Math help - rolling average Wind Direction

[skimask]

Maybe I'm thinking linearly too, but doesn't that just move the problem from North to South?
At a little SE its 179 or less, and move a little SW and it's -179 or more, which averages to 0 or North.
Maybe there's another trick in that hint?
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What I'm thinkin...(without colons this time )
1st reading is 30 degrees, 2nd reading is 330 degrees, average wanted should be 0 degrees, obviously the AVERAGE is 180...no good...but...
if you take the sin and cos of 30, you get .5 and .866...
if you take the sin and cos of 330, you get -.5 and .866...
the average of the .5 and -.5 is 0, sin(0)=0 (average of .5 & -.5)...
but I'm stuck at what to do with the 2 cosine values. Maybe this example is just a 'special case' where the result is right one 'axis' lines of a graph...
More thought needed...I know the answer is right here...just have to put my finger on it...

Or maybe you could treat the first 30 degree reading as actually 390 (360 + 30), then do the averaging, which would give you the right answer of 360, subtract 360 from the result and you get the answer...

[Melanie]

Another prod...

You have two values, if the sum of the two values (each in the range 0-180, and ignoring if they're positive or negative) are equal to or greater than 180, then simply change the reference point.

Alternatively, how about this simplest solution... forget North, South or whatever, your reference point 'floats' - it's one of your two values (each 0-359)... ValueB is X degrees away from ValueA. So simply halve the difference and always add to ValueA. Note, that your choice of which value is ValueA is important... but then I'd be giving it away... once again it's something about if the difference is greater than 180...

Well there's two potential solutions... need any more?

[Melanie]

Melanie taps her fingers impatiently whilst eyeing up her whip coiled on the wall, then scribbles...

	ValueA var Word
	ValueB var Word
	ValueC var Word
	Result var Word
	If ValueA < ValueB then Swap ValueA,ValueB
	ValueC=ValueA-ValueB
	If ValueC < 180 then
		ValueC=ValueC/2
		Result=ValueB+ValueC
		else
		ValueC=360-ValueC
		ValueC=ValueC/2
		Result=ValueA+ValueC
		endif
	If Result = > 360 then Result=Result-360

There is only one 'anomaly' which could be adjusted whichever way you choose (although there is no true definitive answer). If ValueA and ValueB are exactly 180 degrees apart, you have two potential answers both of which are correct... eg ValueA is Zero and ValueB is 180, you have two answers - one being 90 and the other being 270. My little ditty above always selects the higher - 270 in this case. But what the hell do you expect for 30 seconds work?

[Darrel Taylor]

Or, here's another possibility...

Direction  VAR WORD      ; Holds the Averaged wind direction
Sample     VAR WORD       
AvgCount   CON 3         ; Number of samples to average

; Get New Sample here 0-359

If (Direction > 270) AND (Sample < 90) then 
    Sample = Sample + 360 
else
    if (Direction < 90) AND (Sample > 270) then
        Direction = Direction + 360
    endif
endif

Direction = ((Direction * (AvgCount-1) + Sample) / AvgCount) // 360

Which treats anything in the range of 0-90 as if it's 360-450. But only if the other value is on the other side of the zero point.

Then after averaging, it gets the Modulus of the result, which brings it back into the 0-359 range.

[DavidK]

If it is any help. The boys from Berkeley have an routine for averaging the anemometer direction. It is somewhat specific to their product, but the explanation may be of help. Has links to the math routines.

http://www.emesystems.com/OL2wind.htm#Wind_vane

[Melanie]

ooooh... I don't like bashing other folks work, but that is very amateurish... "Hello Mr Customer. Here's a fine wind direction guage for your airfield - oh, by the way, (at best) it doesn't work 3% of the time..."

[RussMartin]

I had to deal with aspects of this when I was working with weather and water instrumentation back in 1976. At the time, I was with the Wyoming Water Resources Research Institute. We dealt with an enormous volume of meteorological and hydrological data.

A side note for/about the "Boys from Berkeley"--there's no reason for a "dead band" on a weather vane and hasn't been for several decades. (That isn't "bashing", is it, Melanie?)

For clarity, let's agree that a vane tells us direction and an anemometer tells us that air is moving, and how fast--that there is in fact wind!

A wind direction change of exactly 180 degrees in that short a period (3 seconds) is frankly unlikely unless the vane is spinning wildly and sampled at the wrong instant, or just turning lazily, moving almost at random. Most occurrences will be in very light, almost insignificant breezes (hence little or no wind speed), so direction doesn't really obtain.

I would "trap" such a condition by looking at the absolute (i.e., unsigned) value of the difference of every successive pair of readings. If it is 180, it's a null (i.e., useless and/or anomalous); discard the earliest value and use the second against the incoming third, and so forth--truly "rolling" the values through:

If |oldest - next| = 180, then oldest = next and next = newest, then start over; else, average the data and continue. (Melanie can probably write the code in about 15 seconds!)

What is overlooked in this discussion is the fact that wind "direction" means nothing if there is no air movement, i.e., no wind speed. Much more important than a possible but unlikely 180-degree anomaly is if the vane is simply sitting there in absolutely still air. It will continue to report a direction! This would be like your television weather reporter saying, "The wind is coming from the east at zero miles per hour." Huh?