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.
- 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.
- 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.