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

General discussion of map projections.
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

Atarimaster wrote:
Scroll through the list of d3 projections. If one them requires a backwards formula, they obviously accept them.
Thanks--Will do.

Michael Ossipoff
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

But that multiplication by 1.1574 wouldn't work...only for a cylindrical.

I mean, there is a multiplier that will make the map conformal at (lat 30, lon 0), but the approach that I described, that led to 1.1574, was something that would change the conformal point from 21.3 to 30 only for a cylindrical equal-area projection.

I thought that the value (.821454415) for k that I gave, would result in conformality at (lat 30, lon 0). If that doesn't do it, then I'll find the error, and the right values for k, for the two versions that I suggest.

Will find the error and fix scale.

Michael Ossipoff
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

I expected it to be a copying-mistake, but it looks as if it might have been something more obvious:

From what I wrote on one page, it appears as if I might have been looking at:

(d lat / d y) / (d lon / d x)

...instead of (1/cos lat) times that.

If so, the error was easy to find, but it's an error that I didn't make when I was posting scale-disproportions in the linear version.

Michael Ossipoff
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

Yes, that's evidently what the error was.

In my formula for the EW/NS scale-disproportion, on the central-meridian, at a certain latitude, I left out a factor of 1/(cos lat).

So, if k = 1, EW/NS scale-disproportion, on the central meridian, at latitude lat(y) =

[2/pi] [(p+1)/p] [(1-(abs(y))^p)/cos lat(y)]^2

Michael Ossipoff
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

Values for k for Equal-Area PF8.32:

--------------------------------------------------
For conformality at (lat 30, lon 0):

k = .948534
--------------------------------------------------
For equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator:

k = .986147

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

Michael Ossipoff
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

I've edited my post before this one, to say:

For equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator:

k = .986147

MIchael Ossipoff
daan
Site Admin
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Re: Equal-Area PF8.32, power-function world map

Post by daan »

RogerOwens wrote:Values for k for Equal-Area PF8.32:

--------------------------------------------------
For conformality at (lat 30, lon 0):

k = .948534
--------------------------------------------------
For equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator:

k = .986147
Those values are correct.

— daan
RogerOwens
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Re: Equal-Area PF8.32, power-function world map

Post by RogerOwens »

Good to hear it's right this time; it's a problem with so many opportunities for an error, that it wasn't possible for me to really be sure that there still wasn't an error that I'd missed somewhere.. Thanks for checking it.

Michael Ossipoff
RogerOwens
Posts: 403
Joined: Sun Feb 02, 2014 8:24 pm

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

Post by RogerOwens »

daan--

First I want to emphasize that, when I first posted them, I really wasn't entirely sure whether my k values for Equal-Area PF8.32 were error-free. So, as I said, thanks for checking.

And, even if I'd been sure, of course validity-confirmation by someone else still makes all the difference when someone proposes something like a map-projection, especially one whose derivation has lots of opportunities for error.

Could you post images of Equal-Area PF8.32, with both k values?

(...the k value for conformality at (lat 30, lon 0), and the k value for equal and opposite EW/NS scale-disproportion at (lat 45, lon 0) and at the equator)

Michael Ossipoff
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