Talk:Longitude

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[edit] Talk

Are Dava Sobels' books "Longitude" (which I thought was history wrapped in a historical novel) and "A True Story...." the same book? DJ Clayworth 15:07, 20 Oct 2003 (UTC)

I have Longitude, and I can assure you it is purely non-fiction. You may be thinking about the television programme, which was a dramatized and fictionalized (several characters were merged into one etc. similar liberties taken) version of the events covered in the book. -- Cimon Avaro on a pogostick 15:21, Oct 20, 2003 (UTC)

According to Dana Sobel, John Harrison didn't actually win the Longituded prize. The rules were changed under him so that he no longer qualified for it. Instead, parliament granted him a sum of money in lieu of the prize. --Ezra Wax 16:53, 25 Jul 2004 (UTC)

I thought her first name was Dava, not Dana. Was I mistaken? Michael Hardy 02:35, 3 Jan 2005 (UTC)

[edit] Positive and negative longitudes?

This article says longitude runs between +180° and −180°, but it does not say whether east or west is positive. I had always thought that it was standard that west is positive, never suspecting until a couple of hours ago that anyone used the other convention. Then I saw Wikipedia's list of earthquakes, which does use the opposite convention. As I stated on that article's discussion page, I once heard the head of the math department at MIT (David Vogan, who has since been succeeded as department head), speaking before about 150 undergraduates, state that it's appropriate that longitudes in Europe are negative since Europe is a cultural cesspool. No one responded that east longitudes are positive and west negative. (Nor did anyone complain about that characterization of Europe.) So I'm not the only one to think that's standard. Michael Hardy 02:35, 3 Jan 2005 (UTC).

According to the IAU/IAG Working Group On Cartographic Coordinates and Rotational Elements of the Planets and Satellites:
The planetographic longitude of the central meridian, as observed from a direction fixed with respect to an inertial system, will increase with time. The range of longitudes shall extend from 0° to 360°.
Thus, west longitudes (i.e., longitudes measured positively to the west) will be used when the rotation is prograde and east longitudes (i.e., longitudes measured positively to the east) when the rotation is retrograde. The origin is the centre of mass. Also because of tradition, the Earth, Sun, and Moon do not conform with this definition. Their rotations are prograde and longitudes run both east and west 180° instead of the usual 360°.
Thus, Earth longitude should really run westward from 0° to 360°, but because of tradition it runs westward from 0° to +180° and eastward from 0° to -180° (which makes no difference from a geometric point of view).
Urhixidur 04:21, 2005 Jan 3 (UTC)

So whoever wrote list of earthquakes had it wrong, then. Michael Hardy 03:21, 4 Jan 2005 (UTC)

Wait --- the web page you're citing does say longitude is to be measured in an easterly direction. Michael Hardy 03:55, 4 Jan 2005 (UTC)
It is standard for west longitude to be negative;
Standard in what communities? Among professionals in certain fields? Which ones? And do the standards differ from one discipline to another? Michael Hardy 04:05, 4 Jan 2005 (UTC)
I have an inclination that this may be an Eastern Hemisphere vs Western Hemisphere convention, with the sign chosen so that longitude is positive within your "home hemisphere". That way, you will tend to minimise the use of negative numbers. Wardog (talk) 13:50, 25 February 2008 (UTC)
I'm yet to see a single computer application that defaults to west positive. It's possible that west positive is standard among some scientific disciplines, but I've never seen it, and I work with a lot of them.
When giving coordinates, it is normal to describe the coordinate system precisely--e.g. LL84 (lat-long World Geodetic System 1984) or LL84-WP (lat-long World Geodetic System 1984 west positive) -- as well as whether the coordinates are in decimal degrees, degrees minutes seconds, or something else. Antandrus 04:16, 4 Jan 2005 (UTC)
however a lot of applications and hardware (such as GPS units) allow LL-WP (west positive) as a preference. Think of a mapping exercise with the x axis going E-W and the y axis going N-S: with west longitude negative, numbers get larger as you go east and larger as you go north, which conforms to Cartesian geometry (with north at the top). Speaking as a GIS professional we never use west positive. HTH! Antandrus 03:37, 4 Jan 2005 (UTC)
What in the world is "HTH"? Michael Hardy 04:05, 4 Jan 2005 (UTC)
Hope that helps. Antandrus 04:16, 4 Jan 2005 (UTC)
Going back to the IAU/IAG paper (a PDF version is found here), now I'm confused. If « the planetographic longitude of the central meridian, as observed from a direction fixed with respect to an inertial system, will increase with time », then this means the longitude is positive eastward, because that's the direction in which the meridian will be seen to be moving (for a prograde planet). So why then say, in the next paragraph, that « west longitudes (i.e., longitudes measured positively to the west) will be used when the rotation is prograde and east longitudes (i.e., longitudes measured positively to the east) when the rotation is retrograde »? It seems a contradiction.
If the aim is to convert the planetographic coords into celestial coords, by using the equation for W (the celestial longitude of the prime meridian, more or less) for the specified time, and then adding the planetographic longitude, then the latter ought to run eastward in all cases (prograde and retrograde). So I guess that's not the aim, then. It also seems to me that the planetographic longitude of the prime meridian is, by definition, always zero --it is only its celestial longitude that spins around. I'll have to sleep on this.
Urhixidur 04:33, 2005 Jan 4 (UTC)
Ah, now I understand. I was confusing prime meridian with central meridian. The central meridian is the meridian above which the observer is hovering --this changes with time as the object rotates underneath the observer, obviously, and that is what is meant in the first quote. Prograde objects have their longitudes labeled from east to west. Obvious, now.
Urhixidur 14:04, 2005 Jan 4 (UTC)

