I understand that inside any sphere you can place a tetrahedron which
touches the surface of the sphere at each of its four vertices. Now,
if that sphere is the planet earth, I'd like to think there's a way (a
formula, a spreadsheet, a simple piece of software) to calculate the
latitude and longitude of each of the four points of the tetrahedron.
My A-level maths just isn't up to the job. Of course, because there
are an infinite number of different ways to position a tetrahedron
inside a sphere I would need to input a starting point, e.g. the
latitude and longitude of one of the vertices (and some other
parameter?), and the formula/spreadsheet/program would provide the lat
and long of the other three points.

Now, my question is: Can Google Answers provide me with a
formula/spreadsheet/whatever to do this calculation ($50.00)? If not, can you
put me in contact with someone who is knowledgeable enough and
kind enough to help me out ($25.00)?

---

Clarification of Question by onlynick-ga on 25 Aug 2005 11:45 PDT

Yes, I meant a regular tetrahedron, i.e. one where all four faces are
identical equilateral triangles.

---

Clarification of Question by onlynick-ga on 30 Aug 2005 15:19 PDT

Yes, I meant to assume the earth to be a perfect sphere with a single, known radius.

---

Clarification of Question by onlynick-ga on 06 Sep 2005 11:25 PDT

Wow, rak1-ga, it sounds as though we're getting somewhere! Since I'd
like to play around with the location of the starting point I'd like
the spreadsheet/formula to allow me to input the lat/long of the first
point (and one other parameter, an angle probably), then calculate the
other three lat/longs based on what I have input. More complex, I
know. I can only apologize for my neediness.

---

Clarification of Question by onlynick-ga on 08 Sep 2005 12:50 PDT

I can't think of a better use for the $50 - but forgive me if I have a
play with the spreadsheet prior to making the donation. I'll add more
when I've had a go ...

---

Clarification of Question by onlynick-ga on 08 Sep 2005 15:08 PDT

rak1-ga,

Classy spreadsheet. I can't imagine a more complete solution. The Red
Cross get $100. I hope it helps.

Nick

Actually, I think you'd need more than one vertex.  The most you could
get back with just one vertex is a circle of possible tangents that is
109.5 degrees away from the one vertex that you list.

Oh, and you should consider talking to the programmers at
Cambridgesoft.  Their ChemDraw 3D software uses tetrahedrons to
represent carbon atoms.  The user is able to rotate these tetrahedrons
in a sphere.  I bet they could help you out.

Any 4 points on the surface of the earth are the vertices of a
tetrahedron.  Or do you mean a regular tetrahedron?

The Earth is not a perfect sphere. The shape of the Earth surface is
that of an oblate spheroid. It is flattened at the poles. The polar
radius is about 25 km less than the equatorial radius. The oblateness
is a consequence principally of the Earth's axial rotation.

To answer the question we need to simlify the shape of the earth and
assume that it is a perfect sphere with an average radius (we could
use a maximum or minimum instead).

Would this simplification OK?

Simplifying the shape of the earth to a sphere is fine and dandy, but
then the points you wind up with won't be correct. Just be aware.

I can't do the trigonometry either, but it should be possible to
create a formula using Lat./Long. definition of one point (a/b) which
then calculates:  if a second point is at Lat./Long. (c/d) then the
L/L of the other two points will be w/x and y/z which are eguidistant
from the two given points.
Elementary  - in principle  ;)

Do you want A solution for a given starting point, or ALL solutions? 
I've done a spreadsheet that would do the first - the second is a good
deal more complicated

rak1-ga,
Then proudly post your spread sheet.  We commenters can only prove our competence.

OK, needy onlynick, I have your comprehensive answer!

I have posted at robkenny.50megs.com a spreadsheet which calculates
the answers in cartesian and lat/long co-ordinates, and plots the
results.   There's also an explanatory powerpoint (in case you're
interested in the methodology) .  Apologies for the rudimentary HTML.

There's a fair bit of work here.  Being only a commentator, I
obviously shan't ask for a fee, but if you feel it's worth the $50 you
originally offered, perhaps you could make a donation to one of the
Katrina charities, such as the Red Cross?

Let me know if you have any questions, thanks

rak1-ga,
That is a great suggestion  - also for any other questioners
"answered" by comments, an incentive for us all.

onlynick -

Of course, no problem.  Let me know if you spot needed improvements,
and I'll be happy to make edits

Nick -

Thanks for the positive feedback, and on behalf of the bedraggled,
thanks very much for the $100