Ravi,

Good to see you getting into the CPL adventure.

A bit hard to comment without the whole question.  Might you be able to provide any data additional to that contained in your post, please ?

Hi Ravi,

Thanks for raising this — you have actually picked up an important difference between the flight computers.

Your wind conversion is correct:

120°T with 10°E variation = 110°M at 20 kt.

Using a CR-type computer, it is quite reasonable to obtain about a 7 kt headwind component outbound, giving approximately 173 kt, and then treat that as a 7 kt tailwind component on the reciprocal, which leads you towards 187 kt.

The example on page 114, however, has been worked using the E6B-type flight computer method, which is the method CASA uses for these examination calculations. The E6B solves the complete wind triangle, including the effect of the crosswind and the crab angle required to maintain track. This produces approximately:

Outbound: 173 kt
Return: 185 kt

So the difference you have found is not because your magnetic conversion is wrong. It is due to the slightly different way the CR and E6B computers handle the wind triangle.

For CASA exam purposes, I would recommend using the E6B method for these heading and ground-speed calculations.

It is a good pickup, and we will add a clarification to the text so that it is clear that the example is based on an E6B-type computer.

Kind regards,
Paul

Paul,

Your answer confuses me a tad (although I suspect that the story is very simple and will fall out as soon as I can work the specific problem).   

Both the CR (Jeppesen CR and clones) and the Dalton (E-6B and E6B/E6-B clones) must give the same answer (within reasonable reading/plotting tolerances - which should be only trivially different) if they be used correctly.   They have to as they are both solving the same triangle using mathematically correct and equivalent techniques.  The CR does the bulk of it by resolving components along the TR/GS vector, while the Dalton does the whole thing somewhat more simply and easily by a graphical solution using the basic triangle.

To resolve this, as I don't have the source document, might I ask you to post the whole question for readers' consideration, please ?  Then I can comment with my engineering pilot's hat on.

Asides: 

(a) For what it's worth, one can work the sums backwards and get, near enough, TAS 180kt, GS out 172, GS back 186 for the CR and, for the Dalton, TAS 180, GS out 172, GS back 186.   I see no substantive difference between the CR and the Dalton ?

If I run the question in Excel (ie do the trig sums), I get TAS 180, GS out 172.2, GS home 185.9 which shows a reasonable agreement with both my CR and Dalton solutions ?

(b) One needs to be aware, and keep in mind generally, that the pilot nav computers are manufactured to a price and are not always as good as they ought to be.   Certainly, the average computer is not up to the build quality of the (somewhat more expensive) engineering/scientific slide rules of yesteryear.   Generally, they remain fit for purpose but such manufacturing errors can introduce unexpected errors in use.

Indeed, EBJ related a tale that, when he first took over the rights to Lahr's/United's device, the first few manufacturing runs were so bad that he and family spent numerous evenings around the kitchen table running detailed scale error checks.

(c) The other consideration is ETAS.  In Lahr's workup, he didn't coin the term, preferring just to refer to the trig equation in his patent application, and, most certainly, didn't introduce the 10 degrees nonsense which EBJ did in his user guides.  The Jeppesen approximation process is fine for in-flight use but makes absolutely no practical sense at all and, for the exams, just puts the candidate closer to a point where the CASA computer assigns a "wrong" grade for the question.   

One must set TAS on the outer CD scale against 10 (the TAS index) on the inner CD scale to permit the trig multiplications.  Initially, you must run the sum X/W = TAS x sin (drift angle) to get the drift angle.  You now can, immediately, swing your eyes around to the immediate left of the TAS index to figure the sum ETAS = TAS x cos (drift angle).   This takes all of one second (OK, two seconds if you are drinking coffee at the same time) - why ever would anyone intentionally bother approximating ETAS by TAS ?  As the drift angle gets down to below, say, 5 degrees, you will do so simply because you can't read the ETAS value and, for practical purposes, the two are functionally the same value.

For those who have been mischievously seduced by the user guides and use the ETAS=TAS approximation for drift < 10 degrees, you are deliberately introducing a small, and quite unnecessary, error into the G/S calculation.   

