I'd like to set up a UPS based on a distance of 60 seconds and am having a difficult time of it.   It's probably because I've been working at it too long and it's late!

Here are my assumptions;

- The scale for minutes in the middle of the UPS can also be considered to be just 60 seconds.
- In setting the correct meridian distance for say, N 35°, is done exactly as one would set the meridian distance for 35°.

Any ideas about this?

Thanks,

Craig

Dear member of the forum,
Does anybody know where in the USA I can send my sextant mirrors to be re-silvered?
My mirrors are starting to look a little sad.
Regards,
Steamburn

I'm at a loss to understand how to calculate LHA while sailing in Eastern longitudes.

What I know;

LHA in Eastern longitudes=  GHA + ApL (Assumed position longitude) (minus 360° if necessary)


Example;

Date-  November 25, 2022
GMT- 10:10:59

ApL=  E 025° 45'

GHA= 336° 00.7'

LHA=  336° 00.7' + 25° 45'= 361°45.7' - 360° = 1° 45.7'

But, here's the confusion, LHA in Eastern longitudes is supposed to be rounded up or something like that and I'm not really sure if it is supposed to be rounded up.  Information on the web is vague and not explained well enough making too many assumptions.  Another possibility is that very few who speak English are sailing, doing CN, in the Eastern hemisphere.

So, my estimate to solve the problem is simply this (tell me if I'm wrong)

Eastern longitudes, LHA= GHA (whole degree's) + ApL (whole degrees) + 1

The problem above could be easily solved like this-  LHA= 336° + 25° + 1° (minus 360°) = 2°

Any ideas about this?

Gentlemen, 

maybe something of an interest for you. I  kinda wrote a formula  which  maybe  useful.  MAybe;>  It can  be used to get a running  fix from bow  and beam relative bearings. It is well-known  procedure.  But  to speed up  calculations,   there are so-called method  of double angles and 0.7, and also distance travels equalls distance off.  Plus a few so-called special  angles , like 36 deg  bearing1 and 69 deg.  bearing 2.  In the last,  particular case distance travelled is equal  distance oiff the beam to  the object you are  taking bearing on.

Anyway, instead of using  just a few angles  why not  to use any angle at the first  bearing  and calculate what  angle should be  at the second bearing  if  you want  to  have distance travelled between bearings to  be equal  to  distance offf beam of that specific  object you took  bearings at.

a very simple drrawing is following.( see attachment).

Again,  nothin  revolutionary,  just maybe helpful.  Thank  you

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Hi
 
I'm interested in the history of star finder models as an aid to celestial navigation and have a small selection of plan models (list below). My favourite is the Italian Sferoscopio del Pino (image enclosed). Earliest found to date by Rude 1921 leading to the 2102-x models, the 2102-D and British Admiralty NP323 still current.
 
I've recently extended to researching the celestial starfinder globes which predate the plan models  by some margin into the 19thC. Russian and Chinese models made up until the 70s at least come up on eBay. Freiberger also still offer one new. Image of my Russian 1978 example shown.
 
Although there are plenty of resources describing current use of the plan models, notably the 2102-D with David Burch's 'The Star Finder Book' recently into a 3rd edition 2019, have found no reports of the globes being used for navigation, past or present. The globes also have many more stars than the almanac tabulates (the Russian has 160, almanacs typically 57) and unclear how useful these extra stars would be without supporting tables.
 
Can anyone comment on how widely used the globes were or are being used.
 
Thanks.
 
David

Star finder models owned.
2102-B
Sferoscopio del Pino 1971 facsimiles of the 1937+ model (inserts in pocket of book 'La Navigazioni Astronomica' by Mario Sacchetti (1971) to support exercises)
USAF CP-300/U
2102-D vintage and current Weems and Plath
British Admiralty current NP323
Russian celestial star finder globe model 3Г 6.6 inches diameter.

April 30, 2022

Here is a link to a popular and reputable supplier of celestial navigation equipment having a sale on the simple but remarkable Davis Mk3 Lifeboat sextant.  I have no affiliation with the seller or Davis whatsoever other than to have purchased a few items from each of them over the years.

