Wind Triangle Vectors & Curved Trajectories

[Badmachine]

Can Wind Triangle vector calculations apply to curved trajectories or turns?

[ImbracableCrunk]

This sounds like something for Newton and Leibniz.

What would be the application? Are you wanting one calculation for a curved path? A unified howgozit with one entry?

I'll gladly defer to the smarter people.

[5LY]

Why would you need to? Just turn to the calculated heading required to fly your next track. You could do a seperate calculation for each degree of the turn but it would be meaningless. You'd turn through them as you went.

[Badmachine]

For example:

Plane 1 attempts to fly from point A to point B, but drifts laterally "X" feet due to crosswind by the time it was to have reached point B.

Plane 2 then attempts to fly an RF leg from point A to point B that would reach point B under zero crosswind, but experiences the same crosswind as Plane 1. Does Plane 2 drift laterally by the same distance as Plane 1?

[PBL]

The answer to "do wind vector calculations work for curved tracks?" is: it depends on the calculation.

The straight answer to your question
Plane 1 attempts to fly from point A to point B, but drifts laterally "X" feet due to crosswind by the time it was to have reached point B.

Plane 2 then attempts to fly an RF leg from point A to point B that would reach point B under zero crosswind, but experiences the same crosswind as Plane 1. Does Plane 2 drift laterally by the same distance as Plane 1?

The crucial parameter is the time taken, not the position of A and B. If Plane 2 aims to fly its RF leg in exactly the same time as Plane 1, the drift will be the same. If not, the drift will be different.

Basic analytic geometry, of the sort needed for basic calculus, yields the following.

Suppose your position at time t in zero wind is given as <x(t),y(t)> in rectilinear coordinates, i.e. for a flat plane, not lat/long.

Suppose the wind is coming at w kts from z°N, where z for simplicity is between 0 and 90. Then the position at which you need to aim in this wind condition in order to fly the track <x(t),y(t)> is given by <x1(t),y1(t)>, where

x1(t) = x(t) + t.w.sin(z)
y1(t) = y(t) + t.w.cos(z)

If you've done basic calculus, which often includes trajectory calculations, then this is simply a reminder. If you haven't done basic calculus, then this might look like mumbo-jumbo, but I can assure you it's the math you need.

It follows from this that at time t1, your aimed position <x1,y1> is displaced from your desired position <x,y> by <t.w.sin(z),t.w.cos(z)>, which is, expressed in vectors, just

desired position = aimed position + wind-vector.t.

In that sense the wind-vector calculation "works" for a curved track.

PBL

[BOAC]

BadMachine - not having a clue what an 'RF leg' is, I would answer #4 with yes, the same. Maybe if I knew what you meant by 'RF' I might change that answer. :)

As IC said, perhaps it would help us to give you a sensible answer if you could explain what you were seeking to do/prove/develop/solve - or, even what you actually do/what your interest in aviation is?

[PBL]

not having a clue what an 'RF leg' is

Radius-to-Fix leg. Part of RNP. E.g. Jeppesen Briefing Bulletin 05-03 (http://www.jeppesen.com/documents/aviation/notices-alerts/hubwatch/BriefingBullentins/bulletin_jep05-03_RNP.pdf)

I would answer #4 with yes,

And you would mostly be wrong, wouldn't you? Make that Theory 1, Practice O. Or have I misunderstood what you meant?

PBL

[BOAC]

And you would mostly be wrong, wouldn't you? Make that Theory 1, Practice O. Or have I misunderstood what you meant? yes - selective part-quoting again (quite a talent) - you did not read my post. Try the first line again?

New answer = probably not. (New question. Why on earth would anyone want to fly around an arc to go from A to B and increase route distance by 3.142xxx when we know there is a straight line route??)

Too may variables.

BadMachine at what level is your understanding of basic navigation? Do you understand that cross-track error due to drift is dependent on flight time?As IC said, perhaps it would help us to give you a sensible answer if you could explain what you were seeking to do/prove/develop/solve - or, even what you actually do/what your interest in aviation is?

[BOAC]

I do hope you have not scared the OP away.

