Hello stevek Originally Posted by

**stevek** Hi all,

Hoping that someone can come up with a simple explanation or solution to my problem ... yes, I have Googled extensively and haven't come across anything that helps.

Basically, I'm looking for a way to calculate the TOTAL height of an object that includes the height component that can be seen above your personal horizon PLUS the height component that cannot be seen because it lies below your personal horizon.

Picture a very high tower that is located 100's of kilometres away. Naturally the base of the tower and a percentage of it's total height will not be visible to you because of the Earth's curvature. Obviously it's trivial to measure the height component that you can see above the horizon ... but how do you calculate the height component that lies below the horizon ?

Thanks in advance to all those who may be able to shed some light on this.

In the attached diagram:$\displaystyle O$ is the centre of the earth, whose radius is $\displaystyle r$

$\displaystyle PS = h_1 =$ height of observer above surface of earth

$\displaystyle QT = h_2 =$ height of part of tower not visible below the horizon. It is $\displaystyle h_2$ that we need to calculate.

$\displaystyle R$ is a point on the horizon visible from $\displaystyle P$; therefore $\displaystyle PRQ$ is a tangent to the circle at $\displaystyle R$, and $\displaystyle \angle PRO = \angle QRO = 90^o$

$\displaystyle \theta_1$ and $\displaystyle \theta_2$ are the angles (in radians) subtended at $\displaystyle O$ by the arcs $\displaystyle SR, RT$; therefore the arc lengths $\displaystyle SR$ and $\displaystyle RT$ are $\displaystyle r\theta_1$ and $\displaystyle r\theta_2$ respectively. If s is the total length of the arc ST, then: $\displaystyle s = r\theta_1+r\theta_2$

$\displaystyle \Rightarrow \theta_2 = \frac{s}{r}-\theta_1$ ...(1)

Now I am assuming that we are given$\displaystyle r$ (the radius of the earth),

$\displaystyle h_1$ (the height of the observer) and

$\displaystyle s$ (the distance measured around the surface of the earth from the foot of the observer to the foot of the tower).

Therefore, in $\displaystyle \triangle PRO$:$\displaystyle \cos\theta_1 = \frac{r}{r+h_1}$

$\displaystyle \Rightarrow\theta_1 = \arccos\left(\frac{r}{r+h_1}\right)$...(2)

In $\displaystyle \triangle QRO$:$\displaystyle \sec\theta_2 = \frac{r+h_2}{r}$

$\displaystyle \Rightarrow h_2 = r(\sec\theta_2-1)$

This, then, together with equations (1) and (2) will give you the height of the part of the tower below the visible horizon.

I've done a calculation on a spreadsheet (see second attachment) given: $\displaystyle r = 6378000$ (radius of earth in m)

$\displaystyle s = 100000$ (distance 100 km away)

$\displaystyle h_1 = 50$ (height of observer in m)

which gives $\displaystyle h_2\approx 438$ m.

Seems OK to me.

Grandad