POF Sample Question

[Port Strobe]

Hi, I'm not doing my ATPLs just now, a few years off yet, but I downloaded a sample POF paper for the ATPL from either the CAA or JAA website, can't remember which exactly and there is a question that I have half an idea how to figure out but not 100% sure, so I thought I'd ask someone for help please if you can spare a moment (someone who has done ATPLs or someone that knows a thing or two will breeze through it I think). Here goes;

A 50 ton twin engine aeroplane performs a straight, steady, wings level climb. If the lift/drag ratio is 12 and the thrust is 60kN per engine, the climb gradient is (assuming g=10m/s sq'd)

a) 3.7% b) 15.7% c) 12% d) 24%

Where I come unstuck is whether or not to factor in the vertical component of thrust and if so how? Is there a component of drag downwards as well then?

I'd be greatfull for any help thanks.

[FlyingForFun]

Ok, I'll give it a go. But this is way beyond anything I remember learning for the ATPL exams. :eek:

The first part is fairly simple. Yes, you do need to factor a vertical component of thrust, and a downward component of drag. The easiest way of doing this is to resolve all the forces relative to the aircraft (rather than the ground) - if you do it this way, T, L and D are all at right-angles to each other, and W is the only one that isn't. So you need to break W into the bit that oposes L, which is WcosA, and the bit that oposes T, which is WsinA, where A is the climb angle. All of this is fairly basic stuff, and is covered in PPL books such as Trevor Thom - although it makes a lot more sense with a diagram!

Now it starts getting more difficult.

We know that T=120kN.
We know that W=500kN.
We know that L=12D

The only force oposing L is the previously-calculated component of W:

L = WcosA = 500cosA

All the other forces are at right-angles to this one, along the thrust/drag line:

T = D + WsinA, or 120 = D + 500sinA. Re-arranging this gives us D = 120 - 500sinA

Since L = 12D, and L = 500cosA, we know that:

12D = 500cosA, or D = (500/12)cosA

We know have two formulae that give us D, so they must be equal:

120 - 500sinA = (500/12)cosA

Now, it's simply a case of solving this. But this is where my A-Level maths falls over, and I don't know where to go from here.

Fortunately, since it's a multi-choice question, we can simply put in each of the 4 values of A and see which one is the closest. The answers are given in terms of a gradient, so convert them to an angle by dividing by 100 then taking the inverse tangent. For answer b, 15.7% comes out as 8.9 degrees - and this fits the equation pretty much exactly. Answer = B.

I think.

Now I'm waiting for someone to come along with the easy method that I'd completely missed!

FFF
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[Mintflavour]

This exact question came up in my exam. fortunately I looked at this question the night before. had ago but could not do it. I had the answer only, in my feed back and just remembered it was 15.7 not sure if they will change the numbers one day but I was luck they were exactly the same.

Job Job, easy 2 Marks.

Mint

[FlyingForFun]

Remembered it was 15.7<FFF checks back to see what answer he put> Whew, that's a relief!

If that had come up on my exam, it would have been one that I'd have left to the end and come back to, no question at all. Whether I'd have managed to do it in the time available I don't know - I know it took me quite a while this morning, but I was doing bits of it in between bits of work, so I don't know exactly how long.

FFF
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[Dick Whittingham]

you can solve the final equation by asuming that for small angles Cos is 1 and Sin is the same as the angle in radians. Thus the equation becomes:

120 - 500A = (500/12)

This works out as 0.156radians, 8.9deg.

I think!

Dick W

[Keith.Williams.]

To answer this type of question we must ignore the fact that lift will be slightly less than weight in a steady climb. This is because we are effectively being asked to calculate the climb angle (Gamma) and the lift in such a climb is equal to the weight x COS Gamma. So if we don't know Gamma we cannot calculate the lift.

If we assume that lift = weight we can use the given weight (50 Tons) and the Lift to Drag ratio (12:1) to calulate the drag.

50tons = (50 x 1000 Kg x g)N

Which using g = 10 gives weight = 500000N


Lift to Drag = 12 so if lift = weight, D = W/12

Which is 500000/12N = 41667N


We can now use the standard formula:

SIN Gamma = (Thrust - Drag) / Weight

Using the given values of 120000N thrust and 500000N weight and our calculated drag value of 41667N gives:

SIN Gamma = (120000 - 41667) / 500000 = 0.156


This can be converted into %gradient using the standard equation:

% Gradient = SIN Gamma x 100%

Which in this case is 0.156 x 100% = 15.6%

Ths equates to about 9 degrees.

At such small angles it is also reasonably accurate to say that % Gradient = TAN Gamma x 100% which in this case would give 15.8%

This type of question is not difficult provided you learn the principles and standard equations then practice, practice, practice.

If you just learn the answers to feedback questions you can pass the exams provided you are lucky enough to get the right exam paper.

But if you learn how to answer the questions from first principles you will be able to pass the exams even when you are unlucky.

