An elevator motor must provide enough force to overcome gravity, resistance, and any required acceleration.
Mechanical power is the rate at which the motor transfers energy:
\[
P=Fv.
\]
For an elevator, the relevant force is the drive force or cable tension, and \(v\) is the elevator speed.
1. Elevator mass
The total mass on the elevator car side is
\[
M=m_{\mathrm{car}}+m_{\mathrm{pass}}.
\]
This mass includes the empty car plus passengers or cargo.
2. Direct-lift model without counterweight
For upward motion, the cable tension must overcome weight, resistance, and any upward acceleration:
\[
T=M(g+a)+f.
\]
At constant speed, \(a=0\), so
\[
T=Mg+f.
\]
The mechanical power delivered to the elevator car is
\[
P=Tv.
\]
3. Counterweight model
A counterweight reduces the net gravitational load the motor must lift. In an ideal counterweighted elevator,
the approximate drive force is
\[
F_{\mathrm{motor}}
=
(M-m_{\mathrm{cw}})g+(M+m_{\mathrm{cw}})a_{\uparrow}+d f.
\]
Here \(m_{\mathrm{cw}}\) is the counterweight mass, \(a_{\uparrow}\) is acceleration measured positive upward for the car,
and \(d=+1\) for upward car motion or \(d=-1\) for downward car motion.
The term \((M-m_{\mathrm{cw}})g\) is the gravitational imbalance. The term
\((M+m_{\mathrm{cw}})a_{\uparrow}\) appears because both the car side and counterweight side have inertia.
4. Constant speed, acceleration, and deceleration
The calculator estimates three useful power levels:
| Phase |
Acceleration |
Meaning |
| Speeding up |
\(a_{\uparrow}=d a\) |
The car accelerates in the direction it is moving. |
| Constant speed |
\(a_{\uparrow}=0\) |
The car moves at constant velocity. |
| Slowing down |
\(a_{\uparrow}=-d a\) |
The car accelerates opposite its motion direction. |
Since \(P=Fv\), the acceleration phase often has the largest positive motor power when the elevator is moving upward.
During downward motion or with a heavy counterweight, power can become negative. Negative power means the motor must brake
or regenerate energy instead of supplying energy.
5. Motor efficiency
If the motor has efficiency \(\eta\), the approximate electrical input power required during positive-power operation is
\[
P_{\mathrm{input}}=\frac{P_{\mathrm{mechanical}}}{\eta}.
\]
For example, a mechanical output of \(50\ \mathrm{kW}\) at \(90\%\) efficiency requires
\[
P_{\mathrm{input}}=\frac{50\ \mathrm{kW}}{0.90}\approx55.6\ \mathrm{kW}.
\]
6. Worked example
Suppose a \(1500\ \mathrm{kg}\) elevator accelerates upward at \(1.2\ \mathrm{m/s^2}\), with \(g=9.81\ \mathrm{m/s^2}\)
and negligible friction. The required tension is
\[
T=M(g+a).
\]
Substitute:
\[
T=(1500)(9.81+1.2)=16515\ \mathrm{N}.
\]
If the elevator is moving at \(2.5\ \mathrm{m/s}\), then
\[
P=Tv=(16515)(2.5)=41287.5\ \mathrm{W}\approx41.3\ \mathrm{kW}.
\]
At constant speed with the same load,
\[
T=Mg=(1500)(9.81)=14715\ \mathrm{N},
\]
\[
P=(14715)(2.5)=36787.5\ \mathrm{W}\approx36.8\ \mathrm{kW}.
\]
Formula summary
| Concept |
Formula |
Use |
| Total elevator-side mass |
\(M=m_{\mathrm{car}}+m_{\mathrm{pass}}\) |
Combines car and load masses. |
| Direct lift, accelerating upward |
\(T=M(g+a)+f\) |
Required cable tension without counterweight. |
| Direct lift, constant speed upward |
\(T=Mg+f\) |
Required cable tension at \(a=0\). |
| Counterweight drive force |
\(F_{\mathrm{motor}}=(M-m_{\mathrm{cw}})g+(M+m_{\mathrm{cw}})a_{\uparrow}+df\) |
Idealized drive-force estimate with counterweight. |
| Mechanical motor power |
\(P=Fv\) |
Power from drive force and elevator speed. |
| Electrical input estimate |
\(P_{\mathrm{input}}=P_{\mathrm{mechanical}}/\eta\) |
Input power required for positive mechanical output. |
A counterweight can greatly reduce constant-speed lifting power, but acceleration still requires power because both the car and counterweight have inertia.
