Professional Pilot Training (includes ground studies)A forum for those on the steep path to that coveted professional licence. Whether studying for the written exams, training for the flight tests or building experience here's where you can hang out.
Use the following formulas for Altimeter calculations
TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))
where TA = true altitude, IA = indicated altitude, Subscale = altimeter subscale setting, ISA Dev = ISA temperature deviation (I assume you can calculate this). This will work for the vast majority of all altimetry questions, except PPL questions use 30ft (not 27).
This will give you the exact answers to questions 1, and 3. I think you mean the indicated altitude at an MSA (true) of 12000ft in Q1.
Question 2 is slightly flawed in that radio altimeters operate only up to 2500ft, and actually read height above the surface directly below rather than altitude. If you put this aside you can determine that the ISA dev is -ve, ie colder using the equation above.
The following formula
Height change per hPa(ft)= 96 x T/P
where T= temperature in Kelvin (C + 273), P = Pressure in hPa
In Q4, T= 252K, ISA in C is -21, as 5500m is approximately 18000ft, P = 500hPa
That gives Height change per hPa = 96x252/500 = 48.4ft
Q. You intend to overfly a mountain ridge at an altitude of 15000ft AMSL.The average air temperature (=ISA Deviation) is 15 degree lower than ISA. The Sealevel Pressure (=QNH)1003Hpa. Which altimeter indication is needed?
15 x 15 x 4 = 900 1013 - 1003 = 10 x 27 = 270
900 + 270 = 1170
15000 + 1170 = 16170
Read Angleofattacks post.. and use the formulae TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))
I am currently studying for my cdn atpl and I came across a MET question I cannot seem to grasp. The concept of port/starboard drift increasing/decreasing true altitude respectively. Now the in the spirit of trying to understand the subject rather than simply memorizing the port drift=increasing true alt and starboard drift=decreasing true altitude, I am trying to attempt to understand this concept. Now where I get mixed up is how do you know whats happening because you could get either port or starboard drift depending on where you sit around the low or the high right? Isn't it a matter of perspective? eg port drift: I could be to the east of a low or west of a high if i was on a west east trip right???
The theory is that you can determine whether you are flying from/to a higher or lower pressure or temperature based on the direction of crosswind. This is Buys Ballot's law. In the Northern Hemisphere, if you stand with your back to the wind, the low pressure system will be on your left. The reverse is true in the Southern Hemisphere. Therefore if you have a crosswind from the left (ie. starboard drift), you are flying towards a lower pressure. Then remember the phrase: "high to low, watch out below!", which will remind you that true altitude decreases as you fly towards a lower temperature or pressure.
Last edited by Yara-ma-yha-who; 8th Oct 2011 at 23:19.
Friends, i am stuck with one of a question.can some plz shed some light in solving this question.thanks for the help
Q.Air at T = +16° C and DP = +4° C is forced from sea level over a 10.000 ft mountain range and descends back to sea level on the other side. If the leeward condensation level is observed to be 8.000 ft, what will be the final temperature?
Hi everyone A mass and balance question I need help with!
At a given mass the CG position is at 15% MAC. If the leading edge of MAC is at a position 625.6 inches aft of the datum and the MAC is given as 134.5 inches determine the position of the CG in relation to the datum.
Withn the answer it states -
The MAC is 134.5" long. The CG is 15% of this distance back from the leading edge. 15% of 134.5" is 20.17".
The leading edge of the MAC is 625.6" aft of the datum, the CG is 625.6 + 20.17 = 645.77" aft.
I'm looking for recent ATPL questions regarding Pitot blockages. There seems to be some discrepancies between our schools studyguides and real life. As I am a teacher in the subject, it would be interesting to know which stance the current QB has.
Our studyguide maintains that a blockage of the ram air opening of a pitot tube will cause the ASI to freeze at current speed whereas in real life, such a blocking would cause the ASI to read 0 because of pressure escaping through the waterdrain.
The altimeter will freeze if a blockage occurs but that would be in the static system. The VSI would read zero. If the pitot gets blocked, the ASI behaves like an altimeter and will increase its readings as you climb. If the static gets blocked, the ASI error will reverse, i.e. it will under-read as you climb.
At least that's what the JAA questions expect - real life has nothing to do with it!
The water drain is usually operated with a spring (every 200 hours on some aircraft), so that implies some sort of seal. It would have to be sealed in normal circumstances (like the alternate static) otherwise the readings would never be correct.
One it did pretty much as the JAA exams expect - froze at the original value (I never went above about 2000ft that trip, so the altimeter thang wasn't really there).
The other did something I'd never have predicted - it started reading almost in direct proportion to the RPM gauge. During take-off of-course, that actually seems about right - after five minutes I realised that something odd was happening.
Fortunately both totally VFR trips, so all flown on attitude back to an uneventful landing.
The "blow & block" test I've learned now to do as part of my pre-flight whenever I'm doing the first flight of a new-build aeroplane ! It's also why I know that this cobblers about water drains is just that - if it starts going down, there's a system leak and it needs sorting before flight.
It's a problem with "teachers" who have no real knowledge or experience - but aviation suffers from that a lot (as do many schools of-course, so they're in good company).