January exams, please read

[OneIn60rule]

Good luck to everyone who's doing their January ATPL exams.

Now I have a question from Performance.

You descend at a constant mach, what is going to happen to your angle of pitch?

a) increase
B) decrease
c) remain the same
d) depends on temp.

Since you descend at M then I assume you'll have a TAS increase and that's why I went with the angle of pitch as increasing.

L=CL x speed x 1/2p

Since speed goes up, CL has to increase to remain constant but then I didn't factor in the density, so if P and S increase then there's more a reason for CL to decrease, am I nuts????

Maybe someone can help me see what I'm not seeing.

[Keith.Williams.]

As with (far too many) JAR questions, you must make a number of assumptions. Firstly you must assume that you are gliding, so speed control is by means of changes in pitch atitude only (you cannot simply open or close the throttles).

Having made the above assumption you can find out what happens to CAS by using the C-T-M sketches. When descending at constant mach, the M line will be vertical, with the C and T lines converging on it as altitude decreases. This means that as you descend at constant Mach, your CAS will increase.

But increasing CAS means increasing dynamic pressure. If your angle of attack remains constant, increasing dynamic pressure will give increasing drag, so the aircraft will decelerate. But to have constant Mach, the aircraft must accelerate to increase its CAS. So you must push the nose down in order to reduce angle of attack and drag. This will enable you to increase CAS in order to maintain constant Mach.

The next problem comes in deciding whether pushing the nose down means that pitch angle is increasing or decreasing. There is no universally agreed answer on this point. Some schools teach that pushing the nose down below the horizon is increasing pitch, while others teach that it is decreasing pitch.

At the end of the day you must choose one or the other. But you should also appeal on the basis that there is no universally agreed definition of whether pitch is increasing or decreasing in this situation. This problem has been going on for a very long time but the CAA appear to be unable to recognise it.

The problem could of course be solved by a single e-mail message from the CAA to all the schools, stating what the JAA believe happens to pitch angle when the nose is pushed further down or pulled up when it is already pointing below the horizon.

[QNH 1013]

Keith,

Please could you explain the concept of the C-T-M sketches that you mentioned in your post?

I'm taking the JAR CPL exams a week on Monday including Principles of Flight and Performance and I'm wondering if I'm missing a trick here.

Wonder if anyone else is sitting the CPLs at Gatwick this month? I was on my own for some of the exams in November, and I think January may be even less popular.

[Just another student]

QNH 1013, the CTM sketches are a quick way of remembering the relationship between speeds in a descent or climb.

In a Climb Cas Tas Mach

You might be asked, what happens if you maintain a constant Mach no during the climb? You look at C T M as above, and see that for a constant Mach, Tas would decrease. You can see this by noting that Tas is on the left hand side of Mach, the same with Cas.
For a constant Tas in a climb, Cas would decrease and Mach would increase (because Mach is on the right hand side) etc

For a descent the effects are reversed I believe.

With regard to the Jan exams (next week), I too will be taking Performance, and i'm also taking Principles of Flight. I'm not exactly looking forward to taking them, but i've worked damn hard over the last weeks, so hopefully the work will pay off.

According to the feedback I have regarding your question, the pitch angle would decrease. I would think that this is the case as your Tas is increasing, so you would want to raise the nose, decreasing the pitch.But as Keith rightly says, what is a decrease in pitch? Is it nose down or what?

[Dean Johnston]

QNH 1013,

I would have thought that your FTO would have made you aware of the CTM sketch. If not shown to you then that is quite scary, considering the exams are so close. Which FTO are you with?

[High Wing Drifter]

Sounds like the one that we at Bristol know of as ERTM (Equivelent Rectified True Mach).

[OneIn60rule]

And yes this question has dubious choices and that's why I asked a fellow student first, his teacher told him that an increase in pitch angle relative to the horizontal. So if you are already pitching nose down then you'll have to increase pitch (Pitching down in his lang.).
If you are nose up you will have to decrease pitch.

I've seen this question 5 times in my feedback and I really don't like to see it again!


the CTM btw isn't hard to memorize since all you have to do is remember what happens when you are in ISA, Isothermal and Inversion.

ISA> CTM (If C constant in climb then TM increase, if T increase then C decrease and M increase)
Iso> C decrease in climb and MT same.

Inversion> CTM is now CMT!
SO if climbing at constant Cas then M and T go up

The sketches were taught in my school but you can simply use the THREE finger method to help out.

WARNING if a question comes up asking about you maintaining constant MACH and overall temp decreases then the CAS will stay the same. I cannot explain why this is so but the question pops up 10 times in my fback so it's one of the cases where CTM is invalid for this type of question. Otherwise CTM(ISA)/CMT(Inversion) works for climbs and descents.


