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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!
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?
If we restrict ourselves to the "classic" sub-tropical jets then they are westerlies because the thermal equator is more or less at the geographic equator - except in one case. In the northern hemisphere summer, July, August or thereabouts, the tremendous heating of the Tibetan plateau drags the thermal equator well north of the geographic. Now the upper air flow out of the upper high is subject to geostrophic forces opposite to normal and the winds turn easterly
This upper easterly flow can reach jet speeds right up near the trop in the area between N India and W Africa. You can find them on the upper winds/temps charts or sometimes on regional sig wx chats
There are other forms of "jet", for example the low level cold front jets that occur equally in both hemispheres
Last edited by Dick Whittingham; 26th Jul 2012 at 17:28.
Hi all, just came across this (apparently) weird question, can someone explain how to figure out the correct answer or am I missing something?
At maximum landing mass, the structure of the aircraft is designed for a rate of descent:
- 250 fpm
- 600 fpm - 200 fpm (this would be the correct answer, but how is this figured out, is there a formula or something )
- 220 fpm
Hi all, I have a question about easterly waves. Since in summer prevalent upper winds in area above ITCZ are westerlies, and movement of storms is dictated by upper winds, is the westward movement of easterly waves and corresponding storms consequence of Tropical Easterly jet?