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

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From: Tom Sult
Date: 2012 Jul 14, 15:54 -0500

Thomas A. Sult, MD
Sent from iPhone

Indeed yes.  I had it in my head to construct a sine table using methods similar to what you have provided.  Similarly, I expected to construct a simple sun's declination table.  Using these and a simple gravity protractor, I envisioned getting latitude.

Maybe not so accurate.  Maybe not so "easy", but certainly do-able.

Longitude, however, defies these types of methods!

Best Regards

On Jul 14, 2012 1:55 PM, "Örjan Sandström" <pokerbacken@yahoo.com> wrote:

I have on several occasions showed persons how to accurately divide a circle on a piece of sheet metal, usually into 15° segments but on occasions 3 and 1°, this with fairly good precision (1/10 of a degree average). I have shown this using a piece of wire with four loops and two sharp points or dividers.
if you are interested here it is below (messy description) or you can look on the net.

first make a line as straight as you can manage using the stretched wire.
now use a point as near centre of that line you can find and use wire and two points as divider to make a circle that line passes through.
now using same setting make two arcs with "divider at point where line cross circle and let them start and end mm or two inside the circle.
now using these new points make four arcs without changing setting letting them start and end just inside circle.
basic plane trigonometry tell us that a triangle with all sides equal has an angle of 60° at each point and two such triangles with one side in common will point 180° away from each other thus form four 90° segments of said circle if the tips furthest apart are joined.
so we next use points where circle and line meet each side and make two such with as large radius as wil fit, this dome we get two "x" one above and one below the center of circle, draw line between "x's". we now have four 90° segments ans several arcs crossing.
now we use the points where circle is crossed by the two lines with "divider" set at original distance and draw four arcs, these will cross the circle in 8 places and other arcs in useful angles, now using the points where something crosses circle or two arcs intersect we can easily divide circle into 15, 30, 45, 60, 75... by simply making more arcs the tricky part is geting 15 degree divisions divided accurately into parts.

we can then divide 15 into 2 segments but that is 7.5°if we instead divide into 3 we get 5° MUCH better, how to divide an angle into 3? answer, near impossible with simple devices like this.
there are cheats (more below)
dividing the circle into 5 gives us 72°.
draw a line between 30 and 120 and 60 and 330 then 120 to 240 then 240 and 330, now draw an arc with intersection of these lines and the original four straight lines cross as center and divider set to touch where line 90° of crosses circle.
Using the above that divider setting we can then use with the original 0, 15, 30, 45, 60, 90 105, 120, 135
150, 165, 180... by "adding ot subtracting" 72 forming
6, 9, 12, 18, 21, 33, 48, 63, 78, 87, 93, 108, 123, 138, 153, 168, 183...
bisecting these wil give 3, 6, 24, 39, 51, 54, 66, 81, 96, 111...
bisecting them gives among others 3, 27, 33.
now using divider we can check the distance between 3 and 6 with distance between 30 and 33...
6 with distance between 6 and 12 or 12 and 18...
and 12 between 60 and 72...

End result of this exercise, if done with care, is circle divided into 120 parts and within say 1/10 of a degree average deviation.
dividing further to single degrees requires dividing into 3 which is hard to do accurately, one way below that is a fair approximation.
below is what I hope is a description of a kind of "neusis construction" (uncertain about name I last did basic trig 20+ years ago, but this name was used when dividing angles into thirds)
make an angle CAB (centre of circle and two lines forming 3°) then draw CD perpendicular to AB to cut it at D. Complete the rectangle CDAF. Extend FC parallel to AB to point E now line is drawn from A to point E. Have the point E chosen so that HE = 2AC. angle EAB is 1/3 of angle CAB.
did this make sense?
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