Thanks for the response... What I was getting at is I have a calculated density altitude of 5206 ft. Under that condition what is the correction factor to sea level? 13.05 @ 107.57 mph
120 feet for each degree from standard temp for the pressure altitude.
Start with sea l evel pressure diff between it and 1013mb . 30 feet per mb. So if MSL pressure is 1020 them 1013 is above that so 7x 30=210’, subtract that from elevation and you have pressure height.
Then work out temp for that pressure altitude starting with 15 degs celcius. You have to work out what it is a cording tomstd temp so 15. - (altiude in thousands x2degs) = std temp. Temp changes 2 deg C every 1000’. If the actual temp is hotter than std you add the height other wise if colder you subtract it.
For example;
Elv 3000 avtual temp is 25 deg C
Msl press 1020mb
1013-1020= -7 X30’ = -210
3000-210=2790’
15-(2.79x2)=9.42 deg C std temp
Actual temp=25 deg so its hotter by 15.58 deg
Pressure height 2790 + (15.58x120’)=4669.56’ density height
You will have to work it out going backwards.
I bet your confused now lol
