View Full Version : Celestial Navigation with RFB / TMO mods?
Wildcat
06-09-17, 08:30 AM
Does anyone have some hints on how to do this and get it working right?
Thread links / download links would be appreciated.
propbeanie
06-09-17, 10:03 AM
I haven't kept track of them, and the "newest" I could find says for v1.3 on it's release thread:
http://www.subsim.com/radioroom/showthread.php?p=607248#post607248
but for v1.4 on its download post:
http://www.subsim.com/radioroom/downloads.php?do=file&id=525
... if they're the same thing... There's probably others that would be easier to incorporate, since if you read closely, may87th mentions messing with the menu_1024_768.ini file, which always overly-complicates things if you're trying to "merge" mods together...
I'm sure there's others, but... :salute:
Rockin Robbins
06-09-17, 10:41 AM
To make it work you have to cheat somehow, because there are no in-game tools to make it happen. The sky is wrong. Your clock is wrong. You don't have a sextant. In order to make an external program work you have to feed your position into the astronomy program to get the position with the "sextant" which is just chasing your own tail. Not only that, but the earth in SH4 is a cylinder, not a sphere or oblate spheroid. How are you going to get real celestial navigation stuff to fix your position on a cylindrical world? It can't.
SH4 is not a planetarium or a celestial navigation simulator. Any attempt to make it one kills the functionality of the game.
Wildcat
06-10-17, 10:01 AM
That sucks, but oh well I guess.
Is anyone out there doing the cheat method using stellarium?
While I suppose it's sort-of pointless to do it, it's still an interesting exercise in using the sextant or star/sunsighting I suppose.
Does anybody know if the star placements are correct, even if the times are not?
I am a big fan of polynesian navigation techniques and have completed several long distance navigations in FSX and other sims using nothing but star-courses. I would really love to try this in SH4, but the spacing between the stars must be correct. The timing does not matter much, but if the stars are in the wrong places of the sky relative to eachother it is useless.
propbeanie
06-10-17, 11:05 AM
I haven't a clue if this will give you any satisfaction:
http://www.subsim.com/radioroom/showthread.php?t=117216
and
http://www.subsim.com/radioroom/showthread.php?t=116170
Some of the posts overlap what I listed above. Some of the folks back then seemed to think that the stars were "accurate" in SH4. I have no idea what their idea of "accurate" is... The moon isn't, the sun isn't, time isn't, the world is "flat" on a cylinder, but ya never know until you try 'em. I'd make a "spare" copy of the game to experiment in though...
vdr1981
06-10-17, 11:15 AM
SH4 is not a planetarium or a celestial navigation simulator. Any attempt to make it one kills the functionality of the game.
Any SH game should be realistic simulation of submarine operations during the WW2 period. I really don't see how could real navigation mod (optional if you like) in any way "kill the functionality" of the game ? :hmmm:
To be honest, the main reason why I went over from SH3/4 to SH5 is "GPS device" on WW2 subs...Seems pretty unconvincing and game braking to me.
I understand that real celestial navigation mod is not possible in SH3/4 but if there was one, it certainly wouldn't hurt...
Cheers!
Rockin Robbins
06-11-17, 01:50 PM
Any SH game should be realistic simulation of submarine operations during the WW2 period. I really don't see how could real navigation mod (optional if you like) in any way "kill the functionality" of the game ? :hmmm:
To be honest, the main reason why I went over from SH3/4 to SH5 is "GPS device" on WW2 subs...Seems pretty unconvincing and game braking to me.
I understand that real celestial navigation mod is not possible in SH3/4 but if there was one, it certainly wouldn't hurt...
Cheers!
Because of the way that the nav map is designed you give up all targeting and tracking with the nav map in order to get celestial navigation. Not only that but you have to feed where you are (and that's what you're trying to work out) to your external planetarium program in order to work the azimuth and altitude of the object you want to use to yield......the same thing you had to feed the astronomy program with in order for it to work.
Not only that but the Earth in Silent Hunter 4 is a cylinder. However all geometry used in celestial navigation is spherical. Could there be some conflict here with wrong results? NAW!!!
No, in order for celestial navigation to work in SH3/SH4 the entire mapping system would have to be reworked, the sky would have to be useful and accurate, you'd have to write a sextant and chronometer into the game, nautical almanacs, proper charts and tools, a rework of the plotting board, radar and sonar would all be involved.
I don't know about SH5 and don't want to know, but I'll bet not much is truly different. The game is meant to be tactical simulation, Not navigational simulation or plumbing simulation or sonar simulation or radar simulation, or even TDC simulation. There is a long way to go in diesel-electric submarine simulation and I'm not sure anybody really wants to go there.
Bubblehead1980
06-11-17, 05:32 PM
Yes, I wish celestial navigation was a "realism" option and properly simulated along with many other things in SH 4 but it is NOT and likely never will be. Having said that, SH 4 with the proper mods is a great sim. My hope is we get to the point where someone with the skill and knowledge can deal with hard codded issues. Look how far this sim has come in 10 years and is very much alive, no telling what is next.
BigWalleye
06-12-17, 06:31 AM
I don't know about SH5 and don't want to know, but I'll bet not much is truly different.
That's a pity, because SH5, with TWoS, really gets it right. The navigator - not the captain - takes 5 celestial fixes a day, conditions permitting. (He will do additional sights if you tell him to.) He works them up - takes about half an hour - and marks the boat's calculated position on the nav chart. Depending on his skill rating, this mark may be close to the boat's true position. Or not. :) Unless you're in sight of a recognizable land feature, you don't know. He also does DR estimates periodically. These take less time and don't require a horizon, but the errors are greater and compound over time. Kind of like the way it happens on a real boat.
Of course, if the player has doubts about the Nav's ability and wants to take a cel fix himself, then SH5 becomes probably no better than SH3 or SH4. But hopefully, that's a pretty unlikely scenario anyway.
The game is meant to be tactical simulation, Not navigational simulation or plumbing simulation or sonar simulation or radar simulation, or even TDC simulation.
Someone who gets it! Amen, brother. Amen.
vdr1981
06-12-17, 09:29 AM
Of course, if the player has doubts about the Nav's ability and wants to take a cel fix himself, then SH5 becomes probably no better than SH3 or SH4.
I think you're wrong because, according to people who actually know how to do real nav stuff, SH5 has accurate celestial sphere (surprisingly, developers did at least something right)...
There's even a tutorial by sjizzle...http://www.subsim.com/radioroom/showthread.php?t=206103
BigWalleye
06-12-17, 10:30 AM
I think you're wrong because, according to people who actually know how to do real nav stuff, SH5 has accurate celestial sphere (surprisingly, developers did at least something right)...
There's even a tutorial by sjizzle...http://www.subsim.com/radioroom/showthread.php?t=206103
I stand corrected. I have never tried to do cel nav in SH5 or any Silent Hunter game. Not the captain's job. For me, doing "the real nav stuff" is OK when I'm really on a boat, but not appropriate when playing a tac simulation game. YMMV.
I may get yelled at again by Mr know it all but so be it.
In real life the Earth is a bumpy oblate spheroid, in SH4 the Earth is a cylinder.
The first thing I do each patrol is draw vertical lines 7.5 degrees on each side of 180d, the International Date Line. Next I draw vertical lines every 15 degrees from these first two vertical lines; going East and West, every 15 degrees is a vertical line.
Because every 15 degrees is one hour it is easy to find out what time it is at my location. If I am based at Pearl, that is Base Time and is the time of day displayed in my Boat the entire patrol.
Once I cross one of the vertical lines I drew every 15 degrees, I subtract one hour from Base Time to find out what time it is at my location. By the time I reach Empire waters, about four hours need to be subtracted from Base Time to find out what time it is at my location. The line at Western Japan is about 5 hours less than Base Time.
There are charts that list Sunrise, Sunset, Moon-rise and Moon-set, plus the Moon phases by date. I edited in these three pages into my in-game F1 Help file.
Since this is a game on cylinder your distance North & South of the equator don't matter that much for Sunrise, Sunset, etc.
I find that knowing what the 'rough' time is at my location found by subtracting one hour every 15 degrees heading West from Pearl, and the date, can be used with my in-game celestial charts to find out; will it be a Full Moon, Quarter Moon, or New Moon etc. Knowing the time tells me what time the Sun and Moon rise and set. This knowledge helps me plan Special Missions, and plan my attacks to my advantage.
