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JAA ATP - GNAV - Dead Reckoning

Old 29th Oct 2011, 22:44
  #1 (permalink)  
Thread Starter
Join Date: Jan 2006
Location: Europe
Posts: 403
JAA ATP - GNAV - Dead Reckoning

I noticed that using the CR-3/5 for solving the following type of questions (wind triangles - Dead Reckoning) is not really that accurate, since the given answers are somewhat close together. What's the best way to find the correct answer?

TAS = 170 kt
HDG (T) = 100
W/V = 350/30 kt
Track (T) and GS?

1. 109 - 182 kt (correct answer)
2. 110 - 180 k
3. 103 - 178 kt
4. 091 - 183 kt

Thank you!
Transsonic2000 is offline  
Old 30th Oct 2011, 08:31
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Join Date: Feb 2002
Location: Sunny Solihull
Age: 65
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Have you actually done this problem with a small pencil dot and the correct method? Using my 35 year old CRP-5 I have no problem in getting the right answer, mind you that's with 40 years of experience!

Remember the question is asking TK & GS not HDG & GS.

1. Put in wind DOWN.
2. TAS of 170 on centre dot (grommet).
3. Turn to put heading on 100 at the top.
4. Pencil dot should now be on 8/9 right drift line at about 183 kts.

109 at 182 no question.

Check your CRP-5 is not too slack and the grommet remains on the centre line at all times when moving up and down. If not it could explain some of your problems.

Also remember wind blows from heading to track answer 4 can't possibly be right before you even start.

You could of course draw a vector diagram to scale, but you won't have time in an exam for that. Get used to using your CRP-5 accurately then quickly.
RichardH is offline  
Old 30th Oct 2011, 17:10
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Join Date: Apr 2009
Location: down south
Age: 75
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If you consult the navigation tutor at your JAA approved service provider I am sure you will receive the correct method.

Why are you asking this question on this thread?

Lightning Mate is offline  
Old 30th Oct 2011, 17:18
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Thread Starter
Join Date: Jan 2006
Location: Europe
Posts: 403
Thanks Richard! Well the problem is not how to work it out with the CR3, but that the given multiple choice answers are sometimes so close together that one can easily be mistaken. So I was wondering if there was a mathematical way/formula to get this type of question(s) figured out?

@ Lightning Mate:
actually I'm doing the conversion from FAA to JAA and I'm exempt from ground school, which is on one hand an advantage (saves money) but on the other it can be a disadvantage (sometimes).

Last edited by Transsonic2000; 30th Oct 2011 at 17:38.
Transsonic2000 is offline  
Old 1st Nov 2011, 00:11
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Join Date: Jun 2007
Location: Scotland
Posts: 15
Rough Formula

You ask for a mathematical method to get round the whizz wheel, but this involves working sin and cos values for angles in your head, which is not an easy thing to do if they are non standard angles. That is why the whizz wheel is so good for the exam but not necessary for real life. There is a simple way to do it in your head, but it will only be of limited help in your exam. The examiner no doubt knows that you can work the answer out approximately in your head and that is perhaps why the answers are so close, to force you to use the whizz wheel (and to do so properly.) Working it out in your head does at least help you to rule out the dross answers however. The formula to use is: Max drift = 60/TAS x wind velocity. From your example: TAS = 170 kt HDG (T) = 100 W/V = 350/30 kt Track (T) and GS? In your head you can work out your example by rounding TAS to 180 which gives - 60/180 x 30 = 1/3 x 30 = 10 max drift. Using the old rule of 60, your heading is 100 which gives a wind angle of 110 i.e. the wind is blowing from your left and slightly behind you. This gives you a tail wind. The rule of 60 says crosswind is max at 60 or more. Your crosswind is 70 so is at a max of 10. Thus, your track will be 110. If you deduct your crosswind component from 90 this gives you your head/tailwind component. 90 - 70 gives 20. Using the rule of 60 (i.e. 20/60) this equates to 1/3 or one third of the wind speed which is 10 kts and we are dealing with a tail wind here. This gives a ground speed of 180kts. This all sounds complicated, but with practice it takes only ten or fifteen seconds to work out. The answer using this rough rule of 60 method is 110 at 180kts. However, this only narrows the answers down to two options. Unless you have a slide rule in your head or have memorised your sin/cos tables for every 5 to 10 degrees, you will have to turn to the whizz wheel. (You could use vector diagrams as an alternative...) Hope that is of some help.
eninem is offline  
Old 1st Nov 2011, 10:40
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Join Date: Oct 2000
Location: Bristol
Posts: 461
You can solve a triangle if you have the length of two sides and one angle. This is O level maths' Label the angles A, B and C and the sides oposite to them a, b and c.

The formula is a/SinA = b/SinB = c/SinC

You have two sides, TAS 170 and Wind 30 and one angle - between hdg and wind direction

Good luck

Dick Whittingham is offline  

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