Wind Triangle Vectors & Curved Trajectories
Per Ardua ad Astraeus
Join Date: Mar 2000
Location: UK
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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!
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!
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Join Date: Jul 2008
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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
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
Per Ardua ad Astraeus
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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.
If you want any Excel stuff PM me an email address.
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RNAV turn trajectory
I am long time pprune reader, but I've never took enough courage o register.
But today I've found this topic, when I've been searching whole day for the similar answer
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.
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.
But today I've found this topic, when I've been searching whole day for the similar answer
Can Wind Triangle vector calculations apply to curved trajectories or turns?
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.
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.