Author

Topic: Lunars, how sensitive for the heights and how to find heights.

navi

posted December 06, 2017 06:31 PM
Data: UTC 03h43m32s 6 December 2017 Lat 41d 52.7' N lon 87d 37.8' W (Chicago)
Measuring the distance between Betelguese in Orion to the far side of the moon I get sextant distance of 25d 6'
To get the heights I use 2102D and plot the moon at the time of the measurement > GHA moon= 20d 11.1' Dec 19d 45' Then SHA moon= GHA moon GHA aries = 20d 11.1'  131 d 0.3' = 110 d 49.2' adding 360 I get 249d 10.8' Then Rim scale on red disc = 360  249d 10.8' = 110 49.2' I plot the moon through the slot of the red disc at Dec 19d 45'
I put on the blue template 45N and turn it to LHA aries = GHA aries  DR lon = 130d 60.3'87d 37.8' = 43d 22.5'
From that I then read height of moon = 26d height of Betelguese 36d
Using the tables in Starks book I get Ma 25d 7.3' Sa 36d 1.3'
Furtmermore using stark at UTC 3 I get LD 24d UTC 4 I get LD 24d 35.2'
Cleared distance I get 24d 20.4' which give me an UTC of 3h 34m 46s
That is 7m 46s slow.
That is 4.6' error in lunar distance measurment which I think is too high consideriong I use a metal sextant with a 6 magnification scope sitting stable on a chair on land.
Can the error come from me calculating the heights using the 2102D?
What is the alternative way of getting the heights using the nautical almanac?
From: Chi


Capt Steve Miller

posted December 07, 2017 10:55 AM
First of all the Star Finder is not accurate enough for determining the Hc of the Moon and the other body. You should do a sight reduction to compute the Hc. This can be done using the work form as usual dealing only with the Hc computation (ignoring the Hs to Hc computations). You would use all the normal inputs. You could also us a calculator/Computer program to Precompute the body's altitude for the time of your Lunar. This is the method that I use.
Betelgeuse is not a normal star to use for a Lunar  it is not on or near the ecliptic. I ran into the same problem with Sirius.
Your error in Lunar Distance is not due to the sextant you used.
Using 2102D is part of the problem, using Betelgeuse is another part due to the relative motions of the Moon and Betelgeuse.
You might have had some errors in the Stark solution. I am looking into this as well and will make another post in a while.
From: Starpath


navi

posted December 08, 2017 10:03 AM
I will post two pictures of Stark calculations. The first is based on heights using the graphical solution with the 2102D, the second using 249.
From: Chi


navi

posted December 08, 2017 10:09 AM
This calculation is done with heights using sight reduction with 249.
The result with this is WORSE than using the 2102D!
You say Betelgeuse is a bad star for this. I get in the geoecentric LD 24d 00.0' at UTC3 Dec 6 24d 35.2' at UTC4 Dec 6. (Sextant distance is 25d 6.0' at 034332 Dec 6 (Sextant distance is 38d 35' at 034332 Dec 7 Wouldn't that suggest the movement is good?
Where do you think the error is from?!
From: Chi


Capt Steve Miller

posted December 08, 2017 12:11 PM
I have attached my Stark Form using the TI89 calculator and the StarPilot 89 software to determine the altitudes of the Moon (29*48.7') and Betelgeuse (37*04.3')
Note that I got the following: Ma 28* 57', Sa 36* 05.6', UTC 3 LD 24* 03.0', UTC 4 LD 24* 35.1', Cleared Distance 24* 26.6', UTC of 3h 44m 6s, Calculated Time 34s slow, 8.5' Longitude Error
From: Starpath


navi

posted December 08, 2017 01:11 PM
Hi Capten Miller,
The positive things is that my old french sextant must be pretty dang good, and my hands stable! 34 sec and 8.5 miles I am happy about! Shouldn't I be? Time to throw not only the GPS but also the chronometer overboard :) and due true celestials.
Negative is that I do not get same results as you in the calculations.
Please see below:
Note that I got the following: Ma 28* 57', I get 21d 18.6' (big difference) Sa 36* 05.6', I get 36d 5.7' (small difference) UTC 3 LD 24* 03.0', (I get 24d 0.0' about 10% difference) UTC 4 LD 24* 35.1',(I get 24 35.2' (small difference)
Cleared Distance 24* 26.6', (I get 24d 13.3 big difference)
UTC of 3h 44m 6s, Calculated Time 34s slow, 8.5' Longitude Error
Can you please look at my calculated height for the moon (reduction form + how I used wwp and wwr corrections). Calculated heights before corrections are attached here and the wwp and wwr corrections are marked in red in the picture where I clear and get 24 13.3'.
From: Chi


