tmk@io.ATT.COM (59481[rjb]-t.m.ko) (07/21/88)
In article <892@wucs2.UUCP> jps@wucs2.UUCP (James Sterbenz) writes: >In article <1550@tekirl.TEK.COM> jackg@tekirl.TEK.COM (Jack Gjovaag) writes: >>In article <303@btree.uucp> rfarris@btree.UCSD.EDU (rick farris) writes: >>>Yes, and why the heck are manholes round? >>Round is the only shape for the cover that guarantees that the cover can't >>fall into the hole. What do I win? > >Nothing. I suspect that any regular polygon with an ODD number of sides >couldn't fall into the hole. Think about trying to rotate a triangle or >pentagon to fit! A circular cover is just easier to manufacture >(kinda like a polygon with an infinite number of sides :-). > >-- Regular polygon with an ODD number of sides can also fall into the hole unless the ledge is wide enough. Consider a regular pentagon ABCDE. Denote midpoint of CD to be M. Note that AM is less than AC or BD, etc. However circle is *NOT* the only shape that can't fall into the hole. Any shape with a constant breadth (i.e. constant width when measuring in any direction) will do the job. Circle is an example of a constant breadth figure. Another example is as follows: 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. In addition, such a triangular constant breadth cover can be painted yellow to indicate the traffic direction or the flow direction of whatever underneath. (Note: an "equilateral triangular cover" is being used for this purpose in some places) Also, the triangular constant breadth shape fits human better than a circle. Do I win the "Best Manhole Design" award? ****************************************************************************** Tsz-Mei Ko ARPA: bentley!tmk@att.ARPA AT&T Bell Labs UUCP: tmk@bentley.UUCP LC 3S-D20 184 Liberty Corner Road {att-ih,decwrl,amdahl,linus}!ihnp4!bentley!tmk Warren, NJ 07060-0908 *******************************************************************************
rsd@sei.cmu.edu (Richard S D'Ippolito) (07/22/88)
In article <745@io.ATT.COM> >Tsz-Mei Ko asks:
Do I win the "Best Manhole Design" award?
Nope! The woman's libbers (and real men) on the committee will never accept
the name of the thing.
Richpatch@nscpdc.NSC.COM (Pat Chewning) (07/26/88)
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. Pat Chewning NSC Portland Development Center 15201 NW Greenbriar Pkwy Beaverton, OR 97006
g-rh@cca.CCA.COM (Richard Harter) (07/26/88)
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.