Sign in to follow this  
Guest LEP

Wind Correction Question

Recommended Posts

I've bought an E6B, both manual and digital ones, and I realised that the usual wind correction cal. only gives Wind Correction Angle and Ground Speed (assuming we know True Course, TAS, Wind Course/Speed). My question is, when I draw out the vectors and solve using trigonometry, it says I should turn my aircraft xx degrees from true course AND to fly the new heading using a HIGHER TAS. But this higher TAS isn't given by the flight computer. So does it mean that pilots over the world would just turn to the wind corrected heading AND just continue to fly at the same original TAS? If so, isn't that wrong?Hope my question is clear.Lep

Share this post


Link to post
Share on other sites
Help AVSIM continue to serve you!
Please donate today!

I assume that you are using a Pythagoras to solve. If that is the case I would assign the adjacent the a/c TAS (say 100kts) and the opposite the crosswind speed component (say full 90deg @ 20kts). Drift = tan(20/100)^-1 = 11deg. Are you looking at the length of the hypotenuse (in this case 100.2). This does not mean a TAS of 100.2kts, I think this means that because you change your heading (11deg in this case) to account for drift, your ground speed will reduce by 0.2kts, even without any TRUE headwind component. However, 0.2kts would have a nugetory affect on the ETA and as such can be completely ignored.I hope that answers your question.================================================================The standard rules of thumb to get a more than good enough result without using a whizz wheel (coloquial name for a CRP or EB6) is:Calculate Max Drift = Wind Speed * 60 / TAS (say Max Drift = 12deg = 20kts*60/100kts)Calculate Wind Offset = Difference between Wind Dir and Track. For example a Track of 090 and a Wind of 060 = a difference of 030.Use the clockface method: On your watch 15 is 1/4 of the clock face, 30 is half, 45 is 3/4 and 60 = full or 1.So if the wind offset = 30 then that is half of the Max Drift. So your crosswind component is 12deg then your drift is 6deg so your heading will be 84deg to remain on the track of 90deg. You will see that any wind offset of 60 or more will be the full max drift.Similar for ground speed. If the wind offset is withing 30deg of your track then the full wind speed is used. So your ground speed in this case would be 80kts (100kts - 20kts). If the wind were 030 then the ground speed would be 90kts - half of 20kts, geddit?Even though this is a simple rule of thumb that the pilot can easily apply mentally in the air to say plot an unplanned diversion, it, in reality produces very accurate results so long as you fly accurately.

Share this post


Link to post
Share on other sites

We fly the same TAS. The ground speed is what gets affected and is reflected in the calculations.----------------------------------------------------------------John MorganReal World: KGEG, UND Aerospace Spokane Satillite, Private ASEL 141.2 hrs, 314 landings, 46 inst. apprs.Virtual: MSFS 2004"There is a feeling about an airport that no other piece of ground can have. No matter what the name of the country on whose land it lies, an airport is a place you can see and touch that leads to a reality that can only be thought and felt." - The Bridge Across Forever: A Love Story by Richard Bach

Share this post


Link to post
Share on other sites

>>the world would just turn to the wind corrected heading AND>>just continue to fly at the same original TAS? >>Correct. Nobody in his/her right mind would use speed>adjustment to counter wind drift. You adjust heading only.>Anybody in their right mind, would follow a GPS "track", and bag thoughts of wind corrections altogether! :D But if you're bored, and don't feel the need of extra time to watch for traffic, then what the heck... :( L.Adamson

Share this post


Link to post
Share on other sites

Thanks everyone, particularly Adverse Yawn's helpful answers.However, I need to point out that we may be missing something (pls correct me if I'm wrong!).To Yawn : The hypotenuse in your case is 102 KTAS, not 100.2.Pls allow me to use this example. Say We would like to fly True Course 0 degree at 100 KTAS. Wind blows from 45 degree at 20 knots. Using vectors and Cosine rule (a sq = b sq + c sq - 2abCos X), the Wind Correction Angle can be found to be 8.1 degree. But the long-side of triangle (corrected course) here works out to be 115 knots! Now if we turn to new heading and fly at 115 KTAS, yes then we still get 100 KTAS at 0 degree True Course, which will work out to be 86 knots Ground Speed (head-wind component calculated to be 14 knots). BUT if we adjust heading by 8 degree and still fly at 100 KTAS, the intended True Course (0 degree) speed component WILL NOT be 100 KTAS any more, but somewhere LESS, which means less Ground Speed, lesser than the 85 or 86 knots we get from E6B.This doesn't seem that negligible, especially so when we fly high up at higher speed in jets or fast prop, where the winds aloft can be like 50 knots in strength or so.Would really appreciate if anyone of you could help enlighten on this paradox. Thanks.

