# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Can I Navigate Without an Assumed Position?
From: Bill Lionheart
Date: 2017 Sep 24, 18:34 +0100

```For the case of three circles of position (generally not great
circles) I wonder if there is an equivalent geometric construction to
the symmedian point giving the least squares solution. I would expect
it to have the same tendency to be closer to the shorter side of the
"triangle".

Bill

On 24 September 2017 at 17:38, Frank Reed  wrote:
> John Howard, you wrote:
> "Just as an exercise, I used a globe and string to get an AP at 60° N and
> 115° W."
>
> But that's way off! May I suggest that you try this again? With a common
> globe, you should expect to get a position this way within one or two
> degrees. Even if you mark up a common ball (like a 10-inch diameter ball
> that you might find in the toy department of Walmart) with a bit of care,
> you can get a position this way within 3-5°.
>
> And by the way, for others following along who may not realize this, this is
> not technically an "AP". It's a fix. The globe is serving as an analog
> computer, and the result, when done right, is your position. Given that the
> accuracy is rather low, you could then feed that into a standard celestial
> calculation and use this as a starting point or "assumed position".
>
> To reiterate, in the real world, you're never this lost. These are game
> problems, not navigation. Also in the real world, a sight consists of the
> altitude as well as an approximate azimuth, which really ought to be given
> in the statement of the problem. You know what direction you were facing
> within 10 or 20 degrees, and this significantly constrains the calculation,
> whether you're working by direct computation or by analog on a globe. If you
> know that Spica, e.g. was in the SSE at some altitude, h, then you measure
> off a distance of 90°-h away from Spica's GHA and dec (its "GP") and draw an
> arc of a circle. That much is obvious. But you don't need to draw the whole
> circle; you just draw that portion that "sees" Spica in the south-southeast,
> or equivalently, that portion of the circle which is perpendicular to SSE.
>
> Frank Reed
>
> View and reply to this message

--
Professor of Applied Mathematics
University of Manchester
http://www.maths.manchester.ac.uk/bl
```
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