GPS

[Fuji Abound]

Its a 400 mile journey with a front in the middle, as you cross the front the wind changes direction by 180 degrees from a cross wind to the right to a cross wind to the left. Which is more efficent, following the magenta line, or some other way?

[Genghis the Engineer]

In 2D, the great circle route - in other words the magenta line, should always be most efficient.

In 3D however, I'd be looking at the winds at different altitudes on a leg like that to see where I can take some advantage by minimising the head/cross wind components.

G

[what next]

Hello!

Intuitively, I would say: Fly the constant heading computed for zero wind condition all the time (assuming that the cross winds are of equal strength and change direction in the middle). The first half of your journey, you will drift in one direction, the second half, the drift will be reversed and you will arrive exactly at your destination. Your velocity vector will point in the direction of your destination all the way and you will not lose effective speed compensating for the crosswind.

max

[BackPacker]

In 2D, the great circle route - in other words the magenta line, should always be most efficient.

Don't agree. If the front is in the middle, and perpendicular to your 400 mile journey, and the crosswind component from the right before passing the front is as strong as the crosswind component from the left after the front, it's better to simply let yourself blow away to the left first, as this will be perfectly compensated later on.

When sailing at sea, you do the same thing. Let yourself wash one way with the tide, and the other way six hours later. Trying to compensate for the tide only costs in distance traveled.

However, practically speaking:
- Winds aloft forecasts are notoriously unreliable. (Tidal flow is much more predictable.)
- The perfect situation as above will almost never appear, which makes the calculation on how much to compensate much harder. It's far easier to follow the magenta line, or calculate your plog so that it follows the imaginary magenta line.
- Airspace, intersections, reporting points, placement of navigation beacons and general predictability of your (filed) route will usually be more important than saving a few liters of fuel.

[Fuji Abound]

;) Bp for the sake of discussion it is the perfect situation you describe and they really do reverse midway on the course equal cross wind from the right to an equal cross wind from the left.

Its an air race point to point and who gets there first wins -oh and yes theoretically speaking same two aircraft, same performance, airspace and weather not a consideration, but you cant change level, cant change the crosswind, you can only change how you navigate.

- yes and I know not very real world!

[dublinpilot]

Damn it's a tough question! I keep changing my mind!

Sounds like a piece of paper and a calculator is called for!

[peterh337]

The fact is that any crosswind component means you fly a longer distance, unless there is a tailwind component to it which is sufficient to compensate. So even a perfect 90 degree crosswind means you have to fly further than the GC route.

It thus follows that if you get a perfect 90 deg c/w from the right and then get a perfect 90 deg c/w from the left, you end up flying a longer distance to get there.

I don't know the answer, and I am too lazy to try some trig, but I think just following the magenta line (and thus crabbing the wind) will be most efficient because the aircraft is flying the shortest route and thus the wind has the least opportunity to "work" on it.

WN's suggestion of a constant zero-wind heading looks tempting and could well be right, but I feel that the aircraft will be subjected to the effect of the wind for longer.

[IanSeager]

There's a bell ringing somewhere in my head and it is saying something about 'pressure pattern navigation'

I can't recall the exact details, but I'm pretty sure it was about this kind of thing.

Ian

[Torque Tonight]

Fly a constant heading (equal to the track from the start to the end point) and accept the resultant drift. If you correct heading for drift to maintain the track line, you will have a permanent headwind component and get there slower.

[fujii]

This is a GPS question and has nothing to do with drift. The magenta line is your track over the ground. It's that simple. The heading on the compass/DG will change as you pass through the front but provided you stay on the line, your track remains the same. It's a good idea to always keep an eye on the compass/DG heading so if the GPS dies, you have a reference to maintain track.

[peterh337]

If you correct heading for drift to maintain the track line, you will have a permanent headwind component and get there slower.That's true, but if you fly a constant heading all the way you will end up flying for longer, because the wind blows you increasingly off track on the first leg, and the second leg does not compensate for that; it merely puts you back where you should be, after having spent more time in the air.

The answer to this puzzle will be interesting.

A google finds e.g. this (http://williams.best.vwh.net/smxgigpdf/smx2001c.pdf) which makes sense but I am not sure it is applicable to the question as originally posted, which is a different scenario. It is obvious that if you are crossing a high or a low then flying to one side of it will help pick up a bit of tailwind.

