Question

I am studying celestial navigation and have come across various ways to solve the navigational triangle.

Formulas, tables etc

My knowledge is as yet too fragile to be able to evaluate different methods or formulas so here are a couple I have come across if someone else could have a look at them for me-------------------------------Damon!!

Distance and bearing between 2 points on earth.

Distance=60.ACOS(sin(latA)x sin(latB)+cos(latA)x cos(latB) x cos(longB - LongA))

Bearing=ATAN sin(longB-LongA) DIVIDED BY

sin(latA)xcos(longB-longA)-TAN(latB)xcos(latA)

Hope you can follow that.

Mike

Answer 1

Distance=60.ACOS(sin(latA)x sin(latB)+cos(latA)x cos(latB) x cos(longB - LongA))

Remember the trig identity
cos (x-y) = sin x sin y + cos x cos y
that means
cos (Lat A -LatB)= sin Lat A sin Lat B + cos Lat A cos Lat B)
It looks like that was used here

Answer 2

Do you mean

Distance=60.ACOS( {sin(latA)x sin(latB)+cos(latA)x cos(latB)} x cos(longB - LongA))

If so that is the first stab I took at the later question that I replied to before I saw this one. I believe it would only work at latitudes near the equator where the longB-LongA which is the LHA is close to the distance/60. That is the error I caught myself in.

If you mean it as posted, without the brackets, then I do not know if it is right or not. I have not figured out if that would be like my corrected formula.

Answer 3

Yes, it is correct without the brackets