Before I describe my construction, I must commend the more sophicated
method used by Bractals; the error is actually less than this construction,
in fact, considerably less.
I have noted the similarities of the two methods below
Construction
Apologies, but with a 'little' Gestalt closure, please allow this model
to be an ellipse ... and then, with my description below, this will make sense.
Y'
--|--
G M H
|
/ | \
X' -------A---o---B------- X"
\ | /
|
I N J
--|--
Y"
1. Draw a horizontal line with the straight edge,X'-X".
2. Construct a circle (radius X) centred at A.
3. Construct a second circle (same radius) whose centre, B,
is at the intersection of the the circumference of the
circle whose centre is A and line X'-X".
4. Reset the compass radius to 2 * X (the diameter of one of
the circles; the only reason for the straight edge).
5. With this radius draw 2 arcs, one above and one below the
two linked circles and centred at M and N.
Areas
In comparing the outlines of these two entities, I found it easiest to
look at the x-axis (Major radius of the ellipse) as having 24 divisions.
Drawing this up in a CAD program the lowest integers available to me
were R = 8 for my shorter compass radius, with Mj = 12 for the Major
and Mr = 9 for the Minor radii of the 4:3 ellipse.
Construction
There are:-
2 sectors: GIA and HJB, each of area (ΠR^2)/3
and 2 sectors: GHN and IJM, each of area (Π(2*R)^2)/6
which has a superfluous overlap,MBNA; area of 2* (R*R√3/4) or (R*R√3/2)
yielding an enclosure of
Area = 2*((Π R^2)/3 + (Π (2*R)^2)/6) - (R*R√3/2)
= 2Π((R^2)/3 + (4R^2)/6)- (R*R√3/2)
= 2Π(2(R^2)/6 + (4R^2)/6)- (R*R√3/2)
= 2ΠR^2 - (R*R√3/2)
= 2Π8*8 - (8*8√3/2)
= 346.70
Ellipse
Area = Π Mj * Mr
= 108Π
= 339.29
Comparison:
The construction actually contains the ellipse within two
very fine 'crescents'; an error of about 2.14%.
The ellipse is 97.86% in area of the constructed model.
Note: Values are rounded to 2 dec. pl.
Bractal's Solution:
His diagram would look rather like this, but with arcs drawn in.
Y'
--B--
| M E
|
| \
X' -----------o---C------A X"
| /
|
D
--|--
Y"
Points E, C and D are collinear.
His centre C corresponds in function to my centre B in that we both
use that location to construct small circles or arcs.
His centre D provides the same function for my centre N to create an
arc of larger radius than the smaller circle.
My construction really only requires the straight edge so that you have
a place of reference upon which to locate two circles, and then for
'accuracy', gauge the diameter for the larger arcs.
Note that with the method I have listed, I could have chosen to demand
construction by a pair of compasses only.
Both methods have their merits, Bractals, well done.
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