How to use this calculator

Put the origin at the point where the three cables meet, with z pointing up. Enter the load as a mass or a weight, then the x, y and z coordinates of each cable’s anchor measured from that point. If your problem puts the origin somewhere else, subtract the meeting point’s coordinates from each anchor first.

Formulas

Unit vector along a cableu = r|r| = (x, y, z)√(x² + y² + z²)

Equilibrium of the meeting pointT_A u_A + T_B u_B + T_C u_C + (0, 0, −W) = 0

Writing the vector equation one component at a time gives three linear equations in the three tensions, which the calculator solves exactly.

Worked example

A 180 kg crate on cables to A(−2, 3, 6), B(4, −2, 4) and C(−3, −4, 2) m (the default)

W = 180 × 9.81 = 1765.8 N.
|r_A| = 7 m, |r_B| = 6 m, |r_C| = √29 = 5.385 m, so u_A = (−0.2857, 0.4286, 0.8571), u_B = (0.6667, −0.3333, 0.6667), u_C = (−0.5571, −0.7428, 0.3714).
x: −0.2857 T_A + 0.6667 T_B − 0.5571 T_C = 0
y: 0.4286 T_A − 0.3333 T_B − 0.7428 T_C = 0
z: 0.8571 T_A + 0.6667 T_B + 0.3714 T_C = 1765.8
Solving: T_A = 1259.0 N, T_B = 833.9 N, T_C = 352.2 N.

Assumptions and limits

The cables are straight and weightless, and all three meet at one point with the load.
A cable can only pull. A negative answer means that anchor layout cannot hold the load with cables alone.
If the three cables lie in one plane the problem has no 3D solution; for two cables in a vertical plane use the cable tension calculator.

Frequently asked questions

What if the problem uses y as the vertical axis?

Relabel the coordinates so the vertical one goes in the z boxes. Swap the other two the same way for every anchor and the tensions come out the same.

The load hangs below the knot on a fourth cable. Does that change anything?

No. The vertical cable just carries the weight W up to the knot, so use the knot as the origin.