Mixing
How to model an aviation holding pattern mathematically?
$begingroup$
A standard aviation holding pattern has four sections, each of which, in windless conditions, takes one minute to fly. The first is the inbound leg, which is the only one for which precise navigational information is available. On this leg, the pilot can determine the wind correction angle (henceforth $theta_{in}$) needed to produce a straight ground track.
The pilot then flies the outbound turn; the outbound leg, during which the pilot chooses some wind correction angle $theta_{out}$; and then the inbound turn. There is no possibility of wind correction during turns, because they are timed turns: The pilot applies a bank angle that will produce 180 degrees of rotation in one minute, and then holds that bank for one minute. If the pilot tried to use a different bank angle, there would be no way to know when the turn had proceeded through 180 degrees.
Here is a diagram of the holding pattern, courtesy of pilotlist.org:
In this diagram, the wind is coming from the south (assuming north is up). On the inbound leg, if the pilot pointed the aircraft due east, it would actually travel north of east. So the pilot has to point the aircraft south of east, to achieve a due-east ground track. The wind then shortens the outbound turn and expands the inbound turn, so the ground track of the outbound leg must be flown south of west, to connect the end of the outbound turn with the beginning of the inbound turn. Also note that the pilot must point the aircraft even further south, to correct for wind and fly the desired ground track. (If the wind was from the north, this would be reversed.)
Pilots use some rule of thumb to choose $theta_{out}$. The diagram shows the pilot doubling $theta_{in}$; many pilots also triple it. After flying the outbound leg and inbound turn, the pilot will discover that they are too far north or south, and will fly back to the correct position (which they can do because precise navigation is available on the inbound leg).
My question is: What is the value of $theta_{out}$ in terms of $theta_{in}$?
Edit: j.k. points out that the length of the legs must be considered. The outbound leg is the only one that should change. The two turns are necessarily one minute each, and the goal is to fly an inbound leg of exactly one minute. In practice, this is done by timing the inbound leg the first couple times around. So for example, if you flew a one minute outbound leg and find that the inbound leg takes 45 seconds, it means you should fly a longer outbound leg - maybe 1:20 or so. So the revised question is: Given $theta_{in}$ and $t_{in}$, can we determine $theta_{out}$ and $t_{out}$?
trigonometry mathematical-modeling
edited Jan 25 at 7:52
Glorfindel
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
$endgroup$
add a comment |
$begingroup$
A standard aviation holding pattern has four sections, each of which, in windless conditions, takes one minute to fly. The first is the inbound leg, which is the only one for which precise navigational information is available. On this leg, the pilot can determine the wind correction angle (henceforth $theta_{in}$) needed to produce a straight ground track.
The pilot then flies the outbound turn; the outbound leg, during which the pilot chooses some wind correction angle $theta_{out}$; and then the inbound turn. There is no possibility of wind correction during turns, because they are timed turns: The pilot applies a bank angle that will produce 180 degrees of rotation in one minute, and then holds that bank for one minute. If the pilot tried to use a different bank angle, there would be no way to know when the turn had proceeded through 180 degrees.
Here is a diagram of the holding pattern, courtesy of pilotlist.org:
In this diagram, the wind is coming from the south (assuming north is up). On the inbound leg, if the pilot pointed the aircraft due east, it would actually travel north of east. So the pilot has to point the aircraft south of east, to achieve a due-east ground track. The wind then shortens the outbound turn and expands the inbound turn, so the ground track of the outbound leg must be flown south of west, to connect the end of the outbound turn with the beginning of the inbound turn. Also note that the pilot must point the aircraft even further south, to correct for wind and fly the desired ground track. (If the wind was from the north, this would be reversed.)
Pilots use some rule of thumb to choose $theta_{out}$. The diagram shows the pilot doubling $theta_{in}$; many pilots also triple it. After flying the outbound leg and inbound turn, the pilot will discover that they are too far north or south, and will fly back to the correct position (which they can do because precise navigation is available on the inbound leg).
My question is: What is the value of $theta_{out}$ in terms of $theta_{in}$?
Edit: j.k. points out that the length of the legs must be considered. The outbound leg is the only one that should change. The two turns are necessarily one minute each, and the goal is to fly an inbound leg of exactly one minute. In practice, this is done by timing the inbound leg the first couple times around. So for example, if you flew a one minute outbound leg and find that the inbound leg takes 45 seconds, it means you should fly a longer outbound leg - maybe 1:20 or so. So the revised question is: Given $theta_{in}$ and $t_{in}$, can we determine $theta_{out}$ and $t_{out}$?
trigonometry mathematical-modeling
edited Jan 25 at 7:52
Glorfindel
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
$endgroup$
$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43
add a comment |
$begingroup$
A standard aviation holding pattern has four sections, each of which, in windless conditions, takes one minute to fly. The first is the inbound leg, which is the only one for which precise navigational information is available. On this leg, the pilot can determine the wind correction angle (henceforth $theta_{in}$) needed to produce a straight ground track.
