Equal-Area PF8.32, power-function world map

[RogerOwens]

(Edit, February25, 2017: The values of the constant, k, have been replaced with the correct values, in an edit.)

Tobias--

(I think the parentheses are probably fixed now, but in my reply to daan's post (farther down on this page), I wrote the formulas with fewer parentheses, with numbers replacing (p+1) and (p+1)/p. That version with fewer parentheses would probably be easier to use anyway.)

Could you post images for this Equal-Area PF8.32, with the two values for the k constant?

Here are formulas for Equal-Area PF8.32:

For some equal-area projections, including this one, the only formulas that can be written are the "backwards formulas", in which lat is given as a function of Y, and lon is given as a function of X.
----------------------------------------
For longitude, in degrees:

lon = 180 * kx /(2 - 2 * (abs(y))^p)
----------------------------------------
For latitude in degrees:

lat = arcsin ( ((p+1)/p) * (abs(y) - ((abs(y))^(p+1))/(p+1)))

For negative values of y, then put a minus sign in front of the latitude formula.

If d3's arcsin function is given in radians, then put (180/pi) in front of the latitude formula, to get degrees.

p = 8.32

k = .948534 for conformality at (lat 30, lon 0)

k = .986147 for equal and opposite EW/NS scale disproportion at (lat 45, lon 0) and at equator.
--------------------------------------------------

For anyone who hasn't read previous discussion about PF8.32, "PF" stands for power-function. The map's boundaries are power-functions where x is a power-function of y. ...a function of y such that x = K1 - K2*(y^p), where p and K1 and K2 are constants.

The choice of that function, and for the value of p, is to make a large map (for the rectangular space that it occupies) with point poles, and consequently better topological accuracy and more accurate Arctic.

In this case, it's Equal-Area PF8.32

(the "8.32" denotes the value of p)

If the non-arctic part of the version conformal at (lat 30, lon 0) looks a lot like the CEA conformal at lat 30, then there's no error in the formula.

If the non-arctic part of the version with equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator looks a lot like CEA with equal and opposite EW/NS scale-disproportion at lat 45 and at the equator, then there's no error in the formula.

Michael Ossipoff

[RogerOwens]

Correction to formula:

I've changed the name of the constant from "f" to "k".

That constant should be in the numerator of the longitude formula, instead of in the latitude formula.

(I've made that correction in the post containing the formulas.)

Without the constant, the east-west scale would a little smaller than desired, compared to the north-south scale.

Therefore, in the longitude formula, x is multiplied by the constant, k.

...with the result that the longitude is smaller for a given x, meaning that x is larger for a given longitude, which is the desired result.

Michael Ossipoff

[daan]

----------------------------------------
For longitude, in degrees:

lon = 180 * kx /(2 - 2 * (abs(y))^p))
----------------------------------------
For latitude in degrees:

lat = arcsin ( ((p+1)/p) * (abs(y) - ((abs(y))^(p+1))/(p+1))

The parentheses are mismatched in both of these equations.

— daan

[RogerOwens]

(Edit, February 25, 2017: The values of the constant, k, have been replaced with the correct values, in an edit.)

daan--

Sorry about that. I can't say that the regular text-characters are my favorite way to write formulas.

First I'll write it in a way that will be easier for me--with (p+1) and 1/(p+1) written as numbers, so that there will be fewer parentheses.

Then I'll try to fix the parentheses in the original formulas, because I'd like to also write it in terms of p.

But the formulas in this post will be easier to use anyway, as well as being something that I can write more quickly than figuring-out the parentheses.

So the formulas with fewer parentheses, and probably correct parentheses this time, are in this post below, replacing the original formulas:

Tobias--

Could you post images for this Equal-Area PF8.32, with the two values for the k constant?

Here are formulas for Equal-Area PF8.32:

For some equal-area projections, including this one, the only formulas that can be written are the "backwards formulas", in which lat is given as a function of Y, and lon is given as a function of X.
----------------------------------------
For longitude, in degrees:

lon = 180 * kx /(2 - 2 * (abs(y))^8.32)
----------------------------------------
For latitude in degrees:

lat = arcsin (1.12019 * (abs(y) - (abs(y)^9.32)/9.32))

For negative values of y, then put a minus sign in front of the latitude formula.

If d3's arcsin function is given in radians, then put (180/pi) in front of the latitude formula, to get degrees.

p = 8.32

k = .948534 for conformality at (lat 30, lon 0)

k = .986147 for equal and opposite EW/NS scale disproportion at (lat 45, lon 0) and at equator.
--------------------------------------------------

For anyone who hasn't read previous discussion about PF8.32, "PF" stands for power-function. The map's boundaries are power-functions where x is a power-function of y. ...a function of y such that x = K1 - K2*(y^p), where p and K1 and K2 are constants.

The choice of that function, and for the value of p, is to make a large map (for the rectangular space that it occupies) with point poles, and consequently better topological accuracy and more accurate Arctic.

In this case, it's Equal-Area PF8.32

(the "8.32" denotes the value of p)

If the non-arctic part of the version conformal at (lat 30, lon 0) looks a lot like the CEA conformal at lat 30, then there's no error in the formula.

If the non-arctic part of the version with equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator looks a lot like CEA with equal and opposite EW/NS scale-disproportion at lat 45 and at the equator, then there's no error in the formula.

Michael Ossipoff

[RogerOwens]

I've just noticed that, in my post before this one, I wrote "2.12019" when I meant "1.12019"

That error has been corrected in an edit.

I meant 1.12019 in the formula for latitude, instead of 2.12019

Michael Ossipoff

[daan]

For longitude, in degrees:

lon = 180 * kx /(2 - 2 * (abs(y))^8.32)
----------------------------------------
For latitude in degrees:

lat = arcsin (1.12019 * (abs(y) - (abs(y)^9.32)/9.32))

That does yield an equal-area projection, but its latitude of correct scale is 21°18′, not 30.°

— daan

[Atarimaster]

I must be doing something wrong again, because at the moment, the d3 scripts won’t render an image for the new projection.
Maybe I’ll try again in a few days, but for now I’ve got to attend to other things.

Regards,
Tobias

[RogerOwens]

daan--

Ok, thanks. I'll re-check and re-do how I was determining the EW/NS scale-disproportion for particular latitudes.

----------------------------------------------------------------------------------

Tobias--

Does d3 accept backwards formulas?
-----------------------------------------------------------------------------------
Michael Ossipoff

[RogerOwens]

But, until I check and re-do my determination of scale-disproportions for various latitudes:

If the map is conformal at lat 21.3, then shouldn't I multiply the longitude formula by the ratio of the squares of the cosines of 30 degrees and 21.3 degrees, to make it conformal at lat 30?

If it's now conformal at lat 21.3, then the map is too wide. To narrow it, I should multiply the longitude formula by a factor greater than 1.0

So shouldn't I multiply the longitude formula by cos^2(21.3)/cos^2(30)?

So shouldn't I multiply the longitude formula by 1.1574?

But, as I said, I'll check and find out what's wrong with the way I was finding the EW/NS scale-disproportion for latitudes, and fix the formula.

Michael Ossipoff

[Atarimaster]

Does d3 accept backwards formulas?

Hell if I know!
I don’t even know what they are (except for what you said earlier in this thread).

Scroll through the list of d3 projections. If one them requires a backwards formula, they obviously accept them.

Regards,
Tobias