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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/s² 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.
OAA Flight Performance & Planning. It's likely to be a bogus answer if you don't agree with it, it wouldn't be the first time...
On another note, I have stumbled upon a couple of questions involving the dumping of fuel to ensure level-off altitude isn't below obstacle clearance altitude during a drift down procedure. Is this common practice, as the only thing I have read about fuel dumping so far is regarding the procedure of doing so prior to landing to ensure not exceeding MSLM?
The OAA answer to question 3 is incorrect. Hopefully you should be able to see that is the case by applying common sense. If the aircraft gets heavier it won't climb so well. So if we are happy to be limited to a lower climb gradient we can be heavier.
I'm not an airline pilot, so I cannot comment on what is common practice. But if I knew that keeping any unnecessary fuel on board would cause me to hit the mountain tops, I would certanly dump any fuel that I didn't need.
I agree with you regarding question 3. It's just that as with the case of questions 1 & 2, it's easy to miss corrections, that's why it's seems like a better bet to run it by here, rather than just assume that I'm right and they're wrong.
The fuel dumping questions were:
If the level-off altitude is below the obstacle clearance altitude during a drift down procedure?
a. Fuel jettisoning should be started at the beginning of drift down. b. The recommended drift down speed should be disregarded and it should be flown at the stall speed plus 10 kt. c. Fuel jettisoning should be started when the obstacle clearance altitude clearance altitude is reached. d. The drift down should be flown with flaps in the approach configuration.
During a drift down following engine failure, what would be the correct procedure to follow?
a. Begin fuel jettison immediately, commensurate with having required reserves at destination. b. Do not commence fuel jettison until en-route obstacles have been cleared. c. Descend in the approach configuration. d. Disregard the flight manual and descend at Vs + 10 kts to the destination.
Correct answer for is a. for both questions. Looking at them in hindsight I realize that those are the only reasonable answers, it's just that when I came across them there had been no mention of fuel dumping in the previous chapters that I had read, other than for the purpose of ensuring MSLM, and I suppose that the intention after engine failure is to land as soon as possible, thus fuel dumping should be considered pertinent. However there is no mention in the question of what phase of flight is being referred to, and dumping of fuel doesn't seem to be advisory unless absolutely necessary. What had been brought to attention in the previous chapter however, was the use of drift down profile graphs, where one was to determine if the desired altitude, with regard to obstacle clearance, could be met with the current mass, and if not, (to my understanding) a second graph should be used to determine whether or not vertical clearance could be achieved using horisontal distance instead. Thus the reason for my inquery.
You are however absolutely right, Keith, it definately seems like a better idea, if need be, to dump some fuel rather than run in to a mountain top.
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a. Confirmation bias. b. Action slip. c. Environmental capture. d. Frequency bias.
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Question: Which combination of answers of the following parameters give an increase or decrease of the take off ground run: 1 decreasing take off mass 2 increasing take off mass 3 increasing density 4 decreasing density 5 increasing flap setting 6 decreasing flap setting 7 increasing pressure altitude 8 decreasing pressure altitude
I think it would have to be B because it's the only one which lists conditions which all affect ground run in the same direction, as in, they all cause it to decrease. All the rest contain some conditions which cause and increase and some which cause a decrease.