The pilot then flies the outbound turn; the outbound leg, during which the pilot chooses some wind correction angle $theta_{out}$; and then the inbound turn. There is no possibility of wind correction during turns, because they are timed turns: The pilot applies a bank angle that will produce 180 degrees of rotation in one minute, and then holds that bank for one minute. If the pilot tried to use a different bank angle, there would be no way to know when the turn had proceeded through 180 degrees.
Here is a diagram of the holding pattern, courtesy of pilotlist.org:
In this diagram, the wind is coming from the south (assuming north is up). On the inbound leg, if the pilot pointed the aircraft due east, it would actually travel north of east. So the pilot has to point the aircraft south of east, to achieve a due-east ground track. The wind then shortens the outbound turn and expands the inbound turn, so the ground track of the outbound leg must be flown south of west, to connect the end of the outbound turn with the beginning of the inbound turn. Also note that the pilot must point the aircraft even further south, to correct for wind and fly the desired ground track. (If the wind was from the north, this would be reversed.)
Pilots use some rule of thumb to choose $theta_{out}$. The diagram shows the pilot doubling $theta_{in}$; many pilots also triple it. After flying the outbound leg and inbound turn, the pilot will discover that they are too far north or south, and will fly back to the correct position (which they can do because precise navigation is available on the inbound leg).
My question is: What is the value of $theta_{out}$ in terms of $theta_{in}$?
Edit: j.k. points out that the length of the legs must be considered. The outbound leg is the only one that should change. The two turns are necessarily one minute each, and the goal is to fly an inbound leg of exactly one minute. In practice, this is done by timing the inbound leg the first couple times around. So for example, if you flew a one minute outbound leg and find that the inbound leg takes 45 seconds, it means you should fly a longer outbound leg - maybe 1:20 or so. So the revised question is: Given $theta_{in}$ and $t_{in}$, can we determine $theta_{out}$ and $t_{out}$?
trigonometry mathematical-modeling
edited Jan 25 at 7:52
Glorfindel
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
$endgroup$
A standard aviation holding pattern has four sections, each of which, in windless conditions, takes one minute to fly. The first is the inbound leg, which is the only one for which precise navigational information is available. On this leg, the pilot can determine the wind correction angle (henceforth $theta_{in}$) needed to produce a straight ground track.
The pilot then flies the outbound turn; the outbound leg, during which the pilot chooses some wind correction angle $theta_{out}$; and then the inbound turn. There is no possibility of wind correction during turns, because they are timed turns: The pilot applies a bank angle that will produce 180 degrees of rotation in one minute, and then holds that bank for one minute. If the pilot tried to use a different bank angle, there would be no way to know when the turn had proceeded through 180 degrees.
Here is a diagram of the holding pattern, courtesy of pilotlist.org:
In this diagram, the wind is coming from the south (assuming north is up). On the inbound leg, if the pilot pointed the aircraft due east, it would actually travel north of east. So the pilot has to point the aircraft south of east, to achieve a due-east ground track. The wind then shortens the outbound turn and expands the inbound turn, so the ground track of the outbound leg must be flown south of west, to connect the end of the outbound turn with the beginning of the inbound turn. Also note that the pilot must point the aircraft even further south, to correct for wind and fly the desired ground track. (If the wind was from the north, this would be reversed.)
Pilots use some rule of thumb to choose $theta_{out}$. The diagram shows the pilot doubling $theta_{in}$; many pilots also triple it. After flying the outbound leg and inbound turn, the pilot will discover that they are too far north or south, and will fly back to the correct position (which they can do because precise navigation is available on the inbound leg).
My question is: What is the value of $theta_{out}$ in terms of $theta_{in}$?
Edit: j.k. points out that the length of the legs must be considered. The outbound leg is the only one that should change. The two turns are necessarily one minute each, and the goal is to fly an inbound leg of exactly one minute. In practice, this is done by timing the inbound leg the first couple times around. So for example, if you flew a one minute outbound leg and find that the inbound leg takes 45 seconds, it means you should fly a longer outbound leg - maybe 1:20 or so. So the revised question is: Given $theta_{in}$ and $t_{in}$, can we determine $theta_{out}$ and $t_{out}$?