[Port Strobe]

Thanks very much indeed for the replies, I think I understand it. I don't have a pen and paper to hand right now so as soon as I get the opportunity I'll go through the explainations thoroughly with a diagram. Re FFF A level maths bit, I reckon it might involve dividing each side by cosA which eliminates the cosA on the RHS and creates tanA on the left in one of the terms (sinA/cosA=tanA as I'm sure you all remembered :p ), but that kinda creates a mess with the 120 term, I'll explore that one with my pen and paper again! Thanks KW as well, looks like you've cracked it from where I'm standing, I'll let you know if I've any further probs or questions. Thanks again all of you.

[Port Strobe]

I've read through these explainations now and I understand the reasoning behind them. I have one question regarding Keith Williams' reply. I know that lift is less than weight in a steady climb as you clearly stated. but how reasonable an assumption is it to make that lift equals weight in order to solve this problem? Is lift say around 99% of the weight in a typical climb (I know this would depend on the climb gradient but just in general) or is it a lot less. Without knocking your method FFF's method works as well without having to make assumptions in it but then you get to the tricky maths bit and there we hit a brick wall whereas KW's method is very simple to solve, and I suppose as long as the assumption is reasonably accurate then it'd be the better of the two to go with. Is that the way ATPLs are taught using the assumption L=W? I'll state again I'm only enquiring as to how you guys are taught, I've only got a PPL so I'm not going to preach to those who have the know-how way beyond what I have just now but if you can see where I'm coming from it does raise the question is there a limit where the assumption of L=W is no longer that accurate. Anyway, I did get the question answered so thanks to all that took the time to answer it for me, much appreciated.

[Keith.Williams.]

The method to be used depends upon how accurate you need to be.

To be absolutely accurate we must use %Gradient = TAN x 100%. But this leads to the kind of mathematical problem
illustrated by FFF. The majority of the ATPL students that I have met (and I have met quite a few) would be unable to solve it. More importantly they would have absolutely no chance of solving it in the time provided in an ATPL exam.

Dick's solution is similar to mine in that it uses an approximation for small angles. If however you are going to use FFF's method to get an accurate equation, the effort is wasted if you then use an approximation to solve it.

Students (and instructors) must also take account of how the examiners expect us to solve problems. When working out the gradient required to clear an obstacle after take-off, they typically require the use of distance to obstacle and height of obstacle above the runway. This clearly uses the TAN method.

But the CAP 698 Performance Manual issued by the JAA for use in their exams includes the equation:

Still air %Gradient = ROC/TAS x 6000/6080

This clearly uses the SIN x100% method

The ROC must be in feet/min and the TAS in Kts.

The 6000 is taking account of 60 seconds in a minute then multiplying by 100%. The 6080 is converting Kts into feet/min.

Worse still, to calulate %Gradient in a headwind or tailwind the above equation is modified by adding tailwind to TAS or subtracting headwind from TAS. The result is then used in place of TAS in the above equation. This process combines horizontal windspeed with TAS up (or down) the slope, so we are using something between the SIN and TAN!!!

Quite naturally the students are usually in a state of revolt by this stage in the training.

So to do the JAR exams you must be able to jump from one method to another to match the question you are solving. At the end of the day the information provided in the question will determine which method you must use.

In the POF exam however you are rarely asked about % Gradients, but frequently asked about the relationship between lift and weight (or load factor) in a steady climb. For these questions you must use:

SIN Gamma = (Thrust - Drag) / Weight.

Lift = W COS Gamma.

And Thust = Drag + WSIN Gamma

To test the accuracy of using L=W instead of L= WCOS Gamma we need to compare a few figures. If we use 1 to represent weight, a Lift to Drag ratio of 12 and a climb angle of 20 degrees we get:

Using L = W and drag = W/12 gives
Drag = 1/12 = 0.0833

COS20 = 0.940
Using L = COSW and drag = L /12 gives
Drag = 0.0783

The difference between these two values is 0.005

At the end of the day you must pick the method to match the problem being solved (or subject being examined) at the time.

ATPL students reading this post should note the use of the term Gamma for climb angle. I have yet to find a set of FTO notes using this term (I haven't seen them all but I have seen quite a few), but the examiners have started to use it. They tell me that the new questions also expain what it means, but some of the students who have taken recent exams did not appear to think that this was the case.

[Port Strobe]

Thanks very much for the explaination KW, that's cleared things up now. Thanks again to all who have replied to my question.

[My names Turkish]

Doh! Wish I had read this for my December exam, it came up again. Oh well theres always next time:(

[Keith.Williams.]

TURKISH,

As I keep saying, do not learn the answers but DO learn the method. (Even) A CAA examiner could change the numbers without too much effort!!!!!

The strange thing is that they do not to do it very often. Now why is that????????? Some would say inertia, but it is probably more to do with the bureaucracy of the JAA system.

[My names Turkish]

Well I didnt know the meothod either but if I had looked at the sample questions a bit more it might have prompted me to look up the method:rolleyes: Never mind I enjoy doing POF so much I just cant wait to do it again anyway.