WGS84 has longitude positive eastward, negative westward. Urhixidur 14:11, 2005 Jan 4 (UTC)

From a strictly mathematical point of view, our usual standard is to use a right-hand coordinate system, in which angles are measured in a counterclockwise fashion. --Jacobolus 00:10, 15 Feb 2005 (UTC)

And counterclockwise viewed from above the north pole equal to clockwise viewed from above the south pole, so that doesn't help unless you say which pole is preferred. Michael Hardy 01:26, 15 Feb 2005 (UTC)
I added an explanation of the east-is-positive convention. Also note that the apparent annual motion of the Sun is eastward along the ecliptic, and many of the planets (Earth, Mars, Jupiter, Saturn) rotate west to east. I think maybe Neptune, too. It might be worth adding a remark about the Sun's motion along the ecliptic.

[edit] Fuss with clocks

I am curious, however about all the fuss with clocks. Yes, clocks are great and solve the longitude problem. I have taught that. But the motion of the Moon in relation to the Sun and to the stars can be used to tell time, too. Just occurred to me that though it's a bit complicated and not quite so accurate, it might work to low accuracy, but good enough for simple navigation, say within a few degrees longitude.

Pdn 05:19, 25 Feb 2005 (UTC)

If you're not using clocks and only using astronomical objects to find your location, you're doing astronomy instead of navigation. After measuring angles to the Sun, Moon, and stars, you then have to go solve simultaneous equations to find at what time all those objects are at those locations (unless you have astronomical location-time tables) and at what location all the angles match your readings. Moon rise alone doesn't tell you where you are if you don't know the time. If you know what time moonrise was at Greenwich and you know your time you can compare the two times. If you don't know the time then you have to measure at least one other thing which varies on a useful time scale (the movement of Venus is fast enough that you might be able to figure out within several days what the difference is between you and a standard, while the rise time of Rigel changes slowly enough that you might have to watch it for months in order to have enough data to compare it to moonrise and find your location). If you have a chronometer with only a minute and second hand then you'd be able to compare times between events and get better data than only comparing relative angles. -- SEWilco (talk) 15:19, 25 February 2008 (UTC)
Navigation is navigation regardless of how it's done; it doesn't become astronomy just because you are measuring the positions of celestial objects. As far as solely using celestial objects for navigation, see Lunar distance method. --Michael Daly (talk) 17:38, 25 February 2008 (UTC)
You're right. I was actually referring to the amount and type of work needed rather than the purpose. With no timekeeping instrument the task of locating one's position is much more difficult. I've considered astronomical navigation methods on the past, and although they can be used in some situations (primary requirements are suitable measuring tools and being able to wait while getting enough data), a clock and accurate astronomical info greatly reduce the location effort. Having only astronomical info generally requires using that to solve for the local time/position of the Earth. The Lunar distance method depends upon an astronomer having earlier calculated all the necessary tables based upon the astronomer's clock. I was referring to the mentioned problem of calculations from the motion of astronomical objects. -- SEWilco (talk) 17:54, 25 February 2008 (UTC)

