1 = \sqrt{ ( 6371 \textrm{ km} )^2 + ( 21.73 \textrm{ km} )^2 } - 6371 \textrm{ km} = 0.03705 \textrm{ km} = 37.05 \textrm{ m}$$ to stay visible to the observer $A$.