If we mark up the previous graphic with the various quantities, it becomes a bit easier to follow –
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
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
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.
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.
Additionally, I'm seeing formulas for solving ETAS equations that involve COS.
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.
How on earth do we employ COS on a non-scientific calculator, which is all we can take into the exam....?
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.
I came across a practice question today which asked "If your TAS is 150kt and the crosswind component is 35kt what is the ETAS?"
This is intended for the CR as below.
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.
As you can see you get the same answer, viz., ETAS = 146 kts.
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.
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.
