[Proj] Re: "Double ellipsoid" case?
ndzinn at comcast.net
ndzinn at comcast.net
Wed Dec 3 15:59:47 EST 2008
Thanks for your revised assessment of maximum angular distortion in the Google Maps Projection (GMP).
Are you aware that the scale distortion in GMP is very large and unstated? In the vicinity of the Equator and it's about 1.0067 (that's 67cm per 100m). You can't compensate for it with the point scale factor equations of the Mercator projection in the second stage of the GMP (your alternative A) because it's embedded in the first stage of the GMP, the non-conformal mapping from WGS84 to the Google Sphere (GS). Here's how you can demonstrate this to yourself.
Choose any meridian. We're going to traverse north up that meridian (a great circle geodesic) from the Equator to 1 degree North on the Google Sphere (GS) and on the GMP. Consequently, the Eastings are irrelevant; they're constant. Here's the raw data.
The inverse geodesic distance along the meridian on the GS from the Equator to 1 degree North is 111319.4908m. The GMP Northing on the Equator is 0m and the point scale is 1 (unity). The GMP Northing at 30 minutes North is 55,660.4519m and the point scale factor is 1.00003808. The GMP Northing at 1 degree North is 111,325.1429m and the point scale factor is 1.00015233.
To convert from grid distances (111,325.1429m in this case) to geodesic distances surveyors use Simpson's Approximation to compute the line scale factor. That's the sum of the end-point point scale factors plus 4 times the mid-point point scale factor divided by 6, or 1.000050775 in this case. Divide the grid distance by line scale factor and you'll see that we have a sub-millimeter tie with the geodesic distance.
If there were a physical monument on the Equator and one at 1 degree North (same meridian), what would be the distance between them? Well, that would be the geodesic distance on the WGS84 ellipsoid. This is the WGS84 datum, isn't it? Well, that distance is 110,574.3886m. The ratio of these distances is the hidden (and, therefore, insidious) scale distortion of the GMP in this vicinity of the world.
An underlying and yet unmentioned distinction in this thread is that geodesists think in terms of meters on the ground and GIS cartographers think in terms of pixels on a monitor. It's a different frame of reference, a different tolerance for error. If I surveyed football pitch (about 100m), a 67cm error would concern me. My question for you is: How long a distance would it take for the hidden GMP scale distortion to move your image to the wrong pixel on your monitor?
----- Original Message -----
From: "Mikael Rittri" <Mikael.Rittri at carmenta.com>
To: "PROJ.4 and general Projections Discussions" <proj at lists.maptools.org>
Sent: Tuesday, December 2, 2008 10:13:22 AM GMT -06:00 US/Canada Central
Subject: RE: [Proj] Re: "Double ellipsoid" case?
Richard Greenwood wrote:
> I for one, am finding this an interesting, and pertinent thread.
> Let's not quash it, and let's keep it civil.
I thought daan explained the matter clearly, and I have not much to add.
daan Strebe wrote:
> The practical consequence is that they do not have a conformal projection;
> hence local distances measured from a point are not quite uniform in all
> directions, and neither are azimuths quite equally spaced radially from
> that point. So?
I would add that these deviations from conformality of course can be
quantified and bounded. For example, I wrote that "I think the maximal
angle distortion for Sphere Mercator is 0.2 degrees". This was wrong;
I was thinking of the maximal azimuth error, and the maximal angle
distortion is twice that, so make it 0.4 degrees.
It is not surgical precision, and it is 16th century technology,
but I wouldn't say that a 0.4 degree error is "absurd".
(Well, unless my application required exact conformality for some reason.)
The benefit is simpler formulas and faster execution times.
Noel Zinn wrote:
> If I could just switch the WGS84 ellipsoid in the WGS84 datum with
> the Google Sphere (as you suggest), why couldn't I switch
> the International Ellipsoid in ED50 with Clarke 1866? Or any other
> switch for that matter?
Let's see if I understand you correctly. You are asking: if I have data
in the datum ED50, why can't I define a projection which, by definition,
treats (Lon,Lat) from ED50 as if it were (Lon,Lat) on the Clarke 1866
Well, I am forced to say that this is quite possible. The mathematics works,
and it is not illegal. Again, the result would be a map that is not exactly
conformal, relative to the shape of ED50. Because this naive mapping
from (Lon,Lat) on the International Ellipsoid to (Lon,Lat) on the Clarke 1866
ellipsoid - without changing the numerical values - is not conformal.
At least, I don't think so.
But I am just saying that it is possible. I think it would be silly
and meaningless, because I would give up the exact conformality
(relative to the shape of ED50) in return for nothing. I would not
get the benefit of simpler formulas or faster execution times,
since I would have to use ellipsoid formulas to go from Clarke 1866
to the plane.
> In addition to worse fits mathematically
> (since the adjustment was done on the defining ellipsoid),
Yes, spherical formulas give worse maps for large scale maps.
But I would say, only slightly worse. Not absurdly worse.
> we'd open the door to uncertainty and crisis.
> You are proposing (and Google has introduced) the geodetic
> equivalent of sub-prime mortgages in the financial market.
> Don't do it!
Well, we are talking about a maximal 0.4 degree error for angles,
and a variation of the local scale in different direction
through a point, which would be at most 0.5 percent, I think.
If economists could do predictions with 0.5 percent accuracy,
the world economy would be in a better state!
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