spherical trigonometry

So I just finished reading Heavenly Mathematics, by Glen Van Brummelen because, you know, that's the sort of thing I do for fun.  I didn't go full nerd on it, though: I didn't do the exercises.  I'll save that for another time.

Man, did I learn a lot from this book.  Spherical trigonometry was a fairly standard part of the mathematics curriculum until the middle of the 20th century, but it is all but forgotten these days.  Its primary use is for navigation at sea and modern technology has rendered it obsolete.  Well, not obsolete, exactly, but the technicalities have been incorporated into our modern devices that do our navigating for us.  No doubt logarithms will be the next thing to be dropped (and indeed, no student today learns what logarithms are really for).

My favorite things: in the first chapter, we learn how to estimate the Earth's circumference, construct a sine table with our bare hands, and determine the distance from the Earth to the Moon.  Here's the basic idea of the circumference calculation from the medieval Pakistani mathematician Biruni.  First, you need the height of a mountain you can climb.  To do so you make a big square out of wood or something and measure the distance \( E'G\) and the distance from \( F_1' \) to the point where \( BF_1' \) crosses the left side of the square.  Then you use similar triangles to get the height \( AB\).  In Biruni's example, he found the peak to be 305.1 meters.

I've been to the mountaintop...

I've been to the mountaintop...

The next thing to do is to climb the mountain.  If you've stood on top of a big enough hill, you notice that the horizon curves down from where you are.  You can then measure the angle of declination; that is the angle \( \theta \) in the diagram below which is made by taking a tangent to the Earth's surface at the horizon passing through the peak and the horizontal ray from the peak.  Biruni measured this as \( \theta = 34' = 0.56667^\circ \).  Then use the fact that \( \Delta LJB \) is a right triangle to get

the earth, not to scale.

the earth, not to scale.

\[\cos\theta = \frac{r}{r+305.1}\]

where \( r\) is the radius of the Earth.  Provided we have a sine table (which we can build with our bare hands, as the author describes, and which Biruni had available), we know \( \cos\theta \).  Solving for \( r\), we get \( r=\) 6,238 km.  The actual value, accepted by scientists today, is 6,371 km.  Not bad for having only primitive tools.

Once you know the radius of the Earth, it's actually not too difficult to figure out the distance to the moon, provided you can make some measurements in two locations during a lunar eclipse, which Ptolemy was able to do nearly 2,000 years ago.  He missed on the distance to the sun by a factor of 19 or so (too small), but I suppose we can let that slide.

The other thing I really enjoyed was the proof of Euler's Formula \( V - E + F = 2 \) for convex polyhedra.  There are lots of proofs of this, the simplest using graph theory and more high-powered ones via topology.  It turns out that Legendre came up with a wonderful proof using spherical trigonometry.  The idea is this:  take a convex polyhedron, like, say, a cube, and scale things so that the unit sphere fits around it comfortably.  Then project the polyhedron out onto the sphere (imagine a light source in the center of the polyhedron).  You then have a spherical polyhedron.  Now, we know the surface area of the sphere is \(4\pi \).  We need only then add up the areas of the spherical polygons covering the surface.  This is where the spherical trig comes in:  a spherical triangle \(\Delta ABC \) has area \[\frac{\pi}{180}(A + B + C - 180^\circ)\]  where \(A, B, C\) are the triangle's angles measured in degrees.  By dividing a polygon with \( n\) sides into \( n \) triangles we find that the area of a polygon is \[ \frac{\pi}{180} ((\text{sum of polygon's angles}) - (n-2)\cdot 180)\]  Putting all this together we get
\[ \sum \frac{\pi}{180} ((\text{sum of polygon's angles}) - (n-2)\cdot 180) = 4\pi \] and cleaning up a bit we see \[ (\text{sum of all angles}) - \sum n\cdot 180 + 2F\cdot 180 = 720.\]  But the angles go around the vertices, so their sum is \( V\cdot 360\) and since each edge gets counted twice, \( 2E = \sum n\).  So the last equation reduces to \[ V\cdot 360 - 2E\cdot 180 +2F\cdot 180 = 720,\] and dividing by 360 gives us the result.  Cool.

I also learned why Napier was really interested in logarithms and why you used to see tables of logarithms of sines.  It was all about efficient navigation and the resulting cost savings.  Money does indeed make the world go around.

I recommend this book heartily.  You can pick it up and read a chapter in 30 minutes and learn something interesting.