[MetaCRS] mercator scale calculation q
Mikael Rittri
Mikael.Rittri at carmenta.com
Mon Apr 27 03:49:23 EDT 2009
Hello Mark,
> Am I anywhere close to the solution?
Close, but no cigar.
It is not clear to me whether your Mercator uses formulas
for an ellipsoid-shaped Earth, or if you can assume that
the Earth is spherical.
The rule you quote, saying that the Mercator scale is the secant
of the latitude, applies to the spherical Mercator formulas.
For the ellipsoid formulas, I think this rule can give
errors up to 0.5 or 0.67 percent (or something like that).
Of course, that may be enough for your purposes.
A) First, let's assume that's enough.
secant(phi) is defined to be 1/cos(phi). Not as ACOS(phi),
if ACOS means arccos. (You may have been confused by the
exponent notation cos^(-1) for arccos.)
If your scale had been true at the equator, then the local
scale factor at latitude 34° would be 1/cos(34°) = 1.2062179485...
But your map has already been shrunk to make the true scale be
at latitude 31°. This shrinkage amounted to 1/cos(31°) = 1.1666333972...
That is, all distances had to be divided by 1.1666333972...
So, the local scale factor at latitude 34°, in your scale,
equals (1/cos(34°)) / (1/cos(31°)) = cos(31°)/cos(34°) = 1.0339305830...
B) If the spherical formula is not enough, then the proper formula
for local scale factor in ellipsoid Mercator (with true scale at equator)
is
k = sqrt( 1 - e^2 * sin^2(phi)) / cos(phi)
where e^2 is the squared eccentricity of the ellipsoid.
(Source: John P. Snyder, Map Projections: A Working Manual,
page 44, see http://pubs.er.usgs.gov/usgspubs/pp/pp1395 )
That is, e^2 = (2-f)*f, where f is the flattening.
For WGS84, f = 1/298.257223563, which means that e^2 = 0.00669437999014.
So, by the analogous reasoning as under A), we get that the local
scale factor you want is
sqrt( 1 - e^2 * sin^2(34°)) / cos(34°)
--------------------------------------
sqrt( 1 - e^2 * sin^2(31°)) / cos(31°)
which equals
0.9989527964 / 0.8290375726
---------------------------
0.9991117157 / 0.8571673007
or
1.0337661254
(which is only about 0.016 percents smaller than the spherical formula said).
Best regards,
--
Mikael Rittri
Carmenta AB
SWEDEN
www.carmenta.com
-----Original Message-----
From: metacrs-bounces at lists.osgeo.org [mailto:metacrs-bounces at lists.osgeo.org] On Behalf Of Mark Amend
Sent: den 26 april 2009 05:27
To: metacrs at lists.osgeo.org
Subject: [MetaCRS] mercator scale calculation q
Hi there-
This might not be an appropriate post for this list, but I'm struggling to find a solution. Any ideas would be greatly appreciated.
I have a Mercator chart I'm building, at a scale of 1:5000 (Natural scale, at the equator, or in my case he projection has a defined standard parallel of 31N, where the scale factor is 1.00). And yes, I have to use this projection. ;-)
The chart, however, is located north of 31N, somewhere around 34N in the middle of the chart (for clarity of my question, let's just say it is 34N). I have been asked what is the True Scale of my chart at the mid-latitude of that chart. So, it will not be exactly 1:5000 but something ± that, like 1:5043.1 or something (hypothetically speaking).
I need to find an answer.
How do I calculate this "true scale" at the mid-latitude of my chart?
It seems that Mercator scale is defined in various references as the secant of the latitude. If that is the case, then ACOS(34 degrees?
radians?) should give 1.xxx or so, right? But the scale for the
latitude of interest (center of chart) is really 34 - 31 = 3, ..if 31 is where the scale factor is 1. Right? So it should be ACOS(3 degrees)? I then multiply 5000 by this number. Correct?
Am I anywhere close to the solution?
Thanks,
Mark Amend
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