!'!|4

MEN

168

MEN

O

Here - = 4 the half chord.

2

Then (4* + 3 2 )^ = 5, the answer.Or thus, at full length.

4

4

25 ( 5, the answer.

Example 2.—In Stewart’s Ruins of Athens, vol. n. pi. vi.ch. 1. are given, in a section of the columns of the portico ofthe temple of Minerva at Athens, the distance between thechords of two opposite equal and parallel flutes, 6 feet 1.8inches, and the chord of each flute 11.688 inches ; requiredthe diameter of the column, which he has omitted.

6 1.8 2)11.688

2 ) 73.8 diameter reduced to

- inches.

36.9

5.844 half chord.

Inches.

(36.9 s + 5.844’)2 = 37.36 the radius nearly.

Problem V.— The radius of a circle being given, and thechord of a segment of that circle: to find the versed sine ofthe lesser segment.

Rule. — Method 1.—Subtract the square of the half chordfrom the square of the radius, and the square root of thedifference will be the cosine ; subtract the cosine from theradius, and their difference will be the versed sine of the

lesser segment ; that is, v = r — (r* —

Example. —A weir is to be constructed across a river, inthe arc of a circle of 250 feet radius ; the span of the river,or chord, to be 200 feet; what is the versed sine of the lessersegment ?

250 — (250 s — 100 s ) 2 = 20.87122, the versed sine required.

N. B. If the versed sine of the greater segment is required,add the cosine to the radius, and the sum will be the versedsine.

Method 2 .—When the chord is very small, and the diameterlarge.

As the diameter of the circle is to the half chord, so is thehalf chord to the versed sine, nearly.

Example 1.—Suppose a bridge is to be executed, 56 feetin diameter, the breadth or chord of each stone on the faceand on the intrado3 of the arch being 12 inches ; what is theversed sine, or height of the arc of a single stone ?

56

12

672 : 6 :: 66

672)36.000(.053 inches nearly.33 60

2 4002 016

384

Example 2.—Supposing the diameter of the earth to ^7957 miles, how much does the curvature rise in a chord2 miles ? ,

We have 7957 : 1 :: 1 : .000125 of a mile, which reducegives 7.92 inches nearly for the rise of the arc. , g

This would also be the distance that the curvature of 1earth would fall from the tangent in one mile. . 0

N.B. The deflection of the arc from the level is ®s *square of the distance from the point of contact nearly. *. .proportion would give the same result as the method by w “ lCthis example is wrought. .. j

Problem VI. —Given the distance between the two purosides of a trapezoid, and the breadth of the lesser end >' ,find the breadth of the greater end, so that the sides moy ' eto a point at any. given distance.

Note .—The distance from the lesser end to the given P 01is here termed the distance of the point of convergence. .

Rule. —As the given distance is to the sum of * ^distance, and the distance to the point of convergence ; s °the breadth of the lesser end to that of the greater. ^ e \Example 1.— Given the distance, e f, between the pa*® ^sides, A b and d c, of a trapezoid abcb, equal to 3 feet !breadth of the lesser end, d c, 2 feet ; and the distance, E jfrom the lesser end, d c, to the point, g, of the meeting .the inclined sides, a d and b c, 58 feet ; required the breaa b, of the greater end,

58 : 58 + 3

(58 + 3) X 258

,dtb

= 2.103, the brea«

of the greater end. „

Example 2.—A stone weir is to be constructed acr ° 3 0irriver, in the form of the segment of a circle, with the ,, C t j,evex side of the arc towards the stream ; the joints ot ^stones being all to tend to the centre: now the length of e ^,Q

1 foot!

stone in the direction of a radial is 4 feet, the radiusfeet, and the breadth of the lesser end of the stone 1required the breadth of the upper end of the stones.

Foot. In. Sec.

Answer , 1.016 feet, or 1 0 2

This problem is also useful in perspective, in the dratf 1 ’’ 1 ’of two lines towards a vanishing point. . fiiC

Example 3.—A bridge is to be constructed of a i e o(intrados, the section of which is the segment of a cl ~ C f e eti28 feet radius, the length of the archstones are to be ” Jand their breadth in front of the arch and on the intrafoot; required the breadth of the thick end of the ston e>

28 : 28 + 2 :: 1 to the answer.

Or,

28 : 30 : : 11

28)30(1.0714 the breadth of the greater end, as re<l 0 *

28

200

196

120

112

Or, the answer may be reduced to the workman s rUtaking his dimensions, thus: