Solution to yesterdays problem
I think!
I am just beginning to study navigation & on top of that do not have a clue about how things should be expressed mathematically and on top of that have difficulty typing formulas on the computer - they are obviously not designed to do this.
Am I correct in thinking in a triangle the angle should be expressed as a capital letter and the side in small case?
However to the problem:
LHA 194.65 degrees (at North Pole (P))
Latitude of Z 51.74 degrees
Latitude of X 39.24 degrees
LHA = Local hour angle (P)
Z = Observers position
X = Geographical position of a star.
I understand conventionally the LHA is expressed westerly from the observers position to the position of the celestial object. LHA is the difference in longitude between the 2 positions expressed as an angle at the elevated pole. (N pole in this case)
The latitudes given need adjusting in order to create the navigational triangle:
Z 90 - 51.74 = 38.26 degrees
X 90 - 39.24 = 50.76 degrees
So we have a spherical triangle we need to solve.
The formula I think I have understood - at last is:
cosc = cosa x cosb + sina x sinb x cosC
To suit the terminology PZX used in navigation I have adjusted this to:
cosp = cosx x cosz + sinx x sinz x cosP
or
cosp = .6325 x .7852 + .7745 x .6192 x cos (-0.9674)
cosp = .4966 + .4795 x (-0.9674)
cosp = .4966 - 0.4638
cosp = .0328
= 88.12 degrees
x 60 nautical miles in 1 degree
= 5287.22 nautical miles between the 2 points
Please can anyone confirm this
Mike
Yes, that is fine.
It took me forever to confirm the formulas you are using, but they are correct.
I then dug my old books out of the attic and they are in them.
There are by the way programs that you can use to confirm your answers on the web.
Use google to search for "great circle" and navigation and you will finds them.
I did not have to correct the latitudes by doing complements. The equation works as is.
cos^-1[sin 39.24 sin 51.74 + cos 39.24 cos 51.74 cos 194.65]
= cos^-1[.633*.785 + .775* .619 * -.967]
= cos^-1[.497-.464]
= cos^-1[.033]
= 88.1 deg
If you put that problem into a web program with 0 for one longitude and 194.65 for the other of course it will not work.
However putting one point at 100 west and the other at 94.65 east will work.
To solve yesterday's problem, you used a spherical triangle and the formula:
cosp = cosx x cosz + sinx x sinz x cosP
where:
cosp is the cosine of the angle between the two points X and Z,
x is the latitude of point X,
z is the latitude of point Z,
P is the local hour angle at the North Pole.
You substituted the values:
cosx = 0.6325,
cosz = 0.7852,
sinx = 0.7745,
sinz = 0.6192,
cosP = -0.9674.
Using these values, you calculated cosp as follows:
cosp = 0.6325 x 0.7852 + 0.7745 x 0.6192 x cos(-0.9674)
= 0.4966 - 0.4638
= 0.0328.
To convert the angle from decimal degrees to nautical miles, you multiplied it by 60 nautical miles in 1 degree:
88.12 degrees x 60 nautical miles/degree = 5287.2 nautical miles.
Therefore, the distance between the two points X and Z is approximately 5287.2 nautical miles.