These simple MK3 instruments are perfectly adequate for real navigation. They make a great first sextant due to their simplicity and low cost -- yet you can do some real navigating with them and expect perfectly acceptable results. IMHO they are a better choice than the more expensive plastic sextants that are made to look like "real metal sextants." Those fancier plastic sextants are notorious for never holding their adjustment from one shot to the next in a round of sights which only frustrates beginners. The humble Mk3 doesn't suffer from that shortcoming.
 
If later on you decide to treat yourself to a "fancy metal sextant" you will still keep your MK3 forever because The MK3 is also great for backup; emergency; teaching; and for use on days when the spray is flying and you would prefer not to expose your fancy metal sextant to a salt water dousing.

In fact even if you already own a "fancy metal sextnat" you should consider buying a Mk3 as a backup for those reasons stated above. I did and I am very happy I did so.

On the other hand if you decide that celestial navigation is not for  you, you will be able to sell the MK3 used for about what you paid for it. Or you can just give it to a friend or stash it in the ditch bag. It is a rare "fancy metal sextant" that sells as used for about its original purchase price because people are wary of any unknown history of the instrument and possible hidden defects.

As for myself I started out with an (expensive) fancy metal sextant and turned my nose up at the MK3, although it was suggested to me to start with that model.  Many years later (and a few more "fancy metal sextants" in my collection) I bought a MK3 used.  I can't say enough good things about it. It is one of my favorite sextants. Sure, I'm not going to do lunar distances with it, but I can get position accuracy to within 2 nmi fairly regularly -- and as I said, no worries about eager students dropping it or salt spray ruining it.  In fact just recently I was able to bring down Venus in daylight for practice with the MK3 when my fancier Tamaya failed due to a much smaller field of view.

I love this thing even if it is humble and simple. I will keep mine forever for the reasons stated above, and if it gets somehow destroyed I'd buy a replacement in a minute. I think you would love it too if you would give one a fair try.  In fact the biggest down side to the MK3 is the ribbing you are likely to get about your "inadequate" or "toy" sextant from wanna-be navigators and arm-chair pirates who simply do not know any better.  But after you go out and do a round of sights with it you will be convinced  that they are only displaying their lack of experience.

I have seen it stated in reviews that occasionally the non-adjustable horizon mirror may not be perpendicular to the frame. To check for perpendicularity of the horizon mirror cut about 2mm off the corner of a paper business card at a 45° angle and gently present the card to the mirror as you would a carpenter's square. You need to cut the corner off the business card because there is a tiny plastic ridge at the base of that mirror where its mount meets the frame and  you must create clearance for this ridge. If your horizon mirror is not square you must request a replacement instrument -- it can not be adjusted.  Meanwhile expect to see some of the frame in the right hand side of the horizon mirror even if it is properly perpendicular to the frame. That is normal.

https://www.celestaire.com/product/davis...3-sextant/

A digital sextant. Very interesting.
No sight reduction methods or almanac required.

They claim it can get you a line of position from a single shot of the sun. Or it can give you a position fix with two bodies, or a running fix from two sun sights take some time apart.

What I don't' see is an input for a dead reckoning or assumed position, so how can it work?

A built in ephemeris would give it the geographic position of the observed body and the digital observed altitude would give it a zenith distance, so it would have a circle of position -- but it still needs an azimuth or you could be anywhere on that circle of position.

It must have a flux compass to get an azimuth to the observed body but it doesn't mention it.

Link is below. Skip ahead to about 1:52 to see the digital part described.

I would be curious to hear what others think.

https://www.youtube.com/watch?v=2C9-LMNNzH4

Check out the attached illustration and decide for yourself.

Actually all tracks on the surface of a sphere must be curved in some sense or they would not stay in contact with the surface at all, but a great circle is about as straight a track as you can get while staying on the sphere.  It is all a matter of changing  your perspective.

Once again I don't see great circles taught this way elsewhere.  No wonder people find them confusing.

