Re: Spherical Triangle

Charles <cahuffer@xxxxxxxxx> writes:

[snip]

Concerning my programs, I was very precise about the assumptions I
made. When I use your values of 90, 90, and 270 for the three sides,
my program SSSS gives a display of "NO TRIANGLE", which is exactly
what I have programmed it to do. No side can be greater than or equal
to 180. But when I put in 90 for all three sides, SSSS gives 90 for
each of the three desired angles. To get your 270 side, 360 -- 90
does just that.

Yes, this is all what I would expect a calculator program to do.

I have not seen the latest editions of Trigonometry texts. The texts
I have with me are pre-1955. The Shibli book has copyrights of 1928,
1936 and 1949. The Nielsen, Vanlonkhuyzen book has copyrights of
1944, 1946 and 1954. The book by Frank Ayres, Jr. has a copyright of
1954.

What type of problems were solved in your Rothrock book using the
larger spherical triangles? I can see that long-distance sailing
could perhaps need a side greater than 180 but the smaller triangle
could still be used and then subtract from 360. Also, the calculator
inverse trigonometric functions give principal values. How do you get
your equations to provide an output greater than 180? Or do you have
to do additional calculations? But if you do, this is also what you
would have to do using the smaller triangle in the first place. Also,
how many additional cases do you have to deal with using the larger
angles and sides? With me, it is a matter of economy.

As I mentioned in my last post, I didn't use the book to solve your
problem. The Rothrock book probably doesn't use large spherical
triangles--I don't expect any textbook to. In practical problems it
probably comes up about 0% of the time. The only reason I provided a
second solution was to be mathematically complete. I obtained the
extra solution by writing down two answers every time I did an arcsin
or arccos. If y = sin(x) then y = sin(pi-x). Or, if y = cos(x), then
y = cos(-x). I didn't use anything like a "quadrant rule" which would
eliminate answers that I considered valid. There was an inverse trig
function for each of the three unknowns, plus another doubling of
introduced by the solution of a binomial equation, so I produced 16
potential solutions, which I then validated by plugging in to the sin
and cos rules. Only two of them were valid, and one of those was the
angle/side greater than 180 degrees solution. I admit that my methods
were crude, but they were chosen to minimize sources of human error.

The textbook manufacturers would be very happy if all trigonometry
texts were to be re-written!

I said this not because they didn't solve large triangles, but because
the procedures are optimized for someone using pencil and paper. They
were written before the age of the calculator.

Now that we have calculators and computers, what do people use
spherical trig for? For problems involving the surface of the earth,
I don't assume that the earth is a sphere--I have an HP48 program that
calculates geodesics on the WGS-84 ellipsoid. I also have a spherical
harmonic model of the earth's magnetic field, which allows me to
calculate magnetic variation at any point on the surface of the earth.
The combination of these two programs allows me to calculate the
magnetic course between any two points on the earth. I also have a
database which allows me to retrieve by name the latitude and
longitude of any airport, airway intersection, NDB, and VOR in New
England. All of this is in user RPL, by the way.

What formula did you use to calculate C in the original problem I
posed? There was a slight discrepancy in our answers and I would like
to know if you used a different formula.

I used only two formulas:

1) sin(a)/sin(A) = sin(b)/sin(B) = sin(c)/sin(C)
2) cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(C)
cos(b) = ...
cos(a) = ...

I started by solving for b using the sin rule. I then wrote

sin(c) = sin(a)/Sin(A) sin(C)

I then squared both sides of this formula and squared both sides of
the cos rule and added the two results together. This gives
sin^2(c)+cos^2(c) on the left side, reducing to 1. The right side is
a quadratic equation in cos(C), which I solved using the binomial
formula. Finally I solved for c using the cos rule. I did all the
arithmetic using Mathematica, keeping 50 decimal places precision. I
then rounded at the end to the number of digits I used in my post. I
therefore expect that my answers are correct to the number of digits
given. If you have a different answer in the last one or two of your
digits, then I expect that it was caused by accumulated round off
error.




--
Scott Hemphill hemphill@xxxxxxxxxxxxxxxxxx
"This isn't flying. This is falling, with style." -- Buzz Lightyear
.

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