vectors
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Joined: Feb 2011
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From: lyon
vectors
Can anyone help me with a question please, dont know if i can post it here, anyway here it is: if a plane is flying at 124 knots (true airspeed) on a headding of 130 degrees with wind at 225/35. what is the planes true ground speed and track.
Can anyone help me get the angle between the aircraft and the wind to complete the vector triangle, from the book its saying 85 degrees (130-45) but i keep getting 95 degrees.
Can anyone help me get the angle between the aircraft and the wind to complete the vector triangle, from the book its saying 85 degrees (130-45) but i keep getting 95 degrees.
Joined: Jan 2004
Posts: 19
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From: Europe
Wind triangle
Hi,
ok, here is the thing: your wind is FROM 225°, meaning it blows TOWARDS 045°.
So, the angle between 130 (Aircraft) and 045 (Wind) is 85°.
I remember sometimes alos not seeing the obvious from too much staring right at it...
Good luck in your training!
ok, here is the thing: your wind is FROM 225°, meaning it blows TOWARDS 045°.
So, the angle between 130 (Aircraft) and 045 (Wind) is 85°.
I remember sometimes alos not seeing the obvious from too much staring right at it...
Good luck in your training!
Thread Starter
Joined: Feb 2011
Posts: 17
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From: lyon
thanks for that, so would i be right in saying that if plane flying at 124 knots on 130 degrees = |ab| @ 124 knots
wind from 225 degrees = |bc| @ 35 knots
line |ca| = ground speed
and angle abc = 95 degrees?
ive worked it out both by maths and by drawing and this is what i get but the book says angle abc is 85 degrees?
wind from 225 degrees = |bc| @ 35 knots
line |ca| = ground speed
and angle abc = 95 degrees?
ive worked it out both by maths and by drawing and this is what i get but the book says angle abc is 85 degrees?
Joined: Apr 2011
Posts: 14
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From: UK
Flyingmam,
The book is right. You see the wind vector falls below aircraft's 90 degree line.
This is an important question because its needed when you go flying and want to quickly calculate drift angles in your mind, for example in flying holds.
Immagine four quadrants on the aircraft, each 90 degrees. The origin of the quadrants is the aircraft's heading If the aircraft heads 130, the first quadrant is up to 220. The wind vector falls in 225. This means its in the second quadrant.
Now look at the angles subtended in the second quadrant, disregard the first, and its all there. The angle opposite the tailwind component is 5 degrees and the angle opposite the crosswind component is 85 degrees. I made a quick on sketch on paint to make it clearer:
The book is right. You see the wind vector falls below aircraft's 90 degree line.
This is an important question because its needed when you go flying and want to quickly calculate drift angles in your mind, for example in flying holds.
Immagine four quadrants on the aircraft, each 90 degrees. The origin of the quadrants is the aircraft's heading If the aircraft heads 130, the first quadrant is up to 220. The wind vector falls in 225. This means its in the second quadrant.
Now look at the angles subtended in the second quadrant, disregard the first, and its all there. The angle opposite the tailwind component is 5 degrees and the angle opposite the crosswind component is 85 degrees. I made a quick on sketch on paint to make it clearer:
Joined: Dec 2007
Posts: 65
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From: Emerald Isle
Interesting question.
I always thought the Relative Wind Angle was a number between 0 & 180, not 0 & 90.
Is there mention of variation in the question?
Is the wind given as True, not Magnetic? Met reports/forecasts are True, ATS reports are Magnetic.
With a variation of 10E, that would make the RWA 85.
I always thought the Relative Wind Angle was a number between 0 & 180, not 0 & 90.
Is there mention of variation in the question?
Is the wind given as True, not Magnetic? Met reports/forecasts are True, ATS reports are Magnetic.
With a variation of 10E, that would make the RWA 85.
