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If we are on course 360 and we have crosswind from 020 at 20 kts...does the amount of head wind on this leg equal the amount of tailwind on the reciprocal course (i.e. 180)?

To amplify the question with an example:

  • TAS: 260kts
  • Wind velocity: 310/35 kts

Given the above information, calculate ground speed on courses 240T and 060T.

When I use e6b+ aviation calculators, I get the following answers:

  • On a course of 240, the ground speed is 246kts. This is 14kts slower than the true airspeed.
  • On a course of 060, the ground speed is 270kts. This is 10kts faster than the true airspeed.

Since I believe the headwind and tailwind components should be equal, the question is: why is there a difference of 4kts between the computed headwind component and the computed tailwind component

Jimmy
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Joey
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  • Welcome to the site @Joey. Your question is not clear, what are you asking? – GdD Apr 22 '17 at 10:46
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    @GdD I think the question is clear, what's not clear is where Joey needs help. – Federico Apr 22 '17 at 10:51
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    You fly in a mass of air, and that mass is moving 240° SW at 20Kts. If you fly straight north 360° one hour, you'll end roughly 20 miles SW of your theorical destination with no winds. Same applies for a 180° straight south flight : you'll end somewhere 20 miles SW, because your aircraft is drifting within the mass of air. Accuracy is another matter : Your aircraft has a mass, a velocity, is not as "driftable" as a balloon, air being a compressible fluid, and you won't always have 20kts everywhere... You can estimate your ideal course, but you'll have to use references to correct your course. – Karl Stephen Apr 22 '17 at 12:13

3 Answers3

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YES, If the "amount of headwind" was 10 knots due to the crosswind, then the "amount of tailwind" would also be 10 knots on the reciprocal course.

BUT, because the 10 knot disadvantage with a head wind will last for a longer period of time than the 10 knot advantage with a tailwind, the entire trip will take longer and use more fuel than a trip in still air.

EXAMPLE: The headwind leg is 15 minutes longer, but the tailwind leg is only 10 minutes shorter, so you have a net loss of time.

Mike Sowsun
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  • Your answer explains why it will take longer with a headwind that with a tailwind, but does not answer the question of why the OP's computed components are not equal (there is a difference of 4kts). – Jimmy Apr 23 '17 at 20:20
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    My original answer was correct for a wind of 020 at 20 before the OP substantially edited his question. I don't know why his E6B calculations are not equal. My guess is that the increased wind correction angle needed to maintain course effects the ground speed in his 2nd example. – Mike Sowsun Apr 23 '17 at 21:31
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Does the amount of head wind on this leg equal the amount of tailwind on reciprocal course ie 180?

Yes, the intensity of the tailwind will be the same than the intensity of the headwind.

enter image description here

In both case the intensity of the head/tail wind component will be

$\left\| W_l \right\| = \left\| W \times cos\; W_a \right\|$

where $W$ is the wind intensity, $W_a$ is the angle between heading and wind. For a 20 kt wind at 60° ($cos\; 60° = 0.5$), the result is 10 kt.

The cross-wind component is also reversed with the same intensity.

mins
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  • Thanks for your reply.this is what i had calculated but when we use aviation calculators like e6b+ to solve the same problem the effective head wind and tail wind component is coming out to be different.. – Joey Apr 23 '17 at 09:28
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My answer comes from the point of view of a cyclist (also being subject to the effects of headwinds and tailwinds). I would suggest that any form of consistent winds on a round trip can only act to make the net effort required greater than in calm wind conditions.

Consider the effect of two special cases:

  1. Consider a strong cross-wind, blowing at 90 degrees to the path being travelled (say, a westerly wind). In both directions, (north and south), you will face increased head winds. In an aircraft, you would need to aim somewhat west of your intended direction in order to stay on course. Both legs of the journey would be more difficult.

  2. Consider a strong head-wind, blowing parallel to your course (head-wind in one direction and tail-wind on the way home). For the sake easy maths, let's assume that the wind-speed is half that of the cruising airspeed of your aircraft. Going into the wind, your ground speed will be only half that of still conditions and so the journey will take twice at long. A one-hour flight (in still conditions) would take two hours. To make up all that lost time on the way home you would need to be moving very fast. However the tail-wind would only make your ground speed 1.5 times your normal speed and you would return along the 1-hour path in only 40 minutes to give a round trip time of 2hrs 40mins.

In both cases, and therefore all variations of wind directions and for all wind strengths above 'calm', your total round-trip time will always be longer.

If you are only concerned with differences in airspeed (and not how long it will take to cover a certain distance), then a direct tail wind will help in one direction and a direct head wind will hurt in the other direct with commensurate changes to net ground speed. However, as soon as there is any cross-wind component, that will hurt in both directions, hence the 4kt discrepancy mentioned in the question.

David0337
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  • if you have perfect sidewind, I don't see how you can have increased headwind. also, while vaguely related, it does not directly address the question. – Federico Apr 22 '17 at 12:27
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    Because with a side wind, to maintain the intended course, you would need to head slightly into the wind (and the stronger the wind, the greater the deviation) and consequently for a given airspeed, you would cover less ground in the intended direction. – David0337 Apr 22 '17 at 13:05
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    @David0337 I concur, your effort in correcting is wasted as far as desired direction is concerned, with a head or tail wind. Though you gain in speed with a tail wind you would gain even more if ONLY the forward component of the same wind had been directly behind you. – KalleMP Apr 22 '17 at 13:37
  • How did you come up with 40 minutes rather than 30? – Steve V. Apr 22 '17 at 16:02
  • @Steve V My assumption was that the wind speed was half that of the aircraft's airspeed. Let's assume a 60km distance at 60km/h. In still conditions this would take 1hr (t=s/v). With a 30km/h headwind it would be 60/(60-30) = 2hrs. With a 30km/h tailwind, it would be 60/(60+30)=2/3hr=40mins. To do the run in 30 minutes would suggest a 60km/h tail wind but that would make the upwind leg rather more difficult. – David0337 Apr 24 '17 at 05:10
  • You can replicate the results from the E6B calculation with simple triangles to do the vector addition. https://www.triangle-calculator.com/?what=&q=a%3D260+A%3D110+b%3D35&submit=Solve (and then swap the A=110 with A=70 for the return leg). – David0337 Apr 24 '17 at 06:12
  • @David0337 - Oooops, you're exactly right. That'll teach me to try to do E6-b math in my head rather than going and getting the actual E6-b. – Steve V. Apr 24 '17 at 10:51