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.
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
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.
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 17:25.
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?
Obviously I'm missing something, but hopefully some one here can help me out.
Question 1: Given that the characteristics of a three engine turbojet aeroplane are as follows: Trust = 50000 N per engine g= 10 m/sē Drag = 72569 N Minimum gross gradient (2nd segment) = 2.7% The maximum take-off mass under segment 2 conditions in the net take-off flight path conditions is:
a. 101596 kg b. 286781 kg c. 74064 kg d. 209064 kg
Question 2: The determination of the maximum mass on brake release, of a certified turbojet aeroplane with 5°, 15° and 25° flaps angles on take-off, leads to the following values, with wind:
Wind correction: Head wind: + 120 kg per kt OR tail wind: -360 kg per kt Given that the tail wind component is equal to 5 kt, the maximum mass on brake release and corresponding flap angle will be:
a. 67700/15° b. 69000/15° c. 72200/5° d. 69700/25°
Question 3: On a segment of the take-off flight path an obstacle requires a minimum gradient of climb of 2.6% in order to provide an adequate margin of safe clearance. At a mass of 110000 kg the gradient of climb is 2.8%. For the same power and assuming that the angle of climb varies inversely with, at what maximum mass will the aeroplane be able to to achieve the minimum gradient?
a. 121310 kg b. 106425 kg c. 118455 kg d. 102142 kg
It's too late for me to write in my calculations, and for the same reason, please excuse any typos.
Question 1. The first point to note in addressing this question is that calculation of the maximum take-off mass assumes that a single engine failure has occurred. This means that this three engine aircraft effectively has only two engines.
So total thrust available 2 x 50000N = 100000N.
For small angles of climb, the % climb = 100% x Sin angle of climb
This can be rearranged to give: Sin angle of climb = % climb /100
And Sin angle of climb = (Thrust drag) / Weight
Combing the two equations above gives:% climb / 100 = (Thrust Drag) / weight This can be rearranged to give: Max weight = 100 x (Thrust-Drag) / % climb
Inserting the data provided in the question gives:
Max take-off weight = 100 x (100000 72569) / 2.7 = 1015962.963N
This can be converted into Kg by dividing by g = 10 m/s2 to give 101596.2963 Kg or approximately 101596 Kg.
Question 2. To solve this type of problem it must first be noted that winds do not affect the climb limited take-off mass. The wind correction need therefore be applied only to the runway limit (field limited take-off mass).
The next stage of the solution is to calculate and apply the corrections. the question specifies a correction factor of -360 kg/kt tailwind and an actual tailwind of 5 kts. this gives a correction of -360 Kg/kt x 5kt = - 1800 Kg.
Adding this to the figures provided in the question gives:
Yes, but of course. Where I went wrong in question 1 was that I was counting with the thrust from all 3 engines. And in question 2, I was making the wind correction to the climb limit also. Thanks for clearing that up for me. However, when it comes to question 3, you have come to the same solution as I, but the book states that the correct answer should be d. 102142 kg.