GPS
Join Date: Feb 2007
Location: Amsterdam
Posts: 4,598
Likes: 0
Received 0 Likes
on
0 Posts
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.)
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.)
Join Date: Apr 2009
Location: UK
Posts: 816
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Join Date: Aug 2000
Location: Near Stuttgart, Germany
Posts: 1,098
Likes: 0
Received 1 Like
on
1 Post
.
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"
)
..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.
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"
)
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Join Date: Feb 2007
Location: Amsterdam
Posts: 4,598
Likes: 0
Received 0 Likes
on
0 Posts
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.
Thread Starter
Join Date: May 2001
Location: UK
Posts: 4,631
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
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
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
Join Date: Dec 2011
Posts: 2,460
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Thread Starter
Join Date: May 2001
Location: UK
Posts: 4,631
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
Join Date: Jan 2008
Location: London UK
Posts: 517
Likes: 0
Received 0 Likes
on
0 Posts
A footnote on calculating this in the cockpit:
Backpacker wrote:
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.
Backpacker wrote:
So true heading is 012 degrees, while true track is 000 degrees. Ground speed is 100 * cos(12) = 97.81 knots.
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.
Join Date: Aug 2006
Location: Sth Bucks UK
Age: 60
Posts: 927
Likes: 0
Received 0 Likes
on
0 Posts
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.