So, I know what time it is at my location by subtraction and what time the Sun & Moon will rise and set using the charts.
Of course if I was headed East instead, I would add an hour every 15 degrees.
Happy Hunting!
propbeanie
06-12-17, 02:16 PM
Have you found then Art, that the SH4 game is rather "accurate" at the Moon and Sun phases and "times"? What I'm thinking of is if we drew those vertical lines on the game's Nav Chart (would it be easier to put that in the Campaign_LOC.mis file??), using that Blue Chalk font, the player could do the math themselves, using your little Help File add-in?... We'd have "aanker's Sun and Moon Phases Assistance Mod"... ??
"Do you have trouble attacking those Convoys? Is a dead-calm sea the least of your worries? Do the Momi dance like the Kaibokan on your head? Does it seem like the moon is always full when you're trying to do an attack, and you're seen from 9200 yards out? Well, load this puppy up, follow Art's directions, and never more have to find out at 0100 hours that it ain't a cool time to attack! Download now, and we'll throw in a free set of Ginsu knives!"
Have you found then Art, that the SH4 game is rather "accurate" at the Moon and Sun phases and "times"? What I'm thinking of is if we drew those vertical lines on the game's Nav Chart (would it be easier to put that in the Campaign_LOC.mis file??), using that Blue Chalk font, the player could do the math themselves, using your little Help File add-in?... We'd have "aanker's Sun and Moon Phases Assistance Mod"... ??
"Do you have trouble attacking those Convoys? Is a dead-calm sea the least of your worries? Do the Momi dance like the Kaibokan on your head? Does it seem like the moon is always full when you're trying to do an attack, and you're seen from 9200 yards out? Well, load this puppy up, follow Art's directions, and never more have to find out at 0100 hours that it ain't a cool time to attack! Download now, and we'll throw in a free set of Ginsu knives!"
Yes, the Moon phases are accurate; Full Moon, Gibbous, Quarter, etc and Moon-rise, Moon-set are also reasonably accurate using Local time, not Base time found by the 015d equals one hour method. It makes deciding to do a night surface attack and other tasks easier to resolve. Same with Sunrise - Sunset. I don't mind dead calm seas if the Sun is going to set soon.
If the chart says New Moon there is no moon, and despite how good the Japanese lookouts are, they have a harder time in the dark.
I borrowed - stole the Tables from a mod that someone put together called, 'Combat Information Center'
chapter 14: moon phases 1941 to 1945
dates and times of moon phases in the war years.
chapter 15: moon nautical almanac, rise and set times for the moon, during the war year 1941 to 1945
chapter 16: sun nautical almanac, rise and set times for the sun, during the war year 1941 to 1945
and is the only part of the mod I used (I hope the author greyrider is not disappointed ; ) greyrider posted this one mod and I never saw him post again.
Happy Hunting!
propbeanie
06-12-17, 09:17 PM
So, what mod are you usually playing, aanker? I'll see if I can get a straight line to draw in a MIS file, and send you something...
I have modded my own game, fixing and fine tuning many things. The Spyron mod I like to use won't be compatible with FOTRS due to circumstances beyond my control.
If you send me the F1 in-game Help file, it would be easy to add the three Tabs (pages) from greyrider's mod to it.... or you could easily add them yourself - it is a copy paste operation and renumbering the 'chapters' and paragraphs in them with find replace.
Some of the SH4 Help file isn't helpful so I combined some of it and managed to free up 3 pages for the Celestial/Almanac data. That work wasn't necessary but I didn't want Tabs scrolling down the F1 page.
Happy Hunting!
propbeanie
06-13-17, 02:24 PM
...
Some of the SH4 Help file isn't helpful so I combined some of it and managed to free up 3 pages for the Celestial/Almanac data. That work wasn't necessary but I didn't want Tabs scrolling down the F1 page.
Happy Hunting!
:o:o:o Heresey! ~Hare-ess-SEA~!!! :har: - About the only times I've used the Help is when I accidentally hit the F1 key, and it then it takes me a while to figure out how to close it... :lol: - Your idea there is a good one. Actually have a use for hitting that F1 key... :yeah:
Bubblehead1980
06-13-17, 04:49 PM
I have modded my own game, fixing and fine tuning many things. The Spyron mod I like to use won't be compatible with FOTRS due to circumstances beyond my control.
If you send me the F1 in-game Help file, it would be easy to add the three Tabs (pages) from greyrider's mod to it.... or you could easily add them yourself - it is a copy paste operation and renumbering the 'chapters' and paragraphs in them with find replace.
Some of the SH4 Help file isn't helpful so I combined some of it and managed to free up 3 pages for the Celestial/Almanac data. That work wasn't necessary but I didn't want Tabs scrolling down the F1 page.
Happy Hunting!
Would you release this to work with TMO RSRD? This would be a huge help.
propbeanie
06-13-17, 06:00 PM
I'm trying to get his fingers screwed down to the table, but he keeps squirming away... :lol: I haven't gotten into the Help portion yet to see what he did to that, but if aanker doesn't have the time, I'll be trying it, and will see how different FotRSU is from TMO / RSRDC for that portion. That'll be the hard part. The drawing lines is just tedious - unless those can go in the menu_1024_768.ini also, then they become difficult... :har:
Would you release this to work with TMO RSRD? This would be a huge help.
I'll see what TMO & RSRD did with their Help files if anything and attempt to add the three pages of Almanac data. Or, I could send you my Help file. That way you could see what the data looks like before I go through the work of adding and editing them to be compatible with TMO & RSRD. Let me know.
Happy Hunting!
propbeanie
06-13-17, 07:55 PM
I can either open my Google Drive for an upload, or I'll take a link from you, and do the editing, unless you want to dig into that part of it. PM my PM and they'll do lunch... Unfortunately, keltos01 and I filled up my DropBox, and I can't figure out how to get space back... :D some of this new-fangled computer stuff does me in... :har:
I missed this moon/sun rise/set from sh1; find this more of a tactical decision making item, than how long the stupid ship sinks. :doh:
where are these charts?
thanks
:Kaleun_Cheers:
(mod? :06: : [need a smiley looking at watch])
propbeanie
06-24-17, 04:28 PM
They're not directly in the game... That's what we're working on, is a substitute for it for a few mods... It will still involve the user having to do some math :o
check:subsim:
(was looking for and found greywolves download) :yeah:
..... I haven't gotten into the Help portion ...... I'll be trying it, and will see how different FotRSU is from TMO / RSRDC for that portion. That'll be the hard part. The drawing lines is just tedious - unless those can go in the menu_1024_768.ini also, then they become difficult...
Open my Help F1 in-game (maybe in Windowed mode) it is much easier, and edit a copy using a text editor with the FOTRSU Shortcuts and other info you want changed. Then replace mine with your edited Help.cfg file.
Only six or seven lines need to be drawn in a Pearl Pacflt patrol, not tedious at all. From Pearl time it's 4 hours difference from base time to Eastern Japan and 5 hours to Western Japan if that helps put it in perspective.
Draw vertical lines 7.5 degrees E & W of 180d, then every 15 degrees E & W from those 1st two lines. Subtract 1 hour heading W every time you cross a line and add 1 hour heading E. 15d = one hour.
I don't know what keyboard Shortcut commands you are using in FOTRSU. I don't use TMO's or RSRD's keyboard Shortcut commands. You would want to change those in the Help file.
I missed this moon/sun rise/set from sh1; find this more of a tactical decision making item, than how long the stupid ship sinks.
where are these charts?
I missed SH1's calendar's moon/sun rise/set too.
I got them years ago from greyrider's CIC 'Combat Information Center' mod although they are easily found on the internet.
