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An even simpler formula for Lambert's R is
R(phi) = 2 sin((pi-phi)/2).
This is equivalent to
R(phi) = sqrt((2(1+cos phi))
by the trig identity
cos(2theta) = 1 - 2 sin^2(theta),
and relates directly to the diagram. Going into Cartesian coordinates and back is unnecessary.
If phi is measured from S rather than from its antipode, the formula is even simpler:
R(phi) = 2 sin(phi/2).
This works, not because of the diagram, but because it satisfies the differential equation
sin(phi) = R(phi) R'(phi),
which guarantees that R(phi) has the correct amount of compression. The diagram is just a clever way to evaluate the function geometrically.
HuMcCulloch ( talk) 16:26, 4 March 2009 (UTC)
Thanks! I'll leave it to you, since I prefer not to edit Wiki pages if possible -- it would be frustrating to have one's insights revised by untold other editors.
The required differential equation follows from the fact that an annulus at angle phi from S has area 2 pi sin(phi) d phi. Under transformation R(phi), it has area 2 pi R(phi) dR(phi) = 2 pi R(phi) R'(phi) d phi. Equating these gives the differential equation. R(phi) = 2 sin(phi/2) solves it with R(0) = 0, and so must be the solution.
Then, if d is the distance from S to P, d is the base of an isosceles triangle with two unit legs and head angle phi. It follows that d/2 = sin(phi/2), whence d = R(phi), not because it is this distance, but because it is the same as R(phi), which solves the required compression equation. —Preceding unsigned comment added by HuMcCulloch ( talk • contribs) 03:50, 5 March 2009 (UTC)
Lambert is listed as “Swiss” under his biographical page. Digging deeper, he was born in Milhüsa (Mulhouse) and lived there into early adulthood. Milhüsa joined the Swiss confederacy in 1515 as an enclave surrounded by Alsatian territory, long before Lambert’s birth, and remained in the Swiss confederacy until 1798, after Lambert’s death. He lived and studied and worked in Chur, Switzerland, for many years, but worked in Berlin for the last 13 years of his life. It’s hard to say how he would have answered someone who asked about his nationality by the time he died, but “Alsatian” probably was not a candidate. “Swiss” is presumably how he would have answered up until his forties, at least. So… changing the text to “Swiss” from “Alsatian”. Strebe ( talk) 01:42, 6 February 2011 (UTC)
I will be replacing images on the various map projection pages. Presently many are on a satellite composite image from NASA that, while realistic, poorly demonstrates the projections because of dark color and low contrast. I have created a stylization of the same data with much brighter water areas and a light graticule to contrast. See the thumbnail of the example from another article. Some images on some pages are acceptable but differ stylistically from most articles; I will replace these also.
The images will be high resolution and antialiased, with 15° graticules for world projections, red, translucent equator, red tropics, and blue polar circles.
Please discuss agreement or objections over here (not this page). I intend to start these replacements on 13 August. Thank you. Strebe ( talk) 22:42, 6 August 2011 (UTC)
The explanation of how to use the Schmidt Net seems way over-complicated. If the azimuth in use is on the equator (e.g. if plotting the east or west hemisphere of the world), then plotting a map onto the Schmidt Net is a simple process of plotting each grid square into the corresponding distorted square on the Schmidt Net. The complicated procedure currently explained in this section would only be used if the azimuth were one of the Earth's poles. The animated demonstration is simply a case of using the Schmidt Net's vertical axis as a scalar calibration of the radial distance for a polar reprojection. There would be no reason to plot such a projection onto a Schmidt Net anyhow because the Net's lines would not represent parallels or meridians. A polar Lambert azimuthal net should be used instead, with concentric circles representing the parallels of latitude, and radial lines representing meridians converging at the pole. Reprojecting a map would again be a trivial case of re-plotting each grid square. The use of x, y, z coordinates on a sphere is also extremely confusing - in practice, most people would be starting with a latitude-longitude location. I will try to amend this section, but I am not a professional cartographer, so please assist by improving upon my attempts. Ozzorro ( talk) 01:02, 22 July 2012 (UTC)
Thanks - I wasn't aware of the use of the Lambert projection was used in geology - Schmidt nets are probably rarely used today in cartography these days, where paper-based methods have largely been superseded by GIS mapping.
