Every cable cam rig, zip line and lifting sling asks the same question before it goes up: what is the tension in the cable, and is the line strong enough to hold it? The answer is rarely the number people expect. A 30 lb camera rig on a 600 ft span does not put 30 lb into the cable — it puts roughly 450 lb into it, because a nearly flat line has to pull sideways with enormous force to hold a small weight up.
Use the cable tension calculator below to work it out in seconds, then read on for the cable tension formula, the maths behind it, and the rigging angle rules that decide how much line you actually need.
Rigging Load Tool
Cable Tension Calculator
Enter your span, sag and load to find the tension in the cable — then check it against the breaking strength of the line you plan to fly.
Tension in the cable
What Tension in a Cable Actually Means
A cable can do exactly one thing: pull along its own length. It cannot push, and it cannot resist bending. That single limitation is what makes cable calculations so clean — you do not need to know the material, the manufacturing run, the age of the rope or its stretch modulus to solve for tension. You only need geometry and load.
Tension force is measured in pounds (lb) or kilonewtons (kN). If you work in metric, remember that kilograms measure mass, not force. To convert your load to kilonewtons, multiply the mass in kilograms by 9.81 and divide by 1,000. A 14 kg cable cam is therefore about 0.137 kN.
The Cable Tension Formula
For a cable anchored at two level points with a load hanging at mid-span, tension equals the load multiplied by half the cable length, divided by twice the sag:
| Symbol | Meaning |
|---|---|
T | Tension in the cable |
W | Load on the line (rig, camera, rider) |
L | Horizontal span, anchor to anchor |
H | Sag — how far the line drops at the middle |
D | Diagonal length of one half of the cable |
You find D with the Pythagorean theorem, because half the cable, half the span and the sag form a right triangle:
Substitute one into the other and you have the complete equation for tension:
That is the formula the calculator on this page uses.
Worked Example: Calculating Tension for a Cable Cam
A fully loaded FlyLine cable cam weighs about 30 lb. Rig it across a 600 ft span and tension the 8 mm Dyneema until it sags 10 ft at the middle:
- Half span: 600 ÷ 2 = 300 ft
- Diagonal: √(300² + 10²) = √(90,100) = 300.17 ft
- Tension: 30 × 300.17 ÷ (2 × 10) = 450 lb
Now apply the safety factor. Rigging practice is a minimum of 5:1, so 450 × 5 = 2,251 lb. Do not fly that rig on anything rated below 2,250 lb breaking strength — and if the calculated load were 1,000 lb, you would need a cable rated 5,000 lb or better.
How to Calculate Tension in a Cable Using Angles
The trigonometric route gives an identical answer and is often faster in the field, because a sag angle is easier to eyeball than a sag distance.
First, find the angle θ between the cable and horizontal at the anchor:
Then, since the vertical reaction at each anchor is half the load:
Watch your calculator mode — some work in degrees, some in radians. One degree equals π/180 radians.
Running the same numbers: θ = arctan(10 / 300) = 1.91°, and T = 30 / (2 × sin 1.91°) = 450 lb. The two methods always agree, because the triangle formed by the forces is similar to the triangle formed by the distances.
Rigging Angle Calculator: Why Shallow Cables Destroy Hardware
This is the part that catches people out. As a cable flattens, tension does not rise gently — it runs away. The load factor is simply 1 ÷ sin θ, and it is the same figure riggers use for the sling tension formula:
| Angle from horizontal | Load factor | Tension per side (1,000 lb load) |
|---|---|---|
| 90° (straight up) | 1.00× | 500 lb |
| 60° | 1.16× | 578 lb |
| 45° | 1.41× | 707 lb |
| 30° | 2.00× | 1,000 lb |
| 15° | 3.86× | 1,932 lb |
| 10° | 5.76× | 2,879 lb |
| 5° | 11.47× | 5,737 lb |
| 2° | 28.7× | 14,336 lb |
For a two-leg sling, tension in each leg is the load divided by twice the sine of the leg angle — the sling tension formula and the cable tension formula are the same equation wearing different clothes. Below about 30°, small changes in geometry produce large changes in force. This is why you never pull a line dead flat to get rid of sag: you would need a cable an order of magnitude stronger, and the horizontal pull at your anchors goes up just as fast.
How to Find Tension in Two Cables at Different Angles
When the load hangs from two cables that meet at different angles — uneven anchor heights, a rig pulled off-centre, or a two-leg bridle — the two tensions are not equal. Resolve the forces horizontally and vertically and you get:
T₂ = W · cos θ₁ / sin(θ₁ + θ₂)
Where θ₁ and θ₂ are each cable's angle from horizontal. The shallower cable always carries the larger share. Switch the calculator above to Two Cables to determine the tension in each leg, along with the horizontal and vertical components of reaction at both anchors.
How to Measure Cable Tension in the Field
You do not need a load cell to determine tension. You need a tape and a level eye.
- Measure the span. Anchor to anchor, horizontally — not along the cable.
- Load the line. Put the actual rig on it and park it at mid-span. An unloaded sag figure gives you a number that means nothing.
- Measure the sag. Sight a level line between the two anchor points and measure straight down to the cable at its lowest point. A laser level makes this quick over long spans.
- Run the numbers. Feed span, sag and load into the calculator above.
- Compare against rated strength. Use the manufacturer's minimum breaking strength, not the working load limit, and keep at least 5:1.
For reference on Dyneema: 8 mm line will sag roughly 10 ft over a 600 ft span when tensioned appropriately for a FlyLine cable cam. If you are seeing far less sag than that, your tension — and your anchor loads — are much higher than you think.
What These Cable Calculations Do Not Include
The calculations do not account for:
- Knots and splices, which can remove 30–50% of a line's strength
- Shock loading from a rig bouncing, braking, or slamming into an end stop
- Slings and soft shackles at the terminations, and their own angles
- Abrasion, UV exposure and age in a line that has worked three seasons
- Wind load and dynamic side forces on the rig
- Anchor strength — the tree, truss or post is often the weakest part of the system
That gap between theory and reality is exactly what the 5:1 safety factor is for. Do not spend it.
Frequently Asked Questions
What is the tension in the cable if it looks perfectly straight?
Approaching infinity. Tension is inversely proportional to sag, so as sag heads toward zero, tension climbs without limit. A truly straight cable holding any load at all is impossible — there is always sag, and if you cannot see it, tension is dangerously high.
How do you calculate tension in a rope at an angle?
Use T = W / (2 · sin θ), where θ is the angle between the rope and horizontal at the anchor and W is the load hanging from the middle. For a single rope pulling at an angle against one anchor, tension is W / sin θ.
What is tension equal to in a two-leg sling?
Each leg carries W / (2 · sin θ). At 60° from horizontal that is 1.16 times the load share; at 30° it doubles to a full share of the load per leg.
Does the type of cable change the tension?
No. Tension depends only on load and geometry, which is what makes these equations so useful — they hold for steel wire rope, Dyneema, and polyester webbing alike. What the material changes is how much sag you get at a given tension, and how much tension the line can survive.
How much sag should a zip line or cable cam have?
Enough to keep tension sensible. As a rough starting point, 2–3% of the span gives a workable balance between speed, ground clearance and cable load. Tighter than 1% and tension climbs steeply for very little gain.
What safety factor should I use for cable tension?
5:1 as a minimum for cable cam and zip line work — five times the calculated working tension as the required minimum breaking strength. Increase it for anything carrying people, anything over crowds, or any line that sees repeated shock loading.