[Pugilistic Animus]

PBL parametric analysis is not basic calculus:\

:)

[Pugilistic Animus]

I did not like to annoy my teachers--- my teachers thought I was overdoing things:}

I am a teacher:ooh:


as far as this discussion goes---like George Bush the First Na Ganna Duit--not prudent:}:}:}

further I ain't got nothin' against not no one here:)


:8

[Badmachine]

Thanks PBL.

So, if I understand correctly in general, an RF leg (of equal speed to a straight leg) would experience greater drift due to the greater distance traveled and time spent traveling?

[PBL]

You're welcome, Badmachine, glad I was able to help.

an RF leg (of equal speed to a straight leg) would experience greater drift due to the greater distance traveled and time spent traveling?

Yes.

If I had wrote this sentence, though, I would insert the word "therefore" between "and" and "time".

PBL

[DFC]

Why would you try to apply the triangle of velocities to the RF leg on an RNAV procedure?

Does anyone apply the triangle of velocities to a DME arc?

In both cases, the aircraft has positive course guidance and has instant feedback of any errors. All that is required is to correct the error and maintain the required (curved) track.

The simple answer is that on the perfect RF leg or the perfect DME arc there will be no difference between the required track and the track made good.

You could at any point check your heading and compare this to the track of a tangent to the curve at that point. This would be your (instantaneous) drift angle. At the same instant you could compare your TAS and GS. That would give you all the elements of the triangle of velicities for that instant. 1 second later they will be different.

Remember that a RF leg or DME arc are not the same as say the putbound turn in a hold/racetrack because in the hold/ racetrack turns there is no guidance and no allowance for wind can be made until the turn is completed.

if you want to for example calculate a time and fuel burn for a curved track eg DME arc then average the initial and final GS and fuel flow and calculate the length of the arc using basic circle formula. Must be goof enough for the short legs that normally apply.

Alternatively simply make a block allowance for the approach procedure (as most people do).

Why complicate things?

[compressor stall]

Why complicate things? Because it's interesting... :cool:

Recently I had to do some sums to calculate the flight path to do a 90 degree (plus or minus WCA) rolling wings level at an exact lat/lon at an exact heading at a time as accurately as possible.

I calculated the nil wind turn, then backcalculated the entry to the turn. Then I moved this into wind by the appropriate vector to get the entry point.

That's the easy bit.

You have to also consider:

Angle of Bank - in an Airbus you'll get a different angle of bank if you are in HDG or NAV...

Which then follows on to the roll rate into the turn... What is the roll rate? How does this affect the radius for the few seconds it takes to get the normal angle of bank?

It gets very complicated very quickly.

The post flight GPS feed has shown that I was 300 metres out, at 37,000 feet at M.08! The pax were happy (http://www.wired.com/wiredscience/2010/07/eclipse-chasers/) with that :8

[DFC]

I calculated the nil wind turn, then backcalculated the entry to the turn. Then I moved this into wind by the appropriate vector to get the entry point.




One can not use that to complete either an RF turn or a DME arc.

The reason is that while it is possible to apply your technique and end up at the exact end point in the turn, the application of the complete correction at the start of the turn (and then letting the wind do it's thing to put you exactly back on at the end) could cause the aircraft to leave the protected airspace at the start and during the initial parts of the turn.

Imagine a straight run into a 90 degree turn about a point (RF) ending up at a specific point.

If say in zero wind, the aircraft flies nicely round the turn with 20 degrees of bank. Perfect.

With a headwind, the bank angle will be less.

With a tailwind the bank angle will be more.

In all 3 cases thus far the aircraft will follow the exact same ground track from the start to the finish of the turn.

----

Now let's try your idea.

Again in zero wind, the aircraft flies nicele round the turn with 20 degrees of bank. Perfect.

Now with a headwind, you delay the start of the turn which is again completed with 20 degrees of bank and thanks to the late start you end up finishing the turn at the exact end point.

However, by delaying the turn, the ground track will not be the same as the no wind case and thus will not initially follow the required procedure.