Simple stuff, I know. The applied math guys are probably rolling thier eyes, but this helped me "get it."

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I have many books on celestial navigation and none of them explain the simple elegance of great circles as used in c-nav. That is a shame because they are easy to understand without any fancy mathematics whatsoever. I personally didn't “get” celestial until I figured out for myself how these great circles work.  That is why I am posting this with the hope that some of you will find the following helpful – or at least interesting.

Many of us started out in celestial with the analogy of a flag pole and how the top of the flag pole appears at lower angles the further we stand from its base. This is a flat-Earth analogy that can only take you so far and sooner or later runs into limitations.

To begin with we must understand that all of the stars are so far away that the direction in which we see any stars doesn't change with our location on Earth.

For the moment consider Polaris, which seems to stand nearly still as the other stars rotate around it. If the entire Earth were the size of the period at the end of this sentence at that scale Polaris would be many times further away from the ink dot Earth than the Eiffel Tower is from New York City's Central Park.  Clearly taking the bearing from one side of the ink dot in Central Park to the Eiffel Tower and then moving all the way to the complete opposite side of the ink dot and taking it again would not change the bearing any measurable amount.

In fact the real distance to Polaris is so far that even if you wait six months for the Earth to reach the opposite side of its entire orbit around the sun, the direction to Polaris won't change. What causes the change of the angle at which you see Polaris is a change in what you perceive as “horizontal” or “straight up” as you move about on the spherical Earth.

If we could transport ourselves far out into space and look back at Earth to see three people taking simultaneous observations of Polaris we would see that they are all looking in the exact same direction – at a star that appears to us to be directly over the Earth's north pole but which is incredibly far away. From our outer space location we would see that all three would be gazing northward in a direction parallel to Earth's polar axis. But Paul at the north pole sees the star directly over his head; Eddie at the equator sees it low on his horizon; and Matt in the mid latitudes sees it part way up his sky.  What is different for each of them is what they consider to be “straight up” and “horizontal” depending on where they are standing on the sphere that is Earth.

This is true for all the other stars as well. It is only the spinning of the Earth that gives them the appearance of motion. If, from our outer space location, we took a look at several people taking simultaneous observations of Arcturus we would see that they too would all be looking in the exact same direction at that star, but the heights at which they each perceive the star in their local sky would differ by their locations on the sphere that is Earth.

Arcturus travels over the Big Island of Hawaii each day, so let us say we watch a bunch of people taking sights of Arcturus just as it is directly above Hawaii. To get a good mental picture of this imagine that you have a globe in your hands and hold it so that Hawaii is straight up. Now from your perspective all of the observers will be looking “straight up” at your ceiling and far beyond to see Arcturus regardless of where they are standing on the globe. Of course the further they are on the globe from Hawaii the more their personal sense of horizontal will be tipped over.  If a few of them observe Arcturus at the same angle up in their sky we could say that they were all standing on a circle of position determined by the same distance from the star's geographic position.  Alternately we could say that they are all standing on the same circle of position determined by an equal amount of horizontal tilting from the 0° tilt that occurs at the geographic position of the star.


Now on to great circles.

A great circle is a specific concept in spherical geometry. It is the circle that is formed on the surface of the sphere by a flat plane that passes through the exact center of the sphere.  All great circles on any particular sphere have their centers at the center of the sphere. They all have the same diameter as the sphere, and they all have the same circumference, which is the maximum circumference of the sphere.

Think of cutting an orange exactly in half with a sharp straight knife. The outer edge of the peel on each of the two halves is now a circle. If the cut went through the exact center of the orange this circle would be a great circle on the orange.  If instead that sphere were a beach ball it is a great circle on the beach ball; If the sphere were a classroom globe that circle would be a great circle on that globe; If the sphere were the Earth it is a great circle on Earth.

