Counterweight Mass
Calculate an elevator's counterweight mass, M_cw = M_car + factor × Q_max, from the car mass, the balancing factor (typically 0.40 to 0.50) and the maximum load Q_max (kg). The result, in kg, is the mass that balances the car plus a fraction of the payload, so the motor works with the smallest average imbalance. A factor of 0.45 (45%) is common: it fully balances the car and 45% of the rated load, minimizing motor work both with a full and an empty car. Enter the car mass, the balancing factor and the maximum load.
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Counterweight mass
The counterweight is what makes a traction elevator efficient: instead of lifting the entire weight of the car and its load, the motor only has to overcome the imbalance between the two sides of the sheave. The counterweight mass is M_cw = M_car + factor × Q_max, where the balancing factor (typically 0.40 to 0.50) sets what fraction of the rated load it offsets. The classic value of 0.45 (45%) follows an elegant logic: the counterweight fully balances the car itself plus 45% of the rated load. As a result, the imbalance (and therefore the work done by the motor) is the same when the car runs empty (the heavier counterweight pulls it up) and when it carries about 90% of the load (the car is now the heavier side) — spreading the effort symmetrically around the equilibrium condition (car at 45%). That minimizes peak power, energy consumption and wear, since elevators rarely travel completely full. Enter the car mass, the balancing factor and the rated load.
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Hoist Rope Tension
Calculate the resultant force in an elevator's hoist rope, F = (Q + M_car − M_counterweight)·g, from the payload Q, the car mass and the counterweight mass (kg). The result, in newtons, is the unbalanced effort the steel ropes must transmit, already net of the counterweight's balancing effect. It is the basis for sizing the ropes (number, diameter and safety factor, typically ≥ 12 in elevator codes) and the traction sheave. When the load is such that car + load ≈ counterweight, the force tends to zero (balanced system). Enter the load, the car mass and the counterweight mass.
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Calculate the motor power of an elevator, P = m·g·v ÷ η, from the payload m (kg), gravity g (9.81 m/s²), nominal speed v (m/s) and the system efficiency η (motor, gearbox, sheaves). The result, in watts, is the mechanical power needed to hoist the load at nominal speed. In practice, the counterweight (balancing the car plus ~45% of the load) reduces the effective power, and regenerative braking on descent can return some to the system. It is the base calculation for sizing the traction machine. Enter the load, the speed and the efficiency.
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Calculate the power transmitted by a belt, P = (T₁ − T₂)·v, from the tight-side tension T₁ (N), the slack-side tension T₂ (N) and the belt velocity v (m/s). In a belt drive, the driving pulley drags the belt by friction, creating a DIFFERENCE in tension between the two sides: the side that 'pulls' (tight side, T₁) is more tensioned than the side that 'follows' (slack side, T₂). This difference (T₁ − T₂), the effective tension or tangential force, is the net force that actually transmits motion; times the belt velocity, it gives the transmitted POWER. The larger the tension difference the belt can sustain without slipping (depending on friction, wrap angle and, in V-belts, the wedging effect of the pulley walls), the greater the transmissible power. Power also grows with belt velocity — so high-power drives use large pulleys and fast belts (up to a limit, since centrifugal tension reduces available friction at very high speeds). This is central in belt-drive design, present in almost every rotating machine: motors, fans, pumps, compressors, machine tools and vehicles. Enter the tight- and slack-side tensions and the belt velocity.
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Calculate a train's specific movement resistance by the Davis equation, R = A + B·V + C·V², from coefficient A (rolling resistance and mechanical friction, speed-independent), B (resistance proportional to speed, from flange friction and oscillations), C (aerodynamic resistance, proportional to speed squared) and the speed V (km/h). The Davis equation, from the 1920s and still standard in railway engineering, describes the total resistance to motion the locomotive must overcome on straight, level track, per unit weight (N/t or kgf/t). At low speed the constant and linear terms (friction) dominate; at high speed the quadratic aerodynamic term dominates, decisive for high-speed trains (hence their careful streamlining). Davis resistance, plus grade (gravity) and curve resistances, sets the required tractive effort, energy consumption and locomotive traction capacity. It is the basis of traction calculation and train performance. Enter coefficients A, B and C and the speed.
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.