Calculating the distance to the horizon

 

adri
Adri, trying to pose very focused

Last weekend my friend Adri came by to Fuengirola, a nice area of Málaga by the sea where I am currently living. It was Sunday morning and we were chatting on the terrace after breakfast when we noticed a big ship that seemed to be suspended on the horizon. I said “I wonder how far away that ship is. It doens’t look very hard to calculate….” That was the trigger! Adri and I studied Mathematics together at Málaga University and since then we haven’t had the chance to discuss anything related to mathematics whilst having a piece of paper in front of us. This looked like a good opportunity to reconcile that.

 

First the obvious: Due to the curvature of the Earth, the distance to the line of the horizon depends on the height of the observer, and the higher the observer is, the father one can see. It’s helpful to exaggerate things when trying to prove something mathematically, or at least, to have a guess about what may happen. This also applies to drawing: imagine being at least 1000 or 2000 meters high for making a draft of the situation.

 

earth_geogebra

 

 

Before really starting I remember we made a last observation looking at the draft: It is intuitive to see that when h \rightarrow \infty then \beta \rightarrow \frac{\pi}{2}. And the same happens with the left side, which I didn’t draw here. In fact, we can see almost half the moon observing from Earth, right?

And now the real thing: as we can observe, calculating angle \beta as a function of h (and R, of course) we have the problem solved: using \beta and the radius of the Earth it is easy to calculate the arc from the ground where we are to the tangent point.

For that matter we may first try to calculate the value of x in the figure. We are lucky that the line of the horizon means the locus of the tangency points which lines passes the eye of the observer: that give us (in our draft) a square angle in the triangle and we immediately have the value of x:

x=\sqrt{2Rh+h^2}

Now remembering basic trigonometry, specifically the definition of sin:

\frac{R+h}{1}=\frac{x}{\beta}

From where we can obtain that:

\beta = asin(\frac{\sqrt{2Rh+h^2}}{R+h})

From our first floor perspective, about 4 meters from ground, we estimated  that that ship was around 7100 meters from us.

Now that we had a formula we couldn’t stop there… Matlab was the fastest way at that moment. First for home distances (up to Eight-thousander):

R=6371;
h=0:0.1:9000;
beta=asin((sqrt(2*R*h+h.^2))./(R+h));
plot(h,R.*beta)

This simple script give us this beautiful graph, plotting a linear spline every 100 meters:

home

And now going wild, changing h=0:1:100000; we can have a guess of how high we need to be to have the impression of seeing half the Earth:

outerspace

Using much smaller figures for R we could determine fun things, like…. how far you have to be from that orange in your kitchen to see a 48% of it. What about 49.8%?? Please, write your answers below :)

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