[edit] History

I have removed the reference to shipwrecks in Western Australia, since only 5 are known of in early times: English East India Company's, Trial (1622), the Dutch East India Company's Batavia (ship) (1629), Vergulde Draeck (1656), Zuytdorp (1712) and Zeewijk (1727). For these early wrecks I dont know to determine the role of longitude navigation errors as opposed to chart errors (uncharted reefs) and other causes. Anyway ship wrecks were occuring in many parts of the world, motivating the search for a solution. - Op. Deo 20:18, 23 August 2005 (UTC)

The clocks were not so good, but could not the ancient mariners have measured time by the motion of the moon, at least when it's visible at the same time as the sun, so you can get the angle between the two? Pdn

Yes this is the method of Lunar Distances which was the practical competitor to Harrison's chronometer in C18. It was at first limited because the orbit of the moon was not known by ship's officers. But eventually suitable tables were produced which made it easy to measure the moons position relative to certain stars on the ecliptic and determine local time, and therefore longitude. It was then cheaper to provide ships with a set of tables and a sextant, rather than a very expensive chronometer modelled on Harrison's design, So in some ways the impact of Harrison's successful toil was limited. - Op. Deo 22:29, 23 August 2005 (UTC) Thanks. I just saw that Galileo devised a method of using the Moons of Jupiter. see [1] which says: "For Galileo, his admonition marked the beginning of a period of silence. He busied himself with such tasks as using tables of the moons of Jupiter to develop a chronometer for measuring longitude at sea. He endured his rheumatism, ..." Pdn 04:40, 24 August 2005 (UTC)

[edit] Discrepancy between this and Latitude

Latitude says that they are not "exactly" equal to 1 nautical mile whereas this article says that Latitude is "exactly" equal to 1 nortical mile! This seems to be to be something that needs clearing up. —Preceding unsigned comment added by 58.105.29.26 (talk • contribs)

The latitude statement that one minute of latitude is very close to one nautical mile is correct for the reasons it states. Nevertheless, a minute of latitude was indeed originally exactly one nautical mile at the time it was defined, when it was thought that the Earth was a perfect sphere (rather than an oblate spheroid with an undulating geoid). — Joe Kress 01:17, 16 June 2006 (UTC)

The 4th paragraph should be changed, both because it is factually wrong (twice) and because it contradicts the entry under "Latitude." This creates confusion and introduces uncertainty as to which is correct. The differences may seem small but I was led to this entry while trying to resolve an apparent error in the ground track of a satellite orbit simulator. The error was resolved by using the correct definition of latitude as shown by the Wikipedia entry for that parameter (as opposed to the one implied here.) Mennochio 23:38, 30 July 2006 (UTC)

I agree. I also suspect it is worth pointing to nautical miles, statute miles, and the Great Circle Distance Formula in the same discussion. As is, the formula for equivalency in one long. deg. is misleading (i.e., are degrees in radians? do I get km, really?) 71.236.220.71 16:31, 25 October 2006 (UTC)

[edit] Coordinate system

I think the article should emphasize the primary purpose of longitude: that together with the latitude it uniquely identifies any point on the surface of Earth (or other celestial body). This is mentioned only in the third paragraph, saying

A specific longitude may then be combined with a specific latitude to give a precise position on the Earth's surface.

and in the link to Geographic coordinate system in the See also section. I am not sure yet how to change the article to emphasize this more, so I did not edit it. – b_jonas 21:08, 19 June 2006 (UTC)

[edit] Mile conversion

How many miles is 15 degrees (1 timezone) of longitude? --HomfrogTell me a story! 00:41, 30 December 2006 (UTC)

A time zone is similar the section of an orange—it is fat at the equator and slims down to a point at either pole, so its width varies from 1670 km (1037 statute miles) at the equator to 0 km at the poles. Mathematically, 15° of longitude = (1669.800 + 5.595sin²φ)cosφ km, where φ is the latitude, varying from 0° to 90°. Multiply by 0.621 for statute (land) miles, and by 0.540 for nautical miles. Technically, each time zone is a gore, 24 of which can be combined to form a flat world map which can be cut so that it can be glued onto a sphere to create a globe, covering the entire sphere without any gaps or overlaps. The fattest part of a gore touches both neighbors at the equator even when lying flat, but their points all touch at the poles when glued onto a sphere. An example of such a world map/globe with only twelve gores (rather than 24) is here. — Joe Kress 21:19, 31 December 2006 (UTC)
You might want to work that explanation into globe, and link to it from gore. Andy Mabbett 00:45, 17 March 2007 (UTC)

[edit] Measurement of longitude

I suggest a fork of the section ==History of the measurement of longitude== to a new article named measurement of longitude (as a starter).