If you PM me your email I can send you my F1 Help.cfg file but I use mostly Stock commands so you would need to edit your shortcuts page as I described above.
so last night, got distracted by tv and was brought back to the game with a 'warship sighted'. there in the rising first-q moon was an escorted convoy; [on 12th patrol, July 2, 1943 ~22:00 and this is my first convoy, let alone an escorted convoy]. pissed off, decided this was a good spot to experiment.
so just reading off chart in notepad, I got the following: July 2, 1943 - new moon 12:44 ; July 10, 1943 - f-quarter 16:29. (:hmmm:I figured this is saying, over the 8 days, the moon goes from new moon to f-quarter, starting at 12:44 and ending at 16:29. but then, the game only has four phases.)
then pulled out for july 2nd and 3rd:
2- mr: 05:00, ms: 19:03, sr: 05:02, ss: 19:05
3- mr: 05:49, ms: 19:49, sr: 05:03, ss: 19:05
(:oops: oops, I was actually looking at "setting" f-quarter moon)
now the fun part: sub is approximately at 140.5w. add in the 7.5 offset, then (180-148 = 32 and divide by 15 = 2.133 or 2:08 hours. then added in offset hour(?), to get a 3:08 correction. which becomes -
2- mr: 0808, ms: 22:57, sr: 08:10, ss: 22:13
loaded game and sun is totally set by 22:43. moon will not be far behind.
So is sun/moon rise when it first peeks out or fully up? is setting when starting or totally behind horizon?
however, now, moon is 'half full'. saved at radar contact to go to lunch. reloading game, it is still at 'half full. (why?) is math correct for hourly correction? I think this is still very usable for planning and thank you for bringing it up
:Kaleun_Cheers:
thinking about it, I save often, so i'll go back to july 1 and see if it is a new moon.
Not sure if this helps, but in RL celnav the Sun and Moon are considered to be rising or setting when their "upper limb" (the top) is just on the horizon. This occurs when the bodies are actually 0°50' below the horizon. This is because atmospheric refraction* at the horizon amounts to about 34' and the semi-diameter (half of the diameter) is about 16'. Add those together and when the center of the body is calculated to be at an altitude of -0°50', the upper limb should just be touching the horizon.
*Refraction can be calculated using the formula -0.0167°/tan(H+7.32/(H+4.32)) ...where H = altitude in decimal degrees.
so last night, got distracted by tv and was brought back to the game with a 'warship sighted'. there in the rising first-q moon was an escorted convoy; [on 12th patrol, July 2, 1943 ~22:00 and this is my first convoy, let alone an escorted convoy]. pissed off, decided this was a good spot to experiment.
so just reading off chart in notepad, I got the following: July 2, 1943 - new moon 12:44 ; July 10, 1943 - f-quarter 16:29. (:hmmm:I figured this is saying, over the 8 days, the moon goes from new moon to f-quarter, starting at 12:44 and ending at 16:29. but then, the game only has four phases.)
then pulled out for july 2nd and 3rd:
2- mr: 05:00, ms: 19:03, sr: 05:02, ss: 19:05
3- mr: 05:49, ms: 19:49, sr: 05:03, ss: 19:05
(:oops: oops, I was actually looking at "setting" f-quarter moon)
now the fun part: sub is approximately at 140.5w. add in the 7.5 offset, then (180-148 = 32 and divide by 15 = 2.133 or 2:08 hours. then added in offset hour(?), to get a 3:08 correction. which becomes -
2- mr: 0808, ms: 22:57, sr: 08:10, ss: 22:13
loaded game and sun is totally set by 22:43. moon will not be far behind.
So is sun/moon rise when it first peeks out or fully up? is setting when starting or totally behind horizon?
however, now, moon is 'half full'. saved at radar contact to go to lunch. reloading game, it is still at 'half full. (why?) is math correct for hourly correction? I think this is still very usable for planning and thank you for bringing it up
:Kaleun_Cheers:
thinking about it, I save often, so i'll go back to july 1 and see if it is a new moon.
If you're at approx 140d, you are roughly -4 hours if Pearl is your home port. The clock in your boat is home port, Pearl base time and you need to calculate from Pearl 'base time'.
The 180d is not base time unless you're sailing out of Midway. In Pacflt I always sail out of Pearl.
Did I misunderstand your situation? What is your home port?
thx Nathaniel.
home port is midway.
thought of another way. once on station, measure sunrise/set time and calculate offset from that. will try that tonight.
tried several times; all with similar results. currently, being at truk; from midway, I came up with a 2:08 hour difference. date is 20Nov43. sun up is listed 6:27 and measured time was 7:08 or 0:41 difference. sun down is listed at 17:00 and measured time was 18:54 or 1:54 difference. so far, really confusing; :hmmm:, why the differences?
so then checked when the moon rose and it peeked out at 02:30. the listing for moon rise/set is:
Nov19 23:35 12:23
Nov20 ' - ' 12:59
Nov21 00:28 13:33
For Nov20, if the ' - ' means zero, at least closer to the 2:08.
So what does the ' - ' mean? that measured moon rise was at 02:30 on 'Nov21'; but the chart list 00:28 for the 21st. :k_confused:
so far, all the moon phases' have been correct and all though I measured a lot of weird times: still find this all useful for planning. :salute:
The ' - ' means that there is no moonrise on that date.
Rockin Robbins
07-13-17, 06:46 AM
I fear that you can calculate till the cows come home and only (possibly) come up with sunrise times for reality. But the game will stubbornly refuse to cooperate because it is not a solar system simulator, ESPECIALLY where the moon is concerned., I have an SH4 screenshot with a crescent moon less than 10 degrees from a setting (or rising) sun. The phase of the moon is directly related to its rise and set times. Since the screenshot is impossible, the moonrise tme must not be just a little bit, but radically wrong.
And all those calculations just go to waste. SH4 did a TERRIBLE job with solar system astronomy. And just a bad job with deep space astronomy.
The phase of the moon is directly related to its rise and set times.
No, it isn't. The phase of the Moon is a function of its apparent angular separation from the Sun. The percent of illumination can be calculated as the haversine¹ of the separation angle in degrees.
The apparent angular separation can be calculated using the following formula:
θ = acos(sin(Dec1)⋅sin(Dec2)+cos(Dec1)⋅cos(Dec2)⋅cos(G HA2-GHA1)) (Not sure why there's a space inserted into "GHA2". I can't remove it.)
If the Moon was only 10 degrees from the Sun, it should be a crescent. A very thin sliver at 0.7% illumination, but a crescent nonetheless.
¹Haversine = (1-cos)/2
DaveR,
I'd be happy to help you figure this out, but first I need to know a few details:
What, exactly, are you trying to accomplish? Are you trying to figure out the time of Sun/Moon rise/set at Midway or at your ship's actual position or are you trying to find the position of your ship?
Are you able to measure the altitude of the Sun/Moon in game? If so, is the measurement at the lower limb, center or upper limb? What precision can be achieved? (Ideally, you want to be able to measure the altitude to within one tenth of an arc-minute. E.g. 37°12.8'.)
Where are you getting your almanac data? The times you listed for moon rise/set match the Nautical Almanac times for those dates at 30°N. But, the sunrise/set times are a little off.
Is your ship's clock set to Midway time, as I assume it is from reading aanker's post?
Is your ship actually at Midway or some other location? (Still 140°30'W?) What is your DR latitude? Are you underway or stationary?
Rockin Robbins
07-13-17, 12:42 PM
No, it isn't. The phase of the Moon is a function of its apparent angular separation from the Sun. The percent of illumination can be calculated as the haversine¹ of the separation angle in degrees.
You're just repeating my contention that moon phase is dependent on time. For instance, for a last quarter, the moon has to rise 90º before the Sun, and so MUST rise at midnight local time. The last quarter moon can't rise at any other time. And it always sets at noon. (I'm driving myself crazy here trying to visualize this stuff.)
At sunset, the first quarter moon is 90º from the sun, so is at culmination. The first quarter moon, then, MUST set at midnight local time and rise at noon local time. It cannot rise at any other time. And of course, your math verifies that but is meaningless for most humans. Even you, comfortable with the math, haven't made the connection to reality that is vital to understanding whether it is true or false and vital to actually understanding what the numbers mean. Numbers have no meaning in and of themselves. They must be connected to reality.
Just as a check, what time MUST the full moon rise? There is only one time it can rise.
The apparent angular separation can be calculated using the following formula:
θ = acos(sin(Dec1)⋅sin(Dec2)+cos(Dec1)⋅cos(Dec2)⋅cos(G HA2-GHA1)) (Not sure why there's a space inserted into "GHA2". I can't remove it.)
If the Moon was only 10 degrees from the Sun, it should be a crescent. A very thin sliver at 0.7% illumination, but a crescent nonetheless.
