Finding W/V?
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Finding W/V?
Does anyone have a good way of solving this problem:
Track: 315°M
HD: 301°M
VAR: 5°W
TAS: 225 kt
GS: aircraft flies 50 nm in 12 minutes
Calculate W/V (°T) ?
A: 195/63
B: 355/15
C: 195/61
D: 190/63
Track: 315°M
HD: 301°M
VAR: 5°W
TAS: 225 kt
GS: aircraft flies 50 nm in 12 minutes
Calculate W/V (°T) ?
A: 195/63
B: 355/15
C: 195/61
D: 190/63
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Aircraft heading NW, G/S 250kts (50nm/12mins=250kts), ie faster than TAS of 225, therefore tailwind! So wind is from the south. You chaps should know when there are 3 similar answers, the real answer is one of them, not the one on its own!
Londonmet- either use a circular slide rule or draw out a large accurate triangle- you have two sides (250 and 225) and the internal angle- the third side is the wind speed and direction- convert to degrees True.
Londonmet- either use a circular slide rule or draw out a large accurate triangle- you have two sides (250 and 225) and the internal angle- the third side is the wind speed and direction- convert to degrees True.
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Founder,
Firstly, because your wind answer is in degrees true, you need to convert your magnetic track and magnetic heading into true true track and true heading. Variation West, Magnetic Best (i.e. greater than true) therefore you have to take 5 degrees from both 315 and 301 to give 310 degrees true track and 296 degrees true heading.
Next get your CRP-5 out and, using the high speed side, put the centre circle over 225 (TAS), and put the true heading of 296 degrees under the heading index at the very top. Now draw a vertical line DOWN from the centre circle as far as you can. This vertical line represents the true HDG/TAS vector.
You can then see that 310 degrees is 14 degrees to the right of 296 degrees i.e. you have a starboard drift of 14 degrees. Next put your windmark (i.e. a cross) where the 14 degrees starboard line crosses the 250 knots groundspeed line. Draw a line joining the centre circle to this windmark - this line represents the wind vector. As the wind always blows from the centre circle TO the windmark, you can see that the wind is coming from about 190 degrees and that the wind strength is 50- 70 knots in strength.
The last line to draw requires a little bit of imagination. Moving down from the centre circle, try to imagine roughly where the zero speed would be, then draw a line from the windmark to this position. Obviously the line can only go as far as the 150 knot line, but this will suffice. The line you have just drawn represents the true track/Groundspeed line.
Going back to the 50-70 knot windspeed mentioned above - we need greater accuracy than this. We obtain greater accuracy in measuring both the windspeed and direction by rotating the CRP-5 clockwise, to bring the windmark directly underneath the centre circle i.e. the 6 o'clock position. Looking now at the 12 o'clock position, we can read off accurately the wind direction - in this case, 189 degrees. Reading the windspeed from the centre circle to the windmark gives a figure of 63 knots.
189/63, so I would say the closest answer is D.
Hope this helps
VC10 Rib22
Firstly, because your wind answer is in degrees true, you need to convert your magnetic track and magnetic heading into true true track and true heading. Variation West, Magnetic Best (i.e. greater than true) therefore you have to take 5 degrees from both 315 and 301 to give 310 degrees true track and 296 degrees true heading.
Next get your CRP-5 out and, using the high speed side, put the centre circle over 225 (TAS), and put the true heading of 296 degrees under the heading index at the very top. Now draw a vertical line DOWN from the centre circle as far as you can. This vertical line represents the true HDG/TAS vector.
You can then see that 310 degrees is 14 degrees to the right of 296 degrees i.e. you have a starboard drift of 14 degrees. Next put your windmark (i.e. a cross) where the 14 degrees starboard line crosses the 250 knots groundspeed line. Draw a line joining the centre circle to this windmark - this line represents the wind vector. As the wind always blows from the centre circle TO the windmark, you can see that the wind is coming from about 190 degrees and that the wind strength is 50- 70 knots in strength.
The last line to draw requires a little bit of imagination. Moving down from the centre circle, try to imagine roughly where the zero speed would be, then draw a line from the windmark to this position. Obviously the line can only go as far as the 150 knot line, but this will suffice. The line you have just drawn represents the true track/Groundspeed line.
Going back to the 50-70 knot windspeed mentioned above - we need greater accuracy than this. We obtain greater accuracy in measuring both the windspeed and direction by rotating the CRP-5 clockwise, to bring the windmark directly underneath the centre circle i.e. the 6 o'clock position. Looking now at the 12 o'clock position, we can read off accurately the wind direction - in this case, 189 degrees. Reading the windspeed from the centre circle to the windmark gives a figure of 63 knots.
189/63, so I would say the closest answer is D.
Hope this helps
VC10 Rib22
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OK very good, I believe you. Just so we can see your working, would you like to tell us how you got there? I mean I do know...... really.......just like to see your working.....like.......
Actually the answer IS D, as VC10 says.
