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.
I wonder if there are any brain boxes out there who can shed some light on this question? I have (as far as I am aware) followed the correct proceedure of how to calculate a Traffic Load.
ChriSat - thanks for your reply! I have no idea how I arrived at that either. I blame the calculator! I now see that I was working it out correctly, but I must have put in the wrong values or copied down the answer incorrectly.
pilotmike - thank you for your reply also. I see what you're saying, I meant to say 39 200kgs
They are from the Peruvian CPL question bank. I asked Max for help with the question because I am here doing a license conversion I don't have any books with me.
I think this is what it feels like to have mental block! I have just got 99 out of 100 altimetry questions correct, but I just can't work this one out for some reason. Maybe I'm staring too hard trying to find the right answer that I'm missing the obvious solution, but it's something I'd like to get clear in my head either way. You may ask why I'm so bothered when it's only one question out of 100 - well, because it's 'sods law' that it'll come up in the exam! This same question has been answered further up the thread, but it doesn't help solve the confusion in my case. Here goes:
"You plan a flight over a mountain range at a true altitude of 15000ft/AMSL. The air is on average 15C colder than than ISA, the pressure at sea level is 1003 hPa. What approximate altitude should the altimeter read (pressure setting 1013 hPa)?"
Now I'm quite good at remembering and re-arranging mathematical formulas as opposed to rules (just a personal preference), and so I use:
True Altitude = Altitude on QNH + [(ISA Deviation x 4) x (Pressure Altitude/1000)] and I've always got the answer to similar style questions spot on!
With the question above though, I can't get my head around which one of the missing components I'm actually looking for? We're given the True Altitude as 15,000ft, and I know that the ISA Deviation x 4 = -60, and so that means I either need to find the Altitude on the QNH or the pressure altitude, but I can't work out which.
Is anyone able to help explain using my formula? The answer by the way is supposedly 16,230', although I'm aware with these that the answers are quite often 50' out and you just have to pick the closest one.
True Altitude = Altitude on QNH + [(ISA Deviation x 4) x (Pressure Altitude/1000)]
Gives
Alt on QNH = True Altitude - [(ISA Dev x 4) x (Pressure Alt/1000)]
Inserting the data provided gives
Alt on QNH = 15000 – ((-15 x 4 x 15) = 15900
But we are going to use 1013 which is 10 HPA higher than QNH.
Using 27 ft / Hpa this means that the 1013 level is 270 feet below msl
So adding 270 to 15900 gives an indicated altitude of 16170 feet.
At that altitude the estimated lapse rate of of 27 feet per Hpa is a bit too low. It looks as if the examiner has used 33 ft per HPa (30 HPa per 1000 ft) which is probably more realistic.
That's brilliant Keith, I think I've pretty much got it now!
Am I right in thinking that you're effectively working out what the Altitude on the QNH would be (1003 set) and so therefore Pressure Altitude is the same as True Altitude in the initial calculation? And then you're taking in to account the variation between sea level QNH and 1013?
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
Strictly speaking the term "Pressure Altitude" means the altimeter reading when 1013 is set on the subscale. So if we were to use Pressure Altitude instead of indicated altitude in the above equation we would be applying the temperature error from the 1013 level upwards.
My use of TA in ,y previous post was an error. Using the above equation we can solve the problem as follows.
TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))
ISA Deviation = -15 True Altitude = 15000 ft QNH = 1003
TA – (27 x (QNH-Subscale setting) = IA + ( (4 x IA x ISA Dev)/1000 )
Inserting the data provided in the question gives
15000 – ( 27 x (1003 – 1013) = IA + (4 x IA x (–15) ) / 1000 )
15000 + 270 = IA – 0.06 IA
15270 = 0.94 IA
15270 / 0.94 = IA
16244 = IA
So Indicated Altitude = 16244 feet
This is not quite the correct answer of 16230, but it is pretty close to it.
Last edited by keith williams; 19th Feb 2012 at 16:25.
If an alternator is run at below normal frequency, then: A) Electric motors will stop. B) Inductive devices will overheat. C) Lights will become dim. D) Lights will become brighter.
The correct(?) answer is B. I agree that if the frequency is below normal, that would mean that the inductive reactance is lower and thus the current will be higher, but conversely, a lower frequency would also mean that capacative reactance would be higher, thus giving a lower current, which should balance the lower inductive reactance, as there is no mention of one reactance being more dominant than the other, or any one kind of reactance at all for that matter, other than what might possibly be deducted when the answer alternatives are examined. Am I to assume that the other alternatives, being placed in a purely capacative reactant circuit couldn't occur in the case of decreased frequency?
This is a case (just one of many) where the examiners have made some simplfiying assumptons, but have not stated what these assumptions were.
All they really want you to do is to note that reducing frequency reduces inductive reactance so for a given applied voltage the current through inductors will increase.
As you have said the situation is much more complicated in real circuits where different types of component are connected in various ways.
But if you look at the other options, the voltage regulator should prevent these effects unless the RPM is very much lower than standard.