I note that CASA cautions the ATPL candidate against using the approximation but, as a sop to the vast numbers who think it is a "good" idea, goes along with it for drift < 5 degrees as it becomes moot at very small drift angles.

Thank you here is the screen shot of the page. Hence turning at 180 degrees, I would have thought that the magnitude would be the same, just the vector 180 opposite. Kind regards Ravi Mahajani

Respected all, the question asked for the minimum fuel for ballast. You still have, 110 -80 kg so 30 more kg could be added to the wings as hard ballast. I used this method + another 30 kg in the nose ( to max to 55kg) Given the BEW was 2539 in the original, and the limit for ZFW is 2630, you can still add 91kg.

I calculated therefore 30kg into the nose and 30 kg into the wind to make this 55 kg and 110 kg respectively. Then my ZFW would be 2599.

Then to correct with fuel would be 39kg.

Please can you advise the error of my calculation, as the answer ( provided screenshot) does not include weight into the wing compartments

Kind regards

Ravi

Ravi,

First, some observations in respect of the whizz wheel question.

(a) Bob's example (using the Dalton device) plots the W/V reversed which, strictly, is incorrect.   Lots of folks use that approach and the error, generally, is trivial.  However, for vector and geometric correctness, the W/V speed should be plotted down, rather than up, from the grommet and the solution run in the standard manner used by Dalton in his patent.  This alternative approach (and I have never been able to find out who introduced the idea .. but it certainly goes back a long way) does, however, add to the user's difficulty when it comes to running different calculations ie, it is easier, overall, to run the flight planning solution in the correct, standard manner.   It doesn't cause me to get the heebie jeebies at all, but I do counsel my own students against using the alternative approach with the Dalton.   For me, I just can't see any practical advantage conferred by its use, so why use it ?

(b) you have fallen trap to running an inappropriate scalar calculation when doing a vector analysis.   Doesn't quite work that way.   There is no reason why the scalar difference between TAS and GS should be the same in both directions.   You are looking at the vector difference, not the scalar.

(c) with reference to my previous post, do keep in mind that these devices are not precision instruments and carry a bit of error along in their solutions.  In my view, providing that my students come up with answers within, say, a knot and degree or so of the trigonometric solution, I am pretty comfortable.  Indeed, the only way to get the "correct" answer is via a trigonometric (ie mathematical) solution - while the CR runs a trig solution, it is no better/worse than the Dalton.   For those folks who can fly better than to a knot (mind you, we could do that on the Electra - the throttles were welded to the ASI) or degree, I can only sing their praises - I can't fly that accurately overall when one considers all the other stuff that Captain Speaking is doing at the same time, especially if one loses the autoflight gear.

I'll have a looksee at the second problem in the next day or two.

best regards,

John

Ravi,

Ran your result to brief myself.   Not having your detailed solution, I can't comment on your 39 kg other than my sums suggest you would need closer to 46 kg or so.  So your solution approach is "OK" - in that you ended up with a solution but the question now is, as you have posed, why didn't Bob's solution run that way ?   That is to say, does Bob's solution offer some sort of advantage over yours ?

First some background thoughts -

(a) ballast represents an undesirable operational cost.  It comes out of the bottom line profit for the flight, ergo, we do our level best to avoid carting non-essential weight around the place.   That is to say, we keep any necessary ballast to the minimum feasible.

(b) when considering ballast (or any loading, for that matter) always have a background think about what the added/removed load does to the existing (ie present) cg before you do the load change.   This can be summarised, very easily, in the following manner - cg moves toward weight added and away from weight removed - that is to say -

(1) if you add weight forward of the cg, then the revised cg moves forward of the original cg

(2) if you remove weight forward of the cg, then the revised cg moves aft of the original cg

(3) if you add weight aft of the cg, then the revised cg moves aft of the original cg

(4) if you remove weight aft of the cg, then the revised cg moves forward of the original cg

(5) a corollary is that you need less ballast weight if the arm is greater, ie we are concerned with moment changes (weight x arm) not just the weight with which we might be playing.

