I mean if your a perfectionist like moi, I think this formula does the job sin (latitude of a lows center)/sin60. Example sin40/sin60 or 9.8/9.9, equals 0.989, then multiply that by 24 and you get 23.7 mb's rounded to the nearest tenth. Obvioulsy the pressure drop difference is minute but it's there.
Edit: I'm not using the same sin as on a caculator because that would give you a negative anwser, as it is meant for geometery and those problems have both a negative and positive anwser. So respectively at 60 degrees sin= .165 and is .245 at 40 degree's, increasing by .04 every 10 degrees of latitude.
-------------------- "I became insane with horrible periods of sanity"
Edgar Allan Poe
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Re: Pressure drop accounting for latitude
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