Hi Bob, Richard and all!

Once again, great courses and I love seeing things from different angles, but just for the benefit of others unless I am mistaken, there appears to be an error and assumption on Page 3.17 of BT CPL NAV Issue 3.

Specifically in EXAMPLE 3, the example question asks: the "aircraft is cruising at 2500 ft at the position marked A in the figure below. The pilot wishes to commence a continuous climb OCTA to 6500 ft. If the rate of climb is 500 ft/min, and the ground speed during the climb is 100 kt, what is the earliest distance at which the climb may be commenced?"

Now my workings showed the difference between 2500ft and 6500ft to be 4000 ft. Divide 4000 ft by the rate of climb of 500 ft per minute and that would be 8 minutes. 8 minutes at a 100kt GS would be very close to 13.3 nm.

However in the SOLUTION (below) there would appear to be one error, and one assumption.

Firstly the ERROR would appear to be that instead of 6500 be the new climb altitude, this has reduced to only 6000 ft, which of course means that instead of 8 minutes, we are now at 7 minutes for climb to this altitude. 7 minutes on the E6B would be 11.7nm (instead of 13.3nm above).

On top of this I was unsure of how far "A" was from the 35nm boundary, but it would appear the ASSUMPTION was that point A was "right at the aerodrome" since our workings are based on 50nm minus 11.7 nm. This appeared to me to be a bit ambiguous, if stated at all.

Once again I don't mean to pick nits, just verify that I am operating from the same platform as everyone else (i.e. the same QNH datum

Hi Brook,

Thanks for your post. I can see where you are getting a bit confused with that question so I'll try and clear up a few things.

First off, when you look at the vertical profile of the climb (which I'd recommend you do with any of these problems), you will see that the critical factor in the climb is the edge of the 6000ft step at 50 nm. Yes, you are still climbing to 6500 ft but once you are passed that 6000 step, you are in the clear and the rest of the climb is irrelevant. (Note, If you were climbing to say 8500, then you'd need to check clearance of the 8000ft step too).

So, what we are interested in is the climb from 2500 to the last altitude critical to our climb: the 6000 mark and we must ensure we do not reach 6000 before 50nm which is the bottom corner of that 6000ft step.

(6000 - 2500 ) = 3500ft. 3500ft at 500ft per minute = 7 minutes. 7 minutes at 100kt = 11.7 nautical miles.

Those CTA steps are centred on the aerodrome and the distances are measured from there. So, we can commence our climb no earlier than (50 -11.7) or 38.3 nm from THE AERODROME. Where point A actually is, is irrelevant. The answer is given as a distance measured from the aerodrome which is acting as the reference point for the CTA steps. According to the diagram though, point A is pretty much on the aerodrome anyway since it is virtually at the centre of the control zone.

In any case, an exam question will tell you explicitly from where you should measure your answer. The problem on page 3.17 served only as an example of the type of working you need to use.

Cheers,

Rich

Thanks Richard,

Appreciate you checking that out and even replying on a Sunday too!

Thanks for that Clarification on the 6000ft CTA step, yes I see where I went wrong now, it would appear the solution doesn't mention the continued climb to 6500ft so I can see why you pitch it that way given the probability of a CASA exam doing the same!

As far as the A point goes, point taken, CASA exams WILL specify that, just wasn't there in this guess so it left me guessing.

Thanks, I know with all this online stuff it will be easier to explain graphically as well, good luck with it all!

Brook

Hi Richard and Bob,

I hope you both have had and continuing to have a nice Christmas and End of year. Mine has been very laid back, but alas I must provide a boot to my own backside again and here we are

When you do return to the forum can you please provide some advice on the page 4.39 exercise 5 (BTW thanks for the work out its great) since its doing my head in, I have run at it from a number of angles, so I must be missing something I am guessing.

I am going off both the applied use of the E6B and also the calculation for both TE and CA that on their respective legs that d x 60 / D = TE and CA respectively.

The nav exercise is from a Silo 1nm east of Lake Goran to Inverell, and the off track fix is taken from Over a mine 7nm SE of Barraba township.

Now when I measure the track from the Silo to the off track fix I come up with a track length to the fix, of 55nm (D).
When I measure the track deviation from the FPT I come up with something fairly close to 1.5 nm (but lets round it up to say 2, which is (d).

