Eckert IV vs Eckert VI. Other related topics.

[RogerOwens]

I posted the first images that you linked to, by right-clicking the link and selecting "Copy The Location", and pasting copied url within the Img tags (from the top of the edit-screen). But that didn't work this time, where the urls themselves are what the links consist of. For the first set of images, the link was in the form of a word to click on. With that, the way I posted it worked fine.

Michael

[RogerOwens]

When I enclose the url in the img tags, that doesn't work. I don't know why.

2. Cylindrical Equidistant, conformal at lat 30.
Horizontal compression factor in the formula for X: .866025





3. Cylindrical Equal-Area with scale disproportion equal and opposite at latitudes 45 and 0
Horizontal compression factor in the formula for X: .7071






Cylindrical Equal-Area conformal at lat 30
Horizontal Compression-factor in the formula for X: .75
(This is a familiar already-proposed version of Cylindrical Equal-Area, but a posted image would be good, for completeness)




(And I think there’s no harm in stating that the »familiar already-proposed version« is the Behrmann equal-area…)



Linear PF8.32 conformal at lat 30
Horizontal compression-factor in the formula for X: .86612




When I enclose the url in the img tags, that doesn't work. I don't know why.

Michael Ossipoff

[daan]

The URLs that you are embedding between Img tags are not URLs to images. They are URLs to Web pages that execute code to render images.

--daan

[RogerOwens]

Ok, thanks

And of course thanks for hosting what seems to be the only forum on map projections, even if it only has 3 people.

Michael Ossipoff

[Atarimaster]

The URLs that you are embedding between Img tags are not URLs to images. They are URLs to Web pages that execute code to render images.

Exactly.

However, I’ve added a set of buttons which will render some configurations without having to enter the values for p and f before.
If you’re on a halfway decent internet conncection, it should be fast enough to compare the different settings.

Moreover, thanks to the nice guys who write extension scripts for d3.js, I was able to add a "Export to PNG" button. While for this kind of images, SVG is much better (because it can be scaled and edited without any loss), handling a bitmap format like PNG might be easier.
Unfortunately, the PNG export will always save the full canvas (the "paper sheet" that I was talking about earlier in this thread) so you’ll always have a projection image with a huge transparent margin.
So, here’s the link again:
http://tobiasjung.name/mop/mo-projections.php

Regards,
Tobias

[RogerOwens]

Tobias--

Thanks for those improvements.

But especially, thanks for making, and posting or linking to, those projection-images that I requested.

Michael

[RogerOwens]

A few miscellaneous comments, about:


1. Briefer summary of my world-map desiderata

2. What I mean by "easy explanation"

3. Magnification

4. Position in Cylindroid maps


Briefer Summary of My World-Map Desiderata:

To the extent possible:

* Easy explanation

* Large map-size (for available rectangular space)

* Properties
...such as = Area, Conformality, Linearity, and properties of Cylindroid projections

* Low scale-disproportion

What I Mean By Easy Explanation:

I like for it to be possible to tell people how a map was given its conformality or equal-area.

Conformality:

I'd point out to people that when Mercator introduced the Mercator projection, the calculus of infinitessimals hadn't yet been introduced.

Mercator must have used a numerical process. Maybe or likely this obvious process: At lots of fairly closely-spaced latitudes, starting from the equator, make the north-south scale equal to the east-west scale (the exaggerated east-west scale resulting from constant map-width, when the lengths of parallels are really proportional to cosine(lat), instead of being constant).

Nowadays that would be called a numerical integration.

Equal-Area:

Mercator's way of making his map conformal could likewise be used to make any cylindroid map equal-area. Starting at the equator, at closely-spaced latitudes, make the north-south scale vary as the reciprocal of the east-west scale (so that their product is constant). ...so that each tiny latitude-band will have an area on the map that is proportional to its area on the Earth.

If that sort of numerical integration was good enough for Mercator, and all of the many mariners who used his map, then it's good enough.

But there are equal-area projections for which the specifics of why they're equal-area can be easily and briefly completely explained. No calculus required for a brief, easy, complete explanation. That's true (only?) of the Sinusoidal and Cylindrical-Equal-Area (CEA) projections.

