Konstantinos

Today, in Greece, maybe we will have the first fall raining. Thunderstorm, low temperature and winds.

Summer 2007 in Greece. I still remember it. This summer I understood the difference of feeling hot, hot waves ... and ... fire. I remember a fire night without sun, but with 40 -42 degrees.Don't think that we felt hot. You did not have time. All day you were surprised about the fire around you.Now, for me, the 40 - 42 degrees is cold.

What is the highest altitude - elevation that the eye of a cyclone (hurricane - typhoon) has reached? 100 m ASL? 500 m ASL? 1000 m? These wikipedia records did not help me: http://en.wikipedia.org/wiki/List_of_Atlantic_hurricane_records http://en.wikipedia.org/wiki/List_of_tropical_cyclone_records

We have Fujiwhara effect? http://en.wikipedia.org/wiki/Fujiwhara_effect

Ok thank you and sorry. This thread must be closed or deleted.

So, Humberto died and now it is going to resurrect? Is it possible? And if it will resurrect, it would take a new name or the old one?

70% during the next 48 hours and 90% during the next 5 days http://www.nhc.noaa.gov/gtwo_atl.shtml

I believe Ingrid and Manuel have future (maybe together) in Pacific Ocean. (Please, change the title of the thread storm --> hurricane).

So, air temperature and heat index are not influenced by atmospheric pressure?

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?

Can a moderator change the title of the thread because I cannot?

What will the name of this storm be?

80% during the next 48 hours and 90% during the next 5 days.

60% during the next 48 hours and 80% during the next 5 days.

50% during the next 48 hours and 70% during the next 5 days.

40% during the next 48 hours and 70% during the next 5 days.

20% during the next 48 hours and 70% during the next 5 days.

THIS SYSTEM HAS A LOW CHANCE...10PERCENT...OF BECOMING A TROPICAL CYCLONE DURING THE NEXT 48HOURS...AND A HIGH CHANCE...60 PERCENT...OF BECOMING A TROPICALCYCLONE DURING THE NEXT 5 DAYS.http://www.nhc.noaa.gov/gtwo_atl.shtml

What about Invest 92? It is near Gabrielle or it was?

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?

Ok, now everything is clear. Thank you again flyer. I answered on your pm. See it.

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.

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?

If I am in an altitude about 2000 m, after 5-6 hours hiking, then my multimeter will show me my position's current atmospheric pressure. So this time how can I know the current sea level temperature and atmospheric pressure? You meant that I can see it on internet (probably with a mobile internet 3G - 4G) or I can take with me a chart prediction of it, before I begin hiking? flyer, I created a script on matlab (name altitude) using the barometric formula from wikipedia: http://en.wikipedia.org/wiki/Atmospheric_pressure#Altitude_atmospheric_pressure_variation which calculates the altitude. Please tell me your opinion. Code: fprintf ( 1,'Hello, I will calculate your altitude.n' );cslt=input('Please, give me the sea level temperature (degrees C):');fprintf ( 1, 'Thank you.n' );cslap=input('Please, give me the sea level atmospheric pressure (mb):');fprintf ( 1, 'Thank you.n' );cpsap=input('Please, give me the atmospheric pressure of your position (mb):');fprintf ( 1, 'Thank you.n' );alt=8.31447*(cslt+273.15)*log(cslap/cpsap)/(9.80665*0.0289644);altf= alt*3.2808399;fprintf('Your altitude is %d m or %d feetn', alt, altf); And a running of my position now: >> altitudeHello, I will calculate your 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 atmospheric pressure of your position (mb):999.9Thank you.Your altitude is 1.515332e+002 m or 4.971562e+002 feet>>

Thank you johnholmes. Hello again. A simple question: In the Barometric Formula, http://en.wikipedia.org/wiki/Atmospheric_pressure#Altitude_atmospheric_pressure_variation if I replace the sea level standard atmospheric pressure = 101325 Pa, with the current sea level atmospheric pressure, near to my position, and the sea level standard temperature = 288.15 K with the current sea level temperature, then the accuracy will be better or badder?