With a tailwind, you will start the turn early and again maintain 20 degrees angle of bank which thanks to the wind will cause you to end the turn at the exact end point.

in this case, by starting the turn early the ground track will not be the same as the no wind case and thus will not follow the published procedure.

DME arc's and RF parts of a procedure are required ground tracks to be followed as closely as possible i.e. not started early or late but simply followed.

[compressor stall]

One can not use that to complete either an RF turn or a DME arc.


I never made mention of it! My post was an anecdote as to why some people might be interested in delving deeper into the maths.

[Badmachine]

In theory, could one perform a separate vector calulation for each degree of turn traveled, then calculate the average lateral displacement of all degrees of turn traveled to obtain a correct lateral displacement figure?

[BOAC]

If one were a truly anal pilot, yes. We more normal pilots prefer to think that when we are flying in a 40kt for 15 minutes we are displaced 10 miles downwind from where we would have been in still air - so we aim 10 miles the other way. Simple but effective. Very little trigonometry/calculus/transformation required. Even easier for magenta-liners as the software does it for you.

[jonseagull]

Chaps,

soory to dumb things down a bit here, but, I want to use an xl spread sheet to produce an OFP/Plog call it what you will. What are the equations I need to calculate a simple heading, drift and G/S, assuming I know track, TAS and W/V?

[BOAC]

Try it yourself with a piece of paper - it is simply cosines and sines - I assume you understand simple velocity triangles and basic trig? If not PM me or Google them.

REMEMBER HOWEVER:

Correct IAS for altitude to get TAS
Correct W/V for variation (met winds are true) unless you will be using true headings.

Cannot see why you need drift on a plog when you have heading and track?

I have always found it easier to use 'sheet 2' of a spreadsheet to do mathematics on values so that sheet 1 has only the corrected/derived values - far tidier!

[Badmachine]

One example would be:

P (plane): approximate compass bearing 209° (traveling approximately southwest) at 272 f/s (185 mph); W (wind): traveling south at 32 f/s (22 mph).

Plane and wind vector components represented by ordered pairs:

P = [272 f/s cos(241°), 272 f/s sin(241°)] = -131.8, -237.9
W = [32 f/s cos(270°), 32 f/s sin(270°)] = 0, -32

-131.8 + 0 = -136
-237.9 + (-32) = -269.9

Resolved components substituted into Pythagoreans theorem for resultant speed:

||P + W|| = 131.8² + 269.9² = 90,2171/2 = 300.4 f/s (205 mph)

Resolved components substituted for resultant bearing:

tan −1(269.9/131.8) = 64°; (90° - 64°) + 180° = 206°

Deflection angle = 209° - 206° = 3°

Ground track displacement = 3°tan(300.4 f/s) = 15.7 f/s

[BOAC]

I take it you are not a pilot, BM - we start knowing where we want to go (Track) and work out a heading to achieve that. You appear to be starting with a heading and trying to see where the wind eventually blows you (mind you, I've know pilots like that:)).

If you want any Excel stuff PM me an email address.

[litinoveweedle]

I am long time pprune reader, but I've never took enough courage o register. :O
But today I've found this topic, when I've been searching whole day for the similar answer :8

Can Wind Triangle vector calculations apply to curved trajectories or turns?

Is there universal formula to calculate trajectory of ROT affected by constant wind?

I am currently programming GNSS HT1000 simulation and I need to decide when (at which acft position) Fly-by WPT should be marked as overflown and GNSS should start commanding AP to fly next leg.

Here is what I have for fly-by WPT:

- inbound/outbound tracks over WPT
- aircraft tas, hdg, track
- wind vector (speed + direction)
- RNAV turn are designed as ROT(3deg/s) up to max bank of 25deg.

I am able to do calculation based just on ROT, I am able to compensate initial and final track/hdg for wind, but when it come to wind effect to whole trajectory during the turn, I am lost. :ugh:

Please for now forget about RNAV protected areas and RNP Fix tolerance limits, as first I need to know trajectory, then I can check, if it complies with them.

There is 5sec allowance for acft to establish required bank at the beginning of the turn, and then similar time for acft to return to wing level - For simplification please disregard both.

I would welcome any help and/or comment, thank you.