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 29th Jun 2012, 18:07 #101 (permalink) Join Date: Nov 2009 Location: PK Posts: 171 Thanks Keith Hdg 360 CRS 345 TAS 240 GS 285 W/V comes out to be 116/82 With CRS 165 and wind 116/82 the headwind component comes out around -53.8 with this formula: Wind Speed x Cos (Wind Direction - Course) Shouldn't the answers be close together if not exactly the same? Regards
 29th Jun 2012, 19:25 #102 (permalink) Join Date: Jan 2011 Location: England Posts: 657 I think that your wind should be closer to 112/82 And I think that your equation will get you wind component down the track. If you use Windspeed x Cos(Wind direction - Heading) You get 82 x Cos(112 - 149) = 65.5
 29th Jun 2012, 19:52 #103 (permalink) Join Date: Nov 2009 Location: PK Posts: 171 ok, I think more than the wind its the formula that's different I was using Wind Speed x Cos (Wind Direction - Course) Whereas you have mentioned it is: Windspeed x Cos (Wind Direction - Heading) Is (Wind Direction - Course) wrong or is it used somewhere else, I think i read it somewhere
 30th Jun 2012, 15:54 #104 (permalink) Join Date: Jan 2011 Location: England Posts: 657 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.
 3rd Jul 2012, 09:22 #107 (permalink) Join Date: Nov 2009 Location: PK Posts: 171 Thanks BGS
8th Jul 2012, 23:02   #108 (permalink)

Join Date: Jan 2006
Location: Europe
Posts: 370
Performance: Domain

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:

Domain 1/2 width = 60m + 1/2 wing span + 0.125 D
Domain 1/2 width = 60 + 32.2m + (0.125 x 1500m)
Domain 1/2 width = 279.7m

Thanks

 9th Jul 2012, 00:00 #109 (permalink) Join Date: Nov 2009 Location: PK Posts: 171 Hi For aircraft with a wing span of more than 60m*domain starts with a semi-width of 90m.* So it will be 90 + (1500 x 0.125)
 9th Jul 2012, 01:33 #110 (permalink) Join Date: Jan 2006 Location: Europe Posts: 370 thanks for the prompt respond, much appreciated! I guess I was wrong in this case.
 9th Jul 2012, 13:11 #111 (permalink) Join Date: Aug 2010 Location: europe Posts: 7 if a radio altimeter fails the right answer is actually a) height information disappears I know the question is tricky and too type specific, so the only answer which can't be proved wrong is a. cheers
 25th Jul 2012, 11:44 #112 (permalink) Join Date: Sep 2008 Location: NOTTINGHAM Posts: 11 Converging aircraft Hi 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?
 25th Jul 2012, 13:41 #113 (permalink) Join Date: Jan 2011 Location: England Posts: 657 Assuming that they both take-off at the same time: Flight Time to meeting point = Distance / (VA + VB) Where Distance is distance between the two starting points. VA is ground speed of aircraft A. VB is groudn speed of aircraft B. Position of meeting relative to start point A = Time x VA Position of meeting relative to start point B = Time x VB
 25th Jul 2012, 14:25 #114 (permalink) Join Date: Oct 2000 Location: Bristol Posts: 460 Assuming that A and B and the meeting point are in a straight line and that the meeting point lies on the line between A and B
 25th Jul 2012, 15:09 #115 (permalink) Join Date: Jun 2010 Location: Stockholm Posts: 24 Hey guys, Just came across this question: 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?
25th Jul 2012, 16:10   #116 (permalink)

Join Date: Jan 2011
Location: England
Posts: 657
Quote:
 Assuming that A and B and the meeting point are in a straight line and that the meeting point lies on the line between A and B
You are of course correct Dick, but Bradders 147 appeared to be looking for something simple, so I went for the most simple scenario.

If there is a simple equation that covers all possible combinations of track geometries I don't know what it is. Do you?

 26th Jul 2012, 10:03 #117 (permalink) Join Date: Oct 2000 Location: Bristol Posts: 460 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.
 27th Jul 2012, 15:24 #118 (permalink) Join Date: Jan 2006 Location: Europe Posts: 370 Performance - Landing (Class A) 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 Thanks in advance! And I need a brake now!
 27th Jul 2012, 18:09 #119 (permalink) Join Date: May 2001 Posts: 10,858 Its a design function in the relevant certifying specs
 3rd Aug 2012, 08:15 #120 (permalink) Probationary PPRuNer   Join Date: Aug 2012 Location: zagreb Posts: 1 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?

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