Maybe, too, (but not as briefly) for the diamond-shaped projection (Colignon?) in which the outer boundary from the equator to the pole (and every meridian) is a straight line. But, since the Sinusoidal is better, and also much more briefly & easily explained, then there isn't much need for the Colignon, unless one wants a point-pole map and insists on easily-drawn straight-line meridians. (Colignon by itself or grafted to CEA)

For those reasons, I prefer Sinusoidal and CEA, and maps consisting of a grafting of CEA and Sinusoidal, as the equal-area world maps to offer to the public.

(But I propose PF8.32 anyway, as a possibility, even though I prefer Sinusoidal, Cylindrical, and Cylindrical/Sinusoidal grafts, for public offerings.)

But I also like Elliptical-Linear (Apianus II), when it's desired for the map to reflect the Earth's round shape. I feel that Apianus II and Sinusoidal would be a good combination of first maps for early-grade schoolrooms. ...one elliptical and pictorially-good; and the other the most natural and easily-explained equal-area.

...both easily-explained, both linear.

Mollweide has an unnecessary (without high space-efficiency to show for it) skinny Africa, and lacks the easy complete specific explanation of Sinusoidal and CEA.

Position in Cylindroid Maps:

A cylindroid map is a map that is either cylindrical or pseudocylindrical.

Linear maps pretty-much always have a simple construction with easy explanation. Another advantage is that it's always easy to find or determine lat/lon co-ordinates.

But, for easy finding or determination of lat/lon co-ordinates, linearity isn't really necessary.

It can be achieved by making and providing a "latitude-ruler" for a cylindroid map.

A latitude-rule is a ruler that's marked in latitude, for measuring or finding latitudes on a cylindroid map.

Of course, another solution would be to have finely-divided vertical latitude-scales up both sides of the cylindroid map.

Since few maps have those built-in latitude-scales, then latitude-rulers should be made for cylindroid maps that don't have finely-divided latitude-scales up their sides.

I suggest that, on a latitude-ruler, near the edge-side opposite to the edge with the latitude-marks, there should be indications of the magnfication, and the central meridian scale-disproportion, for selected latitudes along the ruler.

Of course, if the map is conformal or equal-area, then only one of those indications, instead of both of them, would be needed.

Michael Ossipoff

[daan]

Mercator's way of making his map conformal could likewise be used to make any cylindroid map equal-area. Starting at the equator, at closely-spaced latitudes, make the north-south scale vary as the reciprocal of the east-west scale (so that their product is constant).

That doesn’t work for pseudocylindrics. If h is scale long parallel and k is scale along meridian, then constant h•k only holds if the angle between h and k is 90°. This is because the flation (areal inflation/deflation) at a mapped point is given by h•k•sin θ′, where θ′ is the angle between parallel and meridian.

But there are equal-area projections for which the specifics of why they're equal-area can be easily and briefly completely explained. No calculus required for a brief, easy, complete explanation. That's true (only?) of the Sinusoidal and Cylindrical-Equal-Area (CEA) projections.

I’d argue that’s true for any cylindric or pseudocylindric projection. All you have to do is set a given parallel at the height on the map at which the area below the parallel is the same as the area below the unmapped parallel on earth’s surface.

— daan

[RogerOwens]

Mercator's way of making his map conformal could likewise be used to make any cylindroid map equal-area. Starting at the equator, at closely-spaced latitudes, make the north-south scale vary as the reciprocal of the east-west scale (so that their product is constant).

That doesn’t work for pseudocylindrics. If h is scale long parallel and k is scale along meridian, then constant h•k only holds if the angle between h and k is 90°. This is because the flation (areal inflation/deflation) at a mapped point is given by h•k•sin θ′, where θ′ is the angle between parallel and meridian.

True. When I said "north-south scale", I meant the ratio of Y distance to latitude-difference (constant at all longitudes on a pseudocylindrical map). ...something that can be measured by the north-south scale on the central-meridian. So, when I said, "north-south scale", I should have said "north-south scale on the central meridian".

