I'm not sure what you mean by 7 min. delay. Please explain where it occurs.

Edit: Oh wait! Are you referring to my post #47 above? Well, I made a big mistake when I posted that. I was trying to use the Sun Almanac to find Latitude. BUT, what the SUN ALMANAC is for is locating your LONGITUDE. So if thats what you meant by 7 min....disregard that post.

Only two bodies come to mind when trying to obtain LATITUDE, those would be, POLARIS, and the, SUN at noon. Trying to get the other 56 navigational stars or moon at meridian passage would require additional pgms such as starrynight or starcalc. Meridian passage provides an east to west line of position (LOP) which intersects the computed vertical Longitude, giving us a fix....but, you know all that and I'm preaching to the choir :)


Naw, you shouldn't have to...just use the above GMT conversion chart. It'll work for every location in the world.

-...-

Tnx for the mod, minimax.

No - allways I was looking for the longitude - maybe therefore I didn't understand some of your postings (;) or :oops: )

(And I still work on a sextant-simulation to get the latitude by the Solaris azimuth.)

My tests show time-delays from 0 to 7 minutes between N.A. and SH4 time at different positions and times for sunrise .


I will make some screenies for explanation tonight.

Don could you explain your Time to Arc conversion. When i do that the 7 min's become:

15 deg/60 min = 0.25 deg / min ~ 1 deg = 4 min

7*0.25 = 1,75 ~ 1 deg 45' (60*0,75)

what am i doing wrong here...

I don't understand that too!

360° / 24h = 15° per h = 0.25° per minute = 0°15' per minute
--> per minute, sunrise is late, we're 0°15' more west.

1 minute delay in SR = 15' east-west difference
7 minutes delay = 7*15' =105' = 1°45' east-west difference

How many miles are 1°45' longitude-difference at 30°N ?
Or otherwise: how long ist 1° longitude at 30°N ?

It seems, Don used a formula... (I don't have one in my hands or brain :( )
11nm?

But 7 minutes means too:
the 11nm for 30°N will be increasing, if we move up to the equator. Am I right?


Don, please explain!

Holy Socks!!!! :o

You're not going to believe what I'm about to show you guys...

But first, let me set this up...


Reminder: I still have my sub icon exposed for testing purposes and comparisons. (Pretend it's not there for this example.)

Date: 11 April 1942.
Time: 0807 (showing on the sub's clock. Time zone (+11))
GMT: 1907
Estimated position: 30N 160E Time zone (-11).
Course: 095° True
Speed: 11 knots

I was too slow and just missed getting sunrise so I decided to use the Observation scope as my sextant, measure the sun's altitude and complete a sight reduction to get a line of position (LOP), using the sun's center and the horizon as the angle of interest.

In the following picture the sun's center is about 2.1° above the horizon:

http://i77.photobucket.com/albums/j4...SUNtestSH4.jpg


In the next pic, using a real life navigation program, "Navigator", I plugged in the data I just explained and it coughed out the solution that the sun's LOP was 54.6 nm from my estimated position on an Azimuth of 81°.

Well...

take a look at the following pic...it says it all!!! It shows my True position and the Celestial Navigation solution to be just ONE (1) nautical mile difference!!!! :rock:

http://i77.photobucket.com/albums/j4...LOPtestSH4.jpg

I don't think I can get any closer than that.

very interesting!

Did you get simular results for other times and locations?

And:
Do you use the free shareware - navigator light? Where did you get the data for 1942? My navigation software (nautic tools) starts with data for 1950.

edit:


sorry:, I see: you're a registered user.

next question:
how many feet/inches/yards/whatever did you calculate as eyelevel?

Height of Eye = 20 feet.

Here its later in the day.

Note: I couldn't get a decent noon sight using the Observation scope due to bobbing action and no horizon to gauge by. So, I waited until later in the day.

Distance = (Speed knots x Time in minutes) / 60

It's 620 minutes after my last sun sight 620 x 11 / 60 = 113.6 nm. I advance my morning position 113.6 nm,This gave me a new Assumed Position of 30°N x 163°E took another sun sight and measured it at 26° 40' altitude......see pic.

http://i77.photobucket.com/albums/j4...eed/LOPSH4.jpg

The Celnav puts me about 2 nm from the sub's icon and about 4 nm from my DR position.

Thanks Don!

So you get your longitude by th sun - did you get your latitude too?

(sun at highnoon = 180°?)

Getting Noon sight with the Observation scope might be difficult. I've not had much luck so far while surfaced. I've got to try it a PD for more stablization. A lot more testing is in order, it seems.

I'm not sure of the elevation limits of the Obs scope. I think it's 90°, but not sure.

Do you know any kind of almanac or software, we can use for the whole community with data usable for 1942 ?
(we can't copy the N.A....)

edit:

Maybe we can use Solaris, but we need free data - and therefore we can't use the 2007 - Data :(

The YF just got home and hankerin' for a dine-out...so, til later..

...and yes, I've got some decent freeware for the war years somewhere on the HD...I'll get back.

Edit: http://www.relex.ru/~zalex/files_eng.htm Starcalc is freeware also.

Disappointment, dang.

Searched everywhere and all I can come up with is freeware beginning 1950.

The only freeware I have that works is "STARCALC".

In order for the average-joe or jill to get LAN is to have a method of measuring the Sun's altitude and a copy of the Sun's Declination (DEC) tables.

For those who don't know there are 3 scenarios for determining LOCAL APPARENT NOON (LAN):

A) 90° - sun's altitude = x,

1) LAT = DEC - x (If the Sun is further away from the equator than your boat. (Both you and Sun in same hemisphere.))

2) LAT = DEC + x (If the Sun is between your boat and the equator. Both you and Sun in same hemisphere.))

3) LAT = x - DEC (If you and Sun are in opposite hemispheres.)

I know of no other way to navigate without using math of some kind.

Using POLARIS to obtain LAT, one would definitely need an Almanac for the war years (3 pages of tables)
LAT = Apparent altitude (corrected for refraction) - 1° + a0 + a1 + a2
...and to get a fix using 3 or more stars.

Its hard cheese for the noob to find out that there's more to nav than pushing a "where-am-I" button on a Garmin. It should give them pause and added respect for those who've gone before.

I don't see any easy way around the math envolved....in actuality, its no harder than balancing one's check book or following a bus schedule.

http://www.nautictools.de/index.html

here you get a free nautical application with data before 1950 - only in unknown accuracy - but it's only in german.

Here : http://www.stargazing.net/astropc/ you can get another free application, but i'm not shure, wether we can use it for nautical solitions.


Another problem:

how do you read an angle using the observation scope, without seeing the horizon?
If we can't read angles above 15°, we will need a sextant.


edit: just found your old posting, Dan:

btw:

I will try this weekend to get the longitude by measure two times the sun declination to get noon - time. And will test the accuracy of the 'nautic tools' (nautical appl., link above).

Edit:

same with latitude using the observation scope down from 90°...

On this site:

http://ftp.linux.org.uk/pub/Navigation/

There is this list of navigational tools - one of those might work

http://ftp.linux.org.uk/pub/Navigation/DIRECTORY

They are dos tools but run just fine in XP