On earth we can divide our day into 24 hours, and 365 days in the year. I am trying to imagine how people living on airships above the cloud-tops would think about time on Venus given that a single day lasts longer than a Venus year 243 earth day, vs 224 earth days for a year. While at the same time the wind blows you around the planet in about 50 hours.

That makes it hard to decide what would be the most practical way of thinking about time and dates on Venus.

  • $\begingroup$ It's largely a solved problem on Mars. Venus would be different, but I don't imagine it would be harder. $\endgroup$
    – user
    Oct 17, 2016 at 9:36
  • $\begingroup$ Since the atmosphere only takes four earth days to circle the planet, airships might have a relatively convenient day/night cycle decoupled from the planet's rotation. $\endgroup$
    – called2voyage
    Oct 17, 2016 at 17:11
  • $\begingroup$ @MichaelKjörling I would expect it to be harder on Venus than Mars. IMHO it is trivial in comparison on Mars, because each day is roughly the same as on earth and you would be living on fixed location on the surface. So you could deal with time roughly the same was as on earth. $\endgroup$ Oct 23, 2016 at 23:30
  • $\begingroup$ @called2voyage agree it would be convenient that it is 50 hours rather than 243 ;-) Still it leaves the problem how do we measure time? Should it be related to this cycle or to the planet's rotation? Having morning happen at the same time each day seems to make sense, but then we would have to have multiple timezones. That seems tricky given that they will not be based on the planet's rotation as on earth but on the fact that you get blown around by the wind. $\endgroup$ Oct 23, 2016 at 23:37
  • $\begingroup$ The wind blows around Venus in about 50 hours in the upper atmosphere. If we, for example, put an air ship in Venus' atmosphere, we'd probably look at the 55 km altitude range. At that height, the atmosphere rotates more slowly. I couldn't find a figure, but it's considerably less, perhaps an orbit in 2 weeks, not 50 hours, but it wouldn't be precise, so not ideal for time. en.wikipedia.org/wiki/Atmosphere_of_Venus#Circulation or, they park a ship over one of the poles perhaps, no circling the planet there. $\endgroup$
    – userLTK
    Jul 23, 2019 at 13:39

1 Answer 1


Short explanation: Treat Venus as though it rotates 224.701 times per one orbit of the Sun (instead of one rotation per 243.025 Earth days, retrograde). Alternatively make Venusian minutes longer? than "Earth minutes" so there are 24*60 Venusian minutes per day with 224.701 Earth days per Venusian year.

That's an attempt to assign a reasonable period of time to each (modified) "Venus day" for the purposes of dividing each day into a reasonable number of chunks.

A standard 12 or 24 hour clock could be used by simply adjusting the speed that it runs at. Once 24 (modified) hours pass it would be called a "day" and 224.701 of those periods would be one orbit around the Sun.

On Earth:

  • One year = 365.256363004 (Earth) days
  • One day = 23 hours, 56 minutes, 4.0916 seconds
  • One second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom.

On Venus:

  • One year = 224.701 Earth days
  • One day = −243.025 Earth days (retrograde)

Earth: 365.256363004 / 365 = 1.00070236439 days, or 365.256363004 / 365.256363004 = 1 day or 365 / 365.256363004 = 0.99929812857 days.

Venus: 224.701 / 224.701 = 1, but 365.256363004 / 224.701 = 1.62552175114 Earth days, or 1.62552175114 * 23 hours, 56 minutes, 4.0916 seconds = 1.62108339 Earth days, alternatively 224.701 / 365.256363004 = 0.61518709257 Earth days, or 0.61518709257 * 23 hours, 56 minutes, 4.0916 second = 14.7241769 Earth hours. [Shhh, don't tell anyone at math.stackexchange.com]. No doubt there's something wrong with that math. :)

So I propose that either 1.62108339 Earth days (1 day, 14 hours, 54 Earth minutes), or for the other different method 14.7241769 Earth hours, equals one Venusion day.

Don't rely on wind speed, rotation, or even solely Venus' orbital period; do a mashup of the math (a mathematical abomination) to derive a reasonable number of "Earth hours" per "Venus day", or change the duration of an hour (relative to "Earth hours") and use "Venus hours" per "Venus day".

Work it out so 224.701 Earth days (one "Venus year") equals 224.701 * one "Venus day".

Rationale: It is probably most useful for one orbit of the Sun to equal one year. It's not particularly useful for either 243.025 Earth days to equal a single day on Venus. Using 1.62552175114 Earth days or instead using 14.7241769 Earth hours to represent a day provides a day of reasonable length.

[Spot an error in the pseudo-math it's OK to edit, as long as the means to derive a reasonable daylength isn't altered.]

I'll scratch my head and take another look at this suggestion in an hour or two.

Interesting background info:

Sidereal time is the angle, measured from the observer's meridian, along the celestial equator, to the great circle that passes through the March equinox and both poles, and is usually expressed in hours, minutes, and seconds.

Common time on a typical clock measures a slightly longer cycle, accounting not only for Earth's axial rotation but also for Earth's annual revolution around the Sun of slightly less than 1 degree per day (in fact to the nearest arcsecond, it takes 365.2422 days to revolve, therefore 360 degrees/365.2422 days = 0.9856° or 59′ 8″ per day, i.e., slightly less than 1 degree per day).

A mean sidereal day is 23 hours, 56 minutes, 4.0916 seconds (23.9344699 hours or 0.99726958 mean solar days), the time it takes Earth to make one rotation relative to the vernal equinox. (Due to nutation, an actual sidereal day is not quite so constant.)

The vernal equinox itself precesses slowly westward relative to the fixed stars, completing one revolution in about 26,000 years, so the misnamed sidereal day ("sidereal" is derived from the Latin sidus meaning "star") is 0.0084 seconds shorter than Earth's period of rotation relative to the fixed stars.

The slightly longer "true" sidereal period is called a stellar day by the International Earth Rotation and Reference Systems Service (IERS). It is also referred to as the sidereal period of rotation, or simply as the period of rotation or the rotational period.

  • $\begingroup$ thanks for the thought you put into this. I still have to try to process this. I should probably re-read this but I don't quite get the rational for the 1.6 earth days being a Venus day. When 224 Venus days have passed 365 earth days will have passed but that is longer than a Venus year. Also there is no easy way of relating a Venus day to any observable cycles on Venus. The sun wont rise and fall in that period. Nor will anything else observable happen. Thus I was contemplating simply ditching the whole day concept on Venus as it makes no sense. Or use earth day due to our bio clock $\endgroup$ Jan 1, 2018 at 23:08
  • $\begingroup$ @AdamSmith - Basically I'm saying 'ditch the concept of a true day' as it's not a particularly useful interval - It's not as though: You need to sleep 1/3 of a day (80 Earth days), or you work on a Farm and your crops require tending in the morning, etc. -- An alternative that occurred to me is to divide the orbital period by 360 then you could say you've traveled one degree around the Sun; also not particularly useful info but the daylength would be a reasonable duration. My rational was simply to not have each 'Venus day' be too long. $\endgroup$
    – Rob
    Jan 2, 2018 at 2:31
  • $\begingroup$ yeah I get that the actual Venus day length is very impractical. Your suggestion gives a time length which is easier to work with. Although I guess whatever we do, there is simply no practical day unit on Venus. $\endgroup$ Jan 3, 2018 at 21:42

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