ELL
369
ELL
Ora* ,
6 c et)t ltle diagonals a c and b d, cutting each other at s,4 B r ® > through s draw e f and g h, respectively parallel
tL
S'J'V
^ AD 5 upon s, with the radius s i, equal to the half« r Ec > describe the quadrant iEt, cutting ef at e ;jjr, c a rc ikl at k, and through k draw mn parallel toln g the diagonal bd at n ; join in, and through bto 0 ■}. • Parallel to it, cutting g h at g, and make s h equal
^tino’ ^°' n N o, and through b draw b p parallel to it,t'vo 5 at p > make s e equal to s f, and e p and g h are
^tho,
Q , ; then the curve may be described by any of theProty s a°wn in Problem 1.
- elh n erri ^—A trapezium, a b c D, being given, to inscribeP r0r j therein,
6 C t 0 ® e the sides b a and CDtoQ; also the sides ad and^ r; ara ' v the diagonals a c and b d, meeting each others? 1 atid r ° u Sh P draw eih, cutting the sides of the trapezium0 f ’ also, through r draw Q e G, cutting the other two
in
<ha*“ e tra P ez ‘ utn ’ n E an d G ; bisect i h at n, and e g at° w ' r ‘ and, bmp; join ip, to which produce k;
H e '7 Qnp
t. o
BMP; join I p, to
tl,g ec fiml to pI ; draw G l parallel to CD, cutting IK at
" u ord; lK is a diameter bisected by p, the centre, and L G ist : °ate. J
Problem might have been constructed, as in theaa ti n g (see Yol. I. Problems xviii. xix. and xx.) byNcli s i 0,16 , °i' the points of contact given ; but it is here18 and reduced into one problem, by which it
f“r 0 ii Uet ter adapted to perspective.i£j e> ? iO.— To find the area of any segment of anh,et t ~ the greater axis,o = the lesser,y = dh, the ordinate,
%
and
*= ad, the abscissa.
Th
11 hy the property of the curve, we have y =. — — X x
(«* i *
V
but
X % (ax — x 2 ) is a fourth proportional
* (ax — x s ) 2 for t : c :: x (ax — X s ) 2 : — X x
* A *
> and since x (ax — x s ) 2 is known to be the
' and;
(a*
fllly. ^ * 2 '
. l ; a, ^ le semi-segment of the circumscribing semi-circlec o' 8 ’ ° r article Segment ;) therefore, as the transverseofiiagatg e diameter of the circumscribing circle, is to thes em' >r ^'ameter of the inscribed circle, so is the area.^ent A ''Segment of the circle to the area of the elliptic® 15 : but the ordinate pd of the circle is to the
^ -wv,!, miv vx vtiiuutv 4.' XJ \J A WIU V/Jl V1V lO l>V
Ing C . I> °P the ellipsis, as the diameter of the circum-d-,■vfor e Clr . c ^ B ' s to the diameter of the inscribing circle :C b << mrcular and elliptic segments upon the same base,dj 8 ^, ty]® en the same parallels, are to one another as their1 ete r of 1 tbe greater axis of the ellipsis is equal to the•tip 1 ' s th tbe circumscribing circle.
t^Ure 0 f I ' e /' ore evident, that, whether we know the specificcf i C ' r curr, ' le greater axis of the ellipsis, or the diameter of
•lie - mg '—-' 1 - -
W! e fc lli
s
‘scribin
"'M, t]? 1 * 0 Se gment, by a ciroular segment; provided it be
g circle, or not, we still can obtain the area
Of
the greater axis of the ellipsis is equal to the
ell - 18 tile
own
circle. In architecture, this circumstance issuppose, for example, that in a groin,segment of a circle, and the other the segmentit follows, from the construction of the groin,ax' 1 ' 6 Ve rtical diameter of the circular side, and the»!>(, '' , 'dtl 1 qp °f the elliptic side, are equal; and therefore, if6 g; v eac ii side of the groin, which is the chord of its*iij e e to () 0 n ’ and the height of the arch, we have nothing‘ a ti(l th tllan to ii n( i tim area of the circular section or
area of the
elliptic5 B
side will be found by
the rule of proportion. For practical use, take the followingrule:
First.— To measure the circular segment .—To two-thirdsof the area of the base, mutiplied by the height, add thecube of the height, divided by twice the base of the segment,and the sum is very nearly the area of the circular segment:then To find the area of the elliptic segment, say, As thechord of the circle is to the chord of the ellipsis, so is thearea of the circular segment to the area of the ellipticsegment.
Example .—What is the area
of the elliptic end of a groin,
which rises 5 feet, and extends at its base 15 feet, supposingthe base of the circular end to be 18 feet ?
2 X 15 = 30, twice the len
gth of the segment.
18
5
15
5
90
25
18
5
3)270
3,0)12,5
90
4.166, &c.
90
180 two-thirds of the product of the base and height.4.166 cube of the height, divided by twice the base.
184.166 area of the circular segment.
Then 18 :15 :: 184.16615
920830
184166
18)2762490(153.471 the area of the elliptic segment.
96
90
62
54
84
72
129
126
.30
18
12
To have wrought this example according to the series,would have been too operose for practical purposes.
The above method for finding the area of the segment ofa circle, was discovered, or invented, by the author, in theyear 1794, and published in his Principles, in 1795.
It is evident, that whatever takes place in the segment,must also occur through the whole curve : therefore, in anellipsis having its greater axis equal to the diameter of acircle, it will be, As the diameter of the circle is to the lesseraxis of the ellipsis, so is the area of the circle to the area ofthe ellipsis.