So, getting back to the question of whether Bob's solution holds any advantage ?

In your solution, you started with a cg of 2712 and you needed to get that moved forward to, at least, 2680.  So you cast about for a suitable loading station up towards the front of the aircraft.  The most forward station available was the nose locker, at 500 mm.  Following (1), you added 30 kg at 500 mm (ie forward of the existing cg of 2712) and the new cg moved forward to 2686 mm.   Still not far enough so we cast about for somewhere else to put some ballasting weight.  Now, you considered the wing lockers.   These are located at 3550 mm which is AFT of the present cg of 2686 mm.   So, if you now added weight to the wing lockers, you are adding it aft of the present cg.  Following (3) this will move the existing cg (2686) AFT to a new cg location of 2696 mm.   That is, you have moved the cg aftwards and all that has done is undo some of the benefit gained by putting the first ballast load in the nose.   The logical conclusion is that ballasting the wing lockers in this situation doesn't help at all and only makes the loading situation worse.

... and that is why Bob didn't choose to load ballast into the wing lockers.

A very easy overlooking mistake to make, which is why I suggest that, when doing any of these sorts of questions, you always need to keep a close eye on where the cg is, where you want it to go, and where you might consider sticking weight - whether ballast or just normal loading.

While I know that you didn't need the detailed explanation, I have added it for the benefit of others who might be reading this and who might have a problem with getting their heads around the details of the storyline.

@john-heddles Thank you again, I agree that adding balasts into the wings confers no real advantage over fuel, but the question was very specific when it  said what is the minimal fuel you need to carry as I understand. Hence why I added ballast to the wing

As an aside, and by way of reminiscing, as a general rule, using fuel to ballast, while mathematically OK, represents an undesirable operational risk for light aircraft.  Indeed, if we were to go back 30-40 years ago, the then DCA (now CASA) airworthiness engineering group made an all out effort to introduce regulatory pressure to ban the practice.  The problem is quite straight forward – should fuel be used for ballast, and the ballast fuel then be burnt, inadvertently, then the aircraft ends up with two problems – the original balance concern, and a critical fuel situation.

A bit different on heavies, where we have multiple crew to look after the in-flight fuel management.

adding ballast into the wings confers no real advantage over fuel   Your suggestion represents an error in logic.  The consideration is only one of loading arm.   Any time you have a cg other than where you want it, the concern is one of moment.   As I suggested before, it is critical to monitor where the cg is to start with, where you want it to go, and what loading arms you might employ to make the change.  If you add/reduce moment inappropriately, then you have to undo that “error” which is going to require the use of more moment change than you really needed.   As you generate moment by weight (strictly mass) for a given loading arm, having to correct a moment loading error requires more weight movement than is minimally necessary.   It follows that adding ballast into the wing lockers results in a very real disadvantage in that you end up having to use more fuel weight than is necessary to fix the original problem.

the question was very specific when it  said what is the minimal fuel you need to carry  That is the problem at hand.   By adding load, inappropriately, to the wing locker, you end up having to use MORE fuel for ballasting than is minimally necessary.                                                                                                                                                                                                                                                                                 

Comparing the two situations –

Original loading ZFW gives ZFW 2539 kg and 2711.6 mm

When I add 30 kg to the nose locker, this improves the situation to 2569 kg and 2685.7 mm.   Still outside the aft limit but pretty close.

When I add fuel ballast of 17 kg, I get 2586 kg and 2679.8 mm, which is just inside the aft limit.

Now, if I start with the original loading and add 30 kg to both the nose and wing lockers, I get 2599 kg and 2695.7 mm and note that this is still somewhat outside the aft limit.

I put in some fuel for ballast and find that I need 46 kg to end up with 2645 kg and 2679.8 mm which is just inside the aft limit.

The conclusion is that, without wing locker ballast, I need 17 kg fuel while, with wing locker ballast, I need 46 kg.   If the question was after minimum fuel to ballast, the use of load in the wing locker has resulted in the minimum fuel ballast increasing from 17 kg to 46 kg so the solution is unsatisfactory.