However, the old plotting did rather miss the whole point of the Schmidt net: it is a gridded sphere projected onto a paper disc, with the meridians running from top to bottom. i.e. the z-axis of the spherical coordinate system has been projected onto the vertical axis of the Schmidt net.
The old explanation assumed the z-axis is perpendicular to the page, and converted the cartesian coordinates (0.321, 0.557, -0.766) into spherical coordinates relative to this axis. It then used the Schmidt net as a glorified protractor and ruler to plot polar coordinates onto the disc.
To demonstrate how it can be done more easily, I'll use the same point as before in terms of its 3-dimensional position relative to the disc, but I'll rotate the coordinate system so that the z-axis points to the top of the Schmidt net (which also points the y-axis to the back). This transforms the point's coordinates to (0.321, 0.766, 0.557) - it's the same point, just in a different coordinate system now.
Converting to spherical coordinates: angle above the x-y plane = arctan(z/sqrt(x*x+y*y)) = 33.8 degrees angle projected on the x-y plane = arctan(y/x) = 67.3 from the x-axis = 22.7 degrees from the y-axis
If you have a look at where the point gets plotted on the Schmidt net in the original animation, you'll see that it is indeed 2.27 grid units to the right, and 3.38 units above the centre of the grid. So if the spherical coordinates are calculated relative to a vertical axis, you get directly to numbers you can plot as a grid position on the Schmidt net. This really is the whole point of a Schmidt net - you don't need to go through all the rigmarole of rotating pieces of tracing paper etc.
Do you think we can turn this into an appropriate explanation & example?
Incidentally, from my reading of a few geology books, the article's claim of the Lambert projection's use in geology seems to be incorrect. A stereonet (ie stereographic projection of a sphere) is required to determine the orientation of faults, because it preserves angles. The Lambert projection is used when evaluating the density of points in a sample, due to its area-equivalency.
Ozzorro ( talk) 18:25, 22 July 2012 (UTC)
Thanks, I've raided my university's library and tracked down two of the books you suggested, plus a few more on structural geology, mineralogy and crystallography. I see now how the Wulff net and Schmidt net are used, and why. The stereonet is really a paper-based graphical technique for solving a whole range of 3D geometry problems - e.g. measuring the angle between two planes and determining the axis of intersection. I've done these sorts of calculations using vector maths, which is quite cumbersome, but I imagine the stereonet would be much simpler to learn, more intuitive to the user, and works entirely with angles. Quite nifty really!!
I think we need to make this very clear - that the stereonet method is not just about projecting information onto a sphere, but it's about using the hemisphere as a virtual mathematical space, plotting the intersection with the sphere of lines and planes of various orientations, and then using it as a tool to answer real-world problems.
I've written up a full explanation of the use of a stereonet (see below), illustrated with examples from structural geology (eg. deriving fold axes etc) - I might need you to check that my applications are valid, and I'd like to make an animation to demonstrate all these operations. I think that the stereonet deserves its own page, since it has a distinct purpose from the simple projection, and is really a mathematical tool that happens to utilise the projection. The Schmidt net could be included as a section within this same page - as you've mentioned, it's not technically a 'stereonet', but is commonly referred to as such in practice. Here's a draft: http://en.wikipedia.org/wiki/Wikipedia_talk:Articles_for_creation/stereonet Ozzorro ( talk) 18:59, 26 July 2012 (UTC)
Hello! This is a note to let the editors of this article know that File:Lambert azimuthal equal-area projection SW.jpg will be appearing as picture of the day on September 7, 2016. You can view and edit the POTD blurb at Template:POTD/2016-09-07. If this article needs any attention or maintenance, it would be preferable if that could be done before its appearance on the Main Page. — Chris Woodrich ( talk) 23:27, 24 August 2016 (UTC)
Can you add Polar version, like in Azimuthal equidistant projection? — Preceding unsigned comment added by 2A01:119F:21D:7900:59D0:EDEC:7FD4:7271 ( talk) 17:12, 27 April 2017 (UTC)
Hello fellow Wikipedians,
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I was very confused to come to this page and see cosine of half the colatitude used instead of the sine of half the colatitude. Without carefully reading, I had been expecting an orientation-preserving (when the globe is seen from outside) map centered at the north pole, instead of an orientation-inverting (or inside-out if you like) map centered at the south pole. Particularly since the colatitude starts from 0 at the north pole; this map is putting 0 colatitude at its outside rim and π colatitude at the origin.