Track is 310o T
Heading is 296o T
Therefore drift angle (DA) is 14o
TAS is 225 Kt
GS is 250 Kt
Using wind triangle and cosine rule wind velocity is:-
WS2 = TAS2 + GS2 – 2 x TAS x GS x Cos DA
WS2 = 2252 + 2502 – 2 x 225 x 250 x Cos 14o
WS2 = 3966.73
Therefore WS = 62.98 = 63 Kt
Now having wind velocity, we can calculate wind direction:-
WV/Sin DA = TAS/Sin X
Therefore Sin X = (TAS x Sin DA) / WV
Sin X = (225 x Sin 14o ) / 63
Sin X = 0.864007
Therefore X = 59.77o
Finally Wind Direction = (Track – 180) + X
Wind Direction = 310 – 180 + 59.77 = 189.77 = 190o
Therefore answer is 190/63 = D
PS Sorry, could not figure out how to do degrees and squared as suffixes!
Track is 310o T
Heading is 296o T
Therefore drift angle (DA) is 14o
TAS is 225 Kt
GS is 250 Kt
Using wind triangle and cosine rule wind velocity is:-
WS2 = TAS2 + GS2 – 2 x TAS x GS x Cos DA
WS2 = 2252 + 2502 – 2 x 225 x 250 x Cos 14o
WS2 = 3966.73
Therefore WS = 62.98 = 63 Kt
Now having wind velocity, we can calculate wind direction:-
WV/Sin DA = TAS/Sin X
Therefore Sin X = (TAS x Sin DA) / WV
Sin X = (225 x Sin 14o ) / 63
Sin X = 0.864007
Therefore X = 59.77o
Finally Wind Direction = (Track – 180) + X
Wind Direction = 310 – 180 + 59.77 = 189.77 = 190o
Therefore answer is 190/63 = D
PS Sorry, could not figure out how to do degrees and squared as suffixes!
Last edited by Groundloop; 19th Dec 2006 at 11:55.
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Look at the answers!
They'll always try to catch you for doing all the working then forgetting to account for the variation, so I'd immediatley be drawn to answer D as it is 5 degrees less than A!
It pays to think like the dastardly examiners do!
It pays to think like the dastardly examiners do!
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W/V = 189.7976154°(T)/62.98198784
That makes none of them correct, but (D) is close enough for government work.
Founder, you could have used Plane Trigonometry, any one of many "Pilot" computers, or a Calculator. This one's a bit basic isn't it?
Regards,
Old Smokey
That makes none of them correct, but (D) is close enough for government work.
Founder, you could have used Plane Trigonometry, any one of many "Pilot" computers, or a Calculator. This one's a bit basic isn't it?
Regards,
Old Smokey
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Tis simple really.........
Drift angle = 14 degrees. 225knots divided by 60 = 3.75.
14 (degrees) by 3.75 = 52.5 knots crosswind component.
12 minutes is a fifth of an hour. 50nm in 12 minutes is 250knots therefore.
TAS 225 knots, means 25 knots tailwind component.
Even a rough guess gives a triangle with one side twice as long as the other. The hypotenuse (can't spell it) will be longer than the longest side by a small amount. Let's guess at - side one=52.5, side two=25, hyp then=60??
A right angled triangle with one side about twice as long as the other gives an angle of about 25 degrees. Since there is more crosswind than tailwind, and the drift is to the right, the wind is from about 25 degrees behind the left wing. Heading is 301 degrees, left wing is pointing at 211 degrees, less 25 gives 186 degrees. So my guess is 186M at 60 knots. Less variation makes
181/60. D is the closest answer.
Sure tis the way we used to do it in the old days. Back when we walked in bare feet uphill both ways for school and kept a piece of coal in our back pocket to keep us warm.
If you want to be more accurate than this brain work, CRP is your only man.
In fact, my way is absolute rubbish for accuracy if the truth be told, but tis handy for real life!
Drift angle = 14 degrees. 225knots divided by 60 = 3.75.
14 (degrees) by 3.75 = 52.5 knots crosswind component.
12 minutes is a fifth of an hour. 50nm in 12 minutes is 250knots therefore.
TAS 225 knots, means 25 knots tailwind component.
Even a rough guess gives a triangle with one side twice as long as the other. The hypotenuse (can't spell it) will be longer than the longest side by a small amount. Let's guess at - side one=52.5, side two=25, hyp then=60??
A right angled triangle with one side about twice as long as the other gives an angle of about 25 degrees. Since there is more crosswind than tailwind, and the drift is to the right, the wind is from about 25 degrees behind the left wing. Heading is 301 degrees, left wing is pointing at 211 degrees, less 25 gives 186 degrees. So my guess is 186M at 60 knots. Less variation makes
181/60. D is the closest answer.
Sure tis the way we used to do it in the old days. Back when we walked in bare feet uphill both ways for school and kept a piece of coal in our back pocket to keep us warm.
If you want to be more accurate than this brain work, CRP is your only man.
In fact, my way is absolute rubbish for accuracy if the truth be told, but tis handy for real life!
Warning Toxic!
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I with you there! KISS (Keep It Simple Stupid) As long as you are not trying to find Norfolk Island after 25 hours in a Gipsy Moth with one arm tied behind your back, it'll do....for want of any other accurate readings!
