[Proj] Standard projection for Mediterranean basin
j.l.h.hartmann at uva.nl
Tue Feb 16 06:01:04 EST 2010
Do you have a programmatic version of the algorithm? It's something that
would be really useful for a global map provider. I am working with the
Royal Tropical Institute here in Amsterdam (www.kit.nl), on an index for
their 10.000s of historical maps from all over the globe. In the long
run we would like to offer some thematic functionality, with an optimal
projection, of course
On 16-2-2010 4:55, strebe wrote:
> Jan & colleagues:
> Regarding the question of optimal standard parallels for Albers, I
> found my notes from 8 years ago. To summarize, there is no good rule,
> so perhaps the exact calculation I describe below has more importance
> than just academic.
> — daan Strebe
> The problem: given two latitudinal extents phi[N] and phi[S], find the
> optimal standard parallels for the Albers projection. Optimal is
> defined to mean that the maximum angular deformation between the two
> standard parallels equals the maximum angular deformation at the two
> latitudinal extents. In this way one is assured that the maximum
> deviation from true remains minimal throughout the region of interest.
> In a breathtakingly brief treatment, Melluish touches on the issue by
> describing how to select standard parallels such that the "error" on
> the extreme parallels is equal in magnitude and opposite in sign to
> the error of the central parallel. The distortion measurement is
> equivalent to mine, but the central parallel (the average of the north
> and south extreme parallels) generally does not carry the greatest
> distortion, so his method does not yield an optimal map. Hinks
> discusses the problem precisely, even noting that the central parallel
> is not the parallel of greatest distortion... but his treatment
> applies to the equidistant conic, not the equal-area, contrary to
> Snyder's comment in USGS P.P 1395.
> We define a family of Albers projections to be all Albers projections
> characterized by a common 'central' latitude at which angular
> deformation reaches a peak between the two standard parallels, or
> sinks to zero in the case of one standard parallel. I have placed
> 'central' between quote marks because it is not any simple average of
> the standard parallels. We can demonstrate that the ratio of the
> angular deformation of any two points on the map remains constant for
> every member of the family. This is the beautiful and useful
> characteristic of the family. The property arises out of a happy
> mathematical coincidence.
> First I declare the member of the family with only one standard
> parallel to be the canonical form. Because of the family's property of
> invariant scale ratios, we can first reduce the problem of choosing an
> optimal Albers projection to the single standard parallel case, which
> is much more tractable than trying to deal with two free parameters.
> This gives us a projection with zero distortion along the canonical
> parallel and identical distortion at the northern and southern
> extremities. Once we have found the canonical projection, we can
> choose the member of its family that produces the same magnitude of
> distortion at the canonical parallel as at the extreme parallels. We
> are guaranteed that the projection that fulfills this condition is one
> from the family of the canonical projection.
> Let phi[S] denote the southern parallel of interest; phi[N] the
> northern parallel of interest; phi the canonical parallel; phi a
> standard parallel; and phi the other standard parallel.
> phi[S] and phi[N] are given. From them we must determine phi,
> phi, and phi, such that the magnitude of the angular deformation
> at phi is the same as the magnitude of the angular deformation at
> phi[N] and phi[S]. Furthermore, there must be no greater angular
> deformation registered anywhere between the phi[S] and phi[N]. Given
> those conditions:
> 1) sin (phi) = A +/- sqrt (A^2 - 1)
> 2) A = (sin (phi[N]) * cos^2 (phi[S]) - sin (phi[S]) * cos^2 (phi[N]))
> / (cos^2 (phi[S] - cos^2 (phi[N]))
> That establishes the canonical parallel of the canonical form, and
> therefore the family, wherein lies the solution.
> If we arbitrarily select some phi, one may ask how to arrive at the
> correct phi within the family, given phi.
> 3) sin (phi) = [sin (phi) - 2 sin (phi[N]) + sin (phi) sin^2
> (phi[N])] / [2 sin (phi[N]) * sin (phi) - sin^2 (phi[N]) - 1]
> However, we do not yet know the proper phi. With much manipulation,
> we discover the iterative equation:
> 4) cos^2 (phi) * sqrt (1 + sin^2 (phi) - 2 sin (phi) * sin
> (phi[S])) - cos (phi[S]) (1 + sin^2 (phi - 2 sin (phi) sin
> (phi) = 0
> We solve for phi with (4) and then use (3) to compute phi. Now
> have the optimal Albers for any circumstance.
> The literature fixates quite a bit on rules of thumb for choosing the
> standard parallels, with some texts recommending inward 1/5th the
> meridional separation of the extreme parallels. Other texts recommend
> 1/7th, and yet others recommend 1/6th. My own calculations based on
> the above criteria show that the correct values range from 5% to 25%
> for reasonable conics, and the symmetry between north and south is not
> necessarily very good. Hence, no rule of thumb is very good. The
> higher the latitude of the northern parallel, the greater the
> asymmetry. The asymmetry remains remarkable even if both latitudes are
> high. For instance, extreme latitudes at 50° and 70° result in
> standard parallels placed optimally at roughly 10% below phi[N] and
> 18% above phi[S].
> Deetz's and Adams's recommendation of 29°30' and 45°30' for the US are
> pretty good, of course. Given maximum extents of 24°34' and 49°25',
> the optimal standard parallels are 28°42' and 46°19'. If we just can't
> bring ourselves to treat Lake of the Woods equitably then the upper
> standard parallel drops to 45°56'. The lower parallel only drops by
> 5'. The former map yield a maximum east-west scale distortion of
> 1.19%; the latter yields 1.15% (excluding Lake of the Woods, of
> course). Deetz's and Adams's recommendation yields 0.98% in the
> heartland; 1.25% at the northern border; 1.45% at the northern tip of
> the Lake of the Woods boundary; and 1.37% at Key West.
> On Feb 15, 2010, at 1:05:19 PM, "Jan Hartmann" <j.l.h.hartmann at uva.nl>
> From: "Jan Hartmann" <j.l.h.hartmann at uva.nl>
> Subject: Re: [Proj] Standard projection for Mediterranean basin
> Date: February 15, 2010 1:05:19 PM PST
> To: "PROJ.4 and general Projections Discussions"
> <proj at lists.maptools.org>
> Cc: strebe <strebe at aol.com>
> What's the 1/6 rule, Daan, and how much does it deviate from the
> optimal formula?
> BTW Nice Java applet to study projection deformations here:
> Pity they didn't make it possible to alter projection parameters
> like standard parallels.
> On 15-2-2010 21:04, strebe wrote:
>> I do not understand Dr. Mugnier's response. Using the authalic
>> latitude does not yield an equal-area projection if you project
>> using a conformal projection. I cannot think of any use for
>> mixing authalic latitude with conformal projection.
>> The purpose of the authalic latitude is to provide an equal-area
>> projection of the ellipsoid onto the sphere. From there you may
>> project to the plane using any spherical equal-area projection,
>> yielding an equal-area projection of the ellipsoid onto the
>> plane. While this technique will not preserve other properties of
>> the spherical projection (such as standard parallels or constant
>> scale along some meridian or somesuch), at least it drastically
>> increases the available projections for ellipsoidal use.
>> I have, by the way, formulæ somewhere that I derived for
>> computing the optimal standard parallels for Albers given the
>> north/south extents of the map. I will post them if I happen to
>> find them. The 1/6th rule would be fine for any real use, of
>> course. The optimal is more of academic interest.
>> — daan Strebe
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