¹Haversine = (1-cos)/2
The problem with trotting out formulas is that they are not intuitively validating. You can use any form of equation you wish and it looks equally valid. Not only that but results of a formula are not intuitively verifiable either. Your calculation could show that the sun rose at midnight, but being a cloud of meaningless numbers, an error picked up in less than a second graphically only yields to calculation by calculation checking of the math. Often the conclusions of pure mathematics are in stark contrast to the phenomenon they describe, and our mental picture of what they describe is entirely bogus.
I apologize, my memory was wrong. The moon, in the SH 4 screenshot I was describing, was rendered a completely wrong size with the crescent moon actually in contact with the disk of the sun. Phase angle would have been less than .001% illumination at those positions.
https://goo.gl/Pse9qK
So here you go, courtesy of Kim Rohnoff from Dutch Harbor and I have the date somewhere but it doesn't matter. This is just a mangling of solar system astronomy so bad it really doesn't matter when and where the shot was taken..... In reality, the sun and moon are essentially the same size in the sky, 1/2 degree, and the moon in that position would not only be invisible, but we would have a partial solar eclipse, which, needless to say did not happen on that day. I believe in this screenie the moon is rendered more accurately to scale than the sun, which is pretty close to double its real size.
SH4 is not an astronomy simulator and any attempt to use it as such is doomed to failure.
tried several times; all with similar results. currently, being at truk; from midway, I came up with a 2:08 hour difference. date is 20Nov43. sun up is listed 6:27 and measured time was 7:08 or 0:41 difference. sun down is listed at 17:00 and measured time was 18:54 or 1:54 difference. so far, really confusing; :hmmm:, why the differences?
so then checked when the moon rose and it peeked out at 02:30. the listing for moon rise/set is:
Nov19 23:35 12:23
Nov20 ' - ' 12:59
Nov21 00:28 13:33
For Nov20, if the ' - ' means zero, at least closer to the 2:08.
So what does the ' - ' mean? that measured moon rise was at 02:30 on 'Nov21'; but the chart list 00:28 for the 21st. :k_confused:
so far, all the moon phases' have been correct and all though I measured a lot of weird times: still find this all useful for planning. :salute:
Nuts! - lol ...... Yeah, a lot of weird times. I used these charts for rough information, a "close enough for Government work" type thing.
It is frustrating not to have precise times though.... plus your in-game times that you posted seem to be closer to a Pearl home port which is interesting. I forget if the Stock game has Midway as a 'Home Base' - maybe that's it?
The time differences do seem to be closer to a Pearl home port even though they still are not precise times, they are close (for a game ; ) Maybe these charts are only 'good' if using/calculating from Pearl as 'Base Time' for PacFlt? Sail out of Midway but calculate from Pearl where the Admirals are - ha!
I wonder how close SoWesPac is, (one way to find out) although those missions are roughly in the same time zone, or close to it.
The phases are 'good' but the times aren't exact, however they're still close enough to plan an attack or Special Mission.
That's all I got.... it's a game - and the Dev's did get it close. I'm going with the use Pearl to calculate from theory.
Happy Hunting!
(I'm driving myself crazy here trying to visualize this stuff.) [...]
And of course, your math verifies that but is meaningless for most humans. Even you, comfortable with the math, haven't made the connection to reality that is vital to understanding whether it is true or false and vital to actually understanding what the numbers mean. Numbers have no meaning in and of themselves. They must be connected to reality.
First of all, I'd like to say that I'm no mathematician. Math was actually my worst subject in school and I absolutely hated it. However, if you're like me and find you have a passion for celestial navigation and devote years to studying it, you will learn to do the spherical trig because that's the heart of celnav.* Whether or not the equations mean anything to you personally is actually irrelevant. The equations I quote are literally the science behind the planning and practice of celnav.
Having said that...I totally understand what you're saying about visualization, because that's also how I understand and integrate new knowledge. So, with that in mind, I have prepared a few pictures to help illustrate why the Moon's phase is only tangentially related to rise or set times and even then only under certain circumstances. (Apologies for only being able to link to the pictures. Apparently, I haven't yet earned the privilege of attaching my own pics in-line in a post.)
Let's take a look at the case you cited as a "check":
Just as a check, what time MUST the full moon rise? There is only one time it can rise.
Many times, it is useful to look at an extreme example of a situation to understand the mechanics behind it. So, let's assume we are almost at the North Pole...our latitude is 89°59'59"N. Only one arc second away from the pole itself. Our longitude is 0°00'00"E. We are on the Greenwich meridian and therefore our local time is exactly equal to UTC. The date is September 6th, 2017 and the time is 07:30 UTC. The Moon is full (as full as it will get at this time at 99.9% illumination). Now, what is the Moon's altitude? If you click on Fig. 1 (https://www.dropbox.com/s/j2kjqfir0ass0z6/FullMoon1.jpg?dl=0), you will see that it is at 8°54'21" below the horizon (represented by the green circle).
Next, we wait six hours until 13:03 UTC. Now what is the Moon's altitude? Fig. 2 (https://www.dropbox.com/s/h1l9fh3t4og36gd/FullMoon2.jpg?dl=0) shows that it is still 7°53'18" below the horizon. Another six hours later at 19:03 UTC and Fig. 3 (https://www.dropbox.com/s/va1to9ny26tsk10/FullMoon3.jpg?dl=0) shows that the Moon is still 6°50'38" below the horizon. So what time will it rise? Well, the bottom left corner of Fig. 3 gives away the fact that the Moon will not rise at all on this date, at this location. In fact it won't rise until two days later, on September 8th, at which point it won't set again until September 20th...another twelve days away.
Suppose our latitude was 45°00'00"N (still on the Greenwich meridian, back on September 6th). Now what time will the moon rise? Fig. 4 (https://www.dropbox.com/s/g1qbxd4xsdm9vv7/FullMoon4.jpg?dl=0) shows that it already rose at 18:22 UTC the previous day. And if we were almost at the South Pole...what then? Fig. 5 (https://www.dropbox.com/s/ucp3fnef5d3pl65/FullMoon5.jpg?dl=0) shows that the Moon never rose on this day. At this latitude and time it is "circumpolar", meaning it stays above the horizon all day.
So, "what time MUST the full moon rise?" The answer is: it depends. But, on what? ... If you answered "the observer's latitude", congratulations! You are correct...partially. Unlike the Sun, the Moon is revolving around the Earth as the Earth itself rotates. The speed of the Moon's revolution is great enough that, the farther away from Greenwich you are, the more the time of Moon rise/set changes from what is listed in the Nautical Almanac.
That is why there is a "Table for Interpolating Sunrise, Moonrise, Etc. (https://www.dropbox.com/s/2dohr4d56ng3sb5/Interpolation%20Table.pdf?dl=0)" included in the N.A. This table allows for the adjusting of sunrise/set times for latitude and Moon rise/set times for both latitude and longitude.Because these are two of the primary factors which govern what time the phenomena occur for a given observer.
As I said: I have spent years intensely studying celestial navigation. I'm not just speaking from some cursory understanding of the subject. I own several sextants and have taken hundreds of sights, including "lunars (https://en.wikipedia.org/wiki/Lunar_distance_(navigation))". I have even generated my own "Time Sight Logarithm Tables" which can be used to navigate the way sailors did back in the 19th century. You may have noticed that my forum nickname is actually the name of the author [or editor, more precisely] of one of the most famous American books on navigation at sea: Nathaniel Bowditch...author of The American Practical Navigator, otherwise known simply as Bowditch.
Anyway, I hope all of that was clear enough. And I hope to be able to help DaveR with his problem...although from my initial calculations, it seems that SH might actually have some inaccuracies. I'll need to investigate further and get the answers to my questions to really know. We'll see.
BTW, for some reason I can't see the picture you posted. All I get is a rectangle containing a circle with a horizontal line through it. (?)
*Actually, learning and/or understanding the math is not required to practice celnav. One can simply use pre-calculated tables and fill in pre-printed forms and do just fine. To be perfectly honest, I still don't completely understand it, myself.
