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Guy Schwarz wrote-

I was working out a problem here are the particluars
DR 40 deg 9 min N
        45 deg W
object Regulus
Dec 11deg 59. 9 min N
LHA 16:25:48
HC 58 Deg 20.4 Min
Using HO 249 the Z is 150 and the ZN is 210 I also checked with the
star section and the ZN was the same.
When I did the calculations using
Cos rasied to -1 (Zn) = sin (dec) - Sin (lat) x sin (Hc) / cos (Hc) x
cos (lat)

The answer came out to 148 deg 11.4 min which is close to the 150 Z
however I though that formula gave me the Zn?


New question, what is the equalvalent function in excel to the Shift
Sin-1
and the Shift Cos-1
Thank you.
Guy


      Hc and Zn by calculation using Casio fx 260 calculator
    Sin-1Hc = sin(dec) x sin(lat) + cos(dec) x cos(lat) x cos(LHA)
     Cos-1 (Zn)= sin(dec) - sin(lat) x sin(Hc) / cos(Hc) x cos(lat)
  (I can't paste in Guy's quoted tables here, presumably being in
HTML.)

========================
Reply from George.

Guy Schwarz is making somewhat heavy weather of his trig. He isn't
alone; there are many whose trig has gone rusty since childhood, or
who managed to evade it at school, only to discover possible uses in
later life.

But Guy is tackling it rather well. He quotes two formulae, as

Sin-1Hc = sin(dec) x sin(lat) + cos(dec) x cos(lat) x cos(LHA)
Cos-1 (Zn)= sin(dec) - sin(lat) x sin(Hc) / cos(Hc) x cos(lat)

but he is stating these wrong. They should be, more simply-

Sin(Hc) = sin(dec) x sin(lat) + cos(dec) x cos(lat) x cos(LHA)
Cos(Zn)= sin(dec) - sin(lat) x sin(Hc) / cos(Hc) x cos(lat)

That is, each formula gives, not the angle required, but its sin or
cos. Then you have to convert that sin or cos into the angle , Hc or
Zn, that is sought.

Doing that final step like we (well, I) used to do at school, using
trig tables, which give for each angle its sin, cos, etc., was very
simple. You had to look in the body of the table for the sin (say),
that you were after, a number between 0 and 1. And then see what was
the corresponding angle (in degrees) at the edges of that table, that
gave rise to it. Just using the same table, but backwards way round to
provide the "inverse sin", in degrees, of that number. Instead of
finding the sin of a particular arc (or angle, in degrees), you were
finding the arc from its sin.

Nowadays, to do the same trick, a scientific calculator has a special
function for the purpose, which it labels (or should label) arcsin,
perhaps shortened to asin or even asn, for finding the arc from its
sin. Unfortunately, another notation, which confuses many, is common,
in which inverse sin of x is written as sin (to the power of -1) x. I
have to write it down like that because I can't do superscripts in
email ASCII, but you will know what I mean. That is easily confused
with a similar expression for sin x (to the power of -1), which
correctly means the reciprocal of  sin x , or 1/sin(x).  In my view
the use of sin (to the power of -1) x is a misleading notation that
should be avoided, but readers will come across it, so they need to
know what it means.

So another way to express the same formulae written above is-

Hc = arcsin [ (dec) x sin(lat) + cos(dec) x cos(lat) x cos(LHA) ]
Zn = arcsin [(dec) - sin(lat) x sin(Hc) / cos(Hc) x cos(lat) ]

=======================

Here's an unfortunate complication. In the early days of computers,
many mathematical programs expected angles to be expressed in radians
rather than in degrees. There are 2 pi radians in a round turn of 360
degrees, so a radian was 180 / pi degrees, or 57 and a bit degrees.
This was convenient for the programmers, and for many mathematical
applications, but damned awkward for everyone else. Nevertheless, that
was how sin, cos, tan, and other trig functions were defined for
computers, and that has continued ever since. To get the sin of an
angle (in degrees) on a computer, you must first turn it into radians,
by multiplying by pi / 180, and then use the sin function. Similarly,
to get the arcsine of a number n, which must be between -1 and +1 as
usual, you use asn (n), then convert that answer in radians into
degrees by multiplying it by 180 / pi. It doesn't affect the
mathematics, just the units of measurement, but it has to be got
right.

Many scientific calculators, on the other hand, were better designed,
for easier use, and often there's a switch or mode-change, to allow
all trig functions to be expressed either in degrees or in radians. I
would always choose degrees, for navigational jobs. That same choice
may be available for certain computer applications.

There's a trick you can often apply, by defining your own functions in
a computer program, such as Basic, including the version of Basic used
in Excel. Say you want to use sin, but in degrees rather than in
radians. Your home-brewed function for doing that job can't be called
"sin", because that name has already been allocated to the internal
sin function working  in radians. A sensible name might be "sind", to
imply that it's working in degrees. Then you have to define it such
that sind(x) is sin(x *pi /180). Then you can express x in degrees.
Similarly for arc sin x, where you can define  asnd(x)  as
asn(x)*180/pi, which will provide an end-relult in degrees. And so on
for all the other trig functions you might need. It only has to be
done once, then those functions can always be available to you. Then
you can use those modified functions just like using a calculator in
degrees-mode.

===========================

Guy quotes the Local Hour Angle of Regulus as 16:25:48, presumably in
degrees, minutes, and seconds. Not sure why, when other angles are in
decimal minutes. But not to worry.

I have checked over Guy's calculations, and they seem perfectly valid,
and to give sensible answers, exactly as quoted at Hc =  58deg 20.4',
Az = 148 deg 11.4'. But an azimuth calculated from its cos in that way
just gives the angle as measured from true North, either clockwise or
anticlockwise. From the cosine, there's just no way of telling,
because the cos of a negative angle is exactly the same as that of a
positive angle. It's an ambiguity, and you have to use some
commonsense with it. If the local hour angle is between 0 and 180, (as
in this case it is, at 16 and a bit degrees) it means that the body
has passed your meridian, and is now descending, so it is somewhere to
your West, wherever you may be on the Earth. In which case, its
azimuth can't possibly be 148 deg 11.4', which would be somewhere to
your Eastwards. Then it must be the corresponding negative angle,
measured anticlockwise from North through 148 deg 11.4', which is
exactly the same as if you measured it clockwise through (360 less
148deg 11.4'), therefore an azimuth of 211deg 48.6'. Does that make
sense?

As for discrepancies with HO 249, I can't comment, not having those
tables.

There's a better way, in my opinion, of doing the job of getting the
azimuth, using the tan function, or even using the special "atan2" or
"pol" functions available with some calculators or computing programs.
But that's another story, that's been dealt with on occasions in the
Nav-l archives, but could have another airing if anyone shows
interest.

George

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

   

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