# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Noonsite using artificial horizon? Did I do it correctly?**

**From:**Peter Hakel

**Date:**2014 Aug 26, 07:13 -0700

Sketching a cartoon also helps (I attach my own humble attempt at “computer art” :-) ). Sight reduction procedures often recommend drawing pictures to visualize the relationships between GHA, LHA, and the various angles within the navigation triangle. We can do the same for meridian passages - after all, those are just a special case, in which this spherical triangle collapses into a single arc.

Peter Hakel

Peter Hakel

**From:**Gary LaPook <NoReply_LaPook@fer3.com>

**To:**pmh099@yahoo.com

**Sent:**

**Subject:**[NavList] Re: Noonsite using artificial horizon? Did I do it correctly?

So for everybody who's latitude is greater than 23° 26' (either north or south) the first method will always apply,

"this produces ZD, zenith distance

then add the declination (if of the same name as latitude or subtract if not)

the result is your latitude at noon."

If your latitude is less then the second method is used some of the time,

"In this situation you subtract the ZD from the sun's declination to determine you latitude."

If your latitude is 5°(either north or south) then 162 days per year; 10°, 132 days; 15°, 102 days; 20°, 60 days.

Gary LaPook

**From:**Gary LaPook <NoReply_LaPook@fer3.com>

**To:**garylapook---.net

**Sent:**Tuesday, August 26, 2014 4:26 AM

**Subject:**[NavList] Re: Noonsite using artificial horizon? Did I do it correctly?

I need to add a clarification.

I wrote:

"this produces ZD, zenith distance

then add the declination (if of the same name as latitude or subtract if not)

the result is your latitude at noon."

This is correct for the common case when you are looking in the direction of
the equator at noon. In
other words, your latitude is of opposite name than the sun's declination OR your latitude is of the same name and is greater than the declination of the sun. In the less common case, which can only occur in the tropics, the sun's declination is of the same name and greater than your latitude and you are looking away from the direction of the equator at noon. In this situation you subtract the
ZD from the sun's declination to determine you latitude.

Gary LaPook

**From:**Gary LaPook <NoReply_LaPook@fer3.com>

**To:**garylapook---.net

**Sent:**Monday, August 25, 2014 9:46 PM

**Subject:**[NavList] Re: Noonsite using artificial horizon? Did I do it correctly?

There are a number of problems with your computation.

You used the wrong method but, even if you were doing it correctly, there is a large math error.

How did you get from 59° 45' to 60° 39'?

59° 45' -1' -4' = 59° 40 not 60° 39'.

Even if you did this subtraction correctly you must have made some other errors, apparently a large error in the measure altitude.

The way to do the calculation correctly is

Measured altitude

+ index correction

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

then divide by 2

which gives you Hs

then subtract refraction

which then gives you Ho (you don't apply SD or dip when you superimpose the images in an AH)

subtract Ho from 90° (it's easier if you write 90° as 89° 60.0')

this produces ZD,
zenith distance

then add the declination (if of the same name as latitude or subtract if not)

the result is your latitude at noon.

Let's do this using your numbers

119° 30'

IC +4' (IE with reversed sign)

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

119° 34'

119° 34' /2 = 59° 47'

59° 47'

ref - 1'

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

59° 46'

89° 60.0'

- 59° 46.0'

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

30° 14.0'

+10° 57.3

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

41 11.3' as the measured latitude so you are in error by 33'.

We can work this backwards to see what you should have measured in the AH

latitude 41° 43.5'

decl. - 10° 57.3'

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

30° 46.2'

89° 60.0'

- 30° 46.2'

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

59° 13.8'

ref + 1.0'

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

59°
14.8'

X 2 118° 29.6'

- IE 4.0'

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

118° 25.6'

you measured 119° 30' so were in error by 1° 4.4'

gl

**From:**Samuel L <NoReply_SamuelL@fer3.com>

**To:**garylapook---.net

**Sent:**Monday, August 25, 2014 4:16 PM

**Subject:**[NavList] Noonsite using artificial horizon? Did I do it correctly?

I just got interested, again, in
Celestial Navigation and bought a Celestaire Astra III-B

I did a noon site yesterday using a pan of water for an artificial horizon and need your help and advice in letting me know if I did it correctly. The result obained for Latitude got me within 4' 10" of our actual location.

In performing the Noon Site I "brought the Sun down" until it was completely superimposed on the Sun reflected from the artificial horizon.

Noon in my Longitude was 17:04:59 GMT

Here are the figures;

Hs using articial horizon- 119 deg 30' (I forgot to read the tenths on the vernier!)

I divided the Hs reading by 2 to get;

Hs- 59 deg 45'

Ie- -4'

Refraction- +1

Ho- 60 deg 39'

Declination -10 deg 57.3"

.....what follows is where I think I either made a mistake or didn't make a mistake....

Sun Semidiameter from the Almanac was 15'.8.....BUT....I wasn't getting the "latitude I wanted" so I divided the SD by 2 and used 7.9'

So....SD used= 7.9'

Final Ho was 41 deg 47' 42"

My actual Latitude was 41 deg 43' 32"

Any ideas on whether I performed the Noon site correctly?

Any other suggestions?

Thank you