Since most of the pictures here show outside-in maps of the earth, as do general references seen in cartography, I think it would be helpful to switch the formulas and description to be centered on the north pole.
Is anyone strongly attached to the current version?
– jacobolus (t) 20:52, 19 March 2021 (UTC)
I've started looking through some of the sources to try to find the Cartesian coordinates listed in the article. It's unclear if the Borradaile citation is just for the spherical coordinates or is also meant to give the cartesian ones. Is this original work? I'm fairly sure it's correct (I can verify that myself) but I'd like to cite it for an short article. I'll be seeing if I can get my hands on Borradaile to verify myself but It would be nice to know if this actually appears in said book before I go through the lengthy library loan process. — Preceding unsigned comment added by 198.109.218.2 ( talk) 17:41, 9 September 2022 (UTC)
I've gone ahead and tried to show the equivalence of and by pushing through the polar coordinates . They are equivalent-ish but maybe it is worth mentioning that they don't have the same setup in the main article. By the half angle formula . The second equality follows from being a colatitude . By this and a right triangle relation on the angle , we get . Since this is on a unit sphere , we get . So by cancelling , we get . The scaling factor of that appears in the article is mentioned in one of the references to make the projection for the whole sphere fit in radius 2. The sign difference on the z is explained by the polar formula picking the center to be the antipodal point S=(0,0,1) of the cartesian projection center S=(0,0,-1). — Preceding unsigned comment added by 73.191.198.100 ( talk) 21:14, 14 September 2022 (UTC)
This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||||||||||||||||||||||
|
An even simpler formula for Lambert's R is
R(phi) = 2 sin((pi-phi)/2).
This is equivalent to
R(phi) = sqrt((2(1+cos phi))
by the trig identity
cos(2theta) = 1 - 2 sin^2(theta),
and relates directly to the diagram. Going into Cartesian coordinates and back is unnecessary.
If phi is measured from S rather than from its antipode, the formula is even simpler:
R(phi) = 2 sin(phi/2).
This works, not because of the diagram, but because it satisfies the differential equation
sin(phi) = R(phi) R'(phi),
which guarantees that R(phi) has the correct amount of compression. The diagram is just a clever way to evaluate the function geometrically.
HuMcCulloch ( talk) 16:26, 4 March 2009 (UTC)
Thanks! I'll leave it to you, since I prefer not to edit Wiki pages if possible -- it would be frustrating to have one's insights revised by untold other editors.
The required differential equation follows from the fact that an annulus at angle phi from S has area 2 pi sin(phi) d phi. Under transformation R(phi), it has area 2 pi R(phi) dR(phi) = 2 pi R(phi) R'(phi) d phi. Equating these gives the differential equation. R(phi) = 2 sin(phi/2) solves it with R(0) = 0, and so must be the solution.
Then, if d is the distance from S to P, d is the base of an isosceles triangle with two unit legs and head angle phi. It follows that d/2 = sin(phi/2), whence d = R(phi), not because it is this distance, but because it is the same as R(phi), which solves the required compression equation. —Preceding unsigned comment added by HuMcCulloch ( talk • contribs) 03:50, 5 March 2009 (UTC)
Lambert is listed as “Swiss” under his biographical page. Digging deeper, he was born in Milhüsa (Mulhouse) and lived there into early adulthood. Milhüsa joined the Swiss confederacy in 1515 as an enclave surrounded by Alsatian territory, long before Lambert’s birth, and remained in the Swiss confederacy until 1798, after Lambert’s death. He lived and studied and worked in Chur, Switzerland, for many years, but worked in Berlin for the last 13 years of his life. It’s hard to say how he would have answered someone who asked about his nationality by the time he died, but “Alsatian” probably was not a candidate. “Swiss” is presumably how he would have answered up until his forties, at least. So… changing the text to “Swiss” from “Alsatian”. Strebe ( talk) 01:42, 6 February 2011 (UTC)
I will be replacing images on the various map projection pages. Presently many are on a satellite composite image from NASA that, while realistic, poorly demonstrates the projections because of dark color and low contrast. I have created a stylization of the same data with much brighter water areas and a light graticule to contrast. See the thumbnail of the example from another article. Some images on some pages are acceptable but differ stylistically from most articles; I will replace these also.
