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Keith234
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(Storm Chaser)
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Fri Oct 01 2004 10:48 PM
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Pressure drop accounting for latitude
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I have heard that you have to correct the pressure drop because of the latitude. How would one represent that, the proportions are not consitent, is there some type of correction factor. Would the correction factor be similar to that of cosine, sin etc? I am a little fuzzy on it, mostly because I not sure why you would need to correct the pressure, isn't pressure and latitude not related? Thanks
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Clark
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(Meteorologist)
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Fri Oct 01 2004 11:29 PM
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Re: Pressure drop accounting for latitude
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Pressure really isn't a function of latitude, not that I have heard of. On such a small scale, I don't even think a latitude correction is needed to track the precise center of a hurricane from a radiosonde (Coriolis effect relatively negligible in the eye), but they probably do that anyway. In any case, however, pressure is above all a function of altitude, with other variables (temperature, etc) included.
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Keith234
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(Storm Chaser)
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Fri Oct 01 2004 11:33 PM
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Re: Pressure drop accounting for latitude
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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.
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Dawn
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(Weather Watcher)
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Sat Oct 02 2004 12:44 AM
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Re: Pressure drop accounting for latitude
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How good it feels to see 2 registered users and 1 non registered users. Hope it says this boring for a long while. You guys are great, got my power back two days ago but could not get my computer or dsl to work till yesterday. Sisters are still with out power, they have our generator now, hope they do not burn it up, given lessons, but they can be air heads at time.
My question is - I am on vacation from work till the 10th, do we take down the window coverings? Do we get our generator back from sisters.
The question " What is you Forecast", do I stay prepared or relax. Looked at lasted maps and not relaxed, am I just overdone when looking at the weather? I wish I had the knowledge you all have, learning, but will take me time to understand.
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Keith234
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(Storm Chaser)
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Sat Oct 02 2004 12:54 AM
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Re: Pressure drop accounting for latitude
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I hate to say it, but this season will have a least one more hurricane making landfall on the Gulf coast. Just be prepared for one more, don't get totally relaxed just yet. Thanks for the compliments, you may say that what I just said in the previous post was complicated, I'm sure you could understand it. So just don't count your chickens before they hatch, as it may be too soon to say the hurricane season is over.
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Dawn
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(Weather Watcher)
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Sat Oct 02 2004 01:20 AM
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Re: Pressure drop accounting for latitude
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Thank You for your reply. I kinda thought your answer is what I would hear.
Please folks do not return your supplies until after the season. Stuff that is being returned today is being sold today with no shippments to replace anytime in the near future.
Please be prepared and do not think you will find that little bottle of propane at Home Depot,, have been zero shipped on all orders since Charlie, along with sheet plastic-tarps and drylock in paint. Hardware - Gabber screws and fasteners-Tapcons- nails-roofing nails-coil nails and any air tool roofing or concert nail,
Weatherstripping, power tools,just plain anything. You may e-mail me pm and I will tell if we are having trouble getting.
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Clark
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(Meteorologist)
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Sat Oct 02 2004 02:59 AM
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Re: Pressure drop accounting for latitude
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I'd like to know where you heard about this and come up with that calculation, as I cannot find anything about it whatsoever. Further, using your calcuations for a second here (more on this in a second) when you talk about something like 950mb, you are talking 10mb difference -- that's significant, and I'd think it'd be more widespread or commonly known if such were the case.
Edit: where did you get your values for sin(40) and sin(60)? You can use a calculator to compute those, just ensure that you are in degrees and not radians. There's nothing weird to it, and calculators weren't built with those functions in them just for geometry. In any case, sin(40°) = .643 and sin(60°) = .866, making the ratio between them .742, or resulting in 1/4 reduction in the pressure. Something smells very fishy here.
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Keith234
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(Storm Chaser)
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Sat Oct 02 2004 12:44 PM
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Re: Pressure drop accounting for latitude
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I agree with you, I got this from a book, Weather Predicting Simplified by Micheal William Carr. On page 23, I'll parapharse in addition a low- pressure system is rapidly intensifying if its central sea-level pressure drops at least 24 mb in 24 hours at 60 degrees latitude. At other latitudes the sea-level pressure drop is multiplied by a correction factor of sin (latitude of a low's center)/ sin60. For example, at 40 degrees north the factor would be sin40/ sin60, or 9.8/9.9, which equals 0.989. Then since there are 24 hours multiply that by a pressure drop of .989 an hour to get 23.7mb(pg. 23, Weather Predicting Simplified). He is saying that the pressure drop is a factor of the latitude in degrees, maybe he thinks, as you travel further away from the equator pressure drops faster but why?
Edit: I found this website on it http://weather.ou.edu/~mbergman/bombcyclogenesis/index2.html
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Keith234
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(Storm Chaser)
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Sat Oct 02 2004 08:21 PM
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Re: Pressure drop accounting for latitude
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I did some more research on it, but found zilch accept the above link. But in my effort I also found some useful info, such as the angular velocity, tangetial velocity, the conservation of angular monmetum and the coriolis effect, etc. I never knew the earth moved at 7.29 x 10 to the negative 5th power and one radian equaled 57.2958. Very interesting. I have come up with another hypothesis, what if the forces and effects metioned above indirectly effect the pressure drop at the surface, making it easier to fall 24 mb in 24 hours at 60 then opposed to the mid-latitudes. Some forces are much greater at 60 then opposed to 45 degrees. Just a suggestion.
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Clark
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(Meteorologist)
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Sat Oct 02 2004 10:57 PM
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Re: Pressure drop accounting for latitude
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Storms can - and often do - intensify more deeply in the higher latitudes. That doesn't mean there is a correction factor, just a different set of standards for defining, in that case, bomb cyclogenesis.
In any case however, sin and cos cannot equal anything greater than 1 (or less than -1), whether in degrees or radians, unless multiplied by something before it (i.e. 2 cos x).
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Keith234
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(Storm Chaser)
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Sat Oct 02 2004 11:19 PM
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Re: Pressure drop accounting for latitude
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I hate things that you just can't figure out, I'm emailing the guy. I"ll post what he say's on the website, that is if he does email me back.
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