Hi ... anyone...

I'm lost and confused by ETAS. I've searched all of the BT resources I can find, and any others within reach, to get some clarity on the topic of ETAS.
Firstly, I'm using an E6B. Numerous resources suggest "Don't worry about ETAS, it's done by the E6B automatically". That's great but it doesn't explain in simple terms (if they exist) what ETAS is.
Additionally, I'm seeing formulas for solving ETAS equations that involve COS. How on earth do we employ COS on a non-scientific calculator, which is all we can take into the exam....?
I came across a practice question today which asked "If your TAS is 150kt and the crosswind component is 35kt what is the ETAS?"
I'm more than a bit concerned that after having completed the BT Performance text I have no idea how to answer what I assume is a simple question.
This is my final CPL subject and with my exam looming I'm more than feeling the pressure.

Any help would be greatly appreciated.

Simon

We can sort out your concerns dead easy.

However, just to set off on the correct heading, can you precisely define what computer you are using. References to ETAS only apply to the CR but we have had great confusion in the past with folks talking at cross purposes with Dalton and CR devices.

How do you tell the difference ? The Dalton has a slide with fan shaped drift lines while the CR doesn't have any slide at all.

Then we can sort out your problems easy peasy.

Hi John,

Im assuming it's a Dalton, it has a rectangular sliding panel (pic attached).



quote="Simathawk77 post=15323 userid=6435"]Hi ... anyone...

I'm lost and confused by ETAS. I've searched all of the BT resources I can find, and any others within reach, to get some clarity on the topic of ETAS.
Firstly, I'm using an E6B. Numerous resources suggest "Don't worry about ETAS, it's done by the E6B automatically". That's great but it doesn't explain in simple terms (if they exist) what ETAS is.
Additionally, I'm seeing formulas for solving ETAS equations that involve COS. How on earth do we employ COS on a non-scientific calculator, which is all we can take into the exam....?
I came across a practice question today which asked "If your TAS is 150kt and the crosswind component is 35kt what is the ETAS?"
I'm more than a bit concerned that after having completed the BT Performance text I have no idea how to answer what I assume is a simple question.
This is my final CPL subject and with my exam looming I'm more than feeling the pressure.

Any help would be greatly appreciated.

Simon[/quote]

Indeed, a Dalton.

Quick history - The original device (with a rolling slide rather than the simple fixed side in your picture) was designed by Philip Dalton in the 1930s. It was adopted by the US forces and its original USAAC designation was E-6B. When it moved to civilian use, the designation became the now-familiar E6B or E6-B.

In respect of your concern with ETAS, may I first suggest an analogy ?

Imagine we have two target shooters, one an archer with a bow and arrow and the other using a rifle. Both weapons do much the same thing - you can target shoot with them. However, each functions in different ways. The archer may be interested in bolt actions, magazines and the like but these are only of relevance to the rifle, and of absolutely no relevance to the bow and arrow.

In your case, you are using a Dalton while your colleague is using a CR. Both do the same thing in that they solve the navigation wind triangle. However, they do the exercise rather differently. The Dalton solves the basic navigation triangle graphically while the CR solves it using a combination of graphical and trigonometric (a particular branch of mathematics) techniques. In your case, you are like the archer; you have an interest in ETAS, which is of vital interest to the CR user, but absolutely of no value to you with your Dalton

Quick history - The CR is marketed by Jeppesen and, with the patent's expiry, a number of clone manufacturers. Jeppesen didn't design the device; they bought the rights from Ray Lahr and his employer, UAL, in the mid-50s. Lahr's device, in turn, can be traced back to a German design by Siegfried Knemeyer in the 1930s. His device was termed "dreieckrechner", which basically means "triangle calculator", and was given the designation DR1, DR2 and so on. The DR2 became the device of preference for the Axis forces in WW2. It was quite sophisticated, using a totally trigonometric triangle solution.

Your question centres around ETAS. You can, if you want, figure out ETAS on the Dalton (graphically) but it is of no use to the Dalton's triangle solution. Rest assured that you will not see such a question in the exam as the examiner is well aware that it only applies to the CR and would incur outrage among the Dalton fraternity.

So, in response to your first post, my thought is

(a) unless you have a specific interest in how the CR solution works in the background, just ignore and forget all about ETAS. It is absolutely irrelevant to the Dalton's triangle solution. Any wild tales you hear to the effect that "The E6B figures out ETAS in the background", or similar, are arrant nonsense and just indicate a total lack of knowledge on the part of whomever might make such a suggestion. The test question you pose, also, has no relevance to the Dalton. You certainly can figure out what the ETAS is using the Dalton but that is only a matter of sideline interest.

(b) If ETAS is your interest, we can continue with the thread and describe that for you.

I appreciate the reply John but I'm not sure how I can just ignore and forget all about ETAS. Every practice exam question I encounter, either BT or other sources, all use ETAS (notably PNR/ETP questions) and my answers are invariably incorrect because of it. Given the exam is both multiple choice and insert the answer, you can understand my concern.

Your reply doesn't make all that much sense unless you are being required to use a CR and that is NOT the case for the CASA exams.

The absolute fact is that, for the Dalton, you don't need ETAS, you don't use ETAS, it is TOTALLY valueless. It only is involved with the CR solution as that device calculates stuff along the track vector and you need to resolve both TAS and WV to make that work.

Maybe if you post a couple of the problem questions (scan them - and any solutions - and post, ie no paraphrasing) so I can see the original words and then I will be able to make some sense of what might be going on here.

Every practice exam question I encounter, either BT or other sources, all use ETAS (notably PNR/ETP questions)

That makes no sense to me, at all.

