MEN
170
MEN
I
of the latter. Then the root of the fourth term so found willhe the ordinate required.
Example. —In the parabola, a b c a, are given the base, ordouble ordinate, A c, 18 feet, the height, or abscissa, B D,16 feet, and the abscissa, b e, 8 feet; required theordinate, e p.
18
—= 9, the ordinate,
A
and 9 s = 81.
81 V 8
16 : 8 : : 81 : - ttt~ = 40.5 the square of e f.
1
Therefore 40.5 2 = 6.364 = ef, the ordinate required.
In this manner any number of ordinates, and consequentlyof points, may be found in the curve, so as to construct thefigure. This method is the most appropriate for use in theformation of the curve, when the abscissa is greater thanthe ordinate, or equal to it.
Problem V.— To find any point in the curve of a parabolaby means of an abscissal parallel, or transverse ordinate ; theheight, or abscissa, the ordinate, and the distance of thetransverse ordinate from the abscissa, being given.
Note. —A transverse ordinate, or abscissal parallel, is aright line terminated by the curve and by the base parallelto the abscissa, commonly called a diameter, in this particularcurve.
Rule. —Multiply the difference of the squares of the ordi-nate and of the distance of the abscissal parallel, or transverseordinate, from the abscissa, by the height ; divide the pro-duct by the square of the ordinate, and the quotient will givethe height of the transverse ordinate.
Or, thus : Take any two numbers, in the proportion ofthe ordinate, or half base, and the distance between theabscissa and transverse ordinate ; call these numbers respec-tively a and b : multiply the difference of the squares of Aand b by the height ; divide the product by the square of A ;and the quotient will give the transverse ordinate.
Example .—In the parabola, abc, are given the abscissa,b d, 8 feet, the ordinate, A d, 40 feet, and the distance, d e,16 feet ; to find the abscissal parallel, or the transverseordinate, e f.
(40.-16-) X 8 = the
40 2
Or, because 5 and 2 are in the same proportion as 40 and
16, we have _ on v 8
as before. - J — - = 6.72 feet,
O
This method, as it avoids the square root, will not only beeasier than the last, of finding a point in the curve of theparabola, but will be much more accurate when the abscissais less than the ordinate.
Problem XI.— Given the abscissa and ordinate of a para-bola, to find any number of equidistant diameters, or abscissalparallels, and consequently points, in the curve, at theextremities.
Take the simple arithmetical progression of the numericalscale 1, 2, 3, 4, 5, &c. till the last contain as many units asthe number of equal parts intended to be contained in theordinate. Then if the last number represent the ordinate,the preceding numbers, 1, 2, 3, &c., will represent the res-pective distances that each transverse ordinate is from theabscissa. Proceed with these numbers as distances, andcalculate as in the last Problem, and the several results willgive the ordinates.
Or thus : If a double ordinate be given, and the numberof transverse ordinates be even, or the equal parts of the
double ordinate odd, take the odd numbers of the aritl)® jpgprogression 1, 3, 5, 7, &c., and proceed as above : andfound the transverse ordinates, let the real base, or ^ordinate, be divided into the equal parts requireperpendiculars erected at the points of division, and ^respectively equal to the results on each side of the at> ^^ywill give the transverse ordinates, and consequently aspoints in the curve. . , g of
This method applies to the construction of a parao o(ievery great extent, as the extrados of a bridge, not 1° ,^ earch only, but for the upper line of a series of arches. ^parabolic curve is well adapted to bridges, as it gets q ^towards the middle, and therefore the contrast with tl>is not so violent as the circular arc, which has the sain® e rS&vature at the ends as in the middle. The use of tra ° x: ped>'ordinates in constructing the curve, is not only more e r ^tious in calculation, but more accurate in ascertainingcurve at the vertex.
Synopsis of the Principal Rules of Menscka 1
Theorem I. Rule 1.—The circumference, c, of abeing given, to find the diameter, d.
/»
d =
TJOtf'
c ir cle
3.1416
Theorem II. Rule 2.d — 22
The Rectificaton or Development of Ce bveS ‘
•fOllO 1 '
Theorem I. Rule 1.—To rectify or develop the ®ference of a circle.
c = 3.1416 d.
Theorem II. Rule 2.
22 dc - f
Theorem III. Rule 1.—To rectify the arc of *the radius, r, and the sine, s, of half the arc,
Let z be the length of the arc required, in all tking cases; then f
„ 1.1 q 3.3 q 5.5 q , a
z = 2 s + —^A + -~B + -~-c, &c. where ?
2.5 4.5 6.7
and A, b, c, the preceding terms. , 0 ( ^
Theorem IV. Rule 2. —Let h equal the chorhalf arc, and c equal the chord of the whole arc.
8 h — c
Theorem V. Rule 3. —The diameter, d,sine, v, and the chord, c, being given : | ^
6d ~ v = 2 X
4 vc
5 ZZ C *4**_ ZZ C X
6 d — 5v 6d — 5v
6 d — v
ve>-
J) k
6 d — 5 d
Theorem VI. Rule 4 . —The chord, c, and ^sine, v, being given : f
8 t> s c 3c'+J°S.
Z ~ C "** 3 c‘ + 2 if 01 ° 3 c a + 2 *
ye)' s