Climb at constant C in ISA, T and M both increase
Climb at constant T in ISA, C decrease and M increases
Climb at constant M in ISA, C and T both decrease

*Climb at constant C in ISOTHERMAL, T+M increase (This is the *odd one)
*Climb at constant M in Iso, C would decrease and T would stay *the same. (again this is the odd one so be carefull with it)*


Inversion>
This one is where CTM becomes CMT (yes only T and M switch their positions)
Climb constant C in Inversion, T and M increase
Climb constant T in Inversion, C decrease and Mach DECREASE.
Climb constant M in inversion, T increase and C decrease.



It's very easy to use after a bit of practice and helps me to find out what I have to do with angle of attack if descending at constant mach and such. They might not ask you what happens with TAs, they are more likely to ask what happens with angle of attack or the dreaded PITCH angle. (In which case I just write down the L=Clx etc..)


Hope this helped a bit. :ok:

[Keith.Williams.]

The C-T-M lines are a simple way of predicting how CAS TAS and Mach number vary in a climb or descent.

Draw a vertical line (Y axis) at the left of your page to represent altitude. Moving up this Y axis means that altitude is increasing. Moving down means that altitude is decreasing.

From the base of the Y axis, draw a horizontal (X axis) to the right to represent speeds along the base of the page. Moving left to right means that the speeds are increasing. Moving right to left means that they are decreasing.

The C-T-M lines are three straight lines fanning out as they go upwards from the X axis. Below the tropopause in the ISA they always fan outwards as they go up the graph. The order of these lines is C for CAS on the left, T for TAS in the centre, and M for Mach at the right.

To find out what happens (for example) in a constant Mach climb (or descent) just draw the Mach line straight up the graph. Now draw the other two lines in the correct order, fanning away from it. Moving up the graph you will see that both the CAS and TAS lines are sloping to the left. This means that CAS and TAS decrease as altitude increases in a constant mach climb.

To see what happens in a constant Mach descent just observe what happens as you move down the graph. CAS and TAS both slope to the right, meaning that they increase as altitude decreases at constant Mach.

The method for constant CAS or TAS is the same. Just draw a vertical line to represent whatever is being held constant. Then draw the other two lines in the correct order, fanning away from it. To see what happens in a climb move up the graph. To see what happens in descent move down the graph.

The straight line C-T-M sytem is only accurate for ISA conditions below the tropopause, but can easily be modified to take account of inversions and isothermal layers.

As we climb in an isothermal layer the pressure continues to decrease, but the temperature is constant. The relationship between CAS and TAS is determined by density (and hence pressure), so the CAS and TAS lines still move apart in an isothermal.

But the TAS at any given mach number is related to temperature. Temperature is constant in an isothermal layer, so the mach number at any given TAS is constant. This means that the Mach and TAS lines are parallel to each other in an isothermal layer. It is also worth noting that the biggest isothermal layer in the world is (the stratosphere) above the tropopause.

So to modify the CTM lines to account for an isothermal layer just draw the TAS and Mach lines parallel to each other and the CAS line fanning away to the left as you move up the graph.

In an inversion the temperature increases as altitude increases. This causes the TAS at any given Mach number to increase. So in an inversion the TAS and Mach lines move together with increasing altitrude, as the CAS line fans away to the left as before.

It probably all sounds a bit complicated without drawings but is really very simple and reliable given a bit of practice.




Now getting back to the original question in this string, doing a quick CTM indicates that as altitude decreases in a constant mach glide, the TAS increases. This means that if we want to maintain constant mach in our glide we must make the aircraft go faster. We do this by pushing the nose down. This reduces the angle of attack, thereby reducing drag. But more importantly it makes the aircraft go down a steeper slope. The force making the aircaft fly down the slope is the weight x the sine of the angle of descent. As angle of descent increases, so does the force down the slope. It is this which makes it go faster.

It is important to fully understand these concepts because other questions include such things as "What happens to descent gradient (or glide angkle) in a constant mach glide"?

Any student who is not thoroughly up to speed with things like CTM will be seriously disadvantaged in the POF and PERF exams.

Those students who have done lots of practice using the standard feedback lists might like to try these questions. None are direct feedback, but they are all inverted versions of real feedback. If the original feedback question asked about a climb for example, I have changed it into a question about a descent.
I have produced them to help students to fend of the temptation to simply learn the answers and ignore the concepts. Those of you who get consistently high scores using standard feedback might find the results of this test rather surprising.

1. Given the following data calculate the maximum allowable take-off mass for the aircraft in a 5 kt headwind?

LIMIT FLAP 5 FLAP 15 FLAP 25
Field limited TOM 66000 Kg 69500 Kg 71500 Kg
Climb limited TOM 73300 Kg 68900 Kg 65600 Kg

Wind corrections: +120 Kg/Kt headwind -360 Kg/Kt tailwind

a. 66600 Kg.
b. 70100 Kg.
c. 72100 Kg.
d. 68900 Kg.

