Comprehension question regarding Gott-Mugnolo azimuthal

General discussion of map projections.
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Atarimaster
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Joined: Fri Nov 07, 2014 2:43 am

Comprehension question regarding Gott-Mugnolo azimuthal

Post by Atarimaster »

Hello,

we've probably all seen it:
Somebody wants to know the distance between two points, e.g. New York and Moscow, so he holds a ruler against a world map and then he multiplies the centimeters/inches he measured by the nominated scale of the map. And we all know that he won't get the actual result in kilometers/miles unless it's a special kind of map projection – equidistant azimuthal centered to either New York or Moscow, two-point equidistant with at least one of the anchor points being New York or Moscow; or a cylindric projection with equally spaced parallels and he decides that he'd rather know the distance between JFK International and Concepción, Chile. ;-)

Some time ago, I read that "the 'Gott-Mugnolo azimuthal' has the lowest distance errors of any map".
Does that mean that if you use a Gott-Mugnolo azimuthal map to measure distances between two arbitrary points using a ruler, you'll still get wrong results, but on the average the result will deviate less from the actual distance than on other projections?

If the answer is "No": So what does it mean?
If the answer is "Yes": Is this really true for a world map or only for maps that cover smaller areas (e.g. a single continent)?

Kind regards,
Tobias
daan
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Re: Comprehension question regarding Gott-Mugnolo azimuthal

Post by daan »

Hello Tobias.

I looked over Gott-Mugnolo again to refresh my memory.
Atarimaster wrote:Some time ago, I read that "the 'Gott-Mugnolo azimuthal' has the lowest distance errors of any map".
Does that mean that if you use a Gott-Mugnolo azimuthal map to measure distances between two arbitrary points using a ruler, you'll still get wrong results, but on the average the result will deviate less from the actual distance than on other projections?
That’s right; that is the intention. However, one could measure “least average error” many different ways. The particular way chosen by the authors yields an azimuthal projection that measures very slightly better than Lambert azimuthal equal-area, which came in second. Other metrics would yield other rankings, however. And, with the measure so close to Lambert’s, using the new projection hardly seems worth the effort. It’s not as if most distances would be even remotely accurate anyway!

Regards,
— daan
Atarimaster
Posts: 446
Joined: Fri Nov 07, 2014 2:43 am

Re: Comprehension question regarding Gott-Mugnolo azimuthal

Post by Atarimaster »

Once again, daan, thanks four your explanation.
And since I love the Old Masters, I think it's nice that an 18th century projection can't be improved significantly (at least, not in this regard) even with contemporary technology. :-)

Regards,
Tobias
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