Forum Replies Created
-
Posts
-
Hi Terence,
A crosswind at 90 degrees to the track such as in this case always has a negative effect on the ground speed due to the division of the TAS vector into 2 smaller vectors, one of which is along the track (GS) and the other which is across the track to correct for the wind (Drift rate).
Using the whizzwheel or cx2 you should get a GS of 120 knots for both home and out.
This should give you a distance to the PET of 135 nm (exactly half way) and a time of 68 minutes.
The database does seem to have an error where it has marked 150 nm and 65 minutes as correct which definitely cannot be the case. I will make the correction and try to have it included in next Friday’s update.
Hi Marle,
Unlike the calculations for take-off distance and ground roll, the instructions in CAP 698 5.1.1 “Field Length Requirements” for landing only looks at the regulatory requirements for the landing distance. In addition, SA CARS 135.08.7 (1)(a)(ii) states:
“Subject to subregulation (3), no person shall dispatch or conduct a take-off in an aeroplane unless the mass of the aeroplane on landing at the destination aerodrome will allow a full-stop landing in the case of a large propeller-driven aeroplane, within 70 percent of the LDA.”
As you can see the regulations here are only concerned with LDA and not the ground roll. I have seen questions asking for a landing ground roll but I haven’t ever seen a question asking for a landing ground roll required as the regulations have no specifications in this regard.
Hi Marle,
This exact question is explained in the CPL Flight Planning, PET and PNR lesson, slide number 8.
If the question asks for the PET you would use the 2 engine TAS to find the distance and time. This question is asking for an engine failure PET. From the PET the aircraft would either continue on or turn back with the one engine inoperative (OEI) TAS but to get to the point where the engine failed the aircraft must have flown there with all engines operating.
Once you know the position of the PET (distance from A) you will use the 2 engine TAS, corrected for wind component to find GS to find the time it would take to fly on 2 engines to the PET where the engine failure is assumed to occur.
Hi Jonathan,
Using the E6B or CRP5 I get a TAS of 380 knots. With a CX2 I get 381.6 knots. For both the “trick” to this question is the temperature as 39000 feet is above the tropopause in ISA so you should use a temperature of -51.5 and not -58 if you had forgotten about the tropopause.
Hi Ahmed,
I cannot find any guidance for this phraseology in AIC 005-2018 which leads me to believe this is a question the CAA may have bought from the EASA database.
In the South African AIC 005-2018 the only guidance given is for the phraseology to be used for an ATC to stop a takeoff after the aircraft has commenced the takeoff roll: “callsign ABORT TAKE-OFF (repeat callsign) I SAY AGAIN ABORT TAKE-OFF” to which the pilot should reply “ABORTING”.
The phrases “ROGER”, “WILCO” and “STOPPING” no longer appear in the AIC.
If you get this question in the exam I think it should be appealed as none of the answers comply with the SA circular.
In order to answer this question though the guidance from UK CAP 413 (on which the EASA questions is based) may be helpful.
Paragraph 4.42 (page 18) states that the pilot should use the phrases “STOPPING” and the ATC should acknowledge with the callsign only.
For your first question I would answer A instead of C as I suspect the CAA may still have the EASA answer. For your second question I would answer A. Technically none of the answers are correct as the proper response should be callsign only (for both South Africa and EASA) but the phrase “ROGER” is still allowed in EASA and defined as “I have received all your last transmission.”
I see what you mean, that question is not nicely worded but that is believed to be exactly how the CAA asks it.
When they ask about a “2 axis gyro” we should interpret that as 2 degrees of freedom with a third spin axis. This should lead you to the right answer – in this question they’re asking about the AH.
Thanks Jonathan,
The essential components of a flight director are :
1- a computer
2- an automatic pilot
3- an autothrottle
4- command barsI think I see why you’re asking this question as it was showing the correct answer as 1 and 2 which cannot be true. I have corrected the error in the database.
An autopilot is not required for a flight director, it merely shows the pilot what an autopilot would be doing if it was installed /engaged which is useful guidance to the pilot when flying manually. An autothrottle is not required as the flight director only gives pitch and bank angle advice to the pilot to achieve the desired performance. The energy must be managed to the pilot using manual thrust or an autothrottle but this does not affect the functioning of the flight director.
Hi Jonathan,
“Localiser ARM” means that the vertical or lateral mode of the flight director is armed and ready to provide guidance to track the localiser once the the aircraft intercepts the localiser course. There is a question in the database on this which had the incorrect answer but has now been corrected.
Hi Yaseen,
Jonathan is on the right track there.
A gyro can be assessed by looking at its freedom to spin on its axis (the plane of rotation) and whether they have freedom in the 2 other axes which are 90 degrees to the plane of rotation and to each other.
The planes of freedom includes the plane of rotation.
The degrees of freedom does not include the plane of rotation.
Exam questions can use either of these terms so be sure to read the question carefully to decide whether or not to include the plane of rotation in your answer.
Good day Jonathan,
That’s quite a broad question, I need a little bit more information to go on. Could you please provide me with the question reference so that I can assist you?
Hi Mark,
To answer this question we will use the formula:
change of mass / new total mass = change of CG / distance from mass to old CG
re-arranged we get:
distance from mass to old CG = (new total mass x change of CG) / change of mass
= (250000 kg x +2 m) / 10000 kg
= 50 m.
By convention, arms aft of the datum are positive and arms forward of the datum are negative.
The CG moved aft (or in the +ve direction) when the mass was added, so this tells us that the mass must have been added aft of the original CG, or put another way, the original CG is forward (or in the -ve direction) of the mass that was added.
From the arm of where the mass was added (+24) we must move forward (-ve) 50 m, so the original CG must have been -26 m, or 26 meters forward of the datum.
