Hi,
I am just trying to figure out how to calculate the crosswind as accurate as possible without using the whistwheel. Does anyone know how to do this? From my understanding, if the wind is 30 degree off, the crosswind is 1/2, 45 degree off is 2/3, 60 degree is 9/10, and 90 degree is all crosswind. In this question the wind is 73 degree off, so how can I get a good approximate on this one without whistwheel?
Here is the question:

An airplane along a track of 093 degrees, wind 020/20, cruises speed (TAS) 110 kt. Fuel-endurance 6 hours (4 hours 15 min plus reserve).

1. The heading to steer out is
a. 073 degrees
b. 083 degrees
c. 093 degrees
d. 103 degrees

2. The heading to steer home is
a. 273 degrees
b. 283 degrees
c. 293 degrees
d. 303 degrees

Hi,

There's a few rules of thumb floating about just like the one you mentioned. Another option is to use the headwind/crosswind tables in ERSA Gen-Con-6 which gives a quick look up for the wind components.

To answer the question you described you actually want the drift angle to calculate a heading correction. A few rules of thumb exist but you'd be best off asking your instructor which one he/she recommends.

Here's one that might be useful:

First, find your Maximum drift i.e. the drift you would expect if the wind was all crosswind:

Max Drift (in degrees) = Windspeed x TAS / 60

Then use the clock code to work out how much of that max drift actually applies. The clock code can be used by imagining a clock, taking the number of degrees the wind is off your nose and pretending these degrees are minutes on the clock face.

If the wind is 15 degrees off the nose look at the 15 minute mark on your clock face: 15 minutes is quarter of an hour so the drift you can expect is a quarter of the max drift you calculated before.

If the wind is 30 degrees off the nose, look at the 30 minute mark on your clock face: 30 minutes is half an hour so the drift you can expect is half the max drift you calculated before.

Similarly:
20 degrees off -> 20 minutes -> 1/3 of an hour -> drift = 1/3 max drift.
45 degrees off -> 45 minutes -> 3/4 of an hour -> drift = 3/4 max drift.

Again, that's just one suggestion but there are others. Please remember, rules of thumb are there to help you make an educated guess! I'd not recommend using a rule of thumb over the flight computer in the exam. You may well find the resulting loss of precision lands you too close to a wrong answer option.

Cheers,

Rich

Your method works like a charm. In this question, the wind is 73 degree off so it is 73 minutes on the clock and the maximum drift is (20*110)/60 = 36.6 or 37 degree.
73 minutes on the clock is about a quarter, so 37*1/4 = 9.25 or 9 degree.
So the correct answer is 93-9 degree is 84 degree to steer out, did I do this correctly?
Do I have to use the same method to fine the heading to steer home?
Thank you very much

Ah, the dangers of a rule of thumb You can't wrap past the hour mark. If you did, a wind 61 degrees of the nose would mean 1 minute or 1/60th of the max drift. In other words virtually no drift correction needed at all! With this rule of thumb, you say anything over 60 degrees off the nose means the equivalent of max drift. Yes, you can use this for the return journey in just the same way. Just remember the direction of drift is reversed.

Getting back to your example, that means you need to kick in 37 degrees of drift correction. This unfortunately doesn't match any of the exam question answers you had.

Check your drift correction needed using the whizz wheel and see what you get. You'll see the rule of thumb is not accurate enough for the exact science of theory exams. If this was the first step in a longer question you'd find your final answer would be way out.

I'm afraid for your exams you'll not be able to escape using your whizz wheel. It is part of the curriculum.

Cheers,

Rich

I am actually preparing for airline selection in the future so I have to be able to figure out how to get the estimate of wind as accurate as possible without the wheel. I just want to confirm that the wind is less than 60 degree of the nose, this rule of thumb would not work. For some reasons, I applied this rule for the questions and the answer is only one degree off. Maybe It was a fluke.

maximum drift is (20*110)/60 = 36.6 or 37 degree.
73 minutes on the clock is 13 minutes after one hour so what I did was I rounded off to 15 which is a quarter, so 37*1/4 = 9.25 or 9 degree.

I have one more question to ask, could you please guide me to the right direction?
It is the same question but this time I have to find the radius of action.

An airplane along a track of 093 degrees, wind 020/20, cruises speed (TAS) 110 kt. Fuel-endurance 6 hours (4 hours 15 min plus reserve)
The formula for this is EOH/O+H, so what I did was
(4.15*ground speed on*ground speed home)/(ground speed on+ground speed home)
The correct answer is 226 NM. The only thing that I am stuck with is the wind.

Hi Richard and Bob,

Just on BT CPL Nav Issue 3 Progress Test 4 Question 3 there is a question pertaining to a crosswind component:

If the Wind is 220 deg magnetic at 20 kt and the runway to be used in runway 16 the crosswind component is:

a) 17 kt from the left
b) 17 kt from the right
c) 10 kt from the left
d) 10 kt from the right.

Now the answer specified on page 6.6 says it should be b - 17 kt from the right.

I am trying to discover the most efficient E6B WORKING of this, since when I wizz wheeled it I got a different answer (d - 10 kts from the right).

Firstly I tried setting the wizz wheel to the effective WIND HEADING (like the TMG calc) and put the wind dot straight up 20 kts higher than the ground speed dot on any arbitary line.