Using either the calculator d x 60 / D or the E6B I come up with a TE of 2.2 deg (so we could round it down or up, but in your answers you go to 3 degrees so I will match this for now.

But now it comes to working out the CA, I have d still as 2 nm but my new D (distance between fix and destination (Inverell) is now the distance left to Inverell which I measure as say 44.5nm so I will round it up to 45nm (D for CA)

So by this we should be able to work out the CA by 2x60 / 45 = 2.66 recurring, which should mean the CA rounded up to 3 deg.

However in the Answer on page 4.43 the CA is 5 deg which is effectively double what the measurements would indicate, and about 2.5 degrees off. Now this too me is important as it effects all other "derivative" measurements and answers.

Would you please let me know if it is me making this error (I have remeasured several times) as this one question has been quite a stumbling block to both my progress and motivation!

Kind Regards

Brook

Hi Brook!

Happy New Year and I hope you had a great Christmas. Sorry your post has sat here unanswered for so long.

Before you get too worried about this question, just double check which mine you are measuring from and which aerodrome you are measuring to.

The mine is "7nm ESE of Barraba township" which places it about 3 nm NE of the northern tip of Split Rock Resovoir. Also, from the answers it would appear the "Tracking To" point is the Inverell aerodrome and not Inverell township. The Inverell aerodrome is actually about 7nm south of the town.

If you take your measurements from those two points you'll get a 3nm off-track distance and you'll find there is actually 38nm to run to Inverell airport tracking 024M (as in the answer) and not the 45nm you were getting. The 45 nm will naturally give you a shallower CA than the answer.

Also, when calculating your closing angle look at the distance to run and your off-track distance. You are 3nm off-track and you need to regain track at the destination 38nm away.Try using those end points and see if you still get a different answer. Let us know if you do and we'll take it step by step.

Cheers,

Rich

Thanks Richard,

I'll have a look over the weekend and see if we are on the same page, and might scan my map that I have drawn on. Appreciate that and sorry for my late reply, I have been offline again for a few days apart from email.

Brook

Hi Richard,

I scanned the map with my lines on it, are we singing off the same hymn sheet? Maybe I missed something - when I measure the distance from B to the final destination. I get 45nm.

*** STOP PRESS *** could it be that I used the Inverell AERODROME (Licenced Civilian) rather than INVERELL airPORT (RPT). That would explain the issue.

This is probably what I have done by the looks. Interesting to see how much of a sticking point it was with me. I only got that as a result of you saying that it was 38nm not the 45 I had measured which of course also changed the track etc.

OK time to play the DER song

Yep, from your scan, you are tracking to the wrong aerodrome. You want to go to Inverell (YIVL) not Inverell North (YINO).

Cheers,

Rich

Thanks Rich,

Bugger, and to think I was starting to get upset that the refueller was taking a long time to get here...

Brook

Hi Again Richard - last but not least of the series of questions up to topic 6 anyway

Can you please elaborate on the workings for Question 26 on Page 4.55:

Heading and TAS have been constant from 2110 to 2128. If heading is changed by 12 deg L at time 2128, at what distance from A would you expect to regain flight planned track:

a) 100nm
b) 150m
c) 50nm
d) 80nm

Now this would be an easy one to make an assumption on - If you look at the picture, the first leg A to B is 50nm, and the distance off track is 5 nm. So d=5 whilst D=50. Based on that the TE is 6 deg right.

Now it wants to know if the heading is changed by 12 deg L at time 2128, at what DISTANCE from A would you expect to regain flight planned track.

In the absence of any workings on this specific answer in the answer legend (I initially got this one wrong with 50nm instead of 100nm) Now would I be wrong in saying that how I would have gotten 100nm was like this:

Since the TE and CA must always add up to the track change, so in this case the current TE is 6 deg (Right) and the track change 12 degrees left, so we can safely assume the CA is also 6 degrees, which gives us our course correction of 12 deg left.

However since there is also a relationship RATIO between the TE and CA that the TE is twice the CA, so if the CA is the same size as the TE in this case, then the distance must be double (ie 100nm vs 50nm based on a CA of only 3).

Does this sound right to you?

Brook