In our university there's a posted sign that encourages students to take the stairs instead of the elevator to save the university electricity during the hot NYC summers.

When you climb the stairs you generate something like 8x your mechanical energy in metabolism which is dissipated as heat and must be cooled by the air conditioning in our cooled stairwells.

What are plausible assumptions about the energy efficiency limits of air conditioning and mechanical lifting of an elevator? What (fundamental) limits are there that prevent 100% efficiency in the elevator case? Are there other important factors for this analysis?

Is it plausible that the university is wrong? Should we take the elevator instead of the stairs to save the university energy? Do the humans of an energy-efficient future look like Wall-E?

The source of this question is a bathroom physics problem posted above university physics department toilets near grad offices. Problems are submitted by department members. To the best of my knowledge, none of these were ever homework questions. This question is adapted by my toilet memory from a problem that was provided by Andrei Gruzinov.

  • 4
    $\begingroup$ Don't forget that the elevator's motor is lifting much more than just a person, and the energy used (plus heat) is correspondingly larger. $\endgroup$ – Javier Jun 7 '18 at 16:42
  • 2
    $\begingroup$ @Javier Elevators usually have a counterweight equal to that of the elevator-car itself. So it really should only need to expend the energy to lift / slow-the-falling-of the people in the car, and overcome friction. $\endgroup$ – Glurth Jun 7 '18 at 16:57
  • 2
    $\begingroup$ The secret real reason for the counterweight on a traction elevator is to keep the hoist rope under tension so that it does not slip on the drum. A traction elevator can not function without a substantial counterweight, no matter how powerful you make the motor. If it also makes the system more energy efficient, that's a side benefit. $\endgroup$ – Solomon Slow Jun 7 '18 at 19:44
  • 2
    $\begingroup$ As fun as this is to ponder one must also consider long term effects, not just a comparison of work done for the single event of being elevated. Using stairs will make the human a more efficient machine, capable of doing more and consuming less. Healthy people can tolerate moderate temperatures without turning on the AC in their rooms. They require less medical care over time (although the unhealthy may die sooner). It's hard to say which path is better based on speculation. You need to include all these factors and get a real simulation going. I'm eager to see the results. $\endgroup$ – ggcg Jun 7 '18 at 21:26
  • 3
    $\begingroup$ A quick google search shows that this question has been asked and answered many times. Many times, and with multiple conflicting answers. The health benefits are indisputable, ignoring the risks of falling down the stairs (definitely non-zero; this is one of the a most common work-related injuries). I sometimes use the stairs to get to my fifth floor office -- oftentimes during regular business hours, but never when I'm at work during off-hours. $\endgroup$ – David Hammen Jun 8 '18 at 1:27

I wondered the same thing and so did some empirical testing with my elevator.


The elevator uses a huge amount of energy to move just the elevator.

For 5 floor ascent using my single piston hydraulic lift elevator on the day I tested...

Elevator: ~550KWs

Stairs: 92KWs / (22 calories)

If the elevator is going up anyway, then you should hop on since your additional weight has almost no effect on the total power used.

If you are considering taking the elevator alone, you should take the stairs.

Also keep in mind that most the energy used by the elevator is turned into heat and that heat is generated inside the envelope of the building, so must be rejected by the AC system.

More explanation and data here...


  • $\begingroup$ I'm glad to see an actual measurement rather than an estimate. However, according to wikipedia, hydraulic elevators are less power efficient that traction elevators because traction elevators have counterweights and hydraulic elevators don't. Wikipedia mentions that traction elevators can use up to 80% less energy than hydraulic elevators, so then each elevator ride would be about 110 KWs, which is pretty comparable to taking the stairs. $\endgroup$ – lnmaurer Jun 18 '18 at 16:15
  • $\begingroup$ I should add that the elevator in the original question is not a hydraulic elevator since the building is at least 10 stories tall (poster mentioned this in the comments), and hydraulic elevators are not used for such tall buildings. $\endgroup$ – lnmaurer Jun 18 '18 at 16:24

Accepting that climbing stairs generates heat equal to 7 times the mechanical energy (one part goes into you lifting yourself), how much energy does it take to remove that heat?

The 2nd Law of Thermodynamics says the best you can do is

$W = Q \frac {T_h - T_c}{T_c} \sim Q \frac{\Delta T}{T}$

Where $\Delta T$ is the difference between inside and outside temperature, and T is the outside temperature. For ease of calculation, taking T around room temperature of 280K, you get that the air conditioner is better (uses less energy, i,e, W/Q is better than 1/7) if it’s working into a temperature difference less than $\Delta T = 40\rm{K}$.

Large-building air-conditioners routinely even better than this for electrical power usage by using e.g. evaporative chillers to do most of the thermodynamics.

Unless it’s really really hot outside, take the stairs.

  • $\begingroup$ @sammygerbil if the AC is more efficient, you save energy by using the stairs. The goal of the sign mentioned in the question was to save energy for the University. $\endgroup$ – Bob Jacobsen Jun 9 '18 at 2:08
  • $\begingroup$ @gphys If taking the stairs wins for perfect elevator efficiency, you don’t need to consider the elevator further. $\endgroup$ – Bob Jacobsen Jun 9 '18 at 2:21
  • $\begingroup$ @GPhys I look forward to reading your answer. Actual power efficiencies for buildings are complicated: I work in one where most of the year all the thermodynamic cooling is done by evaporation, and electrical power is just used for fans. It doesn’t even have a closed-cycle refrigeration unit. So, it works even better (for power) than my indicative calculation. $\endgroup$ – Bob Jacobsen Jun 9 '18 at 2:39
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Bob Jacobsen Jun 9 '18 at 3:07
  • $\begingroup$ @BobJacobsen Ah yes, sorry my mistake. I got confused. $\endgroup$ – sammy gerbil Jun 9 '18 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.