2. How do VX and VY compare?

a. Vx is always greater than or equal to VY.
b. VX is always less than VY.
c. VY is always greater than or equal to VX.
d. VY is always less than VX.

3. The reduced thrust take-off procedure?

a. Improves engine life.
b. Cannot be used in low ambient temperatures.
c. Requires at least 10 Kts headwind component.
d. Can be used if TOM is greater than performance limited TOM.

4. VS will not be decreased by?

a. Increasing flap angle.
b. Decreasing weight.
c. Increasing altitude.
d. Moving the C of G aft within the allowable range.

5. The TOD at ISA msl on a flat hard dry runway with no wind is calculated to be 800m. What would it be in the following conditions?

Assume + or – 20m/1000 ft elevation, + 10m/Kt tailwind,
+ or – 5m/10C ISA deviation and the standard slope adjustments.

2000 ft airfield Elevation. QNH = 1013.25 mb. 5 Kts tailwind.
210C ambient temperature. Dry hard runway. 2% upslope.

a. 832m.
b. 954m.
c. 1034m.
d. 1195m.

6. How can the climb limited TOM be increased?

a. Reduce V2.
b. Reduce VR.
c. Reduce V1.
d. Reduce flap angle and increase V2.

7. Which of the following approximates to VMD for a turbojet aircraft?

a. 1.32 VS.
b. 1.35 VS.
c. 1.5 VS.
d. 1.6 VS.

8. A jet aircraft is climbing at Vx and constant powers setting. If speed is increased while power setting held constant?

a. Climb gradient and ROC will increase.
b. Gradient and ROC will decrease.
c. Gradient will increase and ROC will decrease.
d. Gradient will decrease and ROC will increase.

9. An aircraft is gliding at its best range glide speed. If angle of attack is decreased?

a. Glide speed will decrease.
b. Glide endurance will increase.
c. Glide range will decrease.
d. Glide range will increase.

10. How do IAS and drag vary as a flight progresses for a jet aircraft flying at its maximum range cruise speed?

a. Increase, Increase.
b. Increase, Decrease.
c. Decrease, Decrease.
d. Decrease, Increase.

11. What is the tyre speed limit?

a. Max VLOF in ground speed.
b. Max V1 in TAS.
c. Max V1 in ground speed.
d. Max VLOF in TAS.

12. The long range cruise speed for a jet aircraft gives?

a. 1% increase in TAS.
b. 1% increase in IAS.
c. 1% increase in ground speed.
d. 99% of maximum cruise range with an increase in IAS.

13. An aircraft is gliding at its best range glide speed. If angle of attack is increased?

a. Glide speed and range will decrease.
b. Glide speed and endurance will increase.
c. Glide speed will decrease and range will increase.
d. Glide speed will increase and range will decrease.

14. When descending at constant mach number IAS…… and the margin to low speed buffet ………?

a. Increases, Increases.
b. Increases, Decreases.
c. Decreases, Decreases.
d. Decreases, Increases.

15. A jet aircraft is climbing at Vx and constant powers setting. If speed is reduced while power setting held constant?

a. Climb gradient and ROC will increase.
b. Gradient and ROC will decrease.
c. Gradient will increase and ROC will decrease.
d. Gradient will decrease and ROC will increase.

16. Distance available for a jet aircraft planning to land on a wet runway?

a. Must be at least 15% greater than the dry landing distance.
b. May be less than 15% greater than the dry landing distance, provided such data is included in the flight manual.
c. May be less than 15% greater than the dry landing distance, provided the airport authority gives approval for the landing.
d. May be less than 15% greater than the dry landing distance provided the operating company gives approval for the landing.

17. Maximum endurance may be achieved by?

a. Carrying out a steady climb followed by a steady descent.
b. Flying at the absolute ceiling for as long as possible.
c. Flying at constant altitude and constant speed using minimum fuel flow.
d. Flying at VMRC in straight and level flight.

18. Which of the following statements is true?

a. Best range speed is lower than best endurance speed.
b. Best prop range speed is equal to best glide range speed.
c. Best range speed and best endurance speed are the same thing.
d. Best glide speed is approximately 1.32 VMD.

19. Which of the following balance thrust in a steady climb?

a. Weight.
b. Drag.
c. W sin Gamma.
d. Drag + W sin Gamma.

20. How will variations in C of G position within authorised limits affect the fuel consumption in terms of ANM/Kg?

a. Forward movement will increase ANM/Kg.
b. Forward movement will decrease ANM/Kg.
c. There is no relationship between C of G position and ANM/Kg.
d. Rearward movement will decrease ANM/Kg

21. In order to achieve maximum glide range an aircraft must be flown at?

a. VIMP.
b. VIMD.
c. VX.
d. VY.

