flight planning
If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, endurance 3 hours and actual time of departure is 1105 UTC. The distance from departure to Point of Equal Time (PET)
9000ft/ISA -10⁰C (OAT= -13⁰C) TAS =212 kt D = 350 NM … O = 176 kt … H =246kt Distance to PET = D x H / (O + H)…………………. = 350 x 246 / (176 + 246) = 204 NM
Answer 203nm
I am unsure about the temprature corrrection as i get is -7 { 15 - (9 x 2) }= 3 and -10 =-7
can someone just explain this -10 or +10 isa and correction made accordingly please
thank you in advance.
Last edited by clkorm3; 6th February 2014 at 04:59.