3

If we assume that it is possible to dig a hole through the earth, how can we calculate exactly where the hole would exit the earth if we know .... 1) The point of entry (gps coords) 2) The angle of entry 3) The direction of entry

So if a hole was dug with an entry point of 51.50360, -0.11959, in the direction of 0 degrees (north) at an angle of 45 degrees (straight down = 90 degrees), how can I figure out where on the planet the exit to the hole will be?

Any ideas?

If anyone can help with even a better way of asking the question, key words to use on Google, or related reading that might help me solve this I would appreciate it. So far I have found a few sites which talk about chords through the earth and euclidean distance. But they all seem to start from the basis of knowing the point of entry and exit to work out the euclidean distance. I only know the point of entry, angle of entry and direction of entry and need to calculate the exit (..and distance would be a nice extra).

user99443
  • 33

2 Answers2

4

Assume for simplicity that you are drilling at the point $P$ with geographical coordinates $(\phi_0,\theta_0)$ on the northern hemisphere. Here $\phi=0$ corresponds to the meridian of Greenwich, $\theta=0$ to the equator, and $\theta={\pi\over2}$ to the north pole $N$. Assume that you are drilling the hole in the direction $\alpha$ measured counterclockwise from due north and at an angle $\delta>0$ downwards with respect to the local horizontal plane. The drill will reappear at some point $Q$ whose spherical coordinates $(\phi_1,\theta_1)$ we now have to compute.

The three points $P$, $N$, $Q$ form a spherical triangle $\Delta$ on the unit sphere $S^2$, see the figure below. About $\Delta$ we know the following: The side $PN$ has length ${\pi\over2}-\theta_0$, the angle at $P$ is $\alpha$ (supposed $<\pi$ in the following), and the side $PQ$ has length $2\delta$ (convince yourself of this!). The formulas of spherical trigonometry then allow to compute the angle $\nu$ at $N$, as well as the length $\sigma$ of the side $NQ$: Using $\cos\bigl({\pi\over2}-\theta_0\bigr)=\sin\theta_0$ and vice versa one obtains $$\cos\sigma=\cos(2\delta)\sin\theta_0+\sin(2\delta)\cos\theta_0\cos\alpha\ ,$$ which gives $\sigma$. Furthermore one has $$\cos\nu={\cos(2\delta)-\cos\sigma\sin\theta_0\over \sin\sigma\cos\theta_0}\ ,$$ which gives $\nu$. It is then easy to see that $$\phi_1=\phi_0-\nu,\quad \theta_1={\pi\over2}-\sigma\ .$$

enter image description here

Christian Blatter
  • 226,825
  • Thanks Christian. It looks like you have maybe given me the answer I need....now to try and understand it :-) – user99443 Oct 08 '13 at 12:13
  • OK, I have spent a while looking at this and cannot figure it out. I am afraid the maths that I needed for my comp sci degree did not prepare me for this. Is there any chance someone could help me out by plugging some numbers into this so that I can understand what is happening? – user99443 Oct 10 '13 at 00:19
  • That is brilliant!! You are a legend Christian. Thanks ever so much :-) – user99443 Oct 10 '13 at 10:30
  • I am feeling really thick now. I have spent a few days on this trying it for P(0,0) at a 90 degree angle of entry, due north (360 degrees). The direction shouldn't matter as it will be straight down and should result in the antipode (0,180). I get as far as calculating cosv and it returns -8.16 which must be wrong as it should be a value between 1 and -1. I am converting everything to radians (coords, angles and direction) and am failing to get an answer that is even close to 0,180. Any ideas? – user99443 Oct 15 '13 at 01:03
  • There was a typo in the second formula; sorry for that. – When $\theta_0=0$ and $delta={\pi\over2}$ the first formula gives $\cos\sigma=0$, or $\sigma={\pi\over2}$. Then the second formula gives $\cos\nu=-1$, or $\nu=\pi$, as it should be. – Christian Blatter Oct 15 '13 at 08:33
  • Thanks Christian. I am now getting the results I expect from my antipode tests – user99443 Oct 17 '13 at 11:24
1

A simple approach would be to identify the surface of the earth with the surface of a unit sphere in$~\Bbb R^3$ (making the somewhat false but hardly avoidable assumption that the surface is a sphere to begin with, and choosing the earth's radius as unit of measure), in such a way that spherical coordinates match longitude/lattitude coordinates. Then the entry point defines a point $P$ on the unit sphere, and the digging directions can be translated into a nonzero vector$~v\in\Bbb R^3$. Then just compute the second intersection of the line $P+\langle v\rangle$ with the unit sphere, and take its spherical coordinates. Probably a messy computation, but an entirely straightforward one.

Marc van Leeuwen
  • 115,048
  • Thanks Marc. If only I could use the simpler approach :-) Your answer is much appreciated nonetheless.....I may end up having to use it when I fail to get my head round Christian's answer :-) – user99443 Oct 08 '13 at 12:17