22. If a pilot elects to land with flap 35 instead of flap 25 the field limited landing mass will……. and the climb limited landing mass will………?

a. Increase, increase.
b. Increase, decrease.
c. Decrease, decreases.
d. Decrease, increase.

23. Which of the following statements is true of two identical aircraft of different masses when descending at idle power setting at any given angle of attack?

a. The heavier aircraft will have a greater forward and greater vertical speed than the light aircraft.
b. The heavier aircraft will have a lower forward speed but greater vertical speed than the light aircraft.
c. The lighter aircraft will have a lower vertical speed but greater forward speed than the heavier aircraft.
d. The lighter aircraft will have a greater forward speed but a lower vertical speed than the heavier aircraft.

24. To maintain a lower speed greater than VS, when flying at the back of the drag curve?

a. More flap is required.
b. Less thrust is required.
c. Less flap is required.
d. More thrust is required.

25. Which of the following conditions might cause V2 to be limited by VMCA?

a. High ambient pressures.
b. Low ambient temperatures.
c. High flap setting.
d. All of the above.

26. What happens to the descent angle as an aircraft descends from FL370 to FL250 at constant Mach number then from FL250 to FL 100 at constant CAS?

a. Increase, increase.
b. Increase, constant.
c. Decrease, decrease.
d. Decrease, constant.

27. How is SFC affected by C of G movement?

a. SFC is not affected by C of G position.
b. Forward movement increases SFC.
c. Rearward movement increases SFC.
d. Rearward movement decreases SFC.

28. If a pilot elects to land with flap 35 instead of 25 flap the landing distance will……. And the go-around performance will………?

a. Increase, increase.
b. Increase, decrease.
c. Decrease, decreases.
d. Decrease, increase.

29. When carrying out certification test flying to establish VMCG, why is nose-wheel steering considered to be inoperative?

a. Because nose-wheel steering is ineffective after VR.
b. Because nose-wheel steering would not be used in the event of an engine failure during the take-of run.
c. Because VMCG must be valid for both dry and wet runway conditions.
d. Because aircraft may be operated with defective nose-wheel steering.

30. What would be the obstacle clearance in a 5% gradient take-of climb given the following data?

Obstacle height 160m above the airfield elevation.
Obstacle 5000m from the screen.
Screen height 50 ft.

a. 90m.
b. 105m.
c. 200m.
d. 250m.

1. d
2. c
3. a
4. c
5. c
6. d
7. d
8. d
9. c
10. c
11. a
12. d
13. a
14. a
15. b
16. b
17. c
18. b
19. d
20. c
21. c
22. c
23. a
24. d
25. d
26. b
27. a
28. c
29. c
30. b

[OneIn60rule]

you made one mistake when making the answers.

in question 27. The correct answer is given as CG will not affect SFC, which is answer A).

I disagree with this since a forward CG will make it harder to pitch up and since it's harder you'll need more power and therefore SFC will increase.


So the correct answer should be> SFC will increase with forward CG movement and that's B)

[QNH 1013]

Keith, Many thanks for the explanation. Saves me having to picture the lss in the atmosphere which is what I was doing before.

I'll work through the examples when I can keep my brain working long enough, and I'm not drugged up on Day-Nurse / Night-Nurse or whatever it is Mrs QNH keeps giving me; I'm suffering a stinking cold at present. At least I should be over it by exam day.

Good luck to all of you sitting this month.

[Keith.Williams.]

Onein60rule.

I made no mistake in question 27.

The question asks about SFC. SFC is a measure of how much fuel is required per hour to produce each unit of thrust. SFC depends upon RPM and ambient temperature. It is not affected by C of G position, provided the RPM is not changed sufficiently to require the engines to be moved out of the optimum 85% to 95% RPM range.

Even if we assume that RPM does need to change that much, SFC will increase only if we were already in the optimum RPM range prior to the C of G change and the C of G change caused us to move out of that optimum range. I believe that this question is actually probing the student's knowledge of the difference between SFC and specific Range.

A number of schools have previously taught that C of G affects SFC. Most have now amended their teaching, but some have not. I believe that they are wrong.

I suspect that you are confusing this with question 20, refers to " fuel consumption in ANM/Kg". These unit refer to Specific Range not SFC. It is however true that forward movement of the C of G would reduce specific range and increase gross fuel consumption.

[Just another student]

Keith, thanks for posting the questions. Personally I have found those questions, very similar to the ones I have already worked through, in fact, a few are exactly word for word. I am finding that the more times I work through my feedback, my knowledge and understanding is increasing. 2 or 3 weeks ago, I would not have stood a chance in answering the questions above. However now I am able to go through them, and similar questions with few problems.

Thanks for your help.

Good luck to eveyone sitting exams.