226
Mechanics
DegSine of
3
4
5
6
7
8
9
10
11I 2
IZ
-4
-5
16
>7
18
Parts.
1743
3483
5 2 336975871510452121861 39 1 7-564317364
D.l Parts.19J3255620212223
34202 ^
35 8 36j393746040
. 39°73 4!244°673i4 225:42261143
2643837:44 69465,
27,45399'45;70710.6328 46947 46171933^4190802948480.47,73135165
Parts.
60181
61566
62932
64278
65605
66913
68199
Parts.
81915
82903
83867
84804
85716
86602
8746
8829.
89106
89879
90630
9-354
20791 30,50000487431^66' .
22495!31,5-503 49 75470 67 9205024192,32^52991:50 7660468^92718
25881:33 54463 I 51 777 14699335827563 34 ^ 55919,52 78801 70.9396929237^35 57357 53 79863 7 - 19455 -30901'36 58778,54 80901 72,95105
D.
Parts.
73
95630
74
96126
75
96592
76
97029
77
97437
78
97814
79
98162
80
98480
81
98768
82
99026
83
99254
84
99452
85
99619
86
99756
87
99862
88
99939
89
99984
90
IOOOOO
25. The Use of this Table will be obvious from twoor three Examples. It was observed, that the Power isto the Weight it sustains on any Inclined Plane IC, asthe Height of the Plane IL to the Length thereof IC ythat is, as the Sine of the Plane’s Inclination to the RadiusSuppose the Angle of Inclination ICE = 40 Degrees,then will the Sine IL be equal to 64.278, and the Ra-dius CI — to 100000, which Numbers are as 64 to100 ; therefore 100 Pounds will be sustained on the In-clined Plane by a Power equal to 64 Pounds nearly.
26. Again; since EC = 100000 represents the cen-trifugal Force under the Equator, then will IM = 76604(the Sine of 50 Degrees, and Co-Sine of 40) be as thesaid Force in the Latitude of 40 Degrees : WhichNumbers are as 1000 to 766; and such is the Propor-tion of the Forces in those two Places.
27. In the same Manner, if the Radius CD — 100000express the Force of any <Hr eft Stroke , then will theSine IL—64278 be expressive of the Force of an obliqueStroke in the Direction F C, every Thing else beingequal.
28. Again ; since the Force of a direct Stroke is ex-prefs’d by CD = 100000, if it were required to findthe Angle of Obliquity, such that the Force of the
Stroke