![]() |
perf problems part 2
Hello all,
Having some more performance questions. Also from feedback so anyone sitting this subject soon will find the help and questions useful, thanks, For this question refer to perf manual SEP fig 2.2, With regard to the take off performance chart for the single engine aeroplane determine the take off dist to height of 50ft. OAT = 38C Pressure alt = 4000ft Aeroplane mass = 3400lbs T/W component = 5kts Flaps = approach setting Runway: dry grass correction factor = 1.2 a) approx 4200ft b) approx 5040ft c) approx 3960ft d) approx 3680ft Following a takeoff limited by the 50ft screen height a light twin climbs on a gradient of 5%. It will clear a 160m obstacle in relation to the runway (horizontally), situated at 5000m from the 50ft point with an obstacle clearance margin of: a) it will not clear the obstacle b) 105m c) 90m d) 75m With a TAS of 194kts and a vertical speed of 1000ft/min the climb gradient is about a) 3degrees b) 3% c) 5 degrees d) 8% On a dry runway the accelerated stop distance is increased: a) by uphill slope b) by headwind c) by low outside air temperature d) by a lower take off mass because the aeroplanes faster to V1 During certification flight testing on a four engined turbojet a/c the actual TOD measured are: 3050m with failure of critical engine 2555m with all engines operating and all other things being equal The take off distance adopted for certification file is: a) 2938m b) 3050m c) 3513m d) 2555m Why isnt it A as in 2555m x 1.15? |
Following a takeoff limited by the 50ft screen height a light twin climbs on a gradient of 5%. It will clear a 160m obstacle in relation to the runway (horizontally), situated at 5000m from the 50ft point with an obstacle clearance margin of:
a) it will not clear the obstacle b) 105m c) 90m d) 75m 5% of 5000m = 250m 50 ft = 15.6 m 250m + 15.6m = 265. 6m 265.6m - 160m = 105.6m So B is the right answer. With a TAS of 194kts and a vertical speed of 1000ft/min the climb gradient is about a) 3degrees b) 3% c) 5 degrees d) 8% 1000 ft * 60 = 60000 ft 60000 ft / 6080 = 9.86 NM X * 194/60 = 9.86 X * 3.23 = 9.86 X = 9.86/3.23 X = 3 So A is the right answer. It is only a 1 in 60 rule. During certification flight testing on a four engined turbojet a/c the actual TOD measured are: 3050m with failure of critical engine 2555m with all engines operating and all other things being equal The take off distance adopted for certification file is: a) 2938m b) 3050m c) 3513m d) 2555m 2555 * 1.15 = 2938.25 3050 the right answer is the bigger one for safety reasons. So B About the accelerated stop distance A is the correct one. Sorry that Im not ale to help you with the first question. I dont have the CAP. :ok: |
jnikster,
Your chances of passing the exam will be far greater if you learn how to answer questions, rather than simply learning the answers to them. Q1. From Fig 2.2 I get a graphical distance of about 3300. Multplying this by 1.2 gives 3960. Q2. The % climb gradient is a measure of the height gained divided by the horizontal distance flown. So a 5% gradient means that your height gain is 5% of the horizontal distance flown. In this question the obstacle is 5000 m from RZ, so the height gain is 5% x 5000 m = 250 m. But at RZ you are already at screen height which is 50 ft or 15 m. So your total height at the obstacle is 250 + 15 = 265 m. The obstacle clearance is the aircraft height minus the obstacle weight which is 265 - 160 = 105. Q3. For this type of question go to page 28 of the CAP 698. This gives the equation: Still air % gradient = (ROC/TAS) x (6000/6080) Inserting the given figures gives: Still air % = (1000/194) x (6000/6080) = 5.0868% Converting this into a decimal gives 0.050868 But gradient = height gain / distance so the figure above is also the TAN of the climb angle. Using your calculator will give an angle of 2.912 degrees. So the closest option is 3 degrees. Q4. Go to figure 3.4 which is the ASDA chart for the MEP1. Examination of this chart shows that Option (b) a headwind, option (c) low outside air temperature, and option (d) lower take off mass, all decrease ASDA. Now go to the bottom of page 19 where the final sentence and the note state that ASDA must increase by 5% for each 1% upslope. Q5. Lots of students think that you answer this type of question by simply multiplying the all engines distance by 1.15. In many cases this is true, but it is not always so. Go to page 53 of the CAP 698 where the rules are stated. Note that para a is for TORR, para b is for ASD and para c is for TODR. So for this question we need para c. The sub-paragraphs define the factors to be used for single engine dry and single engine wet. In this case the single engine is greater than 1.15 x the all engines. Although the exam usually includes very few questions which require you to actually work through the graphs in the CAP, many of the questions can be answered by knowing your way through the CAP. Time spent familiarising yourself with the CAP will greatly improve your exam score. |
Your chances of passing the exam will be far greater if you learn how to answer questions, rather than simply learning the answers to them. All. Many thanks for the replies. Jinkster |
Jinkster,
Go easy on Keith, he's only trying to help you out! Try not to get so upset over such a minor comment. This industry requires that you develop a thick skin and a positive attitude to constructive criticism because you'll get lots of it and you can't afford to get upset. Just call it character building! ;) Hopefully Keith will help you next time if you ask nicely!!:p Anyhow, keep up the good work and I hope you get there in the end! |
Apologies Keith and thanks for the great help, I dont want to be "tarred" with the same brush of learning which answers go with what question.
I want to be a proffesional pilot and the only way I am going to be able to do this is by hard work and study! Hopefully no harm done - I do understand your explanations clearly. Thanks, Jinkster |
No worries Jinkster, the important thing is that you understood the explanations.
|
Thanks Keith, great explanations :ok:
|
| All times are GMT. The time now is 20:09. |
Copyright © 2026 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.