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a. No heading is specified so select one at random, 000 degrees for example.
b. Set the high-speed slide with the centre dot at 240 kts.
c. The question specifies a +45 wind component. This means a 45 kt tailwind component. This means that the ground speed is 240 TAS + 45 kts = 285 kts.
d. The question also specifies 150 left drift. So draw a cross where the 15 degrees left drift line crosses the 285 kt arc.
e. With 15 degrees left drift the track is 345 degrees. So the reverse track is 165 degrees.
f. Rotate window to align 165 degrees with true heading index.
g. The cross now indicates 19 degrees right drift. Align the 165 degrees track with 19 degrees right drift. The cross now indicates 15 degrees right drift. Align 165 degrees with 15 degrees right drift. The cross now indicates 16 degrees right drift. Align 165 degrees with 16 degrees right drift. Both the cross now indicates Move and the outer scale now indicate 16 degrees right drift, so the drift is balanced.
h. The cross is now on an arc 65 kts below the centre dot. The wind component was a headwind during the outward leg so it must be a tailwind during the return leg This means that the wind component is -65 kts.
There are quicker ways of doing the job (I saw one once here in PPRuNe but I can't remember what it was), but the above is the method described in the CRP 5 instruction sheet.
Last edited by keith williams; 29th Jun 2012 at 14:28.
Looking again at my description of the solution using a CRP5, I can see that I have made an error in that I assumed that the 240 knot TAS plus the 45 knot wind component would give a ground speed of 285 knots. This would be true if there was no drift, but it is not true with 15 degrees drift, because drift adds a lateral component to the ground speed.
If we split the TAS into one component along the 15 degree drift track, and another component at right angles to it we get 231.8 knots along the track and 62.2 knots across the track. Adding the 45 knot wind component to the along-track speed gives a ground speed of 231.8 + 45 = 276.8 knots.
The modified solution using the CRP5 become:
a. No heading is specified so select one at random, 000 degrees for example.
b. Set the high-speed slide with the centre dot at 240 kts.
c. The ground speed is 276.8 knots.
d. The question also specifies 150 left drift. So draw a cross where the 15 degrees left drift line crosses the 276.8 kt arc.
e. With 15 degrees left drift the track is 345 degrees. So the reverse track is 165 degrees.
f. Rotate window to align 165 degrees with true heading index.
g. The cross now indicates 19 degrees right drift. Align the 165 degrees track with 19 degrees right drift. The cross now indicates 15 degrees right drift. Align 165 degrees with 15 degrees right drift. The cross now indicates 16 degrees right drift. Align 165 degrees with 16 degrees right drift. Both the cross now indicates Move and the outer scale now indicate 16 degrees right drift, so the drift is balanced.
h. The cross is now on an arc at approximately 55 kts below the centre dot. The wind component was a headwind during the outward leg so it must be a tailwind during the return leg This means that the wind component is -55 kts.
This matches the solution that you got with your E6B.
It is curious that my error led me to answer that was marked as being correct. This makes me suspect that the author of this question made the same error.
I think your previous calculations were right. Perhaps I was mixing up two different modes of the E6B.
1) To find the W/V
Hdg 360 CRS 345 TAS 240 GS 285
W/V comes out to be 116/82
2) To find HDG and GS on reciprocal course
W/V 116/82 CRS 165 TAS 240
GS comes out to be 178 HDG comes out to be 150
Thus the wind component comes out to be TAS-GS (240-178) = 62kts Headwind.
3) There is another mode that calculates the X-wind and H-wind
For a W/V of 116/82
If you work it out with Course i.e. 165; H-wind component comes out to be -54 and X-Wind is 62 from the left.
If you work it out with Heading i.e. 150; H-wind component comes out to be -68 and X-Wind is 46 from the left.
Since we are maintaining the track and not letting the aircraft drift, we'll be heading into the wind so more headwind component. If we are not heading into the wind then less headwind component but more x-wind and the aircraft will drift.
