6 ms·
Of course you can average times of day, as well as wind directions. Express them as vectors on a 2-D plane, then separately average the components in each dime
by bluesnowmonkey 16y ago
Of course you can average times of day, as well as wind directions. Express them as vectors on a 2-D plane, then separately average the components in each dimension. Although this is somewhat obvious in the case of wind measurements, it works for times of day as well if you think of them as points on a 24-hour clock face.
Your medical friend was not checked out. She was busy stifling impolite laughter at your unfamiliarity with Cartesian coordinate systems.
- roel_v 16y agoFor time, what would then be the average of let's say 3 and 9 o'clock? Center of the clock I presume, but what does that mean in terms of time? Or would you then need to define the average of several points in time as a non-point in time?
- copper 16y ago0 o'clock could be interpreted as 12 o'clock too.
- roel_v 16y agoThe GP explicitly specified a 24-hour clock face, plus even on a 12-hour clock the question would remain.
- DougBTX 16y agoHand waving: I'm guessing that you could make a new vector from the average x and the average y. The direction would indicate the average time, and the length would indicate the certainty, so a length of zero would mean total uncertainty.
- cscheid 16y agoThat's pretty much exactly right - the variance of the von mises distribution is essentially the length of the average vector. http://en.wikipedia.org/wiki/Von_Mises_distribution http://en.wikipedia.org/wiki/Von_Mises_distribution
- Vivtek 16y agoYou'd have to talk about a time vector, with your case being the zero-length vector, or "no preference". Then a vector with a longer extent would indicate a "strong preference". Something like the standard deviation or something in a linear measure, I guess.
- d0mine 16y agoFor 24-hours clock you should have said 6 and 18 hours.
- roel_v 16y agoHow embarrassing - I initially had written a paragraph here pointing out how wrong you are and I was just seconds away from posting it before realizing you're right ;) So yes you're right, to be consistent with the article and my reply elsewhere in this thread it should've been that. Luckily it seems that most posters managed to derive the underlying question from the circumstances ;)
- eru 16y agoYou can do that for wind, but what does it mean? An equally valid measure for wind would be to take the absolutes, and average their squares. That'll give you a measure of the average energy in the wind.
- hugh3 16y agoIt means the average direction of the wind, which is an important factor in many places. Of course this might need to be combined with an average windspeed to get a better picture of what the wind in a certain location is really like.
- eru 16y agoWhy don't we just look at the distribution in the first place? Make a nice circle chart for all wind vectors and color it according to probability.
- beambot 16y agoThat may be a decent "hack" approach, but more generally you should use distributions meant to operate on circles, eg. von Mises (analagous to Gaussians): http://en.wikipedia.org/wiki/Von_Mises_distribution http://en.wikipedia.org/wiki/Von_Mises_distribution