Any plane that does not go through the exact center of the sphere still makes a circle where it intersects the surface, but that is not a great circle. Instead it is technically a “small circle” even if it is enormous in size.  All lines of latitude save the equator are small circles on Earth. The flat planes that any of them lie upon cut across the Earth's polar axis at some distance north or south of the actual center.  Nearly all circles of position that you plot as Sumner lines or LOP's in celestial are also small circles. Small circles do not share the handy characteristics of great circles, so be sure not to conflate the two.

For our purposes it is useful to alter our perspective a bit. We say that any circle on a sphere that has the same circumference as the maximum circumference of the sphere will lie in a flat plane that cuts the sphere exactly into two halves passing through the center of the sphere, and is therefore a great circle.

Imagine a schoolroom globe of the Earth. Now imagine you have made a bracelet that slips down over this globe so that it sits exactly on the equator. The plane it lies in is the equatorial plane of the globe which we know cuts the globe into two equal halves and passes through the exact three dimensional center of the globe. This bracelet is a great circle on the globe and it represents a great circle on the real Earth.

Next imagine that you can slip and slide this bracelet anywhere you want on your globe provided it stays in full contact with the globe. In other words if you slide the near side up the back side slides down, but the whole bracelet stays in contact with the globe.  The circumference of the globe and the circumference of the bracelet stay the same regardless of where you slide the bracelet because the globe is a sphere. No matter where you slip the bracelet on the globe it still has a circumference equal to the maximum circumference of the globe and so it always remains a great circle on the globe – and it always represents a great circle on Earth.

Realizing that the circumference of the bracelet never changes you mark a scale along it in degrees from 0° to 360° and now you can slide the bracelet anywhere you want on the globe to connect any two points and measure the distance between them in degrees.

Because the real Earth is (almost) a sphere and all of Earth's great circles have the same circumference we adopted the convention that on Earth one degree along a great circle is 60 nautical miles.  That convention won't work on other planets because they don't have the same circumference as Earth, but for the moment we are only concerned with navigation on Earth. This convention was not arrived at by accident or coincidence – it was deliberately contrived for the purposes of navigation. If we can measure the angle between any two points on Earth along a great circle with our globe and bracelet we can now easily convert that measurement into nmi. What follows explains why we take the measure in degrees instead of just directly measuring it in nautical miles.

Now consider the celestial sphere way out beyond the surface of the Earth. The celestial sphere shares the exact same center as Earth's center. Therefore any flat plane that passes through Earth's center and defines a great circle on Earth can be extended out to the celestial sphere where it forms another great circle out there.  In our globe and bracelet model this means that no matter where we slide our bracelet it not only represents a great circle on Earth, it also represents a matched great circle out on the celestial sphere.  These two great circles work as a matched pair.

Consider a classic wagon wheel. Whatever the angle is between any two spokes on the rim it will be the same angle between those spokes at the hub – because they share the same center.

In our model if we take two places on Earth, A and B, and find their zenith points on the celestial sphere, then the angle between those two zenith points when measured along a great circle on the celestial sphere will be the same angle as points A and B are apart when measured along the matched great circle on Earth, provided we make the measurement in degrees of arc and not in linear measurements such as nautical miles.

It turns out this is very handy because in celestial navigation we need to figure out how far we are standing from a star's geographic position (GP).  Usually the GP is very far away from us and below our horizon, so we can't see it to measure it directly – but we can see our zenith and we can see the star sitting directly over its own GP.  The angle measure from the zenith over to the star along a celestial sphere great circle is exactly the same as the angle measure from beneath our feet over to the star's GP along an Earth great circle – and the Earth version is easily converted into nmi.  If we could reliably take the measurement from our zenith over to the star we would know how far we were standing from the GP of the star and our immediate problem would be solved.

It turns out that it is difficult to measure using your precise zenith as a reference, but it is relatively easy to use the horizon, and with a few simple corrections determine the star's height above the true horizontal. (Ho)  Next we apply the knowledge that the entire angle from the true horizontal to our zenith is always 90° and so the distance from the zenith to the star (called the zenith distance) must be whatever is left over from 90° after you subtract Ho.  Zenith distance =  (90° –  Ho).  And this is the angular distance from you to the star's GP which, when converted into nmi, allows you to plot a circle of position centered around the star's GP at a known radius in nautical miles.

