Xref: utzoo sci.electronics:3438 sci.math:4275
Path: utzoo!attcan!uunet!husc6!cca!g-rh
From: g-rh@cca.CCA.COM (Richard Harter)
Newsgroups: sci.electronics,sci.math
Subject: Re: Constant Breadth Manhole(was Re: WHat are those holes for?)
Message-ID: <31374@cca.CCA.COM>
Date: 26 Jul 88 02:04:09 GMT
References: <10790@oberon.USC.EDU> <327@richp1.UUCP> <303@btree.uucp> <745@io.ATT.COM> <1211@nscpdc.NSC.COM>
Reply-To: g-rh@CCA.CCA.COM (Richard Harter)
Organization: Computer Corp. of America, Cambridge, MA
Lines: 23

In article <1211@nscpdc.NSC.COM> patch@nscpdc.NSC.COM (Pat Chewning) writes:
>In article <745@io.ATT.COM>, tmk@io.ATT.COM (59481[rjb]-t.m.ko) writes:

>> Construct an equilateral triangle ABC. Draw arc AB with center C.
>> Draw arc BC with center A. Draw arc AC with center B.
>> The resulting figure bounded by the 3 arcs is a constant breadth figure.

>I created the three arcs as desribed,  I GOT A CIRCLE!
>If there is something I am missing, please let me know.

What you are missing is a good compass (protractor).  If that is not your
problem, reflect on the fact that circles do not generally come with
three centers and do the following:

Find the midpoint of the equilaterial triangle.  Call it O.  Draw a 
circle with center O through A, B, and C.  Compare this with the figure
previously drawn.  

-- 

In the fields of Hell where the grass grows high
Are the graves of dreams allowed to die.
	Richard Harter, SMDS  Inc.