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Jeff Gottfred wrote:

>
>2. Given the following commonly used formulae for computing Z, which is
>the best one to use and why?
>
>      sin Z = (sin t * cos d) / cos Hc
>
>or
>
>      cos Z = [sin d - (sin L * sin Hc)] / (cos L * cos Hc)

I use the formula (as part of an altitude sight reduction program):

       cos Z = -(sin Dec - sin Hc * sin Lat)/cos Hc/cos Lat)*sign(sin GHA)
			
		IF Z<0 THEN Z=Z+360	

This gives an azimuth  0First let me say that I am not a sailor (except dinghies, which don't
>count!), my interest in navigation began when I got a pilot's license
>years ago, and today I am primarily focused on the historical techniques
>used by folks like David Thompson, Alexander MacKenzie, Peter Fidler,
>Lewis & Clark, &c.
>
>You give me my Astra IIIb, a good flat piece of glass,

What do you do with a good flat piece of glass? Use it as an artificial
horizon perhaps? IMHO far too unreliable!

> either a good
>watch (or a lousy watch and a set of lunar distance tables),

A set of lunar distance tables? These are not published anymore as part of
the Nautical Almanac since 1908! I will believe that you are able to
calculate the Most Probable Position from  a single altitude observation,
but to derive the time from a Lunar Distance is a quite different matter,
far, far more complicated.

 a current
>almanac, and a set of log trig tables, and I could map the continent,
>(or, presumably, take you anywhere in the world if someone else drives
>the boat.) Yet, I have never taken a formal (celestial) navigation
>course.
>
For mapping quite different techniques were employed (triangulation), using
different instruments.

Regards,
Jan


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Jan van Puffelen wrote:

>I use the formula (as part of an altitude sight reduction program):
>
>       cos Z = -(sin Dec - sin Hc * sin Lat)/cos Hc/cos Lat)*sign(sin
>GHA)
>			
>		IF Z<0 THEN Z=Z+360	
>
>This gives an azimuth  0What do you do with a good flat piece of glass? Use it as an artificial
>horizon perhaps?

Actually I use it to cover my artificial horizon (a bowl of water) so
that it doesn't blow in the wind...

>A set of lunar distance tables? These are not published anymore as part
>of the Nautical Almanac since 1908! I will believe that you are able to
>calculate the Most Probable Position from  a single altitude
>observation, but to derive the time from a Lunar Distance is a quite
>different matter, far, far more complicated.

Yes, that's why I am quite pleased with getting 10 statute mile accuracy
in longitude using a lunar distance formula of my own computation. I
have since located some formulae for clearing the distance that date to
1786 up to about 1880-- I expect to do much better with them.
(Of course, I do lunars just for fun, and at demonstrations at
historical re-enactments, &c) Remember, lunars are a direct method for
longitude -- I use transits for latitude (200 years ago the noon shot was
the staple of the navigator's diet).

As for the lunar distance tables, you are quite right, they have not
been produced for over 80 years. I generate them myself using the data
in the almanac and a computer program that I wrote...



>For mapping quite different techniques were employed (triangulation),
>using different instruments.

Well, yes and no. The early explorers (circa 1780-1820) basically used
their sextants. David Thompson personally surveyed over one and a half
million square miles of this country using a sextant manufactured by
Dolland. His accuracy is astounding, seeing as how he only had a lousy
watch and a set of lunar distance tables.

The guys with the theodolites and chains came along about 100 years
later....

Cheers!

Jeff,

Calgary.


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Subject: Re: Almanac Question in JN
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Jeff Gottfred (gottfred@agt.net) wrote:

>I believe that "recipe" methods are inherently dangerous. This is
>because any new twist to the problem can leave you uncertain of your
>answer. Please consider the following questions:
>
>1. Are the rules for determining Zn the same using pub 229 and using a
>hand calculator?

The answer is, "It depends." There are two common conventions for
assigning algebraic signs to Latitude and Declination:

        (1) North is + and South is - for both Latitude and Declination.
            This method is taught in USPS JN 89/92.
        (2) Latitude is always +. Declination is + if Latitude and
            Declination have the same name (both North or both South);
            otherwise it is -. This method is taught in USPS N 93.

If you use (1), the answer is "No." If you use (2), the answer is "Yes."

