(15C) Haversine Navigation

[Thomas Klemm]

(11-06-2022 05:14 PM)C.Ret Wrote:  If I understand all this correctly, the final multiplication by 60 gives the displacement in nautical miles.

This is correct.

From Nautical mile:

Quote:Historically, it was defined as the meridian arc length corresponding to one minute (\(\frac{1}{60}\) of a degree) of latitude.

While not exactly 1852m it is probably close enough.



Even though I've never used this calculator the program is quite readable.
Both →REC and →POL appear to use infix notation which is rather unusual.

Here's what I came up with for the HP-48:

Code:

\<< \-> \Gh1 \Gl1 \Gh2 \Gl2
    \<< 1 \Gl2 \Gl1 - 0 ACOS \Gh2 -
        SPHERE \->V3 RECT V\->
        \-> u v w
        \<< w  u \->V2
            SPHERE V\-> \Gh1 + \->V2
            RECT V\-> v SWAP \->V3
            SPHERE V\-> RECT
        \>>
    \>> ROT DROP 60 *
\>>

Since the latitude is measured from the north-pole we have to calculate the complement \(c = 90^{\circ} - \theta_2\).
And then the transformation between spherical and rectangular coordinates seems a bit awkward.
Thus I'm not super happy with it.