Just as soon as the good admin folk sort out my posting problem.   Never had problems previously with graphics ?

I might look into a trigonometry book and see if I can re jig it for pilots

Dead simple.   While most folks have tech calculators and PCs which can be used without much difficulty, the easiest approach is with the CR which has the sine and cosine tables built into the wind side CD scales.

I doubt that you will be able to simplify things further ....

Graphics problem has been referred to the IT guru so, hopefully, we should have a resolution shortly

Ravi, I presume that, by "box charts", you are referring to the old DCA "whizz-around" takeoff and landing charts ?   I have a considerable background with them, having produced more than a few in days now long past.   If these are the subject of your comments, perhaps you could elaborate - I would be surprised were I not able to address your concerns.

Several approaches from which to choose.

(a)  use a squared grid on your nav computer to figure the vector graphics.   Not particularly accurate and takes some time.

(b)  use the AIP wind component table.   Easy, once you dig out the AIP and look up the table.  Accuracy to the knot which is fine.

(c)  use an electronic calculator to figure the trig, which is the same as for (b) and (d).

(d)  if you have a CR navigation computer solve in the same way as you do for the flight planning exercise.  All you have to do is

(1)  figure the angle between the wind and the runway, then

(2)  set the TAS index against the wind speed (the TAS index is just the 10 on the standard CD slide rule scales, such as on the calculation side.   Setting it against the speed sets up for the appropriate trig multiplications.

(3)  look around clockwise to the angle calculated in (1) on the sine scale

(4)  read off the XW as you would for the normal flight planning calculation.

(5)  look around anticlockwise to the angle calculated in (1) on the cosine scale

(6)  read off the TW or HW component as you would for ETAS.

That's quicker, easier and more accurate than the other two ways.

For example, for RWY 29 (assume 290M) and W/V 310M/40 - XW 13.7 on the CR and 14 from the AIP, HW 37.6 on the CR and 37 from the AIP.   Too easy.

Private operation flight and 115L (excluding reserves) 

Means just what it says.  The 115L excludes (ie doesn't have any) reserves so you have to add them into the calculation.

I have elected to address the matter by providing a review of how the navigation triangle analysis might be approached.     We can run the work mathematically, which involves no more than relatively simple trigonometry, through drawing it on paper, or through the various navigation devices which have been in vogue for many a year.  

In Australia, the two principal devices seen are the Dalton (often referred to as the E6B, after its early US military designation) which uses a simple graphic solution similar to what one would do on a sheet of paper, and the Jeppesen CR devices (using a part trigonometric solution), which drew on an earlier 1930s German DR2 device (using a trigonometric solution)

The basic navigation triangle (triangle of velocities, or vectors, are other terms you will come across) can be solved either graphically (sheet of paper or on the Dalton) or the equivalent trigonometric solution (as on the DR2 or CR devices or programmed into an electronic navigation device or PC).  

In the following graphic (for a tailwind case), the basic triangle is shown in red.   On the Dalton, this is simply drawn to scale using the known data and then the data required can be read off to scale.   Super simple and I will show the picture a little later.

Looks like one needs to paste graphics in a different manner since the site was reorganised a while ago.   I will find out the story from admin and then come back and post the graphics into the post.   I can see the upload item but, when the file is selected, nothing happens other than for the filename to be stated in the post ?

We can drop a perpendicular (right angle line) from the intersection of the HDG/TAS and W/V vectors (on the basic triangle in red) onto the TR/GS vector resulting in two internal triangles (shown in blue, at the W/V end, and mauve, at the drift end).   These subsidiary triangles are extracted and shown to the right of the main graphic for clarity.

We can solve the subsidiary triangles without any difficulty but must solve the blue triangle first (which has enough data to start with), then the mauve (after we get some necessary data from the solution of the blue triangle).

In the blue triangle, we can figure the angle A using the known TR and wind direction data.  From this we can determine XW and TW (or HW).

sin(A) = XW/WS  which gives us XW = WS x sin(A)

cos(A) = TW/WS which gives us TW = WS x cos(A)

With XW known, we can move to the mauve triangle.

sin(DR) = XW/TAS  which gives us XW = TAS x sin(DR)  from which we can extract DR

cos(DR) = ETAS/TAS  which gives us ETAS = TAS x cos(DR) – and that is ALL that ETAS is !

With the drift angle, we can figure HDG with reference to the known TR.   

For the tailwind case, GS = ETAS + TW

For the headwind case, GS = ETAS – HW

For those who might be interested, this is the usual approach taken for microprocessor based computers and is easily programmed.    Excel, or similar spreadsheet programs, are the easiest way to go.