The images will be high resolution and antialiased, with 15° graticules for world projections, red, translucent equator, red tropics, and blue polar circles.
Please discuss agreement or objections over here (not this page). I intend to start these replacements on 13 August. Thank you. Strebe ( talk) 22:42, 6 August 2011 (UTC)
The explanation of how to use the Schmidt Net seems way over-complicated. If the azimuth in use is on the equator (e.g. if plotting the east or west hemisphere of the world), then plotting a map onto the Schmidt Net is a simple process of plotting each grid square into the corresponding distorted square on the Schmidt Net. The complicated procedure currently explained in this section would only be used if the azimuth were one of the Earth's poles. The animated demonstration is simply a case of using the Schmidt Net's vertical axis as a scalar calibration of the radial distance for a polar reprojection. There would be no reason to plot such a projection onto a Schmidt Net anyhow because the Net's lines would not represent parallels or meridians. A polar Lambert azimuthal net should be used instead, with concentric circles representing the parallels of latitude, and radial lines representing meridians converging at the pole. Reprojecting a map would again be a trivial case of re-plotting each grid square. The use of x, y, z coordinates on a sphere is also extremely confusing - in practice, most people would be starting with a latitude-longitude location. I will try to amend this section, but I am not a professional cartographer, so please assist by improving upon my attempts. Ozzorro ( talk) 01:02, 22 July 2012 (UTC)
Thanks - I wasn't aware of the use of the Lambert projection was used in geology - Schmidt nets are probably rarely used today in cartography these days, where paper-based methods have largely been superseded by GIS mapping.
However, the old plotting did rather miss the whole point of the Schmidt net: it is a gridded sphere projected onto a paper disc, with the meridians running from top to bottom. i.e. the z-axis of the spherical coordinate system has been projected onto the vertical axis of the Schmidt net.
The old explanation assumed the z-axis is perpendicular to the page, and converted the cartesian coordinates (0.321, 0.557, -0.766) into spherical coordinates relative to this axis. It then used the Schmidt net as a glorified protractor and ruler to plot polar coordinates onto the disc.
To demonstrate how it can be done more easily, I'll use the same point as before in terms of its 3-dimensional position relative to the disc, but I'll rotate the coordinate system so that the z-axis points to the top of the Schmidt net (which also points the y-axis to the back). This transforms the point's coordinates to (0.321, 0.766, 0.557) - it's the same point, just in a different coordinate system now.
Converting to spherical coordinates: angle above the x-y plane = arctan(z/sqrt(x*x+y*y)) = 33.8 degrees angle projected on the x-y plane = arctan(y/x) = 67.3 from the x-axis = 22.7 degrees from the y-axis
If you have a look at where the point gets plotted on the Schmidt net in the original animation, you'll see that it is indeed 2.27 grid units to the right, and 3.38 units above the centre of the grid. So if the spherical coordinates are calculated relative to a vertical axis, you get directly to numbers you can plot as a grid position on the Schmidt net. This really is the whole point of a Schmidt net - you don't need to go through all the rigmarole of rotating pieces of tracing paper etc.
Do you think we can turn this into an appropriate explanation & example?
Incidentally, from my reading of a few geology books, the article's claim of the Lambert projection's use in geology seems to be incorrect. A stereonet (ie stereographic projection of a sphere) is required to determine the orientation of faults, because it preserves angles. The Lambert projection is used when evaluating the density of points in a sample, due to its area-equivalency.
