# NavList:

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

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Re: That darned old cocked hat
From: Gary LaPook
Date: 2010 Dec 9, 21:20 -0800
I see why the 2/3rds length works from this excerpt from Wilipedia (http://en.wikipedia.org/wiki/Centroid) since it locates the centroid:

### Of triangle and tetrahedron

The centroid of a triangle is the point of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side). The centroid divides each of the medians in the ratio 2:1, which is to say it is located ⅓ of the perpendicular distance between each side and the opposing point (see figures at right). Its Cartesian coordinates are the means of the coordinates of the three vertices. That is, if the three vertices are a = (xa,ya), b = (xb,yb), and c = (xc,yc), then the centroid is

$C = \frac13(a+b+c) = \left(\frac13 (x_a+x_b+x_c),\;\; \frac13(y_a+y_b+y_c)\right).$

The centroid is therefore at $\left(\frac13,\frac13,\frac13\right)$ in barycentric coordinates.

The centroid is also the physical center of mass if the triangle is made from a uniform sheet of material; or if all the mass is concentrated at the three vertices, and evenly divided among them. On the other hand, if the mass is distributed along the triangle's perimeter, with uniform linear density, the center of mass may not coincide with the geometric centroid.

--- On Thu, 12/9/10, Gary LaPook <glapook@pacbell.net> wrote:

From: Gary LaPook <glapook@pacbell.net>
Subject: [NavList] Re: That darned old cocked hat
To: NavList@fer3.com
Date: Thursday, December 9, 2010, 3:33 PM

BTW, my "2/3 of the length of the line connecting the midpoint to the opposite angle as another way to determine the fix using the midpoint method" works for triangles of other shapes, not just right triangles.

gl
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