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Re: Another "emergency navigation" sight reduction method
From: Gary LaPook
Date: 2015 Jul 9, 00:44 +0000

An easy way to check the computation on a Bygrave is to do the same computation on a calculator since this allows you to check the intermediate steps.

Just use the standard Bygrave formulas in the three step process following along on the form I have posted.

First calculate co-latitude and save it in a memory in the calculator. If you are using a value for hour angle that is not a whole number of degrees you might want to make the conversion to decimal degrees and save it in a memory since it will we used twice. If you are using whole degrees then this step is not necessary.

Then you calculate "W" using the formula:

tan W = tan D / cos H

and sum it to the memory where you have saved co-latitude which is then X and then make any adjustment necessary to convert X to Y. (If you are just making trials you can avoid this step by your choice of the trial values.) There is no reason to store W itself since it is not used again. You can then convert W to degree and minute format to compare with the Bygrave derived value.

Then you compute azimuth angle using the formula:

tan Az = (cos W / cos Y ) x tan H.

If you want you can also convert Az to degree and minute format to compare with the Bygrave.

The last step is to calculate altitude with the formula:

tan Hc = cos Az x tan Y.

Then convert to degree and minute format to compare with the Bygrave result.

(When entering values in the format of degrees minutes seconds, change decimal minutes to seconds, 6 seconds per tenth of a minute, in your head  before punching in the assumed latitude, declination  and hour angle if necessary.)

Using whole degrees for declination, assumed latitude and hour angle, using a TI-30 with only 3 memory locations the key strokes are:

---------------------------------------------------------------------

(co-latitude = 90 - Assimed latitude)

90
-
Assumed Lat
=
STO 1  (co-latitude stored in memory 1)

---------------------------------------

(tan W = tan D / cos H)

Declination

tan
/
H
cos
=
inv
tan    (computed W)
SUM 1  (X now stored in memory 1)(change X to Y if necessary)

--------------------------------------

(tan Az = (cos W / cos Y ) x tan H)

cos    (of W from prior step)

RCL 1  (recalls Y from memory 1)
cos
x
H
tan
=
inv
tan (computed Azimuth angle)

------------------------------------

(tan Hc = cos Az x tan Y)

cos    (of Az from prior step)

x
RCL 1  (recalls Y from memory 1)
tan
=
inv
tan  (computed altitude, Hc)

2nd
D.D - DMS (changes Hc in decimal degrees to degrees, minutes and seconds)

DONE

From: Bob Goethe <NoReply_Goethe@fer3.com>
To: garylapook@pacbell.net
Sent: Wednesday, July 8, 2015 3:26 PM
Subject: [NavList] Re: Another "emergency navigation" sight reduction method

Greg,
This PDF was just the sort of thing I was looking for, with a concise statement of the formulae for Bygrave.  That said, my first attempt at using the equations came up with nothing like what I got from Pub. 249.  I would appreciate you (or perhaps others) helping me sort out where I am making my errors.
First, I recast the equations slightly to make them fit the way my mind works.
tan(x) = tan(d) ÷ cos(LHA)  I recast as
1. tan(d) ÷ cos(t) = tan(x)
...where d = declination of the GP, and
t = meridian angle
I inserted a line for myself of
2.  lat ~ X = Y
...where I am looking for the difference between the latitude of the AP (i.e. "lat") and the latitude of point X.  This difference I designate as Y.

3.  (tan(t) * cos(X)) ÷ cos(Y) = tan(Az);
I inserted another line:
4.  180 - Az = Z
...where Z is the azimuth angle to the GP (e.g. N 100° W).  From that, I can derive Zn, the azimuth (e.g. 260°)
Finally, where Ronald uses "a" in his final equation, I designate that as Hc.  Hence:
5.  cos(Az) * tan(Y) = tan(Hc);
I *think* I have wrapped my head around the equations...except that I come up with answers that don't match with Pub. 249.  So either I have misunderstood the equations, or I am not using my calculator properly.
Here is the scenario I used.
GP of the sun:   GHA = 48° 03.6'            Declination = N 11° 44.6'
AP = 53° N    114° 03.6' W
Hence, the meridian angle = t = 66°
1.  tan(11.743°) ÷ cos(66°) = tan(X)
0.208 ÷ 0.407 = 0.511
X = arctan(0.511) = 27.1°
2.  53° ~ 27.1° = Y
Y = 25.9°
3. (tan(66°) * cos(27.1°)) ÷ cos(25.9°) = tan(Az);
(2.25 * 0.890) ÷ 0.899 = tan(Az)
2.00 ÷ .899 = 2.22
Az = arctan(2.22) = 65.8°
4.  180 - 65.8 = Z
Z = N 114° E
5. cos(65.8°) * tan(25.9°) = tan(Hc);
0.410 * 0.485 = 0.199
arctan(0.199) = 11.2°
Now, when I do this with Pub. 249, I get an azimuth angle "Z" of N 103° E and an Hc of 23° 44'.
Clearly, I am doing some things very wrong.  Do you observe what they might be?
Thank you for giving thought to this.
Bob

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