Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Kitchen Tabletop


Date: 11/21/2001 at 13:10:49
From: Russell Kopp
Subject: Geometry

Hi:

I have a kitchen table whose functionality involves the use of math.  
First I'll explain the table:

From a crow's eye view, in its folded state the table is a rectangle, 
with adjacent sides of different length. (I suppose it could have been 
square, but I'm not sure).

The tabletop has hinges on one of its longer sides - for reference 
I'll call this its east side. Attached to these hinges is another 
tabletop of equal size that sits on top of the original tabletop when 
the table is in its smaller, folded state.

The bottom of the two tabletops is connected on its bottom or under 
side to a pivoting device that allows the bottom tabletop (and of 
course the top tabletop attached to it) to be rotated clockwise 90 
degrees. Once this rotation is effected, the top of the two tabletops 
may be unfolded from the bottom tabletop, revealing a table twice as 
large in size as the original. In its unfolded state, the overlap of 
the larger table top on its base is such that north and south table 
ends overlap the base by equal amounts, and the west and east ends 
overlap by equal amounts as well. The overlap of adjacent sides need 
not be the same.

Were the table split into four quadrants, the pivot would clearly lie 
in the upper right (northeast) quadrant.

I wish to know how to determine the correct pivot point such that it 
meets all the guidelines discussed above.

I should note that the table's actual dimensions in its folded state 
are 35.4375" by 24.4375". I've done a little research, and should note 
for those with geometry backgrounds that this is not a golden 
rectangle (a rectangle whose adjacent sides are in a ratio of 
approximately 1:1.618.)  I say this because I thought a golden 
spiral's limit might be the pivot point - though it appears not to be 
the correct point.

Thanks,

Russ


Date: 11/21/2001 at 20:20:44
From: Doctor Rick
Subject: Re: Geometry

Hi, Russ, thanks for writing to Ask Dr. Math.

I like questions like this - discovering math in everyday things! I've 
seen the kind of table you're talking about, but I hadn't asked myself 
the question you have. 

It turns out that the answer is simpler than you thought. Here is a 
figure:

               +---------------+
               |               |
     ..........|...............|..........
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     ..........|.......B.......A..........
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     ..........|...............|..........
               |               |
               +---------------+

You have stated that the opened table is centered on the base; you 
didn't state, but I think I can assume it, that the folded table is 
also centered on the base. Thus I've drawn the folded (solid lines) 
and opened (dotted lines) tabletops centered at the same point. The 
point marked A, in the center of the hinged side of the folded table, 
must move to point B, the center of the entire table, when the top is 
rotated 90 degrees.

All we need now, to find the center around which the tabletop is 
rotated, is to locate a point such that lines drawn from it to A and B 
are the same length and form a 90-degree angle. That's easy:

               +---------------+
               |               |
     ..........|...............|..........
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |           O   |         :
     :         |         /   \ |         :
     ..........|.......B.......A..........
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     :         |               |         :
     ..........|...............|..........
               |               |
               +---------------+

We have here a right isosceles triangle OAB; OA and OB are equal, so 
A can rotate to B, and AOB is a right angle, so the tabletop is 
rotated 90 degrees. If the width of the folded table (perpendicular to 
the hinged side) is W, then the pivot point O is W/2 to the east of 
the center, and W/2 north.

The proportions of the table are irrelevant. We just need a base that 
is at least half the width of the folded table.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Practical Geometry
High School Triangles and Other Polygons

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/