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    Re: The magical maths of Google Maps
    From: Frank Reed
    Date: 2015 Oct 6, 10:05 -0700

    David Pike, you wrote:
    "From this I conclude that the Google Maps projection is close to standard Mercator’s."

    Yes, that's right. They call it "Web Mercator" but the difference from a standard Mercator is insignificant. The Mercator projection is a terrible choice for a global map projection, and it was steadily falling out of favor in the late 20th century. You can thank the early developers of "Google Maps" back around 2005 for putting it back on the table. They were not map-makers nor even people with any significant intelligence on the matter --merely programming geeks. They regurgitated the first global map projection that they could think of. And other companies copied Google. It's a habit that has grown like weeds. And now as the garden has grown old (ten years... so old in Internet time!), no one remembers that those weeds weren't planted intentionally in the first place.

    Dave, you continued:
    "You’ll find the distance is 5261km, but the line drawn is a beautiful arc looking very like the great circle. Now, just for fun, go to Google Satellite without changing scale. You’ll find you’ve got a “Globe” presentation, which you can tilt until the line between Boston and London is straight while the distance remains constant. You would appear to have hit the view in with the plane of the great circle."

    Yes, that's right. The only thing you have to double-check with an earth-based mapping product is that they are not being too clever by including ellipsoidal corrections to the distance calculation. If that happens then it's not quite a great circle but awafully close. I haven't checked, but I assume that Google Maps products use the standard spherical great circle calculation.

    You wondered:
    "My question is, is it a simple matter for a computer mathematician to create such presentations, or is it really clever stuff?"

    It's either fairly trivial or entirely trivial depending on how you go about it. First, if you do it the old-fashioned way, you look up great circle distance and calculations, and you code it up. This is a task that has been repeated and described hundreds of times on the Internet so you don't need to do much work yourself as a coder these days. It's all spelled out -- fairly trivial. But if you do it the modern way, you have even less work or at least a very different sort of work. Nearly every device or software package in the modern world that deals with mapping or more generally any type of 3d graphics projection (an enormous number of products) has the great circle calculation, and therefore the solution of the standard spherical triangle, baked right in. And frequently this "baking in" is at the level of hardware with circuits built-in to do the spherical triangle problem in hardware since it is repeated so many hundreds of thousands of times a day in 3d mapping problems. Even GPS chipsets often have the spherical triangle built in (so I have been told, but I haven't confirmed this myself). To get the right answers, you merely ask the environment's "API" the right question. I suspect it's a fair bet that there are more automated devices in existence today that can solve the spherical triangle directly than there have been human beings in all of history who can solve that same problem (without looking it up in a reference).

    You asked:
    "Also, is there any way, just for fun, you could use Google maps to solve the PZX triangle to get Co-Alt, or would it take too much effort to work out the lat & long of X?"

    Yes indeed! Set up the great circle problem with the starting point at the observer's lat/lon. Set the end point at the Dec/GHA of the body. Then do whatever is required to get the great circle distance in any tool you have available. The distance to the body's "GP" is the co-alt (same as "zenith distance"). If you have a GPS device that's displaying your current position properly, you can just enter the GHA and Dec of the other body as "waypoint" or destination. In some tiny fraction of a second, the device will return distance and bearing. The distance, as we know from the basic principle of celestial navigation, is identical to the zenith distance of the body. And the bearing is the usual azimuth.

    For another example, suppose you want the angular distance between two stars. If you have the coordinates in any spherical coordinate system (alt/az, SHA/Dec, ecliptic lat/lon, even galactic lat/lon coordinates), you can enter them as starting and ending points in any earth mapping program. The great circle distance on the globe is identical to the angular distance between the stars.

    Again, for any of these tricks to work, it's important that the underlying software does not attempt to include ellipsoidal corrections to distance. To check for this, you can measure the distance between the north pole and and any point on the equator and then compare with the distance between any two points along the equator separated by 90 degrees of longitude. The great circle distances should match. If they do, then the software is solving a pure spherical geometry problem, not really connected with the Earth in any specific way, and you can safely use it to solve any other spherical geometry problem.

    Frank Reed
    Conanicut Island USA

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