Elevator Tower Mechanism Explained: Mine Headframe Parts, How It Works, and Hoisting Formula

Trace the drum, rope and cage

The side elevation shows a winding drum, a braced headframe, a head sheave and a cage between guides. The rope meets both circular surfaces tangentially and remains connected to the cage roof. As the cage descends, the drawn payout drives both rotating wheels. The cage slows to a stop, pauses, and the illustration restarts.

The side-view angle is the included angle between the vertical downward rope leg and the leg leading toward the drum. Changing it changes the drum position and the force-vector diagram. It is not the out-of-plane fleet angle.

The numeric displacement follows constant-deceleration kinematics. Shaft travel and playback time are normalized to keep the mechanism visible. Wheel rotation follows the drawn payout, so it is not a predicted operating RPM. The illustration does not establish real tower height, shaft depth or clearance.

What the calculator compares

The four controls are static tension, braking deceleration, rope angle and initial speed. Static tension represents suspended weight in this ideal model. The load factor and peak rope tension describe an initially descending load with upward acceleration while it is stopping.

This is an explanatory load comparison rather than a structural design. Rope mass and elasticity, sheave inertia, counterweights, friction, braking delays and tower deformation are not included. A different motion direction or load arrangement requires a different force balance.

Braking tension and rope-force resultant

Let T0=mg be the static suspended-weight tension, and let a be the positive upward deceleration of an initially downward-moving mass. Newton’s law gives T−mg=ma, so T=T0(1+a/g), with g=9.80665 m/s².

If both rope legs carry that same tension and their included angle is θ, their components on the sheave are Fx=−T sin θ and Fy=T(1+cos θ). Their resultant magnitude is R=2T cos(θ/2), directed θ/2 from downward vertical. The bearing support must oppose that resultant. The included angle is measured between the two rope legs in side view.

With initial downward speed v0 and constant positive deceleration a, stopping time is v0/a and stopping distance is v0²/(2a). During braking, v=v0−at and x=v0t−at²/2. At zero deceleration the load continues at constant speed and no finite stopping distance is returned.

Default descending-stop example

For 1 MN static tension and 5 m/s² upward deceleration, the ideal load factor is about 1.510 and peak rope tension about 1.510 MN. At a 45° included rope angle, the resultant on the sheave is about 2.790 MN, tilted 22.5° from vertical.

Starting downward at 18 m/s gives a 3.6 s stopping time and 32.4 m ideal stopping distance. At the illustrated stop, live tension returns to the 1 MN static value while the calculator retains the peak braking result.

Doubling speed quadruples stopping distance at the same deceleration. In this ideal mass model it does not change peak braking tension. Setting deceleration to zero shows a five-second constant-speed sample, not a zero-distance stop.

Limits of an ideal headframe load illustration

The page does not calculate member stress, foundations, rope selection, brake thermal capacity, dynamic amplification from rope stretch or a permissible stopping clearance. Clearance, replacement, alignment and service-life requirements depend on the actual installation and equipment.

The force diagram isolates the rope resultant; it does not show every structural load. Its arrow scale is normalized so the angle remains readable. The reported force values, not the drawn arrow length, carry the magnitude.

Headframe calculator questions

What does the angle slider measure?

The angle between the two rope legs in side view. The resultant of equal tensions bisects that angle.

Why does tension rise while the cage slows?

The cage is descending, so stopping it requires upward acceleration. The upward tension must exceed its weight in this model.

Why is there no stopping distance at zero deceleration?

A positive downward speed persists indefinitely if no opposing acceleration is applied.

Is the drawn shaft to the same scale as the calculated stopping distance?

No. Travel is normalized to keep the mechanism visible, with calculated distance shown numerically.