The German article de:Längenproblem discusses also earlier geometric methods as well as the more curios proposals (firing cannons, injuring dogs and the like), while this article, as well as Board of Longitude and John Harrison, cover much of the same. -- Matthead discuß!     O       09:04, 25 July 2007 (UTC)

I agree - strongly. In fact, I think we need to reorganize the following pages:
Longitude
Board of Longitude
Longitude Prize
John Harrison
Method of lunar distances
and probably a few others. They repeat lots of information and much of it does not agree very well. I've been trying to get them in sync with respect to such things as whether the Longitude Prize was awarded to Harrison (it was not - it was never awarded, though small awards were made to many) and the various dates of events. There has been a non-agreement on the pages as to the relative importance of the chronometer and the lunar distance method.
If all these pages are factored so that they only reference the same history article, then a lot of the discrepancies will disappear and any reader can find a consistent story.
I strongly suggest a different name - since longitude is not measured directly, but calculated based on other measurements, I'd suggest calling it History of longitude determination.
I guess this has to be put up as a proposal, since it's a lot of change to be done. --Michael Daly 21:13, 27 October 2007 (UTC)
The general reader would ignore your distinction that longitude is not measured directly. "History of longitude" or "Longitude history" would be a more appropriate term. Be mindful of Wikipedia:Naming conventions which requires that "Names of Wikipedia articles should be optimized for readers over editors; and for a general audience over specialists." — Joe Kress 05:22, 29 October 2007 (UTC)
Point taken. History of longitude would be my choice. Thanks! --Michael Daly 05:50, 29 October 2007 (UTC)
BTW - I think that any information pertaining to odd proposals involving dogs, cannon etc, should go in the Board of Longitude article, since it had to deal with them. These proposals never contributed to longitude nor represent more than amusing anecdotes relating to the history of longitude. --Michael Daly 19:14, 30 October 2007 (UTC)

[edit] Lack of dates in the "Longitude Act and Harrison's chronometer"

There are very few dates given in this section. I wanted to know when Harrison built his various chronometers, when the Longitude Prize was offered, when the voyages occurred, etc. Could someone with access to reference materials please provide them? --Sapphire Wyvern 01:50, 9 August 2007 (UTC)

All such info was moved to History of longitude and the details on Harrison's chronometers remain in John Harrison. --Michael Daly 19:15, 5 November 2007 (UTC)

[edit] Minutes and Seconds?

Could someone add information on why degrees are divided into "minutes" and "seconds"? Is there some relation to time that I am missing or is 60 simply a useful number to divide from hence a clock analogy? —Preceding unsigned comment added by 142.179.217.154 (talk) 00:13, 8 September 2007 (UTC)

[edit] History of longitude created - history section removed.

As per previous comments above, I've split out the history section and merged it with similar history sections in several other articles. This is now a single article History of longitude. --Michael Daly 19:07, 30 October 2007 (UTC)

[edit] Unspecified meaning of the surface

The following paragraph is found in the section on noting and calculating longitude.

A degree of latitude is always around sixty nautical miles, 111 km or 69 statute miles. This can be expressed exactly as 111.1334 - 0.5594 cos (2φ) + 0.0012 cos (4φ), where φ is latitude. A degree of longitude varies from 0 to 111 km and is 111 km times the cosine of the latitude, when the distance is laid out on a circle of latitude. More precisely, one degree of longitude = (111.320 + 0.373sin²φ)cosφ km, where φ is latitude).

It does not specify anything about the shape of the "Earth" it is using. Obviously it is not a sphere, but exactly what model is referred to is not specified. If someone recognizes the equations and can identify the model used, could they please state what it is? --Michael Daly 19:13, 5 November 2007 (UTC)