Rockin Robbins
07-14-17, 06:48 AM
Let's take a look at the case you cited as a "check":
Many times, it is useful to look at an extreme example of a situation to understand the mechanics behind it. So, let's assume we are almost at the North Pole...our latitude is 89°59'59"N. Only one arc second away from the pole itself. Our longitude is 0°00'00"E. We are on the Greenwich meridian and therefore our local time is exactly equal to UTC. The date is September 6th, 2017 and the time is 07:30 UTC. The Moon is full (as full as it will get at this time at 99.9% illumination). Now, what is the Moon's altitude? If you click on Fig. 1 (https://www.dropbox.com/s/j2kjqfir0ass0z6/FullMoon1.jpg?dl=0), you will see that it is at 8°54'21" below the horizon (represented by the green circle).
Next, we wait six hours until 13:03 UTC. Now what is the Moon's altitude? Fig. 2 (https://www.dropbox.com/s/h1l9fh3t4og36gd/FullMoon2.jpg?dl=0) shows that it is still 7°53'18" below the horizon. Another six hours later at 19:03 UTC and Fig. 3 (https://www.dropbox.com/s/va1to9ny26tsk10/FullMoon3.jpg?dl=0) shows that the Moon is still 6°50'38" below the horizon. So what time will it rise? Well, the bottom left corner of Fig. 3 gives away the fact that the Moon will not rise at all on this date, at this location. In fact it won't rise until two days later, on September 8th, at which point it won't set again until September 20th...another twelve days away.
Suppose our latitude was 45°00'00"N (still on the Greenwich meridian, back on September 6th). Now what time will the moon rise? Fig. 4 (https://www.dropbox.com/s/g1qbxd4xsdm9vv7/FullMoon4.jpg?dl=0) shows that it already rose at 18:22 UTC the previous day. And if we were almost at the South Pole...what then? Fig. 5 (https://www.dropbox.com/s/ucp3fnef5d3pl65/FullMoon5.jpg?dl=0) shows that the Moon never rose on this day. At this latitude and time it is "circumpolar", meaning it stays above the horizon all day.
So, "what time MUST the full moon rise?" The answer is: it depends. But, on what? ... If you answered "the observer's latitude", congratulations! You are correct...partially. Unlike the Sun, the Moon is revolving around the Earth as the Earth itself rotates. The speed of the Moon's revolution is great enough that, the farther away from Greenwich you are, the more the time of Moon rise/set changes from what is listed in the Nautical Almanac.
That is why there is a "Table for Interpolating Sunrise, Moonrise, Etc. (https://www.dropbox.com/s/2dohr4d56ng3sb5/Interpolation%20Table.pdf?dl=0)" included in the N.A. This table allows for the adjusting of sunrise/set times for latitude and Moon rise/set times for both latitude and longitude.Because these are two of the primary factors which govern what time the phenomena occur for a given observer.
As I said: I have spent years intensely studying celestial navigation. I'm not just speaking from some cursory understanding of the subject. I own several sextants and have taken hundreds of sights, including "lunars (https://en.wikipedia.org/wiki/Lunar_distance_(navigation))". I have even generated my own "Time Sight Logarithm Tables" which can be used to navigate the way sailors did back in the 19th century. You may have noticed that my forum nickname is actually the name of the author [or editor, more precisely] of one of the most famous American books on navigation at sea: Nathaniel Bowditch...author of The American Practical Navigator, otherwise known simply as Bowditch.
Anyway, I hope all of that was clear enough. And I hope to be able to help DaveR with his problem...although from my initial calculations, it seems that SH might actually have some inaccuracies. I'll need to investigate further and get the answers to my questions to really know. We'll see.
The answer is that the full moon MUST be 180 degrees opposite the sun. Therefore, if the sun sets at 6:30 pm, the full moon MUST rise at that time. Now we can muddy our thinking by placing ourselves in a place where the sun doesn't set (therefore the full moon would never rise), but just like Einstein's Theories of Relativity and Newtonian Physics, that doesn't invalidate the general rule, it merely clarifies it for extreme situations.
That brings up a really cool situation north of the arctic circle, where in winter, the sun never rises. The full moon, opposite the sun, therefore never sets. You can wake up in the "morning" and see the full moon just run a full circle in the sky for the day. If you're at the pole, the moon's altitude won't change. If you're between the pole and the arctic circle, the moon will incline to the horizon in its circle through the sky at an angle of 90 minus your latitude to the horizontal, but never set. It just appears to orbit you with a period of 24 hours!
Our situation in the game is in the tropics, VERY typical. But even on the north pole, the full moon is found 180 degrees from the Sun. It's interesting that any farmer from 1850 knew more about lunar phases than we do. They would rightly laugh at the proposition that the full moon can rise at any time.
Show me any time in history between plus and minus 45 degrees latitude where the full moon rose at midnight. It can't happen. Two hours before or after sunset. It can't happen. It's a simple mechanical relationship. And don't change the definition of full. Full is 100% illumination, not 99.7%, not 99.9%. The moon moves away from the Sun at about 11 degrees a day. 11 degrees is about 50 minutes time. So a NEARLY full moon naturally changes its rise and set time. The moon, 24 hours after 100% full, rises 50 minutes earlier than it did the previous day! (15º = 1 hour) Changing the terms and claiming that you have somehow invalidated another situation altogether is called a paper tiger argument. We have to use the same terms or there's no effect.
Although Full Moon occurs each month at a specific date and time, the Moon's disk may appear to be full for several nights in a row if it is clear. This is because the percentage of the Moon's disk that appears illuminated changes very slowly around the time of Full Moon (also around New Moon, but the Moon is not visible at all then). The Moon may appear 100% illuminated only on the night closest to the time of exact Full Moon, but on the night before and night after will appear 97-99% illuminated; most people would not notice the difference. Even two days from Full Moon the Moon's disk is 93-97% illuminated.US Naval Observatory (http://aa.usno.navy.mil/faq/docs/moon_phases.php)
Unfortunately, Silent Hunter 4 renders the earth as a cylinder with diameter who knows and a height of the linear distance from pole to pole. Therefore there is no difference in sunrise and sunset with latitude anyway. That alone completely invalidates celestial navigation, which incorporates 3 dimensional trigonometry to fix the ground point of stars and planets. You can't do that on a cylinder. Unfortunately, the ground "point" of any celestial object is a vertical line extending from pole to pole, approximately 12,430 statute miles long! The whole shebang falls flat on its face.
Reality ALWAYS trumps math. Unfortunately, due to a recent act of Congress, reality exists in the computer and we occupy virtual reality!:har:
Oh, my photos link to Google Photos. I'll post a link to the photo in question and lets see if you can get there indirectly: https://photos.app.goo.gl/27C6mQ3TeNue8BO73
Usually I run it through Google IRL shortener, goo.gl (https://goo.gl), and post THAT link, but let's try the raw link from hell here and see if you can see it.
Nathaniel, you're a perfect example to show that people can disagree without being disagreeable. It's a pleasure to swap ideas with you.
Rockin Robbins
07-14-17, 12:00 PM
For the curious, a short, somewhat general description of how celestial navigation works is in order. Nathaniel, back me up here and correct any errors, I'm doing this without any reference materials.
Every star in the sky is at the zenith somewhere on Earth. Now, it might be daytime and you can't see that star, but it's still out there, straight up from a single point on earth. This spot is the ground point.
What if you could have a chart of a small collection of bright stars and their ground points at different times of the night and that chart would cover a year. Shazaam! That's exactly what we have with the US Naval Observatory's Nautical Almanac (http://aa.usno.navy.mil/publications/docs/na.php).
So here's the way it's supposed to work. You see a star up there. The almanac tells you the ground point of that star. But it's not straight up for you. You take your sextant and measure the altitude above the horizon. There are complicated interpolations and correction factors we'll just pretend don't exist right now, pay no attention to the man behind the curtain.
Let's say the altitude of that star is 86º, That means that you are somewhere on a circle drawn around the ground point with the diameter of the linear distance described by that 4º difference from vertical. What good is a circle? We need more information.
So we pick another star, plot the ground point, measure the angle and draw the circle around THAT. Now we have two circles that intersect in probably two places. Errors can have bizarre consequences! Now we know we're at one of the two intersections, usually pretty good.