trigonometry mathematical-modeling
trigonometry mathematical-modeling
edited Jan 25 at 7:52
Glorfindel
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
edited Jan 25 at 7:52
Glorfindel
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
edited Jan 25 at 7:52
Glorfindel
3,42581830
edited Jan 25 at 7:52
Glorfindel
3,42581830
edited Jan 25 at 7:52
Glorfindel
3,42581830
3,42581830
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
asked Nov 11 '13 at 6:00
Graham MainwaringGraham Mainwaring
112
112
$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43
add a comment |
$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43
$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43
add a comment |
1 Answer
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oldest
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$begingroup$
Note that $theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $theta_{out}=theta_{in}$
The case of very long legs is similar to this: $theta_{out}=theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=sin theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (theta_{in} + theta_{out}) / 180$ and
$t_{Tin} = 1 + (theta_{in} + theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(cos theta_{in}, sin theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + cos theta_{in}, 3w_2 - sin theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
alpha_{out} = arctan frac{ 3w_2 - sin theta_{in} }{ 3w_1 + cos theta_{in} }
$$
Note this is not $theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $alpha_{out}$,
and he is able to determine corresponding $theta_{out}$. I am lazy
to do the calculation of the corresponding $theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - cos theta_{out} + w_1, - sin theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = frac { 3w_1 + cos theta_{in} }{ - cos theta_{out} + w_1 }
= - frac { 3w_2 - sin theta_{in} }{ - sin theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $theta_{in}=w_2$ and
$
theta_{out} = alpha_{out} = 2 theta_{in}
$.
answered Nov 11 '13 at 6:18
j.k.j.k.
663
$endgroup$
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
add a comment |
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1 Answer
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$begingroup$
Note that $theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $theta_{out}=theta_{in}$
The case of very long legs is similar to this: $theta_{out}=theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=sin theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (theta_{in} + theta_{out}) / 180$ and
$t_{Tin} = 1 + (theta_{in} + theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(cos theta_{in}, sin theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + cos theta_{in}, 3w_2 - sin theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
alpha_{out} = arctan frac{ 3w_2 - sin theta_{in} }{ 3w_1 + cos theta_{in} }
$$
Note this is not $theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $alpha_{out}$,
and he is able to determine corresponding $theta_{out}$. I am lazy
to do the calculation of the corresponding $theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - cos theta_{out} + w_1, - sin theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = frac { 3w_1 + cos theta_{in} }{ - cos theta_{out} + w_1 }
= - frac { 3w_2 - sin theta_{in} }{ - sin theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $theta_{in}=w_2$ and
$
theta_{out} = alpha_{out} = 2 theta_{in}
$.
answered Nov 11 '13 at 6:18
j.k.j.k.
663
$endgroup$
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
add a comment |
$begingroup$
Note that $theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $theta_{out}=theta_{in}$
The case of very long legs is similar to this: $theta_{out}=theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=sin theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (theta_{in} + theta_{out}) / 180$ and
$t_{Tin} = 1 + (theta_{in} + theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(cos theta_{in}, sin theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + cos theta_{in}, 3w_2 - sin theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
alpha_{out} = arctan frac{ 3w_2 - sin theta_{in} }{ 3w_1 + cos theta_{in} }
$$
Note this is not $theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $alpha_{out}$,
and he is able to determine corresponding $theta_{out}$. I am lazy
to do the calculation of the corresponding $theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - cos theta_{out} + w_1, - sin theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = frac { 3w_1 + cos theta_{in} }{ - cos theta_{out} + w_1 }
= - frac { 3w_2 - sin theta_{in} }{ - sin theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $theta_{in}=w_2$ and
$
theta_{out} = alpha_{out} = 2 theta_{in}
$.
answered Nov 11 '13 at 6:18
j.k.j.k.
663
$endgroup$
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
add a comment |
$begingroup$
Note that $theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $theta_{out}=theta_{in}$
The case of very long legs is similar to this: $theta_{out}=theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=sin theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (theta_{in} + theta_{out}) / 180$ and
$t_{Tin} = 1 + (theta_{in} + theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(cos theta_{in}, sin theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + cos theta_{in}, 3w_2 - sin theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
alpha_{out} = arctan frac{ 3w_2 - sin theta_{in} }{ 3w_1 + cos theta_{in} }
$$
Note this is not $theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $alpha_{out}$,
and he is able to determine corresponding $theta_{out}$. I am lazy
to do the calculation of the corresponding $theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - cos theta_{out} + w_1, - sin theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = frac { 3w_1 + cos theta_{in} }{ - cos theta_{out} + w_1 }
= - frac { 3w_2 - sin theta_{in} }{ - sin theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $theta_{in}=w_2$ and
$
theta_{out} = alpha_{out} = 2 theta_{in}
$.
answered Nov 11 '13 at 6:18
j.k.j.k.
663
$endgroup$
Note that $theta_{out}$ has to depend also on sizes of legs and turns (may be on their ratio). That is because on the outbound leg, you compensate for drift the wind causes on the turns.
