very good, thanks again, note 10
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Hi guys, I also take advantage, as on other occasions, to use your advice and formulas, thanks.
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hi mikeoday, it really just got better now! Is the math from 0 to 40 correct? If you don’t mind and have time without commitment, could you reveal how this math is done? because just looking can not understand how it does! would you like me to somehow repay this help?
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Well I think it is right. I walk you through it and that should confirm ( or otherwise 
)
$#WM#==F?(170-(clamp(#WCT#,32,104)*2.5)):(90-(clamp(#WCT#,0,40)*4.5))$
If the watch/phone weather units are imperial ( ie. #WM#==F ) then the expression is
(170-(clamp(#WCT#,32,104)*2.5))
check:
#WCT# == 32 ( equivalent to 0C )
clamp(32,32,104) will return 32
32*2.5 = 80
170-80 = 90
So, when temp = 32, rotation = 90 degrees
#WCT# == 104 ( equivalent to 40C )
clamp(104,32,104) will return 104
104*2.5 = 260
170-260 = -90
So, when temp = 104, rotation = -90 degrees
If the watch/phone weather units are metric( ie. #WM# not = F ) then the expression is
(90-(clamp(#WCT#,0,40)*4.5))
check:
#WCT# == 0 ( equivalent to 32F )
clamp(0,0,40) will return 0
0*4.5 = 0
90-0 = 90
So, when temp = 0, rotation = 90 degrees
#WCT# == 40 ( equivalent to 104F )
clamp(40,0,40) will return 40
40*4.5 = 180
90-180 = -90
So, when temp = 40, rotation = -90 degrees
I hope that helps.
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very good! now I understand! super explanation, all right here, as soon as I finish revealing how it was and put your credits to the face, I’m very grateful for your help! 
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You are most welcome.
Hi! Do you know how to use these tags for Cº and Fº at the same time? I am using it in degrees Cº I would like it to work in both modes
$ # WCT # <= 10? 100: 0 $
$ # WCT # <= 19 && # WCT #> = 11? 100: 0 $
$ # WCT # <= 25 && # WCT #> = 20? 100: 0 $
$ # WCT # <= 30 && # WCT #> = 26? 100: 0 $
$ # WCT #> = 31? 100: 0 $
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No the results will be different for Celsius and Fahrenheit
You can use the following tag to test for which units are active on the user’s watch:
thanks
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I know this is an old thread by your maths, @mikeoday, helped me greatly so I would like to extend my thanks as well!
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Thanks, are most welcome and I’m glad it helped!