On the DR2, both sets of equations were solved on the device logarithmically.   When Lahr developed the CR, he discarded the trigonometric solution for the wind end to run with a graphical solution.  The backround to this decision was that Jeppeson, who started the exercise, wanted a device which would compete with the numerous Dalton clones then in the marketplace.   The early Daltons included a squared grid section on the slide to facilitate graphical vector work which couldn’t be done on the DR2 and was a significant selling point for the Dalton back in the day.   Most Dalton clones in recent decades have dispensed with the squared grid on the slide which, in this engineer’s view, is a rather sad thing.

The CR solution is pretty straightforward and shown in the next graphic.   The CR wind side outer scales are the same slide rule CD log scales as on the calculation side.  Lahr reworked the inner scale markings to read the sine and cosine values for the angles.  The angles are then marked at the values of those quantities to make things easier for the pilot-user.    This then permitted an easily set up multiplication to figure the previous equations.

On the graphic, the sequence is

1  set the TAS index (which originally was “10”, as on the calculation side) against the TAS value. The index is labelled TAS only to make things easier for pilot recollection.

2  run your eyes clockwise to the XW value,

3  against which we can read the drift angle value (which is actually the sine of the drift angle value)

4  run your eyes across to the cosine scale and the drift angle (which is actually the cosine of the drift angle value), and then

If you want to use this technique for the blue triangle, rather than the (somewhat inferior) graphic version just replace the TAS index by a WS index.  Sine gives you the XW and cosine the TW (or HW).    Also, be aware that this technique is directly applicable to working out runway HW/TW and XW.   The AIP table is just a rounded-off set of trig data.  Using the CR is easier, quicker, and more accurate.    Also, keep in mind that you might need the complement relation: sin(A) = cos(90-A), and vice versa, to cover the problem where Lahr only included part of the angular ranges on the CR.

Back to the Dalton, if we draw the red triangle on the Dalton, the device uses the wind end of the triangle but leaves out the drift angle end as we only need the HDG and TR lines to figure the drift angle.   The next graphic shows this with a tailwind case on one side and a head wind on the other.   If one puts the track vector against the index, it makes plotting the ETAS value a bit easier using a Douglas protractor to draw a nice perpendicular from the intersection of the HDG and WD direction vectors. 

And that’s about it for figuring out ETAS on the Dalton – too easy - even if ETAS is of no value to the Dalton’s graphical triangle solution.  

Note -

While I expect that Ravi will follow the story without any problems, for a lot of students, this post may appear to be a little heretical and confronting as it contradicts a lot of the (erroneous) material you will have read elsewhere or had presented to you by well-meaning instructors who are parroting what they, themselves, incorrectly learned in past times.   However, it is the story, notwithstanding the strange things published in many of your textbooks and user manuals, often as a result of the original tales being written by those not quite understanding what is going on ... and then being copied and re-copied for later documents.   Such is a common problem with which we are routinely faced in the flying game.

If you are interested but find it a bit confusing, ask a question or two and we can sort out any problems which might be causing your difficulties.  

Does it really matter if you don’t understand the correct story ?   Possibly not all that much, in practical terms for the cockpit, but understanding the correct story can only help in your use of the navigation computer devices.

2. In the exam I got confused by one question. It said Example Private operation flight and 115L (excluding reserves) <--- That confused me. Do it mean that I need to add the 30min reserves on that? 

Question 2 – Townsville to Chillagoe

You are absolutely correct!

The Echo has a maximum structural landing weight of 2,725 kg, as stated on page 56.

Although the landing performance chart indicates approximately 2,750 kg, this does not override the aircraft's structural landing weight limitation.

The calculation should therefore be:

Maximum landing weight = 2,725 kg

Enroute fuel burn-off = 82 kg

Maximum take-off weight = 2,725 + 82 = 2,807 kg

Adding the 8 kg start-up and taxi allowance gives a maximum start-up weight of 2,815 kg.

The published solution has incorrectly used 2,750 kg instead of the structural maximum of 2,725 kg.

Thank you for bringing this to our attention. We will note this for correction in the book.

Regards,
Bob Tait's Aviation Theory School

AS for Question 1 it has created much debate with many different answers provided in the same region by the instructional staff, it seems to comes down to accuracy and pencil thickness at this stage. 

Unfortunately, I don't have the current book.   Perhaps you could post the questions ?

1. On page 58 of the book, there is flight and the answer is the landing weight is 2275. I noted in the generation of the answer the top right corner is not exactly on the line but above and to the right. Irrespective I used the CASA booklet and my value for the landing weight was 2330kg. I did this a number of times with protractors etc. Please advise what I am doing incorrectly.

2. On page 59 there is a flight from Brisbane to Chillago. The final answer is the 2750 + the weight of fuel for the flight. I got the answer for the fuel weight without any problem but then wondered where the 2750kg comes from because I thought the maximum landing weight is 2725kg

Can you please assist 

Regards

Ravi