3
$\begingroup$

In common US usage, the heights of tall things are sometimes converted to a "number of stories". The thinking is that people can better compare heights to similar tall buildings that they might have seen.

The US conversion is usually: $$\text{Number of stories} = \text{Round}\Big(\frac{\text{height in feet}}{10}\Big)$$

I assume that something similar is used in metric countries like: $$\text{Number of storeys} = \text{Round}\Big(\frac{\text{height in meters}}{3}\Big)$$

I am assuming that like most things that are common usage, this conversion is not correct in practice.

What is a more accurate height of a building story?
How does this cause confusion when comparing the height of a building in "stories" to the actually number of stories?

$\endgroup$
  • 1
    $\begingroup$ I assume you're talking about in public media (news, blogs, etc.) I honestly haven't seen it, usually the exact or rough measurement is given in metric units. Perhaps because a meter or a kilometer is easy to visualize. Perhaps because dividing a measurement into 3 to give an arbitrary or subjective value is unnecessary. $\endgroup$ – Sam Weston Mar 8 '15 at 0:40
  • 1
    $\begingroup$ @SamWeston Maybe I just assumed incorrectly, and it really is just a US thing. $\endgroup$ – hazzey Mar 8 '15 at 0:42
  • $\begingroup$ California's Capitol Building comes to mind; an annex was added many years after its original construction, which has a different story height, so although the buildings are contiguous you can only move directly between them on levels where the floors are in about the same spot. I looked for a good picture but this was the best I found, looking from the original building into a hallway of the annex on one of the levels that matches up. $\endgroup$ – Air Mar 8 '15 at 4:48
  • $\begingroup$ @EnergyNumbers - Regarding the edit - "story" and "stories" are not misspellings; they're just the American spellings. Not typos. Note the use of "storeys" for non-American countries. $\endgroup$ – HDE 226868 Mar 8 '15 at 13:07
2
$\begingroup$

There's a handy-dandy table here:

                                           Office      | Residential/hotel | Function Unknown 
                                                                             or Mixed-Use 
floor-to-floor height (f)                  3.9m        | 3.1m              | 3.5m 
Entrance lobby level floor-to-floor height 2.0f = 7.8m | 1.5f = 4.65m      | 1.75f = 6.125m 
Number of mechanical floors above ground   s/20        | s/30              | 2/25
(excluding those on the roof)

Height of mechanical floors                2.0f = 7.8m | 1.5f = 4.65m      | 1.75f = 6.125m 
Height of roof-level mechanical            2.0f = 7.8m | 2.0f = 6.2m       | 2.0f = 7.0m 
areas / parapets / screen walls

H = Building height
f = Typical occupied floor-to-floor height
s = Total number of stories

m = meters

These numbers are just averages.

Using this, it is possible to derive the height of an office building: $$H_{\text{office}}=3.9s+11.7+3.9(s/20)$$ The same can be done for a residential building: $$H_{\text{residential}}=3.1s+7.75+1.55(s/30)$$ The page then gives some comparisons of real buildings and the variation from the formula. However, while the "aggregate" variation for residential buildings (for example) comes out to 0.36%, a better indicator of accuracy would be to take the absolute value of the variation.

So, in summary: The average height depends on the function of the building (i.e. office vs. residential). But there are, of course, deviations from these figures.

| improve this answer | |
$\endgroup$
  • $\begingroup$ It is interesting that a group tried to distill it down to a formula. Interesting site. $\endgroup$ – hazzey Mar 8 '15 at 0:51
  • $\begingroup$ @hazzey Indeed. I was skeptical of their legitimacy at first, but some background checks indicate that they're reliable. The formulae, though, do deviate quite a bit in some cases. But after all, there are some things (e.g. spires) that formula can't account for. $\endgroup$ – HDE 226868 Mar 8 '15 at 0:53

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.