>2. Given the following commonly used formulae for computing Z, which is
>the best one to use and why?
>
>      sin Z = (sin t * cos d) / cos Hc
>
>or
>
>      cos Z = [sin d - (sin L * sin Hc)] / (cos L * cos Hc)

If you use the latter, the rules for converting Z to Zn are the same as
for Ho 229. If you use the former, you have to worry about the sign of
t and use different rules for converting Z to Zn. This is just a surmise;
correct me if I am wrong.

>3. What is the following formula?
>
>      For sperical triangle XYZ,
>      cos X = (cos x - cos y * cos z) / (sin y * sin z)

This is a formula for finding an angle (X) of an oblique spherical
triangle given the three sides (x, y, and z) in degrees, where x is
the side opposite the angle X. It is derived from the Law of Cosines
for oblique spherical triangles. See the Chemical Rubber Company (CRC)
Standard Mathematical Tables or Bowditch (1995) Section 2115 or Bowditch
Volume II (1981) Section 142.

>4. Use the formula in #3 above to derive:
>      a) a formula for computing Hc

Let X = LHA (or t), x = Co-Hc, y = Co-Latitude, z = Co-Declination,
where Co-Hc = 90 - Hc, etc.. Observe that cos(90-Hc) = sin(Hc),
cos(90-Lat) = sin(Lat), and sin(90-Dec) = cos(Dec). Then with a
little algebra, you get

        sin Hc = (cos LHA * cos Lat * cos Dec) + (sin Lat * sin Dec)

Find sin Hc, then take the arcsine to get Hc. By happy circumstance,
either LHA or t will give the same answer.

>      b) a formula for computing Z

Let X = Z, x = Co-Declination, y = Co-Latitude, z = Co-Hc and proceed
as before to get

        cos Z = (sin Dec - (sin Lat * sin Hc)) / (cos Lat * cos Hc)

>      c) a formula for computing the great circle distance between any
>two points on the surface of the earth.

Observe that if the starting point is L1, Lo1 and the destination point
is L2, Lo2, then the starting point corresponds to your DR or AP, and
the destination point corresponds to the GP of the body. Co-Hc corresponds
to the great circle distance between these two points. Use the above
formulas with Lat = L1, t = Difference in Longitude = (Lo1 - Lo2),
Dec = L2. Zn corresponds to the initial course angle. To visualize
this relationship, draw the Navigational Triangle and the corresponding
Great Circle Triangle.

>I would argue that if the JN course (which I am not familiar with) does
>not prepare one to answer these questions, then either the course should
>changed,

Sorry, I have to disagree with Jeff on this one. As a teacher of JN and N
for my local squadron, my observation is that only a relatively small
percentage of the JN students I see have enough math background to be
able to handle the above. All of them, however, can learn to be
competent celestial navigators without going into that much of the
mathematical details. It is important that they have a good understanding
of the Navigational Triangle and know at least two methods of solving it,
but they don't need to know how the methods were derived.

> or one should do the extra study required oneself.

I find it rewarding to do so, but to "demand" that others should do so
is not reasonable. People take JN for different reasons. Some just want
to learn to get from point A to point B. To impose such stringent
requirements on JN students would drive away most of them. I submit
that we should try to encourage more people to study celestial
navigation, not discourage them. Let's not be mathematical elitists.
It is hard enough to convince people that celestial navigation is not an
anachronism in this day and age when a GPS can be had for $250 (US),
which is considerably less than the price of a metal sextant.

Having said that, I complement Jeff Gottfred on having the ability and
discipline to work out his own Lunar Distance tables and methods. Jeff,
perhaps you would be willing to share with the rest of us more about the
Lunar Distance method and how you use it.

------------
Chuck Taylor
Everett, WA
ctaylor@eskimo.com

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Chuck Taylor (ctaylor@eskimo.com) has written an excellent response to
the three questions that I posed-- Thanks for responding Chuck!

I have only a few minor comments to add.

>>2. Given the following commonly used formulae for computing Z, which
>>is the best one to use and why?
>>
>>      sin Z = (sin t * cos d) / cos Hc
>>
>>or
>>
>>      cos Z = [sin d - (sin L * sin Hc)] / (cos L * cos Hc)

>If you use the latter, the rules for converting Z to Zn are the same as
>for Ho 229. If you use the former, you have to worry about the sign of
>t and use different rules for converting Z to Zn. This is just a
>surmise;
>correct me if I am wrong.

