[NSRCA-discussion] Correct Geometry on N

[tocdon at netscape.net]

Thanks for the clarification Jim!

Don


-----Original Message-----
From: Stuart Chale <schale at optonline.net>
To: General pattern discussion <nsrca-discussion at lists.nsrca.org>
Sent: Tue, 8 Jul 2008 5:45 pm
Subject: Re: [NSRCA-discussion] Correct Geometry on N


Yes I see what you mean now.?
Strange :)?
Stuart?
?
James Oddino wrote:?
> "The figure is equally tall as it is wide." No it isn't. > Horizontally it is as large as the diameter of the circle. Vertically > it is the distance across the flats which is less.?
>?
> Jim O?
>?
>?
> On Jul 8, 2008, at 2:01 PM, Stuart Chale wrote:?
>?
>> OK lets see if this works.?
>> In the attached drawing there is a hexagon with sharp corners. Drawn >> symmetrically. Superimposed is the same 6 sided figure with radiused >> corners. Around this one is a circle just touching each radius. >> The figure is equally tall as it is wide.?
>>?
>> Stuart?
>>?
>> atwoodm at paragon-inc.com wrote:?
>>> I gotta go with Richard on this. The length of sides of a 30/60/90 >>> triangle are X(short side), 2X (hypotenuse), X * square root of 3 >>> (medium side). If we make X = 1 then the height of the hexagon >>> will always be 3.46 but the width will very as the side "points" >>> are trimmed by the increasing radius. With zero radius, the width >>> in my example would be 4 (thus richards 1.15 ratio). As the radius >>> grows the width narrows.?
>>> Does someone with Cad want to draw this up??
>>>?
>>> Its somewhat irrelevant. If the angles are correct, and there's >>> symmetry, and equal line lengths, the actual shape will be correct >>> regardless of "roundness"?
>>>?
>>> Sent from my BlackBerry Smartphone provided by Alltel?
>>>?
>>>?
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