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). [attachment=2461]20240602_095119.jpg[/attachment] [attachment=2461]20240602_095119.jpg[/attachment]

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

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.

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

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.

[color=blue]I'm using an E6B.[/color]

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

[color=blue]Numerous resources suggest "Don't worry about ETAS, it's done by the E6B automatically"[/color]

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.

[color=blue]That's great but it doesn't explain in simple terms (if they exist) what ETAS is.[/color]

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 –

[attachment=2482]man triangle.jpg[/attachment]

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.

If we mark up the previous graphic with the various quantities, it becomes a bit easier to follow –

[attachment=2483]man triangle marked up.JPG[/attachment]

Next, with the Dalton, it’s just the same, but the instrument makes things a bit easier for you and you can complete the exercise in a very short period of time. Once you have your head around the problem, the Dalton is extremely quick to use for this sort of problem. 

So

(a) Setup the wind vector in the normal manner

(b) Iterate to get the triangle orientation correct

(c) Read off the results

HDG = 044.5
G/S = 158.5

[attachment=2484]Daltonsetup1_2024-07-06-2.jpeg[/attachment]

With the Dalton, we only see the business end of the vector triangle. Note that my solution follows Dalton’s original patent technique. There is another approach used (which dates back to the very early days – indeed, I have a monograph on file for the subject which comes from Weems). This alternative procedure offers a trivial benefit but geometrically it is not quite right. However, the associated errors are minimal and lots of folks prefer it. I don’t see any real value so I stick with the original (and geometrically more logical) way of doing things. Which should you use ? Suit yourself, but the alternative method doesn’t aid in understanding what’s what in the process.

Now, with the CR, we get the following results

R X/W = 38 kt
T/W = 12.5 kt

Drift = 14.5 L

ETAS = 145.5

Giving the same results as for the previous two methods.

HDG = 044.5
G/S = 158.0

[attachment=2486]CR setup 1.jpeg[/attachment]

You see that it doesn’t really matter all that much how you go about the task so long as you are attentive to the housekeeping details. (Normally, we wouldn’t bother with the detail lines on the computers but I have included them here for clarity).

As a side note, routinely you will get a bit of spread with the answers which reflects practical accuracy. So long as the process is done carefully and attentively, then the result will be fine and of acceptable accuracy.

Now, with the CR solution, folks tend to get a bit confused as we don’t see the triangles upon which the solution is based, and the user guide doesn’t help much.

So we might benefit from the CR explanation’s being transferred to a triangle graphic for illustration.

[attachment=2487]CR triangle marked up.jpg[/attachment]

Depending on the orientation of W/V and TR/GS, the final G/S value will be either ETAS + T/W or ETAS – H/W

If you incline to the rather strange 10 deg approximation thing, you will be introducing an unnecessary error. Keep in mind that CASA (I am presuming) does the calculation on the handy desktop PC. While the CASA website suggests that they run the approximation for drift less than 5 deg, that is rather facile as the error at that stage is near zero for all intents and purposes as can be seen, quite easily, on the CR cosine scale.

Lahr’s solution for T/W (H/W) and X/W is run graphically, although it quite easily can be done trigonometrically in the top triangle and is both easier and more accurate.

The solution for ETAS and the drift angle are run trigonometrically in the bottom triangle

(a) X/W = TAS x sin (DR)

(b) ETAS = TAS x cos (DR)

These sums are what you are doing when you set up the wind side CD scales with the CR. You always have to set the X/W sum up (to get the drift angle) so it is easy peasy to read off the ETAS value every time. There is no practical value gained by using the (less correct) approximation (ETAS ~ TAS) for smaller drift angles and it doesn’t save you any time along the way. The correct trigonometric solution for G/S is always G/S = ETAS ± H/W (or T/W), regardless of drift angle, and should be the way you approach the problem every time.

Obviously, you don’t need to know anything at all about the background, behind the scenes, trigonometry. If it is a bit confusing for you, just apply the process via rote learning (providing that you don’t ever make any mistakes in setting up the rote-learned settings). However, for some folks, knowing the background aids understanding of why you are doing what you are doing.