In case of instant turns, the outbound leg is identical to the inbound leg (opposite direction), so $theta_{out}=theta_{in}$
The case of very long legs is similar to this: $theta_{out}=theta_{in}$ approximately.
Edit:
We will use
- length unit equal to 1 minute of flight
- Ground coordinate system (GCS)
- Air coordinate system (ACS)
- wind speed - vector $(w_1, w_2)$
The plane moves in ACS in any direction by a fixed speed v=1 unit per minute (so lengths can be given in minutes).
The trajectory in GCS is pictured in the question.
The trajectory in ACS consists of
- segment of angle $theta_{in}$ southwards, length 1 minute,
- then circle arc of length $t_{Tout}$, [ T stands for Turn ]
- second segment of angle $theta_{out}$ (unknown), also southwards, length $t_{Lout}$ (unknown), [ L stands for Leg ]
- and second circle arc length $t_{Tin}$.
$theta_{in}$ actually corresponds to the wind strength and direction. I think
$w_2=sin theta_{in}$ which can be substituted in final formula.
Times $t_{Tin}$ and $t_{Tout}$ are approximately 1 minute. Precisely they are determined
by the angles (I think $t_{Tout} = 1 - (theta_{in} + theta_{out}) / 180$ and
$t_{Tin} = 1 + (theta_{in} + theta_{out})/180$).
Important is that the two arcs make together the whole circle.
That also implies $t_{in}+t_{out}= 2 $ minutes.
The movements in ACS in changed order are:
- two arcs, together they make no move (but it takes 2 minutes of time)
- first segment (corresponds to inbound leg) - 1 minute, vector $(cos theta_{in}, sin theta_{in})$
- second segment (corresponds to outbound leg).
In GCS, we have to add to this:
- drift caused by wind during the circle and first segment. This is 3 minutes, hence $(3w_1, 3w_2)$
- drift during the second segment.
The circle plays no role for calculations. This is good news. Things about segments are easier to calculate.
The movement in GCS together, with exception of second segment, can be read from the the above, and it is $(3w_1 + cos theta_{in}, 3w_2 - sin theta_{in})$.
This is the vector we have to travel (in the opposite direction) during the second segment. Indeed we have to be on the same point in GCS in the start and end of the maneavour. That means, we have already the direction of the second segment in GCS:
$$
alpha_{out} = arctan frac{ 3w_2 - sin theta_{in} }{ 3w_1 + cos theta_{in} }
$$
Note this is not $theta_{out}$, but it determines it.
Certainly the pilot is able to fly in the direction $alpha_{out}$,
and he is able to determine corresponding $theta_{out}$. I am lazy
to do the calculation of the corresponding $theta_{out}$, and I'm not sure its possible by elementary functions.
As regards the time, the calculation is now straightforward.
The vector speed in GCS is $( - cos theta_{out} + w_1, - sin theta_{out} + w_2 )$
so the time needed is
$$
t_{Lout} = frac { 3w_1 + cos theta_{in} }{ - cos theta_{out} + w_1 }
= - frac { 3w_2 - sin theta_{in} }{ - sin theta_{out} + w_2 }
$$
Please check my calculations. Especially the signs.
Edit: Nota bene, for small winds we get approximately $theta_{in}=w_2$ and
$
theta_{out} = alpha_{out} = 2 theta_{in}
$.
answered Nov 11 '13 at 6:18
j.k.j.k.
663
edited Nov 11 '13 at 8:49
answered Nov 11 '13 at 6:18
j.k.j.k.
663
answered Nov 11 '13 at 6:18
j.k.j.k.
663
answered Nov 11 '13 at 6:18
j.k.j.k.
663
663
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
add a comment |
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
$begingroup$
I edited the OP with a discussion of this. I had initially thought that in a direct crosswind, all four legs would be one minute. But you're right, even if there is no head or tail wind, the outbound leg will be slightly longer than one minute.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:21
add a comment |
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$begingroup$
$theta_{in}$ is the blue angle in your diagram, and $theta_{out}$ the red?
$endgroup$
– shade4159
Nov 11 '13 at 6:13
$begingroup$
Yes, that is correct.
$endgroup$
– Graham Mainwaring
Nov 11 '13 at 6:19
$begingroup$
A recent publication "A Treatise on the Holding Pattern: Expelling the Myths and Misconceptions of Timing and Wind Correction" is on the internet. It is an exact analytic solution of the holding pattern problem under any wind condition. It shows holding patterns that arise under certain wind conditions that have never be discussed in the literature. In addition there are interesting mathematical properties of the holding pattern never previously discussed.
$endgroup$
– Les Glatt
Mar 26 '18 at 3:43