But a good plot is with at least three sightings. That way you get three circles that overlap in one spot. Plus you get more information! Those circles aren't going to intersect at a point, they will describe a triangle. Why? There are observational errors, possible computational errors. This triangle, by its size defines your error envelope. If you're in the US Navy, your error envelope better be smaller than a certain size or you're doing another set of observations!
So if your solution triangle is 100 miles on a side you have junk. If it's 1 mile on a side you did well!
The answer is that the full moon MUST be 180 degrees opposite the sun. Therefore, if the sun sets at 6:30 pm, the full moon MUST rise at that time.
Must it? Take a look at this image (https://www.dropbox.com/s/pztlxr5va5tnjo5/Dec1.jpg?dl=0). The Moon is 100% illuminated, but the elongation (apparent angle from the Sun) is only 177°25'51"...over two and a half degrees less than 180°. How is this possible?
You may have noticed that the equation I posted earlier for finding the separation angle or elongation contained two types of variables. "GHA" or Greenwich Hour Angle, and "Dec" or Declination. GHA is similar to longitude - it's a measure of the distance of a body from the Greenwich meridian in the horizontal axis. Unlike longitude, however, GHA is only measured westward, through 360°. Declination is similar to latitude. It's a measure of the distance of a body from the celestial equator (an extension of the Earth's equator out into space) in the vertical axis. Finding the elongation of a body is not as simple as subtracting one GHA from another. Declination must also be taken into account.
Also apparent in the previous image is the time which the [100%] full Moon will set on this date at this location (July 9th, 2017 at 45°N, 0°E): 04:37 UTC. Now, have a look at this image (https://www.dropbox.com/s/zqr9605rbjlk3o3/Dec2.jpg?dl=0). It shows that the Sun, at the same location on the same date, rose at 04:28 UTC. IOW, the Sun and the full Moon are both in the sky at the same time. There is a nine minute window between when the Sun rises and the Moon sets.
Nine minutes may not seem like much, but consider this: the Earth rotates through 360° in [roughly] 24 hours*. That means that for each second, the Earth rotates 15 arc seconds. That causes celestial bodies to appear to change position by 15" each second. So, when determining longitude from the position of a celestial body, an error of 4 seconds of time equates to an error of 1 nautical mile of longitude. That means that an error of nine minutes of time equals an error of 135 nautical miles of longitude. That's about two thirds of the distance from New York City to Washington D.C. (At the equator. Differences in longitude become smaller with regard to absolute distance as latitude increases toward the poles.)
As for the statement that "Reality ALWAYS trumps math."...well, I almost don't even know where to begin responding to that. Spherical trigonometry is how navigation is actually done, in reality. And not just celestial. Math is a tool which allows us to predict reality with very high precision. It's how the almanacs and sight reduction tables are generated, it's at the heart of how GPS works, and it even helps us understand reality itself through theoretical and experimental physics (a topic which you alluded to in your previous post). You even used it yourself in describing how the Moon changes position relative to the Sun. Without math, things like modern navigation, astronomy and physics would not be possible. Again, just because you haven't taken the time to understand it doesn't invalidate it in any way.
Now, your synopsis of how celestial navigation works is fairly good. Well, that's how "modern" celnav is supposed to work, generally. However, that's not how celestial navigation was actually done for the majority of the time it was used before the introduction of GPS, Loran and other methods of fixing one's position. The concept of "circles of position", or more practically, "lines of position" (LOPs) was a fairly late comer in the history of navigation. Most often, the navigator would stick to the routine of longitude by time sight in the morning, latitude by noon sight and another longitude by time sight in the evening. Longitudes and latitudes would be "run up" to each other to produce a "fix".
We can debate the finer points of celnav until the cows come home. But, although I don't claim to be an "expert" in any field, I'm asking you to trust that I just might be a little more knowledgeable in this one than most people. I'm not saying that you're completely wrong. You seem like a relatively intelligent person. I'm just saying that, whatever is causing DaveR's problem, whether it's SH's model or perhaps a bit of understandable confusion on his part...I might be uniquely suited to ascertain exactly what is causing it and what can be done to remedy the situation. In the end, our collective enjoyment of these sims is what we all have in common, for whatever reason. I'd just like to contribute what I can to that end. Besides that, this is a relatively complex subject and this thread (or even this forum) may not be the place to get into such technicalities.
*We haven't even gotten into the difference between UTC, UT1 and the fact that the Earth's rotation is slowing, causing us to have to insert "leap seconds" into UTC to keep our clocks in step with "actual" solar time.
Rockin Robbins
07-15-17, 06:56 AM
You are so far the most knowledgeable person on the subject that is at Subsim that I've had absolutely nobody to talk to on the subject. What is the time, date and location of your graphics. I don't think the average person can understand them and I think they deserve to. The scale is also too small to read the text.
Let me try from an astronomy program I use called Carte du Ciel (http://ap-i.net/skychart/en/start). I'm not going to be able to pick the same location on Earth, but I can pick one at about 35 degrees latitude. I'll have to download into my Linux installation and I'll do it tonight.
First let's set the observing location:
https://goo.gl/ubQmih
Now let's look at the sky in Carte du Ciel, whole sky view:
https://goo.gl/nkVs4b
A couple of things of note. The altitude is the altitude of the center of the object, not the limb. The moon at zero altitude is half visible. Then look at the altitude at 29'58.5", essentially that's 30 minutes high. Since the moon is 30' wide that means that to the eye, the entire moon is visible with 15' of clear space below. You won't see that without a telescope, if you're lucky.
But look at that thing below the altitude. Nathaniel, you know what that is, but I'm explaining interesting stuff to others. The Geometric altitude is where the moon REALLY is. You see, our atmosphere acts like a lens, bending light that enters it from an angle. For objects on the horizon, and this thing qualifies, you can see that atmospheric lensing raises the moon from only 1 minute above the horizon (half visible) to 30 minutes above the horizon where you can see the entire Moon with half a moon diameter of black space below the moon! Rule of thumb: on the horizon all objects are raised by about half a degree.
https://goo.gl/TP1Egm
And in the far corner of the sky sits the Sun. Its altitude is 1 degree 39' 10.3", but its REAL altitude is 1 degree 19' 13.4". Now the moon isn't perfectly full, but pretty close, and you can see that with the Sun 1 degree 39 minutes above the horizon (you can just see all if it IF you have a perfect horizon plus perfect sky) and the Moon half a degree in the sky (there is a quarter degree of sky under it, not really enough to see even with a telescope unless your viewing conditions are perfect) they are essentially rising and setting as a unit. Considering that every day the moon's rise time is 50 minutes earlier and we're only 3 minutes apart comparing rise time of sun to set time of the moon this is pretty close to an ideal situation. You can possibly measure that they are not EXACTLY 180 degrees apart, but I daresay that no instrument of a submarine or ship would establish that. Measuring positions of objects on the horizon is basically impossible anyway.
And that's why you mentioned shooting noon positions. That's much more accurate. Straight up, you don't have to worry about atmospheric refraction. Since you are looking through the minimum thickness of atmosphere, seeing conditions affect you the least they possibly can for that particular weather condition. BUT the time of culmination, when the Sun is highest, depends on your longitude and YOU DON'T KNOW THAT. So you don't know exactly when to look to find the time of local noon. Of course navigators have sneaky and effective ways to deal with that uncertainty;
Nathaniel, take it away! How about explaining how to do a noon sight and how to use that information to develop your longitude? Latitude is a piece of cake. Just measure the altitude of Polaris. If you're being fussy you can use a polar alignment scope to correct for the offset of Polaris from the actual pole, but that's less than a degree. If you're south of the equator then the bet is off!
Did you know: the rate of continental drift in the Atlantic Basin, the rate the moon is increasing its distance from Earth and the rate your fingernails grow are all just about exactly the same? Coincidence? Or conspiracy?
What is the time, date and location of your graphics.
July 9th, 2017 at 45°N, 0°E; 04:33 UTC
I'm using Stellarium (http://www.stellarium.org/) to produce the graphics. There's a version available for Ubuntu and a link for the source for Linux, so I think you should be able to get it running on your system. It's not as accurate as the program I use for serious calculations: USNO's MICA (http://aa.usno.navy.mil/software/mica/micainfo.php). But, it's great for being able to visualize various scenarios. And, unlike MICA, it's free! (Although, if you're in need of a program which will produce accurate astronomical data, MICA is the way to go. However, the USNO website can be used to produce most of the same data at no cost. Also, NB: MICA only runs on Windows and Mac.)