Anyways, you can see, quite easily, what ETAS is in the triangle. It is just a part of the G/S vector. It is the vector shown, no more, no less, and is, quite definitely, not the bogey man which many pilots perceive it to be.

[color=blue]Additionally, I'm seeing formulas for solving ETAS equations that involve COS.[/color]

That’s because the trigonometric calculation for ETAS involves cosine(drift) as shown above. Nice to know but you don’t need to worry about it if it causes you a problem.

[color=blue]How on earth do we employ COS on a non-scientific calculator, which is all we can take into the exam....?[/color]

You don’t need to take a scientific calculator into the exam. The CR already has both the sine and cosine scales built into the revamped CD scales on the triangle side of the device. If you bother learning how they work, that can be put to your benefit for CPL/ATPL work as some questions can be answered quite nicely using basic trigonometry.

[color=blue]I came across a practice question today which asked "If your TAS is 150kt and the crosswind component is 35kt what is the ETAS?"[/color]

This is intended for the CR as below.

[attachment=2488]CR setup 2.jpeg[/attachment]

Set the TAS. The sine calculation gives you the drift (13.5) and the cosine calculation gives you the ETAS (146), as shown.

It is not intended that you do this on the Dalton but you can do so, with a little bit of fiddling, as shown below.

[attachment=2489]Dalton setup 2.jpeg[/attachment]

As you can see you get the same answer, viz., ETAS = 146 kts.

[color=blue]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.[/color]

This statement just has to be incorrect but, until the OP posts some examples for comment, we can’t help him. I’m presuming that it just relates to some level of confusion ?? ETAS is necessary should you be running the G/S solution via the CR instrument. It doesn’t rate a mention when you are using the Dalton.

Further, just because a question’s published answer uses the CR’s solution (or the Dalton’s, for that matter), doesn’t mean that you have to use the CR, (or the Dalton), yourself. Just run the solution with your preferred gadget using the techniques appropriate for that device. You’ll get the same answer to within the practical limits of the instruments’ accuracy, presuming that you execute the solution with appropriate care and diligence.

It doesn’t matter in the slightest whether you run with the CR or the Dalton, although the CR is a tad more useful for the ATPL courses and exams. This is only the case because Lahr incorporated Huber’s TAS computer on the calculation side - which is what makes that side so off-putting when you first start playing with the CR. Once the penny drops that the presentation is just that of two, quite independent, computer setups, life gets back to some sort of equilibrium. As (probably) none of you has ever heard of Huber, it is worth noting that those scales provide values for TAS appropriate to higher Mach numbers when compressibility starts to become a bit of a nuisance. Using the incompressible equations built into the basic Lahr and Dalton computers, the errors arising in CAS from around M0.25 and up have to be addressed by other means of correction.

Caveat - Just don’t get into the habit of throwing your CR up on the coaming in mid-summer, unless you want to be buying CRs by the dozen

As with any long post, it is difficult to make sure that no errors sneak through. If you think I have got something wrong, please do comment accordingly so I can check it out.

If you have any other questions, please do ask them and I'll do my best to provide some sensible answers.

[attachment=2490]Nav Day 2 copy.mp4[/attachment] [attachment=2490]Nav Day 2 copy.mp4[/attachment]

I hadn't seen that previously, Bob, rather cute. However, not much value unless the student has a go, himself/herself, with pencil etc. and a sheet of paper.

I remember doing that on a WAC before my first solo nav in 1960. I thought it was a bit odd that we had to be so super careful in measuring the angles and distances when, in those days, the forecast winds were not much better than some met man's wild guess. Make sure that you convert his true guess into a magnetic guess or you might get lost....

I was a bit cynical in those days.

Hello,
I think you'll arrive based on your speed and conditions. The E6B handles this automatically, but understanding the concept is crucial. For calculations involving COS on non-scientific calculators, use approximations or simpler methods.