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Konstantinos

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Posts posted by Konstantinos

  1. In Wikipedia, I saw that the most intense (lowest central pressure) record happened in October 12, 1979, during Super Typhoon Tip in northwest Pacific Ocean and it was 870 hPa. And I remembered one day, when I was in 1300 m - 1400 m altitude with about the same atmospheric pressure. I felt very cold compared with sea level. So... Now, I am wondering how cold you feel inside an eye of a cyclone? And how is the atmospheric pressure around the eye?

  2. This is the topic for invest area HS27 and the possible tropical storm or hurricane in the next days. The chance is still 30 % (according national hurricane center)?
     

    THIS SYSTEM HAS A MEDIUM CHANCE...30 PERCENT...OF
    BECOMING A TROPICAL CYCLONE DURING THE NEXT 48 HOURS...AND A MEDIUM
    CHANCE...40 PERCENT...OF BECOMING A TROPICAL CYCLONE DURING THE
    NEXT 5 DAYS.

     

    http://www.nhc.noaa.gov/gtwo_atl.shtml

    From this position, is there chance for this system to come in western Europe?

  3. Hello again flyer. I found my position's declination from this site: http://www.ngdc.noaa.gov/geomag-web/#declination

    I put my Latitude and my Longitude (which I found from Google Earth), and it gave me my declination today: 3° 59' 20" E changing by 5.9' E per year.

    So, every time, when I will go for hiking - climbing, I will find the declination of my destination (probably a summit of a mountain) and I will change the reference in my device.

    Why do you think my figures are drifting a little? For example ESE is in 112.5°, so ESE is 112.5° +- 11.25°. So, ESE is in 101.25° - 123.75°.

    11.25° is derived from 22.5°/ 2.

    22.5° is the angle between two point - names, for example between E (90°) and ESE (112.5°).

    My calculations are not correct?

    Anyway, thank you very much about everything, dear flyer.

  4. Hello again flyer. I studied your link and wikipedia, and I think I understood. So, I created a script on matlab, using wikipedia's formula: http://en.wikipedia.org/wiki/Density_altitude#Calculation

    Code:

    fprintf ( 1,'Hello, I will calculate your density altitude.\n' );ts=input('Please, give me the sea level temperature (degrees C):');fprintf ( 1, 'Thank you.\n' );ps=input('Please, give me the sea level atmospheric pressure (mb):');fprintf ( 1, 'Thank you.\n' );tm=input('Please, give me the temperature of your position (degrees C):');fprintf ( 1, 'Thank you.\n' )pm=input('Please, give me the atmospheric pressure of your position (mb):');fprintf ( 1, 'Thank you.\n' );da=145442.156*(1-((pm/ps)/((tm+273.15)/(ts+273.15)))^0.234969);dam= da/3.2808399;fprintf('Your density altitude is %d m or %d feet\n', dam, da);

    And a simple running:

    >> Density_AltitudeHello, I will calculate your density altitude.Please, give me the sea level temperature (degrees C):17Thank you.Please, give me the sea level atmospheric pressure (mb):1017.9Thank you.Please, give me the temperature of your position (degrees C):23.5Thank you.Please, give me the atmospheric pressure of your position (mb):999.9Thank you.Your density altitude is 4.146678e+002 m or 1.360459e+003 feet>> 

    Is it good? What is your opinion?

    I am trying to understand the difference between "altitude" and "density altitude". I can see only one difference. The altitude does not depend on my position's current air temperature, but the density altitude does. If my position's current air temperature is increased, then the density altitude is also increased, but the altitude remains fixed. If my position's current air temperature is decreased, then the density altitude is also decreased, but the altitude remains fixed.

    I am wondering if I am right.

    Also my multimeter says: The DENSITY ALTITUDE screen is calculated from the absolute values of station pressure, relative
    humidity and temperature. So, what formula it uses?

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