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CPL/ATPL Question

Old 8th Mar 2011, 09:32
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CPL/ATPL Question

Hi,
I have a question that has appeared in DGCA CPL exam, I know the answer is B. 16 o Left – 156Kt. I know it is easy to get this answer using the CX-2.
Can anyone please tell me how to get the answer using a scientific calculator, I know there is a formula but can’t seem to remember it.
Given: TAS = 140Kt, True HDG = 302, W/V 045(T)/45kt, calculate drift angle & GS
Thanks in advance.
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Old 8th Mar 2011, 10:34
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TAS = 140Kt, True HDG = 302, W/V 045(T)/45kt, calculate drift angle & GS

Crosswind Component = W x Sin (diff between Track and W/V)
= 45 x Sin (103)
= 43.8kts
Head wind Component = W x Cos (diff between Track and W/V)
= 45 x Cos (103)
= -`10 kts.......- stands for Tail wind
so Gs will be TAS +(Headwind comp)= Ground speed

140+(10)=150Kts

drift angle
Tan(theta)= Crosswind component/TAS
= 43.8/140=17.13deg

Wind is from around 5 o clock...therefore ur drifting left i.e. 17Left

i don't care what are the options..these are the accurate figures
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Old 8th Mar 2011, 12:59
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Thanks Buddy.
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Old 17th Apr 2018, 19:13
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Originally Posted by planeboy_777
Crosswind Component = W x Sin (diff between Track and W/V)
= 45 x Sin (103)
= 43.8kts
Head wind Component = W x Cos (diff between Track and W/V)
= 45 x Cos (103)
= -`10 kts.......- stands for Tail wind
so Gs will be TAS +(Headwind comp)= Ground speed

140+(10)=150Kts

drift angle
Tan(theta)= Crosswind component/TAS
= 43.8/140=17.13deg

Wind is from around 5 o clock...therefore ur drifting left i.e. 17Left

i don't care what are the options..these are the accurate figures

How we get 103...do explain.
Ty
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Old 18th Apr 2018, 00:08
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360+45 -302
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Old 14th Dec 2018, 06:23
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Could anyone explain to me how does 43.8/140=17.13deg, can't figure it out whatever way I look at it
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Old 14th Dec 2018, 07:56
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Nevermind, finally figured it out, tan−1(43.8/140)=17.13 I hadn't noticed the tan function there
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Old 15th Dec 2018, 12:45
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Can someone explain how to solve this

Aircraft present Hdg 170 (M), ADF 345 relative. Make 450 intercept of the 355 track out bound. What shall be the Hdg to intercept and ADF indication at the time of intercept
– ADF on intercept (Relative) Intercept Hdg (M)
a) 310 175 b) 040 135 c) 045 130
Ans. c
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Old 21st Feb 2020, 00:31
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General nav question

Guys i have a doubt. What is chlat btw these two points....A 71DEG 20MIN north and point B 86deg 45min north (over the north pole). In normal questions i was to just subtract point B from point A but in this question i need to subtract both from 90deg and then add to get ans. Plz explain why so
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