[IanSeager]

That's what I was thinking of Peter.

Ian

[peterh337]

I would put a tenner on the two methods (constant zero-wind heading, and flying the GC track) being exactly the same :)

I reckon both hang on the small angle approximation (sin(x)=x).

[Morrisman1]

the way I see it is we are essentially asking the question:

Is the ground speed advantage by allowing the wind to push us off course sufficient to offset the extra distance we will travel across the ground?

I dont doubt for a second that we wont have a faster ground speed by not accounting for drift, in the end when we account for drift we are flying with more of a headwind (or less of a tailwind) than what we would if we just flew the heading. If we think about the wind triangle, it takes us the same time to travel the adjacent line as it would the hypotenuse. By choosing to stay on track we fly the adjacent line, if we let ourselves drift then we will fly the hypotenuse but the time spent is the same. Hypotenuse is longer and that reflects the extra groundspeed we will have from experiencing less headwind. The second half of the journey mathmatically is exactly the same as the first, just mirrored the situation.

I think that we will end up at the same place in the same time.

[seat 0A]

Very similar to sailing.
As you sail from Holland to England, you encounter drift from the flood- and ebstreams. The direction of the stream changes every 6 hours or so.
The fastest way is not to correct for the drift. Just let the wind (or stream) set you off track. It will blow you back on track during the second half.
Saves you beating into the wind on both sides of the front.

[bookworm]

I would put a tenner on the two methods (constant zero-wind heading, and flying the GC track) being exactly the same

I thought the same for a while but having done the trig I'm now with the sailors!

Constant zero-wind heading gets you there in the zero-wind time (the wind component just cancels). Flying the GC track takes a factor of 1/cos(drift) longer (that's the factor by which your groundspeed is reduced). That means of course that the result is second order in the drift (i.e. if the drift is 0.1 or 6 degrees, the difference is only 1%).

In the real atmosphere of course, nothing is predictable enough to make this a sensible strategy. But a fun question, Fuji, thanks. :)

[Fuji Abound]

Bookworm thank you - anyone want to hazard the difference in arrival times with say a 20 knot x wind component, cruising at 160 knots for the entire course - It is very surprising.

[Torque Tonight]

Although flying the straight ground track gives you the shortest ground distance, flying the heading and accepting drift gives you the fewest air miles. Think about your movement relative to the two airmasses.

Fuji, at a guess I'd say it would be about 5 minutes slower to follow the magenta line. If I have nothing more important to do l'll do the maths later.

[Whopity]

Which is more efficentHow do you define efficiency? In the RAF when confronted with Wind problems, typically crossing the Atlantic, we would look at Specific Fuel Consumption in pounds per mile. Quite often the longer distance at a higher speed gave a better figure however; at the end of the day it depends upon the relative wind speed which is not given. The sailors solution is based upon largely known and predictable conditions.

[Mark1234]

My take: While it's very tempting to concern ones self with ground distance, it's irrelevant. Aeroplanes fly in air, not ground! It's all about frames of reference.

Consider your track through the body of air in which you fly. If you just pick a heading and go, it's a straight line.Conveniently the body moves left and right sufficiently to put you back where you need to be at the end of the line.

If you crab and fly the magenta line, your groundspeed decreases, you fly a longer, squiggly path with reference to the body of air (which is the bit that matters).

[BackPacker]

Okay, here's the math.

Suppose an east-west front oriented along the 'x' axis and passing through the center of your (two-dimensional, carthesian) world. Your journey starts 200 miles south of the front (which is 0, -200) and ends 200 miles north (0, 200). There is a 20-knot x-wind to the left for the first bit, and 20-knot x-wind to the right for the second bit. Airspeed is 100 knots.

If you do not compensate for the x-wind but simply fly a magnetic heading of due north, you end up two hours later crossing the front at the (-40,0) position, and yet another two hours later at your destination (0, 200). The crosswinds have cancelled each other out, and the distance traveled through the air is 400 miles. So a four-hour journey. (Ground speed, by the way, is 102 knots so the distance traveled over the ground is 408 miles.)