navi

posted December 08, 2017 02:07 PM
Please see previous posting as well as these.
I went through it again and I see a few odd things.
First. K for 12d 18.9' in the book gives 1.93911 in your Excelyou get 1.93923 (In my book that is K for 12d 18.8' ) odd! Minor things but still. (Now I am very close for the geocentrics at 3 and 4 UTC. Within one tenth.)
Then I went through my height calculations again and I get calculated distances closer now (For my moon heights I used the wrong declinations in 249!). But still NOT the same as you!
Hc Betelgeuse 36d 44' ww ref 1.3' Sa= 36d 45.3' You have: Sa 37* 05.6'
Hc Moon 29d 48' wwp 56.9' wwr +1.7' Ma 29d 6.6'
You have: Ma 28* 57'
Why do we get different heights?
Please see updated calculation attached here.
From: Chi


David Burch

posted December 08, 2017 05:11 PM
For any given DR Lat LON, and UTC and body, there is a unique Hc and Zn value.
Please use starpath.com/calc or starpath.com/usno to determine the correct values. This should not be an issue in lunars.
From: Starpath, Seattle, WA


David Burch

posted December 08, 2017 05:21 PM
I forgot to mention this one, which we recommend in our online course. So if you are making an error in the use of the tables, this will show you right where the error is. Very nice.
_________________
There is an app for PCs written by Stan Klein for the USPS which you can download and install for various computed solutions. See:
http://www.usps.org/eddept/n/tools.htm
It also provides a unique function of telling you what the intermediate step answers should be using various sight reduction methods.
From: Starpath, Seattle, WA


navi

posted December 08, 2017 09:17 PM
Hello David,
(I agree the basics from the tables should not get wrong!) I get the LD geocentric at UTC 3 and UTC 4 right. Heights are the problem.
Now I used starpath.com/usno for the Hc's. In your form you denote the heights Hs and they are NOT the same as Hc I get from the usno. Should Hc be the same as Hs!? If not what is the difference between Hc and Hs?!
Please see attached picture.
Also you use HD 60.9 when I look in my almanac at UTC 03 to UTC 4 my HD are both 60.8.
From: Chi


navi

posted December 08, 2017 09:43 PM
Please also see previous reply.
So as mentioned in the previous post I get a discrepancy between usno and what you use. (or should Hc be different from Hs as I ask and what is then the difference.)
I once again went through the form and I get the SAME declinations as usno. But i get DIFFERENT Hc, both different from usno and from your Hs using 249. Please see attachment.
This is driving me slightly frustrated. Why would Dec be exactky the same but Hc not?!
From: Chi


David Burch

posted December 09, 2017 07:47 AM
there is just one right answer and the several links given will tell you which one is right. then compare your steps to those to see where the error is.
Use celestial tools to see what each step of the answer is using Pub 249, 229 or any other.
It is far more helpful to you if we guide you to the tools that show you how to find these errors, rather than just track them down individually. celestial tools is your friend.
From: Starpath, Seattle, WA


David Burch

posted December 09, 2017 07:53 AM
Also of course we must not confuse Hs (a number read from the sextant) with Hc that is a computed Height from known location and time.
From: Starpath, Seattle, WA


navi

posted December 09, 2017 08:35 AM
David,
I appreciate your guidance. I have been through the forms many times. I really try to find the errors myself.
I have a few very specific questions that would lead me forward I think:
In your Stark form you put in Hs which in THIS case must be calculated values since I did not made any measurement of the heights. Hence in THIS case is Hs= Hc, is that right or wrong*?
Then I see how you adjust Hs using wwr and wwp from Stark. (I get that part correct.) to get Sa, and Ma.
*What puzzles me is that Hc from usno are NOT the same as the Hs you use in your Stark form!
(I will also now do a sight reduction with NOA and see what I get.)
From: Chi


Capt Steve Miller

posted December 09, 2017 09:56 AM
I should have put Hc in place of Hs on the Stark Form  that is totally my fault. We are calculating Hc (not Hs) and then use the WWP (Wrong Way calculations) and Wwref to take that Hc and convert it to an Hs with corrections for the Lunar.
I did the calculations of the Hc with a few more different calculators and computer and got an Hc for the Moon as 29* 35.4' and for Betelgeuse 37* 01.3. The computed UTC was 3:43:22 with the time error being 20 sec and the Long Error being 2.5'. (I am not sure why the differences between all the results that I got between devices).
From: Starpath

