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Ultra compact sight reduction

Jun 24, 2015
A sight reduction using the Doniol haversine method (latitude and declination contrary in name).  L=latitude; d=declination; hv=naural haversine; LHA=local hour angle; P=product; ZD=zenith distance; Hc=calculated altitude; PD=polar distance; Z=azimuth (LHA > 180° N. Lat.).

A sight reduction using the Doniol haversine method (latitude and declination contrary in name). L=latitude; d=declination; hv=naural haversine; LHA=local hour angle; P=product; ZD=zenith distance; Hc=calculated altitude; PD=polar distance; Z=azimuth (LHA > 180° N. Lat.).

Pat Rossi/Navigator Publishing

Recently, Hanno Ix, a retired engineer and a member of the online NavList navigation forum (www.fer3.com/arc/), presented a challenge to the forum for the development of a manageable means of sight reduction without the use of logs, slide rules or calculators and, if possible, also eliminating special rules and ambiguities. 

The group took up the challenge and we began with the classic Law of Cosines navigational triangle formula:

Sin h = Sin L Sin d + Cos L Cos d Cos t

This formula requires three time-consuming, error-prone multiplication steps. The next effort centered on the Davis haversine/cosine formula:

hv ZD = hv (L ~ d) + hv (t) Cos L Cos d  
(L~d is the same as L +/- d — L - d when latitude and declination are the same name and L+ d when contrary name.)

This formula was almost manageable with two multiplication steps but is still classified as time consuming and error prone.

The breakthrough came with the discovery of the haversine/cosine Doniol formula in the 1977 Volume No. 1 of Bowditch. The beauty of this formula caught my eye as I was leafing through the pages. Very close to perfect, with just a single multiplication step and a few special case rules:

Sin h = n - (n+m)a

n = Cos (L - d) m = Cos (L + d) – same name
n = Cos (L + d) m = Cos (L - d) – contrary name
a = hv (t)

The haversine/cosine Doniol formula was rewritten by forum member Hanno Ix to all haversines that have only one multiplication step and no special rules. Beauty was looking me square in the eye:

Hanno Ix’s azimuth graph. Step 1: Mark dec. on right scale, draw horiz. line. Step 2: Mark LHA on top scale, draw vert. line, it intersects horiz. line at “A.” Step 3: Mark alt. on left scale, draw horiz. line, it intersects curve “C” at “B.” Step 4: draw vert. line through pt. “B” and line intersects bottom axis to indicate azimuth: 44.

Pat Rossi/Navigator Publishing

hv ZD = n + [ 1 - ( n + m)  ] ( a )

n = hv (L - d)   m = hv (L + d)   same name
n = hv (L + d)  m =  hv (L - d)   contrary name
a = hv (t)

ZD = Zenith Distance
Hc = 90 - ZD
L = Latitude
d = Declination
t = Meridian Angle
hv = Haversine

Azimuth is solved with Hanno Ix’s fantastic azimuth graph. I developed an “index card” version of the process see diagram. 

Here’s a summary of benefits for the all haversine Doniol sight reduction longhand method:

1. Full coverage of latitudes and declinations
2. Uses DR as the assumed position 
3. No special rules
4. Ultra compact
5. Quick solution time
6. 1’ precision
7. Not reliant on electronics or mechanical devices. 

Greg Rudzinski is a retired merchant mariner who lives aboard a Bruce Roberts Offshore 38 ketch at Channel Islands Harbor, Oxnard, Calif. 

Edit Module

Sep 15, 2017 05:56 am
 Posted by  Willi Strohl

Dear Greg,
I found this soution accidentally surfing through the net, whe I was looking for a more simple solution of the navigational triangle formula.

I think your solution is ingenious, especially in combination with the azimuth graph. You can solve it within a few minutes, only a (very small) table of halversines necessary. No calculator required. - Very brilliant!

I have one question:
If you assign a negative sign to the declination, when "south" and a positive sign, when "north" (as is common practice), then in my opinion you get along with only one equation:

hv(Zd) = hv(L-d) + (1 - hv(L-d) - hv(L+d)) * hv(t)

No distinction beween "Same name" or "contrary name" is necessary, the pre-sign automatically takes care of it. Do I see this correctly or is there something wrong?

Best regards
Willi Strohl
Beilstein, Germany

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