Trig Table 3Place sines and cosines for problems on this sheet. Refer to mathematical reference book for values not in this table.
Angle 
Sine 
Cosine 
5 
0.087 
0.996 
6 
0.105 
0.995 
7 
0.122 
0.993 
8 
0.139 
0.990 
9 
0.156 
0.988 
10 
0.174 
0.985 
11 
0.191 
0.982 
12 
0.208 
0.978 
13 
0.225 
0.974 
14 
0.242 
0.970 
15 
0.259 
0.966 
16 
0.276 
0.961 
17 
0.292 
0.956 
18 
0.309 
0.951 
19 
0.326 
0.946 
20 
0.342 
0.940 
21 
0.358 
0.934 
22 
0.375 
0.927 
23 
0.391 
0.921 
24 
0.407 
0.914 
25 
0.423 
0.906 
26 
0.438 
0.899 
27 
0.454 
0.981 
28 
0.469 
0.883 
29 
0.485 
0.875 
30 
0.500 
0.866 
31 
0.515 
0.857 
32 
0.530 
0.848 
33 
0.545 
0.839 
34 
0.559 
0.829 
35 
0.574 
0.819 
36 
0.588 
0.809 
37 
0.602 
0.799 
38 
0.616 
0.788 
39 
0.629 
0.777 
40 
0.643 
0.766 
41 
0.656 
0.755 
42 
0.669 
0.743 
43 
0.682 
0.731 
44 
0.695 
0.719 
45 
0.707 
0.707 
46 
0.719 
0.695 
47 
0.731 
0.682 
48 
0.743 
0.669 
49 
0.755 
0.656 
50 
0.766 
0.643 
51 
0.777 
0.629 
52 
0.788 
0.616 
53 
0.799 
0.602 
54 
0.809 
0.588 
55 
0.819 
0.574 
56 
0.829 
0.559 
57 
0.839 
0.545 
58 
0.848 
0.530 
59 
0.857 
0.515 
60 
0.866 
0.500 
61 
0.875 
0.485 
62 
0.883 
0.469 
63 
0.891 
0.454 
64 
0.899 
0.438 
65 
0.906 
0.423 
66 
0.914 
0.407 
67 
0.921 
0.391 
68 
0.927 
0.375 
69 
0.934 
0.358 
70 
0.940 
0.342 
71 
0.946 
0.326 
72 
0.951 
0.309 
73 
0.956 
0.292 
74 
0.961 
0.276 
75 
0.966 
0.259 
76 
0.970 
0.242 
77 
0.974 
0.225 
78 
0.978 
0.208 
79 
0.982 
0.191 
80 
0.985 
0.174 
81 
0.988 
0.156 
82 
0.990 
0.139 
83 
0.993 
0.122 
84 
0.995 
0.105 

Figure 4  Angle of Repose.
As the angle of incline is increased, friction between the load weight and the inclined plane will prevent the weight from sliding downhill until the angle of repose is reached. The coefficient of friction between the two mating surfaces is equal to the tangent of the angle of repose. To experimentally find the coefficient of friction between a load weight and sliding surface, raise one end of the incline until the weight just starts to slide. Measure the angle. Look up the tangent of the angle in a trig table. This is the coefficient of friction between these two surfaces. Example: Suppose the incline can be raised to a 30° angle before the weight starts to slide. The tangent of 30° is .577 and this is the coefficient of friction between these two materials.
Figure 5  Load on an Inclined Plane.
Cylinder and load axes are in line but load is moving on an upward incline. The cylinder must provide enough force to raise the load to a higher elevation against the force of gravity. If the cylinder were mounted in a vertical position, a force of a little over 5000 lbs. would be needed to lift a 5000 lb. load weight. But with the cylinder inclined as in the illustration, the same 5000 lb. load can be lifted with a smaller cylinder force because of mechanical advantage which is similar to a wedging action. Example: Find cylinder force (neglecting friction) to push a 5000 lb. weight up a 28° incline. Solution: Use angle sines as multipliers. Cyl. force = 5000 x .469 (sine of 28°) = 2345 lbs. If there is high friction between load and incline, calculate additional force from cylinder to overcome friction. See opposite side of this sheet. If rapid acceleration of a massive load is required, calculate extra cylinder force required using information in Issue No. 4 of these data sheets.
Figure 6  Cylinder Operating a Rotating Lever.
Cylinder force, F, is horizontal in this figure. Only that portion, T, which is at right angles to the lever axis is effective for turning the lever. The value of T varies with the acute angle "A" between the cylinder axis and the lever axis.
Example: A 4inch bore cylinder working at 750 PSI will develop a 9420 lb. force (12.56 sq. in. area x 750 PSI). Effective force T when working at a 65° angle is: 9420 x .906 (sine of 65°) = 8535 lbs.
Figure 7  Cylinder on a Crane.
Find the lifting capacity of this crane when its members are at the angles shown (capacity will vary as the beam of the crane raises and lowers). A force, F = 15,000 lbs.,is produced by the cylinder and applied to a point 5 feet from the beam pivot. The angle between cylinder and beam axes is 30°. Force F2, the true torque force on the beam= 15,000 x .500 (sine of 30°) = 7500 lbs. Through a 3:1 leverage action, 7500 lbs. at 5 feet from the pivot translates to 2500 lbs., F2, at a point 15 feet from the pivot . To find the vertical lifting force when the beam is at a 45° angle with the ground, F2 must be divided by the sine of the angle with the beam axis and vertical: Lift = 2500 ÷ .707 (sine of 45°) = 3535 lbs.