Close enough. The answer I was looking for was that function #1, derived
from the law of sines, has a quadrant problem, whereas the function #2,
derived form the law of cosines, does not. I.e. the sine of x is equal
to the sine of 180-x, but the cosine of x is not equal to the cosine of
180-x. This means that you are less likely ot make a mistake computing
Zn from Z if you use formula #2.

>>3. What is the following formula?
>>
>>      For sperical triangle XYZ,
>>      cos X = (cos x - cos y * cos z) / (sin y * sin z)
>
>This is a formula for finding an angle (X) of an oblique spherical
>triangle given the three sides (x, y, and z) in degrees, where x is
>the side opposite the angle X. It is derived from the Law of Cosines
>for oblique spherical triangles. See the Chemical Rubber Company (CRC)
>Standard Mathematical Tables or Bowditch (1995) Section 2115 or
>Bowditch Volume II (1981) Section 142.

Actually this is the law of cosines for spherical triangles.
For those who are keen and haven't seen it, I will provide the
derivation of this function from plane trig in a separate message.

>my observation is that only a relatively small
>percentage of the JN students I see have enough math background to be
>able to handle the above.

Yes, I agree that a course my not be able to cover this in depth, but I
would hope that it would instill in the student the idea that the course
does not teach it all... learning never ceases... As an aside, the math
required is taught in grade 12-- remember the infamous "unit circle".
But, I also remember struggling with it, and I am sure that the JN
mandate is not remedial math...

>I find it rewarding to do so, but to "demand" that others should do so
>is not reasonable.

I quite agree. By no means would I "demand" that others learn. But I
would, perhaps, suggest it as good survival training. (I might also add
that a survival situation is not the time to pull out a lifeboat sextant
and start reading the instructions on the side... If the techniques are
not deeply ingrained, then fatigue, panic, &c. can make learning
impossible.)

>perhaps you would be willing to share with the rest of us more about
>the Lunar Distance method and how you use it.

Gee! I thought you'd never ask! I'll send out a sepearte message on the
very subject!

Once again, thanks for your response Chuck!

Cheers!

Jeff.

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Woops.  There is a mistake in the date part of the GHA Aries and sun GHA and
declination calculation forms that I posted a week or two ago.  I slipped up.
The year 2000 is a leap year, so an extra day of 86,400 seconds has to be
inserted after February 29.  I modified the form and a new copy is attached.
If you find any other mistakes, let me know.

For someone who wants to put the formulas a program, there is an easy way to
convert customary time and date to centuries before and after 1200 1 Jan 2000.
It comes from Van Flandern and Pulkkinen, Low Precision Formulae for Planetary
Positions, The Astrophysical Journal Supplement Series, vol 41, p 391, (1979).
It works from March 1900 through February 2100.  They also have another
formula for longer lengths of time.

T = (367*yr-trunc(7*(yr+trunc((mo+9)/12))/4)
     +trunc(275*mo/9)+day+(hr+(min+sec/60)/60)/24
      -730531.5)/36525

The function trunc is just like trunc in Excel.  It drops the fractional part
of the number.
  trunc(8.9) = 8,
  trunc(-8.9) = -8.

Bill Murdoch
Kingsport, Tenn.

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Step 1  Working with the Greenwich time and date, calculate the number of=
 centuries before (-) or after (+) 1200 1 Jan 2000.

Year                    Month
1997 -94,737,600        Jan  0           Jul 15,638,400
1998 -63,201,600        Feb  2,678,400   Aug 18,316,800
1999 -31,665,600        Mar  5,097,600   Sep 20,995,200
2000    -129,600        Apr  7,776,000   Oct 23,587,200
2001  31,579,200        May 10,368,000   Nov 26,265,600
2002  63,115,200        Jun 13,046,400   Dec 28,857,600
2003  94,651,200

Number for the year     ___,___,___      year ____
Number for the month     __,___,___     month ____
Day ___ x 86,400 =3D        _,___,___
Hour ___ x 3,600 =3D           __,___
Minute ___ x 60 =3D             _,___
Second ___                       __
                             __,___for dates in March 2000 and later add =
86,400
Total seconds           ___,___,___ / 3,155,760,000 =3D
Centuries        __.___ ___ ___ ___ =3D T

Step 2  Calculate the GHA of Aries, GHAA, in revolutions.  Remove the int=
eger part of the number leaving the decimal part.  It can be either posit=
ive or negative.  If it is negative, add 1 to it to make it positive.  Mu=
ltiply the revolutions by 360 to convert it to decimal degrees and if des=
ired to degrees and minutes.