I'm guessing here that you need to do a little interpretation when reading the question ? (I'm guessing, also, that I know what the problem is but, until I can have a looksee at the original words, I can't do more than guess.)

I note that Simathawk77 has logged in since I last posted. I really hope that he/she hasn't lost interest in this thread as the subject is very important and equally very misunderstood by a great many pilots.

ETAS is simple but ONLY applies to the CR solution and has nothing to do with the Dalton. So, if you can post some of the problem questions which are causing you angst, we can sort things out for the benefit of all.

Simathawk77
The idea of effective TAS addresses the situation where the wind direction is at an angle to the flight planned track (the wind is not a direct headwind or tailwind). Consider the extreme case when the wind is at right angles to the flight planned track. In that case, there would be no headwind component.

Many would therefore assume that the ground speed would be equal to the aircraft's true airspeed (no headwind component). However that assumption is incorrect, because in this case, to make good the flight planned track it would be necessary to allow for drift by adopting a heading that is off-set into wind.

That means that the aircraft is actually flying into wind to allow it to make good the flight planned track and the ground speed would be slightly less than the true airspeed even though there is no actual headwind component on the flight planned track. If you are using a Jeppesen type flight computer (circular with no slide), this effect can be ignored as insignificant when the drift is less than about 10°.

However, as the drift increases, the effect becomes more significant and it becomes necessary to allow for this 'flying into wind' effect. We do this by adjusting the TAS to obtain an effective TAS which is slightly less than the actual TAS. That is necessary ONLY if you are using a computer (such as the Jeppensen type), that uses wind components to solve heading/ground speed calculations.

However, if you are using the Dalton type computer (the one with the slide), there is no need to consider effective TAS at all. This type of computer solves the heading/ground speed calculations by using vectors and trigonometry and the 'flying into wind' effect is automatically allowed for.

So remember.
ETAS correction is never used for a Dalton (with a slide) type computer.
ETAS is used only for the Jeppesen (no slide) type computer only.
ETAS is considered in determining ground speed only. It has no effect on determining heading.

The OP hasn't been on site for a while. I will need to make up some graphics in the next little while and then I will address such of his questions as I can. The main concern I had was his assertion that all the problems he had seen necessarily involved ETAS and that, really, just has to wrong. However, in the absence of his providing further information, there is not much I can do about solving his dilemma.

As Bob observes, the main takeaway is that ETAS is only relevant to the CR computer and has no relevance to the Dalton.

I would, however, take a slightly different view regarding the observation that the cosine function tends to 1 as the angle approaches zero. The ETAS calculation always applies to the CR's nav solution and is always the correct solution, regardless of drift angle. Using the approximation that ETAS gets closer to TAS as the drift angle reduces introduces an error into the solution that CASA doesn't adopt. Why introduce another source of discrepancy between CASA's answer and yours when that is unnecessary ? More relevant, the way the solution is processed always means that you have only to glance at the cosine scale to read off ETAS so the approximation offers absolutely no useful benefit to the solution, nor does it save any useful time in running the solution.

Some thoughts, pending further post information from the OP.

I'm using an E6B.

First, nothing much wrong with the E6B (Dalton) - a fine machine.

Numerous resources suggest "Don't worry about ETAS, it's done by the E6B automatically"

Such resources are talking utter/arrant nonsense/hogwash and reflect adversely on the knowledge of the authors. ETAS has nothing to do with the Dalton’s navigation solution. It is only relevant for the CR navigation solution. If you wish, though – with a bit of fiddling, you can read off the ETAS value on the Dalton, although that provides no other value.

That's great but it doesn't explain in simple terms (if they exist) what ETAS is.

ETAS is dead simple. It is just one of the triangle vectors used in the routine CR navigation solution. In buzzword speak, it is the resolved component of the HDG/TAS vector on the track/GS vector. No more, no less. An example might be of use ?

Let’s work up to it by considering the basic graphical solution (which is what the Dalton provides).

The first way we might go about the solution (e.g. for the usual flight planning problem) is to run it by hand using no more than a sheet of paper, a sharp pencil, a rule, a protractor, dividers and a compass.

Let’s say the problem is to find some flight plan data in the normal way –

W/V = 140/40
TR = 030
TAS =150

Find HDG and G/S.

The process is -

(a) Draw in a convenient reference direction for north (so we have a datum from which to measure directions).

(b) Pick a suitable scale for speed (if the drawing isn’t to scale, it is pretty meaningless and quite useless).

(c) Draw in a track direction with reference to north using the protractor (if you don’t have a Douglas protractor, please get yourself one). At this stage we don’t know how long to make the track to represent the G/S so just make it adequately long to suit (i.e. longer than you reckon you might need for the problem).

(d) At some convenient point along the track, draw in the W/V vector using your protractor, (remembering that wind will go from the HDG/TR vector to the TR/GS vector). Usually, this is most easily done at the end of the track vector which you have just drawn. Mark off the W/V vector length to suit wind speed with reference to the scale using the compass or rule.

(e) From the end of the W/V vector, mark off the TAS length for the HDG/TAS vector using the compass or rule reference the scale so that it just intersects the track/GS vector. This is most easily done using the compass to draw an arc from the end of the W/V vector to intersect the TR/GS vector.

(f) Measure the drift angle using the protractor. Track and drift angle give the HDG.

(g) Measure the G/S vector length using the dividers or rule.

And we end up with the following results –

HDG = 044.5
G/S = 158.8

The basic drawing described above is shown below –



That’s it. Give it a go and see just how easy it is to do. It is likely that you have never been shown how to do this nor have ever had a go at drawing up the navigation triangle on a sheet of paper. Your first couple of goes might be a bit average but you will pick it up very quickly. Just be disciplined and deliberate with your drawing and measuring actions.