Ozzorro ( talk) 18:25, 22 July 2012 (UTC)
Thanks, I've raided my university's library and tracked down two of the books you suggested, plus a few more on structural geology, mineralogy and crystallography. I see now how the Wulff net and Schmidt net are used, and why. The stereonet is really a paper-based graphical technique for solving a whole range of 3D geometry problems - e.g. measuring the angle between two planes and determining the axis of intersection. I've done these sorts of calculations using vector maths, which is quite cumbersome, but I imagine the stereonet would be much simpler to learn, more intuitive to the user, and works entirely with angles. Quite nifty really!!
I think we need to make this very clear - that the stereonet method is not just about projecting information onto a sphere, but it's about using the hemisphere as a virtual mathematical space, plotting the intersection with the sphere of lines and planes of various orientations, and then using it as a tool to answer real-world problems.
I've written up a full explanation of the use of a stereonet (see below), illustrated with examples from structural geology (eg. deriving fold axes etc) - I might need you to check that my applications are valid, and I'd like to make an animation to demonstrate all these operations. I think that the stereonet deserves its own page, since it has a distinct purpose from the simple projection, and is really a mathematical tool that happens to utilise the projection. The Schmidt net could be included as a section within this same page - as you've mentioned, it's not technically a 'stereonet', but is commonly referred to as such in practice. Here's a draft: http://en.wikipedia.org/wiki/Wikipedia_talk:Articles_for_creation/stereonet Ozzorro ( talk) 18:59, 26 July 2012 (UTC)
Hello! This is a note to let the editors of this article know that File:Lambert azimuthal equal-area projection SW.jpg will be appearing as picture of the day on September 7, 2016. You can view and edit the POTD blurb at Template:POTD/2016-09-07. If this article needs any attention or maintenance, it would be preferable if that could be done before its appearance on the Main Page. — Chris Woodrich ( talk) 23:27, 24 August 2016 (UTC)
Can you add Polar version, like in Azimuthal equidistant projection? — Preceding unsigned comment added by 2A01:119F:21D:7900:59D0:EDEC:7FD4:7271 ( talk) 17:12, 27 April 2017 (UTC)
Hello fellow Wikipedians,
I have just modified one external link on Lambert azimuthal equal-area projection. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
An editor has reviewed this edit and fixed any errors that were found.
Cheers.— InternetArchiveBot ( Report bug) 07:41, 16 December 2017 (UTC)
I was very confused to come to this page and see cosine of half the colatitude used instead of the sine of half the colatitude. Without carefully reading, I had been expecting an orientation-preserving (when the globe is seen from outside) map centered at the north pole, instead of an orientation-inverting (or inside-out if you like) map centered at the south pole. Particularly since the colatitude starts from 0 at the north pole; this map is putting 0 colatitude at its outside rim and π colatitude at the origin.
Since most of the pictures here show outside-in maps of the earth, as do general references seen in cartography, I think it would be helpful to switch the formulas and description to be centered on the north pole.
Is anyone strongly attached to the current version?
– jacobolus (t) 20:52, 19 March 2021 (UTC)
I've started looking through some of the sources to try to find the Cartesian coordinates listed in the article. It's unclear if the Borradaile citation is just for the spherical coordinates or is also meant to give the cartesian ones. Is this original work? I'm fairly sure it's correct (I can verify that myself) but I'd like to cite it for an short article. I'll be seeing if I can get my hands on Borradaile to verify myself but It would be nice to know if this actually appears in said book before I go through the lengthy library loan process. — Preceding unsigned comment added by 198.109.218.2 ( talk) 17:41, 9 September 2022 (UTC)
I've gone ahead and tried to show the equivalence of and by pushing through the polar coordinates . They are equivalent-ish but maybe it is worth mentioning that they don't have the same setup in the main article. By the half angle formula . The second equality follows from being a colatitude . By this and a right triangle relation on the angle , we get . Since this is on a unit sphere , we get . So by cancelling , we get . The scaling factor of that appears in the article is mentioned in one of the references to make the projection for the whole sphere fit in radius 2. The sign difference on the z is explained by the polar formula picking the center to be the antipodal point S=(0,0,1) of the cartesian projection center S=(0,0,-1). — Preceding unsigned comment added by 73.191.198.100 ( talk) 21:14, 14 September 2022 (UTC)