Shears with Winch or Tackle Block: How It Works, Diagram, Formula & Rigging Uses Explained

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Shears are two spars joined at the head and stabilized for lifting, with tackle suspended near the apex and a fall line led to a pulling source. This calculator is an ideal centered vertical-load statics reference, not a rigging plan or capacity rating.

Shears and Tackle Ideal Centered-Load Reference

Explore ideal centered-load statics for selected shears geometry and multipart tackle arithmetic.

0°

Idealized Lift Reference
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Selected Pull Factor
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Ideal Compression per Leg
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Horizontal Reaction per Foot
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Equation Used

W=Pnη; C=W/[2cos(θ/2)]; H=(W/2)tan(θ/2).
Ideal symmetric static reference only; not equipment capacity, safe working load or a lift plan.
  • The apex load is vertical and centered between two equal pin-ended legs.
  • The included apex angle is θ and each leg is θ/2 from vertical.
  • All selected supporting parts share the idealized load through the overall factor η.
  • Guys, anchors, restraint, off-center loads, dynamics, equipment ratings and safety are excluded.

The overall factor η is selected and does not replace manufacturer tackle-efficiency or rating data.

Two-leg frame and multipart tackle

Army rigging guidance determines ideal tackle mechanical advantage by counting supporting rope parts at the moving block and emphasizes rear guys, leg-spread restraint, anchors and proper fall-line routing. The front view shows the equal legs and load. An enlarged tackle view spreads the sheaves into one plane so the continuous rope and selected supporting parts can be followed. Odd supporting-part counts have a moving-block rope attachment; even counts have a fixed-block attachment.

What the calculator determines

Select line pull P, supporting-part count n, a transparent overall factor η and included apex angle θ. The reference reports W = Pnη, factor nη, ideal symmetric axial compression C = W/[2cos(θ/2)] and outward horizontal foot reaction H = (W/2)tan(θ/2).

Ideal centered-load relations

W = Pnη. For equal legs under a centered vertical apex load, vertical equilibrium gives 2C cos(θ/2) = W, so C = W/[2cos(θ/2)]. The horizontal component at each foot is H = C sin(θ/2) = (W/2)tan(θ/2).

Selected-input arithmetic

For P = 500 lb, n = 4, η = 85% and θ = 15°, W = 1,700 lb, factor = 3.4, ideal compression is about 857 lb per leg and horizontal foot reaction about 112 lb.

Centered-load assumptions

For the included apex angle measured between the legs, the vertical component uses cosine. The displayed leg forces exclude the additional forces from the fall line and guys. Real work requires a qualified lift plan, rated equipment and consideration of tackle reeving, per-sheave friction, fall-line direction, guys, anchors, spread restraint, off-center loads, drift, dynamics, shock, wind, ground, inspection and applicable regulations.

Questions

Is W a safe working load?

No.

Does η predict tackle friction?

No; it is selected.

Why cosine for leg compression?

Each leg is θ/2 from vertical, so its vertical component is C cos(θ/2).

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