ADB wrote:
> the rise is sqrt(3)/2 or sin(60), the run is 0.5 (half the side length) +
> sin(30) ==> 1 so atan(sqrt(3)/2)) ~= 40.8934 degrees which makes the upper
> (downward) angle ~= 49.1066 degrees.

and
scott <sco### [at] scottcom> wrote:

> (L/2)^2 = (S/2 + S*sin(30))^2 + (Scos(30))^2
> (L/2)^2 = (S/2 + S/2      )^2 + 3/4*S^2
> (L/2)^2 = 7/4 * S^2
>   L^2    = 7 * S^2
>   L      = sqrt(7) * S


You two seem to have a peculiar (to me) way of going about your trig.
Either they do it a different and, admittedly, interesting way on "The Other
Side of The Pond", than we do here in 'Murica, or -
"We" do it weird, or -
I just never really learned it very well.

I recall going through something like this puzzling out Friederich Lohmueller's
pulley/belt calculations with Jerome.
You pick what, to me, is the least obvious approach.  :D

I'd like to hear and see more about all of this side/2+side*sin(30) business.
I did it what seems to me a more direct way, but got the same answer. {whew!}
:)

See my updated diagram.   Hopefully it's correct - I haven't yet finished coffee
#1.
I'm going to ponder how this gets converted into code, since there's an ever
increasing number of line segments to be drawn at every next level.

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