MEN

169

MEN

1.0714

12

.8568

8

6.8544

4

3.4176

Th*

inch *v ^l ves one f°°t and very nearly seven-eighths of aniiic fi’ es 16 V 101 bein g divided into twelve equal parts, called? e ^om an< eacb ’ ncb into eight equal parts ; the workmeninch. re & ar d anything less than the sixteenth part of an

xi ■

S-Scep^; exarn I J le corresponds to Problem V. Method 2, which^ e * r > or 18 !, 1 * 16 - versed s ’ ne °f tiie arc of the stone of a bridge,'Wse „ b e like, of a given radius and given chord: it like-^ire ct - ert ains the taper that the stones should have in the’Say th n °/ a ra dius. The mould of the section of the stones

V Prohl ° re be f° un< i b y calculation, and the arc described

f^Pass e p* °f tbe article Geometry, almost within the°tirr tbe section itself, without having recourse to the

^S'ent j ance °f a centre, which not only requires an incon-

pr e p^®^. ee °f space, but occasions great loss of time in

.^H.— To find any point in the arc of the seg-

V th e c> a / ClrC ^ e at extremity of an ordinate ; the radius

^incite f' ^ Le chord of the arc, and the distance of thep the middle of the chord, being given.

, r< an b ' n d the versed sine by Problem V. subtract it'’act t| j ad ’»s, and the remainder will be the cosine ; sub-from the square root of the difference lof the1 ' 6 ( -hof ( j 1 le ra< lius and of the distance from the middle ofetl ^h of *| to tbe ordinate, and the difference will give the

S'U '■ uruinaie, anu me uiueience wm jnve me

1 ordi «ate.

ti Je 25o p ® u Ppose the radius, A e, of the segment, see,f 16 rnicjj, eet) tbe chord, g h, 200 feet, and the distance from

om

feet

°f the chord to the ordinate, CD, 50

C D.

ie, will be found to be 20.871, by

Vi^B, OI tlle

V ordinate,

v ed ‘

Co «in e v, > consequently, 250 — 20,871 = 229.129 the

“^Ci^fore (250 2 "— 50 s ) a —229.129 = 15.820 feet,In the *

jj 1 ?® irwI° W ' n S examples, the radius and chord are the^ lri ate f r le ^ast; but the variation is in the distance of the0n i the middle of the chord.

Ex

AMp les for Practice, to be answered.ance from the middle of the chord—Answer,of the chord

to theto the

’he a- ’ re 9uired the ordinate.-> n ate b e “tonce from the middleOfA 61 the a- ’ re< l u ' 1 'ed the ordinate.— Answer,

/*ate bg glf tance from the middle of the chord to thejj. 1>0 tlier ’ required the ordinate.— Answer,

—p- without finding the versed sine, is as follows :

V 'X th ^ le cos ' ne °f G E > the ba lf arc, thus : from^9 lo°f the 6 . S( l uare °f the radius, subtract (100) 2 viz. theof tp Slne > or half chord, 200 ; and the square root,l ’iat 1) ^ 0,i dly i _^L, 1 ! e,ria inder gives A b, the cosine of the arc G e.i^tr, ^ tlin s . P lnd ’lie cosine of the arc D e in the sameOQj ct {5Qy (250) a viz. the square of the radius,

|®ot, 2a >s, ’ m Z ' tbe s 9 ua re of the sine, or half chord, ofl a f „^-949 5 distance from b to C ; and the square

>0t ’ 01 the - - • “

'-It.

remainder gives A I the cosine of

2 x

Thirdly. — Subtract the cosine, a b = 229.129, of thegreater arc G E, from the cosine A i = 244.949 of thelesser arc, de ; and the remainder, bi= 15.82, is equal tothe ordinate C d.

Problem VIII. — Given the chord of a very large segmentof a circle, and the radius of the circle; to find any numberof points in the arc, and thence to describe it.

Rule. —Divide the chord into any number of convenientequal parts each way from the middle, and erect ordinatesupon the points of division ; calculate the length of eachordinate by the last problem ; find the versed sine to a chordnot less than the distance between the two most remoteadjacent points in the ends of the arc ; describe an arc to thischord and versed sine by Problem LII. of the articleGeometry ; then procure a board equal to the length of thechord, and curve its edge to the arc ; with the curved edgedraw an arc between every two points of the extended arc,and the entire arc will be formed.

JV.B. There are several methods of describing segmentsof circles, as shown under the article Geometry ; but noneso eligible, upon a large scale, where accuracy is required,as the above. It will be found convenient to erect the ordi-nates ten feet apart on each side of the middle of the chord,though the last distance at each end, between the extremeordinate and the extreme of the curve should be less than theintermediate 10 feet distances. Boards in general do notexceed 10 or 12 feet; but a twelve-foot board will be suffi-cient for ten-foot distances on the chord, in most cases,even when the versed sine rises high in proportion tothe chord.

Example. —An engineer intending to construct a stoneweir over a river, in the arc, A l m, of a circle; the chord, a m,of which being 200 feet, and the radius 250 feet ; requiredthe length of the ordinates at 10 feet distance from eachother, and the versed sine to a chord of 12 feet, for com-pleting the arc.

n l, the versed sine ..

. 20.871

Problem V.

o k, the first ordinate

. 20.671

Problem VII.

p i, the second ditto..

. 20.07

ditto.

Q h, the third ditto ..

. 19.065

ditto.

H G, the fourth ditto..

. 17.651

ditto.

S F, the fifth ditto.

. 15.82

ditto.

T e, the sixth ditto ...

. 13.565

ditto.

u d, the seventh ditto

. 10.872

ditto.

v C, the eighth ditto...

. 7.726

ditto.

W B, the ninth ditto...

. 4.11

ditto.

These ordinates are respectively calculated at 10, 20, 30,

40, 50, 60, 70, 80, 90 feet distance from the middle of thechord to each ordinate.

It now remains to calculate the versed sine correspondingto a chord of 12 feet of the same arc as that of the weir :This will be found, by Problem V. to be .072 of a foot, orvery nearly seven-eighths of an inch, being, as workmen say,bare measure. Curve the edge of a twelve-foot board to aversed sine of seven-eighths of an inch ; then lay the curvededge, with the concave side towards the chord, successivelybetween every two points, A, b; b, c ; c, d, &c., draw-ing each part until the whole arc is completed to thechord am.

N.B. A boarding is generally fixed to a timber grating,or framing, supported by piles, for the foundation of thestone-work. .

Problem IV. — In a parabola having two abscissas andan ordinate to one of them, to find the ordinate of the other.

Rule. —As the abscissa of the given ordinate is to theabscissa of the required ordinate, so is the square of theordinate of the former abscissa to the square of the ordinate

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