Headwind Component
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Headwind Component
I know the rules of thumb for calculating the x-wind component on T/O ie 1/6 or 1/3 or angle plus 20% but I cant figure out how to reverse it and get the headwind component. Any help!! I know they are related.
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Actually R2000, they're not related. If you know the crosswind component only, you cannot figure out the headwind component. For example, if the wind is 60 degrees off the nose at 10 knots, then the crosswind component is 10*sin(60) = 8.67 knots. However, if all you know is the crosswind component, for example 8.67 knots, then this could equate to zero headwind if the actual wind is a 90 degree crosswind at 8.67 knots, or it could equate to a headwind of 5.00 knots (10*cos(60) = 5 knots) if the actual wind is 60 degrees off the nose at 10 knots. In fact, if all you know is the crosswind component, there is an infinite number of wind/angle possibilities that fit this crosswind component.
Hawk
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R2000
This is a curious source of confusion for many pilots, including (and in my experience, in particular) some very experienced ones.
Here's how to do it:
First of all use the clock-face rule of thumb: 15 degrees = 25%, 30->50% 45->75% etc.
In the same way that one needs to consider how many degrees off the nose the wind is for x-wind component, for headwind, we need to consider the angular distance between the wind and the wingtip.
For instance, consider lining up on runway 27 somewhere or other. If the wind was 240/10, then the wind is 30 degrees (or thereabouts!) from the nose, so 10x0.5=5 kts x-wind.
The wind is 60 degrees from the wingtips, however, which are pointing N/S - 360, 180 degrees. So in this case, the clock-face r.o.t tells us consider all of the wind as h-wind = 10kts. As pilotmike correctly points out, the real number is closer to 90%, but never mind!
Another example, still on rwy 27:
Wind 340/20
X-wind 20 kts (70 degrees from nose, so 1 x 20=20)
H-Wind 6-7 kts (20 degrees from wingtip, so 1/3 x 20 = 6 and a bit)
You'll also notice that in the above example 70+20=90, and in the first one 60+30=90. No big surprise for those who have a recollection of trigonometry!
In my experience it works, more quickly and accurately than the captains calculator. That he's forgotten how to use.
sin/cos on your calculator, are, of course, fine if you have one handy, and the time to use it.
Hope this helps
CC
This is a curious source of confusion for many pilots, including (and in my experience, in particular) some very experienced ones.
Here's how to do it:
First of all use the clock-face rule of thumb: 15 degrees = 25%, 30->50% 45->75% etc.
In the same way that one needs to consider how many degrees off the nose the wind is for x-wind component, for headwind, we need to consider the angular distance between the wind and the wingtip.
For instance, consider lining up on runway 27 somewhere or other. If the wind was 240/10, then the wind is 30 degrees (or thereabouts!) from the nose, so 10x0.5=5 kts x-wind.
The wind is 60 degrees from the wingtips, however, which are pointing N/S - 360, 180 degrees. So in this case, the clock-face r.o.t tells us consider all of the wind as h-wind = 10kts. As pilotmike correctly points out, the real number is closer to 90%, but never mind!
Another example, still on rwy 27:
Wind 340/20
X-wind 20 kts (70 degrees from nose, so 1 x 20=20)
H-Wind 6-7 kts (20 degrees from wingtip, so 1/3 x 20 = 6 and a bit)
You'll also notice that in the above example 70+20=90, and in the first one 60+30=90. No big surprise for those who have a recollection of trigonometry!
In my experience it works, more quickly and accurately than the captains calculator. That he's forgotten how to use.
sin/cos on your calculator, are, of course, fine if you have one handy, and the time to use it.
Hope this helps
CC
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@Chuffer
I'm afraid we both missed the point.
R2000 wants to figure out the headwind component based on the x-wind component, which in fact is not possible as explained by hawk37.
regards
I'm afraid we both missed the point.
R2000 wants to figure out the headwind component based on the x-wind component, which in fact is not possible as explained by hawk37.
regards
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Making all this WAY to difficult!!
Remember for the performace graphs the wind is factored already. HEADWIND component is factored 50% and the TAILWIND component is factored 150%.
So within 45 deg of the nose use ALL, more than 45 deg, use 1/2.
Remember for the performace graphs the wind is factored already. HEADWIND component is factored 50% and the TAILWIND component is factored 150%.
So within 45 deg of the nose use ALL, more than 45 deg, use 1/2.
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When are we going to be given only the x-wind component? We get given a wind velocity from which we can workm out both head and x-wind. Unless it is for theoretical purposes, there is no need to work backwards.