I think you will find that it is being modelled as an oblate spheroid. The radius in the equatorial plane of the Earth is about 0.35% larger than the radius at the two poles. This accounts for the factor 0.373/111.32 increase in the length of the degree of longitude at the pole relative to what would be given by the simple (111.329*cosφ) km which would hold for a perfect sphere. As far as latitude is concerned, one must first ask what definition of latitude is being considered? The geographic latitude is defined as the angle between the normal to the earth's surface and the equatorial plane. The equation given for length of a degree of latitude appears to be consistent with this definition. Note however, the equation is not applicable to other definitions of latitude such as the geocentric latitude, for which the equatorial degrees of latitude and longitude would be virtually the same. Op. Deo 22:48, 5 November 2007 (UTC)
The article states it is geographic latitude. Ooops - I just saw that you added that!
So what would the corresponding equation for longitude be? --Michael Daly 23:19, 5 November 2007 (UTC)
I think the longitude situation is rather simpler, and the equation given seems correct to me as noted above. Basically if one takes a slice of the earth at a given latitude then to high degree of accuracy the earth's circumference at that latitude is a circle. In my attempt to visulise the geometry, a geocentric latitude would be identical to a geographic latitude because the radius at any point will be normal to the circle of latitude. Op. Deo 21:32, 6 November 2007 (UTC)

Both the formula I added for a degree of longitude and the formula added by CielProfond for a degree of latitude represent old ellipsoids and are not exact. The most recent parameters for Earth's ellipsoid from IERS Conventions (2003) (Chp. 1, page 12) are an equatorial radius a = 6378136.6 m and an inverse flattening 1/f = 298.25642. The latter yields e² = 0.0066943980 using e² = 2f−f². An exact expression for the radius of curvature along a meridian (north-south) from The Math Forum is:

M = a(1−e²)/(1−e²sin²φ)3/2

Multiply by 2π/360 to obtain the length of a degree of latitude at a geodetic latitude φ. At the equator (φ = 0°), this reduces to (2π/360)a(1−e²) = 110.57427 km/°. At the poles (φ = 90°), this reduces to (2π/360)a/(1−e²)1/2 = 111.69397 km/°.

As the osculating circle with this radius of curvature moves around the ellipse formed by a meridian and its anti-meridian, its center of curvature moves along the evolute of the ellipse, the astroid, a figure with four concave curves connecting four cusps aligned with the axes of the ellipse, the southern arcs and cusp being the center of curvature for northern latitudes and the north pole, respectively, and vice-versa for southern latitudes.

An exact expression for the radius of curvature along a line of latitude (east-west, but in the prime vertical, the plane perpendicular to the plane of the meridian and also perpendicular to the plane tangent to the surface of the ellipsoid) from The Math Forum is:

N = a/(1 − e²sin²φ)1/2

Multiply by (2π/360)cosφ to obtain the length of a degree of longitude. At the equator (φ = 0°), this reduces to (2π/360)a = 111.31948 km/°. At the poles, of course, this is 0 km/°.

The center of the radius of curvature in the prime vertical is always on the polar diameter, moving from the geocenter at a latitude of 0° to a point on the diameter farthest from the pole at a latitude of 90°. Both radii of curvature have the same value at the poles. These formulae for the radii of curvature also appear on pages 24–25 of Map Projections—A Working Manual by John P. Snyder (1987). — Joe Kress 04:59, 7 November 2007 (UTC)

The lay person is probably not interested in the formulae, but would like some idea of how the length of a degree of longitude varies. Perhaps a table, or a set of examples (e.g. "at the equator, a degree of longitude is 110 km. As you move towards the poles, the length of a degree of latitude get smaller, for example at the latitude of Melbourne it is 90km") would be appropriate in addition to what's already there? —Preceding unsigned comment added by Ac1201 (talk • contribs) 23:34, 2 December 2007 (UTC)

Although most will not be interested in the exact formulae, some request them as in Google Answers (never provided). I'm added a section entitled "Degree length" encompassing most of my previous edit, without mentioning the astroid. But I am including a table for latitudes every 15° per your request. — Joe Kress (talk) 03:36, 25 December 2007 (UTC)

[edit] Image

In my opinion the image is too small to be useful. I would like at least for the prime meridian to be visible. (possibly a higher resolution image?) Also the longitudes shown on the map might be difficult for some to see. jay (talk) 19:38, 29 December 2007 (UTC)

Usually if you click on an image you can get a larger version. -- SEWilco (talk) 15:21, 25 February 2008 (UTC)

[edit] pronunciation

IPA would be helpful. —Preceding unsigned comment added by 81.140.57.113 (talk) 16:38, 27 February 2008 (UTC)

Done and thanks, because I found I have always pronounced it wrong. I said [g] instead of [dʒ]. −Woodstone (talk) 20:11, 27 February 2008 (UTC)
You weren't wrong. The OED gives both, but I've never heard the dʒ version. I think it's an americanism. Algebraist 23:59, 27 March 2008 (UTC)