NavList:

Chris Caswell, you wrote:
"Are you saying John Karl's equations are wrong, or inaccurate?"

Chris, I thought I was pretty clear on all of this, but let me try to re-phrase:
First things first: you don't need this! No one calculates latitude this way, except in quite peculiar circumstances. It is never the case that we know the exact hour angle (or equivalently the exact GMT and longitude) and do not know our latitude. This is just one of those little "idiosyncracies" in John Karl's book. There's plenty of good material in there, of course, but also quite a few oddball bits like this that are not properly explained. If you never saw this calculational trick in your whole life as a navigator, you would not be diminished in any way.

Let me ask you, Chris, do you have any questions about what I have just written in this paragraph? Do you understand what I am saying here? Please feel free to ask for clarification.

As a reminder, when avoiding line of position navigation, latitude is determined almost exclusively by meridian sights, especially "Noon Sun" sights. And then the equation, of course, is beautifully simple and short:
  Lat = z.d. + Dec,
where Dec is the body's declination at the approximate time of meridian passage, and where z.d. is the zenith distance, which is 90° - Ho (in other words, 90 - corrected altitude). There are some rules you need to learn to do it right, but that's the only equation to worry about for latitude. No trig required.

And again, let me ask you here, Chris, does that make sense to you? Do you understand that this is how latitude is calculated (when we are avoiding line of position navigation --which maybe you shouldn't be). The equations you found in Karl's book are not generally useful. And that's why you won't find them in most other books on celestial navigation. They are oddities with almost no practical value. It's not that they are "inaccurate" of anything like that. The problem is that they don't do anything useful for you as a navigator (with very rare and rather exotic exceptions).

You talked about wanting to solve a middle-of-nowhere scenario, where a navigator is "dropped" somewhere in the world with no clue about location. I want to assure you, again, that the equations you previously referenced in John Karl's book are not the answer... and they're not an answer. I am not being disingenuous when I tell you that there's no real use for that particular set of equations in a scenario like this (or, indeed, any practical scenario).

Do you understand what I am saying here? Do you have any specific questions about what I have said here?

I recommend that you set aside those equations described as for "latitude" in Karl's book which you brought up originally in this thread:
   sin(coL+G)=sin H sin G / sin d
   where:  tan G = tan d / cos LHA.
You don't need these.

But just so you know, there is an algorithm in Karl's book which some people cite as a tool for solving exactly this sort of middle-of-nowhere mystery. It's at the end of chapter 7 on "special sights", his formulas 7.5a-f. There are different printings of this book, and the pagination of yours doesn't match mine (as I have alread discovered) so I won't give a page number. I hope that description is enough to find the formulas in question, and you can explore those as you like. They are straight-forward, though a bit tedious.

I also suggest you learn how to do this by drawing circles of position on a ball (or an orange!) as I described in an earlier message. That's not only a good practical approximation to position; it's also a direct application of the fundamentals of celestial navigation. It confirms, reinforces, and enhances your understanding of the primary concept of navigation by the stars.

Frank Reed