MEN
167
MEN
7 a dding 10 to 120, we have 130, the other leg. Andfated\ n ever y instance, each horizontal number will be gene-to en ] addi "g tbe num ber in the vertical left-hand columnc °lum * tW0 ad j acent numbers in the second and third
ns °f the same horizontal line, and by adding 10 to thember expressed in the fourth.
Sides of a right-angled triangle, expressed byany three of the horizontal numbers.
25
27
29
31
33
35
37
39
41
43
45
47
49
61
63
55
57
59
61
63
65
67
69
71
73
75
77
79
81
83
169 . 119
194 . 144
221 . 171
250 . 200
281 . 231
314 264
349 . 299
366 . 336
425 . 375
466 .. 416
509 . 459
654 . 504
601 . 551
650 . 600
701 . 651
754 . 704
809 . 759
866 . 816
925 . 875
986 . 936
1049 . 999
1114 . 1064
1181 .. 1131
1250 . 1200
1321 . 1271
1394 . 1344
1469 . 1419
1546 . 1496
1625 . ' 1575
1706 . 1656
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
Ratio of| the twolegs.
Iris
1^5
1*
14
IJx
m
m
m
2ft2*25%2$2ft
m
2ft2ft2 44StIj
m
8,ft
8ft
3?
3ft
3ft
3®
3®
3ft
4ft
— To find the length of a cylindrical helix.Anther the circumference of the base by the
the Sq 0 revolutions ; to the square of the product adddist aq ar ® the height of the spiral, or the square of theom the beginning to the end ; and theTh e f° 0t tbe sum w * d be tbe l en oth the spiral.
|he bag Brm °*' tbe cylindric helix is a right-angled triangle,height t) wb ’ cb ' s the number of revolutions, and the
strett ’■he spiral ; i. e., if the whole were unwoundf *ght- ant> , C ’ ed upon a plane, the development would be a' ■£*«rrci trian g*e.
fl J ?. ^—Required the length of a screw twisting22 inches in circumference, 3^ times, ande along the axis 16 inches.
22 16
77 base of the development.
256
539
539
5929
square of the base.
LContifMMd—
5929 square of the base.256 square of the altitude.
6185 ( 78.64 inches, the answer.49
148 ) 12851184
1566)10100
9396
15724) 7040062896
7504
Example 2.—Required the number of feet of hand-railingfor a semicircular stair, consisting of 9 winders, each 6^inches high, the diameter of the well-hole being 18 inches.
Problem III.— The chord and versed sine of an arc of acircle being given, to find the diameter.
Rule 1.—Divide the sum of the squares of the sine andthe versed sine by the versed sine itself, and the quotient isthe diameter.
Example 1.—Given, the chord, ab = 48 feet, and theversed sine, d e = 18 feet; required the diameter.
48
— z= 24, the sine, or half chord.
2 -
242 I ] Q2
Then_A- — 50, the diameter required.
18
This arithmetical operation of finding the radius is muchpreferable to the geometrical construction, the calculationbeing so easy, and performed in a very small compass;whereas the other mode requires a floor, or fiat surface, todescribe it upon, which cannot at all times be obtained ;recourse must therefore be had to a temporary floor of roughboarding, which requires an immense time in the preparation,and when done is not much to be depended upon.
Examples for Practice.
A room is to be constructed with a cylindric bow, the planbeing the segment of a circle, whose chord is 18 feet, and theheight of the segment 6 feet; what length of a rod will benecessary to describe the arc ?
A bridge is to be constructed of a cylindric intrados, thesection of which is to be the segment of a circle, to span 100feet, and to rise 33 feet ; what length of a line, or wire, willbe necessary to describe the arc ?
Rule 2.—As the versed sine is to the half chord, so isthe half chord to a fourth proportional ; add this fourth pro-portional to the versed sine, and the sum is the diameter :thus, take the dimensions in the preceding example ; we have
24 x 24 _ 32 ;
18
24
24
18
then 32 + 18 = 50, the diameter.
Problem IV. — :In the segment of a circle are given thechord, and its distance from the centre ; to find the radius ofthe circle.
Rule. —Add the square of the half chord to the square ofthe distance ; and the square root of the sum will be theradius of the circle.
Example 1.—Let the chord, a b, be 8 feet, and the distanceof ab from the centre, CD, 3 feet; required the radius ofthe circle.