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When you're in the sim and the pressure is on and the workload high 
you wont even be able to spell sin or cos, get the simplest method possible 
you wont even be able to spell sin or cos, get the simplest method possible
Last edited by Not Nightowl; 9th Nov 2012 at 18:23.
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I think sin/cos is the simplest, the calculator has to do the job.
sin of (wind direction - runway heading) x wind speed for crosswind
cos of (wind direction - runway heading) x wind speed for head/tail wind
Runway 32 wind from 280 at 15kts
crosswind 9,64kts
headwind 11,5kts
sin of (wind direction - runway heading) x wind speed for crosswind
cos of (wind direction - runway heading) x wind speed for head/tail wind
Runway 32 wind from 280 at 15kts
crosswind 9,64kts
headwind 11,5kts
Last edited by AirGek; 9th Nov 2012 at 18:22.
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Reduce the whole thing to remembering three numbers: 0.5, 0.7 & 0.9. These numbers can be used as the sines/co-sines, as appropriate, for 30, 45 and 60 degrees. Use them as appropriate. They are accurate enough for government work.
PM
PM
If you are taking off to the north ( 000) and the wind is 030/10 then I assume you know that 30 degrees is 0.5 and therefore you have 5kts crosswind.
If the wind was the same but you were departing off 09 I bet you could work out the crosswind too, 60 degrees off therefore 80% = 8kts.
The crosswind in the second example is the headwind in the first example. So all you need to do is subtract the 30 from 90 and then apply the sin rule of thumb you have and you have your headwind.
All you need to learn is the basic sin rules.
30=0.5
40=0.6
50=0.7
60=0.8
Another example to show you;
Departing to the north.
Wind 050/10
Crosswind = 10x0.7 = 7kts
Headwind = (90-50) = 40 now apply the sin rule of thumb to get 0.6
= 0.6x 10 = 6kts.
Simple
the only thing you need to know is the four sin rules and the fact that you have to subtract the "degrees off" from 90 if it is a headwind you are after.
If the wind was the same but you were departing off 09 I bet you could work out the crosswind too, 60 degrees off therefore 80% = 8kts.
The crosswind in the second example is the headwind in the first example. So all you need to do is subtract the 30 from 90 and then apply the sin rule of thumb you have and you have your headwind.
All you need to learn is the basic sin rules.
30=0.5
40=0.6
50=0.7
60=0.8
Another example to show you;
Departing to the north.
Wind 050/10
Crosswind = 10x0.7 = 7kts
Headwind = (90-50) = 40 now apply the sin rule of thumb to get 0.6
= 0.6x 10 = 6kts.
Simple
the only thing you need to know is the four sin rules and the fact that you have to subtract the "degrees off" from 90 if it is a headwind you are after.
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Exactly, Capt Fathom! I still carry one in my pocket.
And for the modern pilot, there are a number of iPhone/iPad apps that do the job well, including the E6B calculator. Pilot Wizz is a free one that is quite handy for most things aeronautical.
Beats all that strenuous mental calculating.
And for the modern pilot, there are a number of iPhone/iPad apps that do the job well, including the E6B calculator. Pilot Wizz is a free one that is quite handy for most things aeronautical.
Beats all that strenuous mental calculating.
Originally Posted by Piltdown Man
Reduce the whole thing to remembering three numbers: 0.5, 0.7 & 0.9. These numbers can be used as the sines/co-sines, as appropriate, for 30, 45 and 60 degrees. Use them as appropriate. They are accurate enough for government work.
Framer, Hefield, 60° is 0.9, not 0.8.
Yeah it's not exactly right but the important thing is that it's simple and quick and very close to being right. If the situation is that critical that you need to use the exact sin of sixty (0.866) , then a mental rule of thumb is not the best method anyway. An FX-82c might be more appropriate.
If you think that the crosswind is going to be a factor then before your approach work out the maximum you can accept for each wind direction. e.g.
Runway 27 (magnetic heading 270 degrees)
Max crosswind allowed 25 knots
Wind 220 degrees, max strength 32 knots
Wind 210 degrees, max strength 28 knots
Wind 200 degrees, max strength 26 knots
Wind 190 degrees, max strength 25 knots
etc...
It'll only take 5 mins to do as part of your approach preparation and it means you won't be caught out when tower gives you that final wind check passing 200ft. It's not the time to be doing maths!
Runway 27 (magnetic heading 270 degrees)
Max crosswind allowed 25 knots
Wind 220 degrees, max strength 32 knots
Wind 210 degrees, max strength 28 knots
Wind 200 degrees, max strength 26 knots
Wind 190 degrees, max strength 25 knots
etc...
It'll only take 5 mins to do as part of your approach preparation and it means you won't be caught out when tower gives you that final wind check passing 200ft. It's not the time to be doing maths!