If you really want to discuss celestial, perhaps we could meet up on Discord for a voice chat. There's a link to the Subsim discord up top ↑. It might save us both a lot of typing. :) I must admit, I'm a total noob when it comes to Discord, but I hear it's similar to Teamspeak, which I've used extensively in the past.
Navigation, celestial in particular, is a passion of mine. So, if you'd like to be bored to death by the minutia of the subject...I'm your man. :D
Rockin Robbins
07-15-17, 05:20 PM
I'm really interested here in getting the Subsim rank and file to understand some of the underlying principles happening in celestial navigation in a way that won't necessarily enable them to do it, but so that they don't feel like idiots whenever someone around them happens to mention it. Doing the chat wouldn't really accomplish that.
Now, is Greenwich Hour Angle the same as the Right Ascension of an astronomical object? I'm inclined to say close but no cigar, because astronomers use the Right Ascension to find an object. You use it for navigation.
So on an astronomical star chart, Right Ascension doesn't move for 25 or 50 years and then everybody changes charts. I imagine the Greenwich Hour Angles are real time adjusted for the very hour of calculation so the Greenwich Hour Angle of a star isn't the same two consecutive days at the same time.
Now there's a way to connect Hour Angles or Right Ascension with Longitude on Earth. Hour lines build to the west, starting at Greenwich. A star overhead at noon, will be 15 degrees away in an hour. So one hour in time is equal to 15 degrees of longitude on Earth. One degree of distance is equal to 4 minutes in time.
Let's say you are looking at the moon in a telescope with the perfect magnification to show the disk of the moon with no black space around. That means that roughly your field of view is half a degree. (notice how daintily I avoid saying 30 minutes, confusing angles and time. That would be bad.....) Now aim the telescope so the moon is just about to touch the field of view on one side. How long until it vanishes from view on the other side of your field of view?
Okay, the moon is half a degree wide. It's going to move half a degree to fill up your field of view and then another half degree to leave. So total movement to travel completely across your eyepiece is one degree. Moving 4 degrees takes one minute, so the moon will pass you completely by in a quarter of that time or 15 seconds! In 15 seconds you will go from seeing nothing to black space to seeing nothing but moon to seeing nothing but black space againi. It's hauling butt! Not really. Actually what you are seeing is the rotation of the Earth moving the aiming point of your telescope.
Rockin Robbins
07-15-17, 05:50 PM
*We haven't even gotten into the difference between UTC, UT1 and the fact that the Earth's rotation is slowing, causing us to have to insert "leap seconds" into UTC to keep our clocks in step with "actual" solar time.
There's that reality trumping math thing biting us on the arse yet again! Math is an imperfect descriptive language like English. It can describe fallacy or truth with equal aplomb, and is only useful if we verify and correct, especially for chaotic change. There is no formula for when to add for a leap second. Yet.
Now, is Greenwich Hour Angle the same as the Right Ascension of an astronomical object? I'm inclined to say close but no cigar, because astronomers use the Right Ascension to find an object. You use it for navigation.
Right ascension is not used in navigation, but a related concept is: "SHA" or Sidereal Hour Angle. To understand both, you need to understand the "first point of Aries", usually referred to as just "Aries"*. This is the point at which the ecliptic (the path that the Sun seems to follow in its apparent course through the stars throughout the year) crosses the celestial equator (the projection of the Earth's equator onto the "celestial sphere") when the Sun is ascending in declination. Ultimately, it's just a completely arbitrary point from which everyone agrees to measure positions.
Right ascension is the apparent arc distance of a body from Aries, measured from west to east. It is usually measured in hours, minutes and seconds. It's most handy for astronomers, because of something called "sidereal time". When Aries is on the observer's meridian, local sidereal time equals 00:00. About 6 hours later*, Aries will be on the observer's western horizon (setting) and sidereal time will be 06:00. Now, if a body has an RA of 6h0m0s* it will be on the observer's meridian - the best time to view it through a telescope. So, if we want to view a body with an RA of say...15h27m43s, we simply watch our local sidereal clock until that time, point our telescope at our meridian at the proper elevation, et voilà! There it is.
Now, SHA is also measured from Aries. But, unlike RA, it is measured from east to west in degrees, arc minutes and arc seconds. In celestial, it is usually only used for stars. The reason for this is to keep the almanac at a reasonable size. You see, there are 57 stars traditionally used in navigation (plus Polaris). They were selected based on their magnitude and distribution in the sky, such that, for an observer anywhere on Earth, the probability would be high that the positions of enough stars visible at twilight would be available for a "fix".
You are correct that GHA (Greenwich Hour Angle) is tabulated for each hour on the daily pages of the almanac, but even that is not enough precision for navigation. The rate of change in GHA varies between stars, planets and the Sun and Moon, but in every case it's fast enough to matter. Therefore, tables are included towards the back of the book which allow for interpolation down to each second of each minute of each hour of every day in a year. However, listing the GHA for all 57 stars, even hourly, would result in a massive tome...much too bulky to be practically usable...let alone even carried on a small vessel, where space and weight are at a premium. So instead, the GHA of Aries is listed, along with the SHAs of the 57 stars. (See here (https://www.dropbox.com/s/fcwfsfy62vrviei/1943Almanac.pdf?dl=0) for a sample "opening" of an almanac, courtesy of Navsoft (http://navsoft.com/astronav.html).) To obtain the GHA of a star, one needs to add the SHA of the star to the GHA of Aries (subtracting 360° when exceeding that value), and then interpolate for minutes and seconds. Longitude is then added (for east longitudes) or subtracted (for west longitudes) to obtain LHA, or Local Hour Angle. (Again, subtracting 360° when necessary.)
LHA along with declination (also listed on the daily pages of the almanac) are the two variables needed to calculate the position a body occupies at a given location. In practice, the position is calculated for an "assumed position" (AP) which is then compared to an actual observation (corrected for various factors) and an "azimuth" (direction) and "intercept" (distance) are obtained, allowing the navigator to draw a "line of position" (LOP) relative to the AP. The result is the intersection (if you're lucky) of the lines at the fix. As you mentioned earlier, the more likely scenario due to errors is a series of lines describing an area (a triangle for three observations) which contains the "most probable position" (MPP).
Since RA and SHA are both measured from Aries, just in different directions, SHA can be obtained from RA by converting RA to degrees (by multiplying by 15) and then subtracting it from 360. And RA can be obtained from SHA by subtracting SHA from 360. (Dividing the resultant degrees by 15 to obtain h:m:s, of course.) It should be noted that this is rarely necessary, due to the aforementioned fact that these two values are seldom used in the same pursuit. The main exception being when a navigator is plotting the positions of the Sun, Moon and planets on a "star finder" for planning purposes.
You may be wondering why the SHAs and declinations are listed on each three-day page of the almanac. It's because they are changing more rapidly than every 25 to 50 years due to "proper motion". Some more than others, but enough that it makes a difference for navigational purposes. However, there are a few pages in the almanac which list the SHAs and declinations for many additional stars for six month periods in case those stars need to be used in an emergency where the traditional stars are not visible due to cloud cover or some other reason.
Finally, see here (https://www.dropbox.com/s/56xhxtfgd0ifc8p/coordinates1.gif?dl=0) for an animation I made which attempts to explain how SHA and declination are measured. As I said, I sympathize with visual learners, being one myself.
...And that is probably way more than you ever wanted to know about that. ;)
*Not to be confused with the Zodiac sign or constellation. Although, Aries gets its name from the constellation, because when it was first described, that's where it resided. Now, due to precession, its apparent position has moved.
*A "sidereal day" is about 4 minutes shorter than a solar day. Therefore, if you look at a particular star each day at the same time, it will appear to have shifted position to the east by about four minutes of time, or about one degree. This is due to the motion of Earth revolving around the Sun.
*Note that when referring to minutes and seconds of time, we use "m" and "s". When referring to arc minutes and seconds, we use ' and ". That's how we avoid confusion.