I was using the course instead of headwind thats why the answers were different.
This is a somewhat lengthy explanation (and using different values) which confirms Keith’s original solution.
You have only been given a fixed numerical value for the TAS, so all other numerical values for the Outbound Track must be based on assumption.
That’s not as ambiguous as it seems at first glance because you are told that the effective Wind Component is +45 kts, making the GS 240 + 45 = 285 kts.
You are also told that the Drift is -15°. Working on the basis that Drift is invariably FROM Heading TO Track, choose some simple numbers to make your calculations easier: I used Heading = 115°T and Track = 100°T.
With these four components we can now establish the Wind Vector.
On the CRP-5 (HI SPEED):
Place centre dot over TAS = 240 kts
Rotate central bezel to place HEADING = 115°T on inner scale under TRUE HEADING index on outer scale
Make a Wind Mark where the straight, vertical Drift line (=15° LEFT) crosses the curved, horizontal speed line at 285 kts
Rotate the central bezel to place the Wind Mark on the central Drift line UNDER the TAS
Read the Wind Direction under the TRUE HEADING index = 230°
Read the Wind Speed by counting down to the Wind Mark from the TAS = 80 kts
Wind Vector = 230° / 80 kts
To calculate the inbound effective wind component, we note that we have a W/V of 230/80, a TAS of 240 kts and a reciprocal Track of 280°T (100°T + 180°)
On the CRP-5 (HI SPEED):
Place centre dot over TAS = 240 kts
Rotate central bezel to place Wind Direction = 230°T on inner scale under TRUE HEADING index on outer scale
Make a Wind Mark 80 kts BELOW the TAS at 160 kts (240 – 80)
Place Track = 280°T on inner scale under TRUE HEADING index on outer scale
Note Wind Mark has moved to indicate 18° of Right Drift
Align 280°T on inner scale with Drift = 18°L on outer scale
Align Track = 280°T on inner scale with Drift = 18°L on outer scale
Note Wind Mark has moved to indicate 14° of Right Drift
Align Track = 280°T on inner scale with Drift = 14°L on outer scale
Note Wind Mark has moved to indicate 15° of Right Drift
Align Track = 280°T on inner scale with Drift = 15°L on outer scale
Note Wind Mark still indicates 15° of Right Drift
The Drift on the outer scale now balances the Drift on the Drift grid, meaning the computation is complete
Read Heading = 265T under the TRUE HEADING Index
Read Ground Speed = 177 kts (ish) on curved speed line
Difference between TAS and GS = 240 kts – 177 kts = 63 kts
GS is slower than TAS, therefore, effective wind component = -63 kts
Closest answer = -65 kts
NOTE that it you are completing these questions under the EASA Learning Objectives, you are required you to use a Navigation Computer and NOT trigonometry. In fact, there ARE some CQB questions where the mathematically more precise trigonometrical solution has been provided as an incorrect option. Bonkers maybe but don't shoot me, I am merely the messenger, I get enough flak for the examiner in class off the students!
Hi there, I'd need some feedback regarding the following question - I believe that the answer provided is incorrect.
Quote:
A Boeing 747-400 has a wing span of 64.4 m. When considering relevant obstacles for the take-off flight path what is the semi-width of the zone at a distance of 1500 m from the TODA?
QDB says that the correct answer is 277.5 m but I come up with 279.7 m applying the following formula:
Can anyone tell me if there is a formula for when 2 aircraft leave different points at different speeds. when will they meet if the distance between the 2 is a certain number of miles?
Where are easterly and westerly jets found? a. Northern hemisphere only. b. Southern hemisphere only. c.Northern and southern hemisphere. d. There are no easterly jets.
The correct answer is a, but I can't seem to find anything to confirm this. Supposedly easterly jets only occur in the northern hemisphere and somewhat rarely(?) at that. I haven't however been able to locate any text stating that this is the fact. I have on the other hand found some indications that easterly jets do in fact occur in the southern hemisphere. Anyone able to shed some light?