Share this post


Link to post
Share on other sites

There is no paradox. You somehow assume that pilots want to fly certain True Course with certain ground speed. This is a WRONG assumption. Pilots definitely want to fly certain True Course which will get them to their destination but they fly with whatever ground speed they will get. In your example the wind very much works against the pilot and this will be reflected in slower ground speed than if there had been no wind at all.You seem to be working with E6B which is (by now a bit obsolete) pilot's well known tool. Are you also studying for PPL and using some textbook? If so please read the chapter on wind correction - you seem to be making assumptions that have no place in real world of flying.And as you observed sometimes a strong wind can seriously diminish a ground speed. Therefore pilots would often look for best altitude to get the most advantageous wind situation. Big commercial aircraft are equipped with FMS that allow pilots to do some sophisticated wind planning while airborne. Their dispatch office can also do some serious pre-planning to come up with best flight plan (may result in a 'curved' path) in the presence of a wind.Michael J.http://www.precisionmanuals.com/images/forum/pmdg_744F.jpghttp://sales.hifisim.com/pub-download/asv6-banner-beta.jpg

Share this post


Link to post
Share on other sites

Thanks Michael. What I meant was, when we turn to the new 8 degree and fly at same original speed of 100 KTAS, by the vector diagram and Cosine Rule, our 0 degree True Course speed would be somewhere LESS than 100 KTAS, meaning our actual ground speed up North would be less than the 86 knots GS that the Flight Comp. gives us. If one still uses 86 knots as GS, there will be serious implication in terms of fuel calculation, since in actual fact we're flying at lesser speed over ground. Hope my message is clear here. Do point out any wrong assumption if you think I've made one.

Share this post


Link to post
Share on other sites

>there will be serious implication in terms of fuel>calculation, since in actual fact we're flying at lesser speed>over ground. Hope my message is clear here. Do point out any>wrong assumption if you think I've made one.Correct. You use the Ground speed to figure out your fuel needs. Yes, the wind and the resulting ground speed may have very serious implications on your fuel requirement.Michael J.http://www.precisionmanuals.com/images/forum/pmdg_744F.jpghttp://sales.hifisim.com/pub-download/asv6-banner-beta.jpg

Share this post


Link to post
Share on other sites

"meaning our actual ground speed up North would be less than the 86 knots GS that the Flight Comp. gives us"I think you may be double counting the wind effect.In your example, the aircraft flies at 100kt TAS at an angle of 8deg to the desired north/south track. The ground speed would be 86kt as you say. This speed has already taken account of the wind effect, so the ground speed would not be reduced further as you suggest.

Share this post


Link to post
Share on other sites

mgh, I'm afraid you may not have gotten what I meant. Flying out with 8 degree correction using 100 KTAS will have to result in a "shorter" KTAS in the 0 degree up North. Draw out the 3 vectors before and after correction and perhaps it may be clearer.Hope someone could help clarify this topic, as the difference is dramatic high up with strong wind and fast airspeed.

Share this post


Link to post
Share on other sites

LEP,Maybe this will help, a picture and a thousand words etc. I believe this is what we are all thinking. A scalene triangle with one egde the wind, the other TAS and other GS. Can you point out from this what you mean?http://forums.avsim.net/user_files/141202.jpgI don't see where TAS increases. The Ground Speed = SQRT( (100^2+20^2) - (2*100*20*cos(45)) ). However, As I mentioned before, in reality the TAS will reduce by a small amount as the aircraft has to change its heading into wind to couteract the drift. But this is ignored as it it practially makes not meaningful difference to the ETA.

Share this post


Link to post
Share on other sites

>result in a "shorter" KTAS in the 0 degree up North. Please, such statement makes no sense and has place in real life. There is only one TAS - TAS doesn't change because of wind. The aircraft knows nothing about the wind and can't sense wind when flying. This is really ABC of aeronautical knowledge. You ought to do some reading - such basic knowledge is available online.Michael J.http://www.precisionmanuals.com/images/forum/pmdg_744F.jpghttp://sales.hifisim.com/pub-download/asv6-banner-beta.jpg

Share this post


Link to post
Share on other sites

Thanks for putting up the sketch. Here's what's on my mind.The 87 knots in your sketch shows the "drifted" course and speed (and it's True Air Speed, not GS, I think).In my sketch, I believe the calculated 115 KTAS is what we should fly out with in the corrected direction of 8 degrees, thereby giving us the 100 KTAS up North still. Had we fly out with 100 KTAS at 8 degrees, the True Air Speed at 0 degree would certainly be less than 100 knots, which when factoring in the headwind of about 14 knots, would result in LESS THAN 86 GS up North. (86 GS is only possible if the 100 KTAS at 0 degrees is maintained).Bearing in mind also that E6B is supposed to give only wind corr angle and GS (GS in the True Course direction --- in this case up North).Not sure if I got my wind correction concept here right or not.

Share this post


Link to post
Share on other sites

you may want to refer to the sketches for what I meant by shorter or longer. By the way, thanks for your advice. I know what TAS means exactly. And by the way, too, isn't this forum for learning from each other, mistakes and knowledge gap and all ??

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this