If you do compensate for the x-wind you've got to compensate by approximately 12 degrees to the right on the first bit, and approximately 12 degrees to the left on the second bit. (Approximately 12 degrees is calculated by the 1:60 method. In reality sin(angle) = 20/100, or angle = 11.53 degrees.)

So true heading is 012 degrees, while true track is 000 degrees. Ground speed is 100 * cos(12) = 97.81 knots.

So you cross the front at (0,0) after 200/97.81 = 2.0447 hours (2 hours and 2.6 minutes). After the front the same calculation applies, but you fly a heading of 348 degrees. And another two hours and 2.6 minutes later you reach your destination (0,200). Total flight time four hours and 5.1 minutes.

So in this idealized situation not compensating for the crosswind is more efficient, though only marginally so. In practice, like I said before, considerations like airspace, navaids, predictable flight paths for ATC and so forth will be more important than shaving the last five minutes of flight time off a four hour journey.

On the other hand, the stronger the crosswind in relation to TAS, the greater the difference. In the ultimate case, suppose the crosswind is equal to your TAS. If you compensate for the crosswind you need to put your nose directly into the wind, and will not get anywhere. You will simply "hover" above your departure airfield until the fuel runs out. Whereas if you just point your nose to true north, you cross the front at the (-200,0) position (crosswind drift to the west being equal to the distance traveled north), and you drift over your destination (0,200) again exactly two hours later. (How you take-off and land in those conditions is an interesting one, but not relevant to the problem.)

[Torque Tonight]

If you swim across a flowing river you point perpendicular to the river's axis and allow yourself to drift downstream. If you aim to swim to the point on the opposite bank... you end up drowning.

Thinking of the situation where the speed of flow of the medium in which you are travelling (ie the windspeed or the river's flow) is a large proportion of your speed or that of your vehicle (airspeed or swimming speed) will help clarify the effects.

[Torque Tonight]

Post 18 I estimated 5 minutes. Backpacker just calculated 5.1 minutes. I thank you.:D

[BackPacker]

Torque, not so fast. Your assumption was based on Fuji's 160 knots airspeed, while my calculation was based on 100 knots.

So you overestimated things by about half...;)

[what next]

...anyone want to hazard the difference in arrival times with say a 20 knot x wind component, cruising at 160 knots for the entire course - It is very surprising.

Surprising indeed, as the difference is really minimal!

For your original problem: Distance=400NM, TAS=160KT, Crosswind of 20KT the groundspeeds (rounded to full knots) will be 161 and 159 for constant heading and constant track, resulting in a time difference of 1 minute and 53 seconds (for a still-air flying time of exactly 2h 30m). Interestingly, allowing yourself to drift will actually shorten your trip time by a few seconds compared to the no-wind case!

To better visualise the effect, you can use the extreme example of an TAS of 100kt into a 100kt crosswind! Trying to follow your magenta line will result in an infinite trip time (limited only by your fuel supply) wheres drifting with the wind will take you to your destination even fasten than in still air...

(BTW: Thanks for this interesting topic, it refreshes some grey memory cells from the CPL course, where the subject was called "effective true airspeed" :) )

[BackPacker]

wheres drifting with the wind will take you to your destination even fasten than in still air...

...as fast as in still air...

Fixed that for you.;)

[Torque Tonight]

Backpacker: :{

[peterh337]

So the difference is just a few seconds?

That would make sense, because it is "obviously" not a first order effect, and the drift for a 20kt x/w against a 160kt TAS is only slightly over 5 degrees.

[BackPacker]

Yes, that's my conclusion. If the crosswind component of the wind is less than about a quarter of your airspeed, the difference between the two different tactics is really low and will probably disappear in the inaccuracies of your flying and the forecast anyway. Only the headwind or tailwind component of the crosswind will make a difference to your ground speed and flight time.

[Fuji Abound]

Thanks for all your input.

Astute of those who mentioned the sailing connection which indeed promted my original question.

As is now clear the higher your speed relative to the "drift" the less impact on the final outcome. For sailors the difference can be significant.

For ease of calculation, imagine crossing Solent to Cherbourg (180 deg) in a boat doing 5 knots. Assume that the tide is doing 4 knots, initially on 90 deg, then 6 hours later it changes to 270 deg. (Yes, I know that tides in the Channel don't reach 4 knots, and that they don't have a step change in direction, but the effects of the real tide will be the same.)

