Question: What is the angle of four single 8metre long chains lifting a 12.186 metre shipping container?
I need to know what the angle would be of four single 8metre long chains lifting a 12.186 metre long shipping container.
Two of the chains and the side of the container form the the three sides of a triangle.
This would be an isosceles triangle, with two equal sides of length l = 8 m, and the third side, or base, of length b = 12.186 m.
Call the angle contained between the two equal sides B. This is the angle you want.
Using the cosine rule, which is:
In any triangle ABC, of sides a,b,c and included angles A,B,C then the sides and angles of this trinagle are related by the following formula.
a^2 = b^2 + c^2 - 2.b.c.cos(A)
So in our triangle we havwe,
b^2 = l^2 + l^2 - 2.l.l.cos(B)
b^2 = 2l^2(1 - cos(B))
1 - cos(B) = (1/2)(b/l)^2
cos(B) = 1 - (1/2)(b/l)^2
Substituting for the values of b = 12.186 and l = 8,
cos(B) = 1 - (1/2)(12.186/8)^2 = 1 - 1.16015 = -0.16015 (negative value means the angle is greater than 90º)
B = 99º 12' 55", or 1.73163 rads
This is the angle between the chains attached to the corners of the side of length 12.186 m.
The angle between the chains along the other side(s) will be different (unless the container is square-shaped), but can be found using the same formula and method.