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On Fri, August 5, 2005 2:55 pm, Nick Kappatos wrote:
> If I remember right, the way we resolved this for real-world surveying problems
> and the like was to convert the degrees 180 < x < 360 into negative values
> relative to 0/north. 350 is -10, 270 is -90, etc.
>
> If x1 > 180
> then x2 = 360 - x1
>
> This makes your numbers relative to 0. You can then convert back if the average
> is a negative number:
>
> If avg1 < 0
> then avg2 = 360 + avg1
Don't you get the same problem at 180, now?
170 + 190 = 170 +(-170) = 0
I think there are several readings that can give you more than one result. For
example, if you take two readings of 90 and 270, what should the answer be? 0?
180?
never mind. I like Derek's answer.
Steve
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Message has 2 Replies:
 | | Re: OT: Math Help
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| (...) One of the things I did pickup from "my" answer is you need to have a sample frequency small enough that you are not going to get wind direction changes of 180 degrees or more. And I say or more, because it's important to the average which (...) (19 years ago, 5-Aug-05, to lugnet.org.ca.rtltoronto)
| | | Re: OT: Math Help
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| (...) Oops, there was a mistake in my first equation above that I've corrected here, in order to give a negative number. (...) That's why I have " > 180", not "> or equal to 180" ;) Since the question was framed in relation to 0, that's what my (...) (19 years ago, 5-Aug-05, to lugnet.org.ca.rtltoronto)
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Message is in Reply To:
| | Re: OT: Math Help
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| (...) If I remember right, the way we resolved this for real-world surveying problems and the like was to convert the degrees 180 < x < 360 into negative values relative to 0/north. 350 is -10, 270 is -90, etc. If x1 > 180 then x2 = 360 - x1 This (...) (19 years ago, 5-Aug-05, to lugnet.org.ca.rtltoronto)
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