If you plot a tidal triangle such that your COG is 180, then you will make the classic 3-4-5 triangle: a 5 knot hypotenuse (boat speed), a 4 knot vector for tidal set, resulting in a 3 knot vector for boat speed over the ground. You will be travelling along a straight line between the Needles and Cherbourg, but your bow will be pointing at about 233 degrees, well off that line. After 6 hours, you will have got 18 miles closer to France.

If instead you point your boat's head at 180, then of course you will be swept up-channel. After 6 hours you will have been swept 24 miles off your direct course, but you will be 30 miles further south.

Now the tide turns. If you have been staying on the direct line, then you will now need to counter the tide in the opposite direction. You'll have to change your heading from about 233 deg to about 127 deg. You'll stay on the line, but you'll still only close France at 3 knots. After another 6 hours, you will have made 36 miles of the 60 mile distance.

What about the boat that was swept up-Channel? She keeps her head on 180 deg, continues to close France at 5 knots, but is now being swept down-Channel. After another 6 hours she has been swept back onto the direct line, but has now made her full 60 miles cross-Channel. She should be just entering Cherbourg Harbour, with the restaurants open and waiting.

Of course the same could be true more or less of slower aircraft - a microlight or a 152 enjoing a cross country with a 30 knot crosswind would notice a significant difference albeit in the aviation world wind reversal along a track are very rare.

Never the less it is interesting I think that what we intuitively believe is happening (in terms of the preferred strategy of following the magenta line) is in fact the wrong strategy even though the difference is small compared with sailinga yacht.

Thank you all again for an interesting debate - I have got another one for you along a similiar theme shortly. :)

[Genghis the Engineer]

As an avid Doctor Who fan of 30+ years standing, I am well aware that there is a fundamental difference between time and space. We all get to travel in both of-course, although time generally in one direction.

The tide issue adds in time, which does not apply to the frontal system which is assumed fixed in space.

So, following the magenta line, time does not affect the wind (apparently nor does height, which is clearly unrealistic), only distance.

However, at sea, time affects the tide. So, you know that the tide will change in a particular way at a particular time. So, you can allow yourself to be swept downstream, knowing that you'll come back this way when the tide changes.


Coming back to the magenta line through a front - constant heading is clearly not the best route and unless you are really clever with your maths, will not get you where you want to be. So, time to destination is infinite. However, constant TRACK, requiring heading changes whenever the wind changes, should still in my opinion be the fastest route from A to B.

G

[peterh337]

A helpful way to work out a lot of puzzles is to consider extreme cases.

Take the case of a 20kt current and a 2kt boat trying to cross the river. It will obviously never get to the opposite point on the other bank.

So the solution must hang on the drift being in some way "less than very significant".

And as soon as it is less than significant, it seems obvious that the two methods are going to yield virtually identical results.

[Fuji Abound]

Genghis - now no wriggling, the difference may be small but it doesnt make it any less real. Whether it is significant is academic as that was not the question, just whether there was a difference. :)

As is so often the case what we do in the real world maybe different for all sorts of reasons - the sailor may follow the magenta line because he couldnt plot a tidal stream if his life depended on it (and yes, they really do exist in this age of GPS), and the aviator may follow the magenta line for other reasons.

[24Carrot]

A footnote on calculating this in the cockpit:

Backpacker wrote:
So true heading is 012 degrees, while true track is 000 degrees. Ground speed is 100 * cos(12) = 97.81 knots.

The quick-and-dirty mental approximation for this, ( 1-cos(angle) ), is:
Percentage Speed reduction = (angle / 8) squared.

So (12/8) x (12/8) = 2.25%

For small numbers the percentage speed reduction is roughly the same as the percentage time increase, so the extra time is:
120 minutes x 2.25% = 2.7 minutes,
versus the 2.6 minutes in Backpacker's exact calculation.

In practice the effect is small, so the calculations can be rounded up to make the maths easier, to get a "worst case" arrival delay which is still typically small.

[stickandrudderman]

Fuji, great post thanks. However, one minor point is that the tide in the channel can reach up to 7kts in places! It's a long while since I did any sailing but it was good to re-awaken the thought process of passage planning.