NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Determing the bearing off a celestial object.
From: Gary LaPook
Date: 2015 Apr 1, 07:38 +0000
From: Bruce J. Pennino <NoReply_Pennino@fer3.com>
To: garylapook@pacbell.net
Sent: Tuesday, March 31, 2015 10:26 AM
Subject: [NavList] Re: Determing the bearing off a celestial object.
Bruce
From: Gary LaPook
Date: 2015 Apr 1, 07:38 +0000
Or you can use the Bygrave formulas since they calculate azimuth prior to the calculation of altitude.
CHECKING YOUR COMPUTATIONS
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
-----------------------------------------------------
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
-----------------------------------------------------
See:
gl
From: Bruce J. Pennino <NoReply_Pennino@fer3.com>
To: garylapook@pacbell.net
Sent: Tuesday, March 31, 2015 10:26 AM
Subject: [NavList] Re: Determing the bearing off a celestial object.
Hello All:
Greg, you raised a point which I never before considered. At any time,
how do you calculate the true bearing of the sun , or any celestial body,
without H? So I went to my trusty “Celestial Navigation in the GPS Age” and
found my answer. Maybe?
Conceptually and visually it seems clear, time is known, your
location is known and GP can be quickly found. I would solve the
triangle using
Tan A = cos LHA/(cosLxtand – sinLxcosLHA)
where A is true bearing, LHA of sun at the time of observation, d is
declination of sun (or moon), L is latitude of observer. Is this correct? An
easier way? When I try to make a sketch of the triangle something seems
incorrect. Is A the true bearing or more correctly I must make a sketch and
allow for quadrant......gotta think about this more.
I’ll try this later today. Thanks in advance.
Bruce
