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ISA Temp deviation display

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Hi all %( 15 (A:Ambient temperature, celsius) - )%!3d!°C% Gets me ISA temp. deviation. I can't get the + and - sign to show correctly. +13 I get 2 should get -2 +20 I get -5 should get +5 and If you use %!+3d! it just reverses the + and - sign. How can it be done.

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Paul,When (A:Ambient temperature, celsius) is 13, you get 2 and you think you should get -2? Well, then I think RPN wins...:-)(15 13 -) equals 2 positive(15 20 -) equals 5 negative (-5)Unless I missed something.Tom

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Jan,Tom I got a way to do it using L:vars:- %((L:ISAtemp,number) 15 >)%{if}%-%((L:ISAtemp,number))%!3d!°C%{else}%+%((L:ISAtemp,number))%!3d!°C%{end} But I still get a minus before the number showing the positive Temp. Is there a way to remove this.

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Hi,A Question:If Temperature = 20 CISA = 15 CWhat must ISA DEV show in your example,+5 or -5 ?If +5 you could try:%(15 (A:Ambient air temperature,celsius) -)%!+5d!

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I should think that "deviation" is most properly expressed as an unsigned value, since it represents the difference between two temperatures, i.e., an "absolute value." ;)

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Hi Jan I tried that and other things and it seems once you go under the 15 you get the minus sign. To explain more +15 degrees C is the temp for International Standard Atmosphere at sea level. They use it for working out aircraft performance and other stuff. So if the real temp is +10 deg C the ISA temp is -5 deg C. You can see the problem already you have a + temp to change to a -. If the real temp was +30 deg C the ISA is +15 so thats the problem.

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>>...So if the real temp is>+10 deg C the ISA temp is -5 deg C. You can see the problem>already you have a + temp to change to a -. If the real temp>was +30 deg C the ISA is +15 so thats the problem.>>> Then why don't you make a simple rest??%((A:Ambient Temperature,celsius)) 15 -)%3d%For example, if Ambient temp is 10, you will get a -5 ISA value. Now, if it is 20, you will get ISA 5 but no plus "+" sign, though it would be easy to add it within the string.Tom

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Hi Tom Well I missed that real good! %((A:Ambient temperature,celsius) 16 >= )%{if}%+%{end}%((A:Ambient temperature,celsius) 15 - )%!3d!%°C%{end} Works OK Thanks to you both. Paul

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