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- Newsgroups: sci.math
- Path: sparky!uunet!stanford.edu!CSD-NewsHost.Stanford.EDU!Sunburn.Stanford.EDU!pratt
- From: pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt)
- Subject: Re: Mercator Projection
- Message-ID: <1992Nov10.060543.25580@CSD-NewsHost.Stanford.EDU>
- Sender: news@CSD-NewsHost.Stanford.EDU
- Organization: Computer Science Department, Stanford University.
- References: <israel.721212129@unixg.ubc.ca> <1992Nov8.214329.27209@CSD-NewsHost.Stanford.EDU> <7868@charon.cwi.nl>
- Date: Tue, 10 Nov 1992 06:05:43 GMT
- Lines: 26
-
- In article <7868@charon.cwi.nl> dik@cwi.nl (Dik T. Winter) writes:
- >In article <1992Nov8.214329.27209@CSD-NewsHost.Stanford.EDU> pratt@Sunburn.Stanford.EDU (Vaughan R. Pratt) writes:
- > > Turns out if you try to calculate this using the 1986 Encyclopedia
- > > Britannica you get sec^3(t). The reason is that EB defines the
- > > Mercator Projection to be the result of projecting the globe from its
- > > center onto the cylinder tangent to the equator. If this were true the
- > > vertical direction would scale not by sec(t) but by the derivative of
- > > tan(t), namely sec^2(t).
- >
- >EB is right.
-
- I think you're alone on this one. Alan Paeth mentioned in an earlier
- post that this was a common but incorrect assumption.
-
- >It is not immediately obvious that the Rand McNall map is conformal;
- >is it? The Mercator projection is. But I do not think that straight
- >lines on the Rand McNall map are lines of constant bearing.
-
- You are right that Mercator is conformal. Since the horizontal
- dimension necessarily scales as sec(t) to keep the meridians vertical,
- an easy check for conformality is to measure the vertical scale and
- verify that it scales as sec(t). If you get anything else your map
- cannot be conformal. Rand McNally measures out to sec(t), surprisingly
- accurately (even more accurate than my Wil Tirion Sky Atlas 2000.0!).
- --
- Vaughan Pratt
-