Celestial Navigation with RFB / TMO mods?

[Rockin Robbins]

Originally Posted by Nathaniel B.

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.

Originally Posted by Nathaniel B.

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.


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.

[Sean C]

Originally Posted by Rockin Robbins

(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":

Originally Posted by Rockin Robbins

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, 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 shows that it is still 7°53'18" below the horizon. Another six hours later at 19:03 UTC and Fig. 3 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 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 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." 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". 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]

Originally Posted by Nathaniel B.

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, 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 shows that it is still 7°53'18" below the horizon. Another six hours later at 19:03 UTC and Fig. 3 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 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 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." 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". 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

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!

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, 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]

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.

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!

[Sean C]

Originally Posted by Rockin Robbins

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. 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. 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]

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. 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:


Now let's look at the sky in Carte du Ciel, whole sky view:


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.


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?

[Sean C]

Originally Posted by Rockin Robbins

What is the time, date and location of your graphics.

Originally Posted by Nathaniel B.

July 9th, 2017 at 45°N, 0°E; 04:33 UTC

I'm using Stellarium 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. 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.

[Rockin Robbins]

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.

[Sean C]

I just noticed the text you added to your previous post. In it, you stated that...

Originally Posted by Rockin Robbins

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:

Originally Posted by Rockin Robbins

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.

[Sean C]

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. (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].)

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]

Originally Posted by Nathaniel B.

*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.