GHAA =3D 0.7790573 + 36,625.0021390T + 0.0000011T^2
       - 0.0000122sin(125.0 - 1934.1T) - 0.0000009sin(200.9 + 72,001.7T) =
=3D
              ____._______ rev.
                _0._______ rev (decimal part)
                 0._______ rev (if negative, decimal part + 1)
                  ___.____ =B0     ___ =B0 __._ '

Step 3  Calculate the Sun's mean anomaly.

g =3D 357.528 + 35,999.051T
    ____.____=B0

Step 4  Calculate the ecliptic longitude of the Sun.

EL =3D 280.466 + 36,000.771T + 1.915sin(g) + 0.020sin(2g)
     ____.____=B0

Step 5  Calculate the obliquity of the ecliptic.

e =3D 23.440 + 0.015T
    __.____=B0

Step 6  Check to see if cos(EL) is positive _  or negative _.  Calculate =
the GHA Sun.  Add 180=B0 to the value if the cosine of EL is negative.  A=
dd or subtract multiples of 360=B0 until the answer is between 0 and 360=
=B0

GHA Sun =3D GHAA - tan-1(cos(e)tan(EL)) (+ 180=B0 if cos(EL) is negative)
          ___.____=B0     ___ =B0 __._ '

Step 7  Calculate the Declination of the Sun.  Negative values are south;=
 positive are north.

Dec Sun =3D sin-1(sin(e)sin(EL))
          ___.____=B0     _ __ =B0 __._ '



Step 1  Working with the Greenwich time and date, calculate the number of=
 centuries before (-) or after (+) 1200 1 Jan 2000.

Year                    Month
1997 -94,737,600        Jan  0           Jul 15,638,400
1998 -63,201,600        Feb  2,678,400   Aug 18,316,800
1999 -31,665,600        Mar  5,097,600   Sep 20,995,200
2000    -129,600        Apr  7,776,000   Oct 23,587,200
2001  31,579,200        May 10,368,000   Nov 26,265,600
2002  63,115,200        Jun 13,046,400   Dec 28,857,600
2003  94,651,200

Number for the year     -94,737,600      year 1997
Number for the month     28,857,600     month Dec
Day ___ x 86,400 =3D        _,864,000
Hour ___ x 3,600 =3D           28,800
Minute ___ x 60 =3D             _,___
Second ___                       __
                             __,___for dates in March 2000 and later add =
86,400
Total seconds           -64,987,200 / 3,155,760,000 =3D
Centuries        -0.020 593 201 =3D T

Step 2  Calculate the GHA of Aries, GHAA, in revolutions.  Remove the int=
eger part of the number leaving the decimal part.  It can be either posit=
ive or negative.  If it is negative, add 1 to it to make it positive.  Mu=
ltiply the revolutions by 360 to convert it to decimal degrees and if des=
ired to degrees and minutes.

GHAA =3D 0.7790573 + 36,625.0021390T + 0.0000011T^2
       - 0.0000122sin(125.0 - 1934.1T) - 0.0000009sin(200.9 + 72,001.7T) =
=3D
              -753.4469769 rev.
                -0.4469769 rev (decimal part)
                 0.5530231 rev (if negative, decimal part + 1)
                  199.0883 =B0     199 =B0  05.3 '

Step 3  Calculate the Sun's mean anomaly.

g =3D 357.528 + 35,999.051T
    -383.8077=B0

Step 4  Calculate the ecliptic longitude of the Sun.

EL =3D 280.466 + 36,000.771T + 1.915sin(g) + 0.020sin(2g)
     _461.6929=B0

Step 5  Calculate the obliquity of the ecliptic.

e =3D 23.440 + 0.015T
    23.4397=B0

Step 6  Check to see if cos(EL) is positive _  or negative _.  Calculate =
the GHA Sun.  Add 180=B0 to the value if the cosine of EL is negative.  A=
dd or subtract multiples of 360=B0 until the answer is between 0 and 360=
=B0

GHA Sun =3D GHAA - tan-1(cos(e)tan(EL)) (+ 180=B0 if cos(EL) is negative)
          301.8001=B0     301 =B0 48.0 '

Step 7  Calculate the Declination of the Sun.  Negative values are south;=
 positive are north.

Dec Sun =3D sin-1(sin(e)sin(EL))
          -22.9252=B0     S 22 =B0 55.5 '



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