(32S II) Position, Distance & Bearing Calculations

[SlideRule]

An extract from Position, Distance & Bearing Calculations, Whitham D. Reeve (©2014 W. Reeve)

Two frequent calculations required in radio propagation work are distance and bearing between two radio terminals. Distance and bearing may be taken from topographical maps for quick analyses but great circle calculations are needed for more accurate work and are described here …
If more than a few sites are being evaluated, it can be more productive to use a programmable calculator or spreadsheet program to find distance and bearings. The program listing in the appendix was written for this purpose on the Hewlett-Packard HP-32S II calculator, but it can be adapted for other scientific calculators that use Reverse Polish Notation (RPN) with minor changes (if any). The program can accept the coordinates in either dd+mm+ss (for example, 66 52' 31.5") or dd.dddd (for example, 66.5407) by setting the input mode, and it automatically resolves any ambiguity.

Excellent documentation w equations, tables, figures, examples etc., …

BEST!
SlideRule

[Jlouis]

It's never too late to say thank you!

Indeed excellent documentation.

I'm always amazed by the excellents professionals and their eager to help!

Thanks again and

Cheers

JL

[Gil]

I just checked, with the given reference,
some of the control results in that paper.

Note that the results are not at all accurate
when reviewed with Vincenty's formulae (error <= 1 mm). :

Point A in deg.mmssss
{ 66.53507 162.35557 }

Point B in deg.mmssss
{ 66.50033 161.02032 }

Correct results:

Bearing forward 1 D.mmss¦ 264.513132284
Backward :2 D.mmss¦ 90: 83.2511696433
effective distance on geoid/Earth surface
[km]: 68.981609066,
or 42.8631846678_mi'
≠indicated 42.7 miles

Besides, Euclidean distance through the Earth (initial altitude 0¦end altitude 0): 68.9812747629 km.

See also, further down, the Post
"Geodesic distance & Earth Euclidean distance calculator, bearing calc", Ver06b.hp for HP50G calculator.

Regards,
Gil

[Thomas Klemm]

This program is a verbatim translation of those for the HP-42S or HP-15C from (15C) Haversine Navigation:

Code:

D01 LBL D
D02 x<>y
D03 R↓
D04 x<>y
D05 -
D06 x<>y
D07 R↑
D08 1
D09 θ,r→y,x
D10 R↑
D11 x<>y
D12 θ,r→y,x
D13 x<>y
D14 R↓
D15 x<>y
D16 y,x→θ,r
D17 R↓
D18 +
D19 R↑
D20 θ,r→y,x
D21 x<>y
D22 R↓
D23 y,x→θ,r
D24 R↑
D25 y,x→θ,r
D26 R↓
D27 60
D28 ×
D29 RTN

Example

Kotzebue: 66°53′50.7″N 162°35′55.7″W
Noorvik: 66°50′03.3″N 161°02′03.2″W

66.53507 →HR
-162.35557 →HR
66.50033 →HR
-161.02032 →HR
XEQ D

37.07535

x<>y
95.14735

The X-register contains the distance in (old) nautical miles, while the Y-register contains the bearing.

For a final result in kilometers, substitute 60 with 111.19.
For a final result in statute miles, substitute 60 with 69.09.

Thus you will get either 68.70681 km or 42.69227 statute miles.

Caveats

• coordinates are negative for west
• only the forward bearing is calculated
• switch the points to calculate the backward bearing
• bearings may be negative
• coordinates are expected to be in HR format
  

Advantages

• no registers are needed
• due to the use of polar/rectangular transformations the program is shorter
• for the same reason it avoids cancelation and handling special cases
  

---

Compared to Gil's program it assumes the Earth is a sphere.
Thus the results are probably not that accurate.

[Gil]

Regarding previous message.

It is written:

"For a final result in kilometers, substitute 60 with 111.19."
Correct should be: "with 111.12“.

And
"For a final result in statute miles, substitute 60 with 69.09."
Correct should be: "with 69.05 or 69.047".

Regards,
Gil

[Thomas Klemm]

(02-06-2023 01:24 PM)Gil Wrote:  Correct should be: (…)

I was just using the values from the original document:

Quote:The above expressions assume the arguments in all trigonometric functions and results are in degrees. If the
inverse cosine function returns a value in radians for the distance calculation, substitute the following
constants:

• For a final result in kilometers, substitute 111.19 with 6370
• For a final result in statute miles, substitute 69.09 with 3,959
  

The idea was to get the same results.

---

How did you come up with your numbers?

[Gil]

A1)
Radius Equator WGS84 (Ra) : 6378137 m
Radius Pole WGS84 (Rb) : 6356752.314245179497563966599633655156798171311085497338848571651283208522086835​1009394057094507133

A2)
Acccoring to
International Union of Geodesy and Geophysics,
Earth radius (2Ra+Rb) / 3 :
(about) 6371008.77141505983252 m
Mean Earth circumference = 2×pi× Mmean Earth radius
= (about) 40030228.7044 m for 360 degrees (circle)
=40030228.7044/360 m for 1 degree (longitude)
=40030228.7044/360/60 for 1/60 of degree (or 1' longitude)
=40030228.7044/360/60 = about 1853.25132891 m for 1' (longitude)

B)
Depending on where you are located, this value (of about 1853...) of "your" nautical mile might vary (for 1' longitudel
—> by convention, nautical mile defined later as = exactly 1852.000000 m.

C)
1 nautical mile —> exactly 1.852 km
60 nautical mile —> 1.852 × 60 = exactly 111.12 km

D)
1 mile —> exactly 1760 yards
1 yard —> exactly .9144 m
Therefore, 1 mile —> 1760 × 0.9144 = exactly 1609.344 m, or 1.609344 km

E)
1 nautical mile —> exactly to 1852/1609.344 mile
= about 1. 15077944802354251173148810944086534637715740077944802354251173148810944086534637​71574007794480235 mile

F)
60 nautical miles —> 60 × 1. 15077944802354251173148810944086534637715740077944802354251173148810944086534637​71574007794480235
= about 69. 04676688141255070388928656645192078262944404676688141255070388928656645192078262​94440467668814100 miles
= about 69.05 or 69.047 miles

Regards,
Gil