Actually for most sight reduction methods we never state the zenith distance explicitly, nevertheless it is there inside the “mathematical black box” of whatever sight reduction method you choose to use.  Ho is related to the observed zenith distance.  Calculated altitude from an an assumed position (Hc) is related to the zenith distance from the assumed position to which we will to compare our Ho. When we combine them to find the intercept the 90° parts from each zenith distance term cancel each other out.
(90° – Hc) – (90° – Ho) =>  (90° – 90°) + (Ho – Hc) => (Ho – Hc)

The last thing we need to do is to convince ourselves that we are measuring the distance up from the horizon to the star along an actual great circle on the celestial sphere.

When we take a shot of a star we face in the exact direction as the light rays coming from the star. Since the star is directly over its own GP that is the exact same direction as from us to the GP.  Next we go to some pains to hold the sextant vertical as we take our measurement.

There are two light rays coming to our eye through the sextant. One is from the star and the other is from the horizon. We rock the sextant to find the spot where the ray from the horizon is straight down beneath the ray from the star and that is when we take our measurement.  One way to define “straight down” is to say directly towards the center of the Earth.  So these two light rays form a flat plane, and that plane can be extended straight down to go through the center of the Earth – and extended all the way out to the celestial sphere.  Because such a plane forms great circles on both spheres we conclude that we are indeed measuring Ho along a great circle on the celestial sphere aligned with the direction of the star, and that it corresponds exactly with the great circle from beneath our feet over to the GP of the star.

I suppose you don't need to know all of this to practice celestial, and some people may find it a bit too esoteric while others may find my math too simple. But it does explain how it actually works. It was a great help to me when I finally figured it out and perhaps it will be a help to others.  I know none of the texts I have explain it like this, which I hope is straightforward enough for most people to understand if they ponder it for a bit.

Other aspects that are germane to great circles also helped me a great deal when I studied the sailings – and that all began with the simple globe-and-bracelet mental model I've outlined above. But that is a discussion for another day.

PeterB

In a much earlier post on this forum titled “A Celestial Navigation Problem I Can't Solve”

There was discussion regarding the determination of the correct Greenwich day and date.  The confusion arises because Chronometer Time (CT) is given as 9:20:05  but this doesn't indicate if this is AM or PM in Greenwich. Apparently this is a common way to state problems on rating exams and it is up to the individual to figure out if it is AM or PM by other information included in the problem.

In this case it could be 9:20 AM in which case it is  09h 20m 05s  ( 2 2/3 hours before noon in Greenwich,) or it could be 9:20 PM which would be 21h 20m 05s ( 9 1/3 hours after noon in Greenwich.) The DR longitude is given as 163° 51.0' E
The solution to the puzzle is to draw a time diagram.

Time diagrams are a schematic representation of the Earth with the locations of pertinent  information sketched in. We think of them these days in terms of angles of longitude, but before about 1940 these were all done in units of hours and only converted to angles at the last.

I like to draw my time diagrams as if you are looking down upon the Earth from above the north pole. This way the motion of the celestial objects progresses clockwise with the passage of time.  A lot of people prefer to draw them as if you are looking up from beneath the south pole in which case the motion of celestial objects is counter-clockwise as time advances.  It is just a matter of personal preference but be sure you pay attention to which convention is in use and to mark your own diagrams with which way is westward.

Fundamental to understanding how to determine the day and date in Greenwich is to realize that the “new” day extends from the International Date Line (IDL) all the way around the world to the midnight demarcation opposite the sun.  As midnight moves westward the new day expands around the world with it wiping the old day away ahead of its advance. When midnight gets back to the IDL the next new day materializes, starting out as a mere sliver, but expanding with the motion of the midnight demarcation.

I simplify my diagrams just to get “the big picture” and fill in the details afterward. If you are near to the real date line it zigs and zags around a great deal, but this method will still get you in the right frame of mind and you can then make adjustments accordingly.

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