When writing that passage, I forgot that the north-south scale on the central meridian isn't the north-south scale elsewhere, on a pseudocylilndrical map. But I didn't forget that when I posted the EW/NS scale-disproportion at various places, with Linear PF8.32

But there are equal-area projections for which the specifics of why they're equal-area can be easily and briefly completely explained. No calculus required for a brief, easy, complete explanation. That's true (only?) of the Sinusoidal and Cylindrical-Equal-Area (CEA) projections.

I’d argue that’s true for any cylindric or pseudocylindric projection. All you have to do is set a given parallel at the height on the map at which the area below the parallel is the same as the area below the unmapped parallel on earth’s surface.

Sure, those areas on the map and on the Earth could be separately numerically integrated, in a plausible, natural way, if the person knows that length of a parallel is proportional to that latitude's cosine, because a length radius drawn parallel to the equator from the axis at that latitude, is.

But it could also be explained just by saying that each next closely-spaced parallel is positioned so that, between it and the previously-mapped parallel, the north-south scale on the central meridian varies inversely with the length of that parallel on the earth, which is proportional to that latitude's cosine.

In that way, it's described as one stepwise process, starting at the equator, instead of having to suggest separate numerical integrations of cos lat, and of area on the map.

...similar to how I'd suggest explaining how Mercator was drawn before the introduction of the calculus--stepwise, adjusting the vertical scale at each next closely-spaced parallel, starting at the equator.

But of course no equal-area map's explanation matches the easiness of that of the Sinusoidal, where one could just point out that the parallels on the map are drawn with lengths proportional to their lengths on the Earth, proportional to cos lat.
...and are equally-spaced, as on the Earth.

Michael Ossipoff

[RogerOwens]

I've just noticed that I left something out in my reply. You'd mentioned cylindric and pseudocylindric, and my reply only addressed cylindric.

For pseudocylindrical maps, I just meant that I felt that the most helpful explanation would be to speak of it as a step-by-step process, starting at the equator, spacing each next closely-spaced parallel so that the vertical scale varies reciprocally with (the map's parallel-length there, divided by the Earth's parallel-length there). ...so as to keep each tiny latitude-strip on the map having area proportional to what it has on the Earth.

More parallel-length than on the globe? Then correspondingly shorter parallel-spacing, to give each little latitude-strip the same area on map and globe.

That's how I like to word the general explanation for cylindroid equal-area maps.

So it's like the way one writes it when one makes an equal-area map analytically, except that it's described numerically. That seems to me the easiest and most helpful explanation.

But it's easier with CEA, because of the situation referred to by Gall's name for it, "Orthographic". It's a projection solvable by a geometric non-calculus trick.

It's easier still with Sinusoidal, because it's only necessary to explain that the parallels are proportional to their actual lengths on the Earth, and are equally-spaced. Proportionally-accurate length X equal spacing, with the natural result of equal-area.

So that's why I say that CEA and Sinusoidal are the ones that are easy to completely explain the specifics of.

A better way to say what I mean: With CEA, by Gall's "orthographic" geometric construction, it's possible to show the derivation of CEA's formulas--to position each latitude-line directly and immediately.

Likewise with Sinusoidal.

...which is why, for equal-area, I favor Sinusoidal, CEA, or a CEA-Sinusoidal graft.

I've derived formulas for Mercator, Conformal Conic, Mollweide, Eckert IV, and now Equal-Area PF8.32 ...but those calculations were far apart in years. Each time I did one of those problems, it was as if I were starting over. That includes now when I'm doing Equal-Area PF8.32 Because it's been so long since I've looked at such a problem, it's a matter of starting from scratch each time.

So I'm not familiar with the problem, because I do it anew each time, and I'm not around those problems a lot.

That's why it's taking me so long to post formulas for Equal-Area PF8.32, with the desired scale-proportions. There are very many opportunities to make a little copying-omission-error, and just one of those would mess-up the map or its scales. So I want to take time to try to be really sure that there isn't an error.

Michael Ossipoff