Rockin Robbins
07-16-17, 07:09 AM
About the first point in Aries, which isn't in Aires any more. It is definitely not an arbitrary point (of course! It has to be measured and confirmed.) The first point in Aries is that point where the Sun's path crosses the celestial equator in its path from south to north. This happens on the first day of spring for us northern hemisphere denizens.:salute:
All this mess with the first point in Aries not being in Aries any more comes from the roots of astronomy in astrology, 3000 years ago. Because the Earth wobbles on its axis, the north pole makes a circle in the sky with a period of 26,000 (roughly) years. This MOVES the first point in Aries, the place where the Sun crossed the celestial equator 3000 years ago to the point where the "first point in Aries" is two constellations away now in Pisces!
For a little fun, go see the total solar eclipse, find an astrology obsessed friend and travel to the eclipse. Before the eclipse, ask them which constellation the Sun is in. Then during the eclipse, have them identify clearly that the Sun is not in its astrological constellation at all! It is two constellations over. But wait! It gets worse. Astrology says that during the year the Sun passes through 12 constellations. That is dead wrong. The Sun passes through 15 different constellations, spending the most time in the largest of them, Ophiuchus, which doesn't rate a seat at the Zodiac table!
So astrology, which as its basis says that the positions of the stars, sun, moon and planets guide our lives, doesn't even bother to track the stars, sun, moon and planets. So much for any credibility. The shock on an astrology enthusiast's face when they see that for themselves, with the Sun two constellations away from where their beloved charts say it is, is priceless.
It's actually a common belief in astrology that, like the first point of Aries, the "houses" were named for the constellations they were in when they were discovered. Don't get me wrong, I don't buy into astrology.
I just noticed the text you added to your previous post. In it, you stated that...
And that's why you mentioned shooting noon positions. That's much more accurate. Straight up, you don't have to worry about atmospheric refraction. Since you are looking through the minimum thickness of atmosphere, seeing conditions affect you the least they possibly can for that particular weather condition. BUT the time of culmination, when the Sun is highest, depends on your longitude and YOU DON'T KNOW THAT. So you don't know exactly when to look to find the time of local noon. Of course navigators have sneaky and effective ways to deal with that uncertainty;
That's not exactly right. While it is true that if a body is directly overhead, at the observer's zenith, you don't have to worry about refraction...this is a rare occurrence. It only happens when the declination of the body exactly matches one's latitude. Any other time, the altitude of a body on the local meridian will be equal to 90° minus the declination plus or minus the observer's latitude (depending on whether they are in the same or different hemispheres.)
You also wrote:
How about explaining how to do a noon sight and how to use that information to develop your longitude? Latitude is a piece of cake. Just measure the altitude of Polaris. If you're being fussy you can use a polar alignment scope to correct for the offset of Polaris from the actual pole, but that's less than a degree. If you're south of the equator then the bet is off!
I'd be happy to. But it'll have to wait. I need to get some rest before work. :)
Also, while I was able to see the screenshot from before when you linked to it, I still cannot see any of the images you are posting in-line. I'm not sure if it's a problem on my end or what.
Alright...back home from work and I've gotten some more rest. Now, for the noon sight:
The noon sight is rarely, if ever used to find longitude. The reason is that, in order to accurately find longitude at noon using a sextant (and chronometer), one needs to determine the exact time at which the Sun transits the local meridian. When the Sun is on the local meridian, it is at its highest altitude for the observer's location. A sextant measures the altitude of heavenly bodies...so that should be easy, right? Not so much. Right around noon, the Sun seems to "hang" at the same altitude for up to several minutes.
Take a look at this graph (https://www.dropbox.com/s/acrwssl9e460uod/NoonSightGraph.jpg?dl=0). (You should be able to zoom it if it is not clear.) The blue line shows the altitude of the Sun every fifteen minutes on July 17th, 2017 at 45°N, 0°E. The red line (which uses the right-hand vertical axis) shows the rate of change in altitude in arc minutes per minute. Notice what happens at local noon: the rate of change drops to nearly zero. That makes it very difficult to determine exactly when the Sun has reached culmination.
There is a "trick" which can be used to try and find the time of local noon: double altitudes. The navigator measures the altitude of the Sun some number of minutes before noon and notes the time. The sextant is left at whatever reading was taken at that time. Then, the navigator waits until after noon, when the Sun drops to the exact same altitude again, and notes the time. splitting the difference between these times should give the time of local noon...if you're stationary...and not on a pitching and rolling ship...and make perfect observations. But, even then, it's tough to get an accurate time.
For the sake of explanation, let's say you did get an accurate time. How does that tell you your longitude? Well, your chronometer would be set to GMT (Greenwich Mean Time or UT [Universal Time], essentially the same thing). Let's say you found that local noon occurred at 14:32:17 GMT on July 17th, 2017. Now, you know that noon in Greenwich occurred at 12:00:00 GMT...or did it? Not necessarily. You have to consider something called the "equation of time". Because the Earth speeds up and slows down in its yearly orbit around the Sun, the Sun appears to race ahead or lag behind "mean" time (the time that a regular clock keeps). On this date, the equation of time is about -6m13s. IOW, the Sun is six minutes and thirteen seconds behind where the mean Sun would be, and noon would occur later. (This information can be found in the Nautical Almanac [see the bottom of page 2 in this example (https://www.dropbox.com/s/fcwfsfy62vrviei/1943Almanac.pdf?dl=0)].)
So, Greenwich noon occurred at 12:06:13 GMT. If we subtract that from the time of our local noon, we get 14:32:17-12:06:13 = 2:26:04. Local noon occurred two hours, twenty-six minutes and four seconds after Greenwich noon. Since we know that the Earth rotates 360° every 24 hours, we know that the Sun appears to move 15° through the sky each hour. So, if we multiply 2:26:04 by 15, we get: 2:26:04∙15 = 36°31'00"...this is our longitude. And since local noon occurred after Greenwich noon, we know our longitude is west: 36°31'00"W.
However, as we have learned, the noon sight is not typically used to find longitude. What it is used for is to find latitude. So, how is this done? Well, in this case, it's kind of handy that the sun "hangs" at the same altitude for several minutes. Because that's what we're after - the maximum altitude. Let's say that, after correcting for index error, height of eye, refraction and semi-diameter, we get a maximum altitude of 66°05.5' on July 17th, 2017 at 14:32:17 UT. We subtract the altitude from 90° to get the "zenith distance" - the distance the Sun is from being directly overhead. (This is equal to the distance we are from the Sun's GP.) 90°-66°05.5' = 23°54.5'. Now, we look in our almanac and find, after interpolation, that the Sun has a declination of N21°05.5'. Since the Sun's declination is in the same hemisphere as our DR, we add the declination to the zenith distance to find our latitude: 23°54.5'+21°05.5' = N45°00.0'.
Finding longitude was usually done by "time sight". But that is another subject altogether.
Rockin Robbins
07-17-17, 05:02 PM
That raises the question of how accurately can you measure the altitude of the Sun, an extended object, on the pitching deck of a submarine? I bet you couldn't duplicate two sightings within 10' of arc. I wouldn't be a bit surprised at an error envelope of a half degree. That would make finding longitude with a noon sight impossible.
It still seems to me the altitude of the north pole is the best way to establish latitude. If you're on land that can be done with incredible precision using a polar alignment scope. Alternately, you could use Polaris itself when the position angle of the pole is parallel with the hofizon. That happens twice a day.
The altitude of the Sun is always going to be a problem. The Sun isn't a point, its a disk. You have to find and align on the exact center of the disk.
Typically, a series of sights are taken around noon. The navigator can choose to use the highest measured altitude directly, or can plot all of the observations onto a graph and rough a curve through them. The top of the curve can then be used as the culmination altitude.
Included in the Nautical Almanac (and also in some sight reduction tables) is a table for correcting the offset of Polaris from the elevated pole. The navigator simply calculates LHA Aries and enters the table to find the needed correction for the time at which the sight was taken. An accurate latitude can be determined fairly easily this way. A corrected azimuth can also be obtained for checking a compass.
It's practically impossible to accurately shoot the center of the Sun or Moon with a sextant. That's why a correction for semi-diameter is included in the calculations. The navigator shoots the lower limb (typically, but sometimes the upper limb for the Moon) and corrects the altitude for half of the diameter to find the altitude of the center. The SD of the Sun is listed at the bottom of each daily page, but the correction is included in the altitude correction tables for both Sun and Moon.
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