Then I rotated the dial to 160 which is the direction of the runway being considered for landing or takeoff. When I do this it shows me a 10 kt crosswind from the right NOT a 17 kt wind - see attached E6B pic.

Now another question, on another part of the E6B there is a cross wind chart (not discussed in the CPL Nav curric that I recall) that was a big part of working out crosswinds at PPL level. Should we use this and compare?

See attachment 2. See if I compare 160 with 220 I have a difference of 60 degrees magnetic. If I feed that into the cross wind calculator attached as attachment 2, entering at 60 degrees (relative wind angle) my question is in this case how do I cross compare, what headwind component do I enter given than I am now confused by the above and the wind given of 20 kts was from 220 not the final heading of 160.

What method are you suggesting for the exam?

Just happens to be a question I got wrong but cant figure out why... I did really well with almost all the others.

Brook

Hi Brook,

Sounds like you're having a brain fart with the E6B. Don't worry, I have 'em all the time

Where are you reading that 10 kts of right of crosswind you mentioned?

You need to use the crosswind table on the reverse side of the E6B or you can even use the Head/Crosswind component table in ERSA GEN-CON (5) on page GEN-CON-6.

You can also use the rule of thumb for crosswinds as a double check on your answer. A wind more than 60 degrees off the nose is nearly all crosswind. A crosswind of only 10 kts was therefore, in this case, impossible. It's only half the windspeed which is the crosswind component you'd expect with a crosswind about 30 degrees off the nose.

Cheers,

Rich

@UnusualAttitude

The Clock-face rule of thumb should work reasonably well for calculating wind correction angle for winds less than 60 kts off the nose. Just remember winds more than 60 degrees off the nose are virtually all crosswind and so you will use all the max drift to correct for it.

You can also use the clock face for a rough estimate of the actual crosswind component:
e.g.
15 degrees off the nose, crosswind is 1/4 the windspeed
30 degrees off the nose, crosswind is half the wind speed
45 degrees off the nose, crosswind is 3/4 wind speed
60+ degrees off the nose, crosswind is the same as the windspeed.

When calculating groundspeeds and crosswind speeds, it might help to remember a few numbers:

Crosswind component = sin (angle off the nose) x Wind speed
30 degrees off the nose: crosswind is 0.5 x Wind Speed
50 degrees off the nose: crosswind is 0.75 x Wind speed
60 degrees off the nose: crosswind is 0.9 x Wind Speed
80 degrees off the nose: crosswind is Wind Speed (all of it!)

Headwind component = cos (angle off the nose) x Wind Speed
30 degrees off the nose: headwind is 0.9 x Wind Speed (about 0.9)
45 degrees off the nose: headwind is 0.7 x Wind Speed
60 degrees off the nose: headwind is 0.5 x Wind Speed
75 degrees off the nose: headwind is 0.3 x Wind Speed

In your example, the wind is 73 degrees off, so the headwind is about a third of 20 kts, say 7 kts. GS on = 110 - 7 = 103 kt. There's a tailwind home so the GS Home = 110 + 7 = 117 kt

A safe endurance of 4 hr 15 minutes is actually 4.25 hours (not 4.15) so watch out for that.

Using the formula for distance to PNR:

Distance to PNR = (4.25 x 103 x 117) / 220 = 233 nm.

Cheers,

Rich

Hi Richard,

Thanks for that - basically what I did was (use the wind whizzwheel and not the table) set the wind direction and velocity on the E6B to the current wind which was (see the attached pic to the previous post and look for the 10 kt wind dot)

I thought I was being smart, but looks like I wasnt. I set the dial to 220, then marked a dot up 20 kts from the GS grommet. Then I rotated to 160 deg, then I saw 10 kt. You know the section on page 3.6 about using the navigation computer to find heading and ground speed - I thought I could be smart and somehow we could use this for crosswind.. (see previously attached pic or try it yourself to see how easy this mistake was

So that makes sense, as always use the wind / crosswind component grid on the E6B (believe it or not on mine its actually on the front!) but just tell me if I am working this correctly then to reach the 17 kts as I may have confused myself: (maybe I am overthinking this) - copy of my E6B wind chart is the second attachment to the last post)

So I take the difference between the two headings (wind at 220M/20 and RWY at 160M - which is a difference of 60 degrees. So I enter the RELATIVE WIND ANGLE section of the chart at 60% (right so far?)

I follow this across diagonally down to the left along the 60 deg mark until I reach the 20 kt arc line (HEADWIND COMPONENT AXIS) - which takes me to the high side of 15 which I can safely assume without greater granularity and resolution is 17kt?

Is this the way of working you were suggesting? It seems to work fine to me, its just the granularity of the answer may vary based on the subscale being in 10 kt increments with a non marked 5 kt sub increment.

Please follow me through on this just one of these to make sure I am not making any incorrect assumptions, just use the attachment 2 on the last post to ensure you are OK with my thought patter and working process, I know most E6B's to be similiar. Does the above make sense to you in terms of my
1) choosing to use the Wind Component Grid, and then the
2)working of that grid as explained above????

Sorry to be high maintenance, I just want to ensure I don't have any flawed logic going on.

I think you were spot on re the brain fart - It's just Nav wrecking havoc with my electricals (PNESD - Pre Navigation Exam Stress Disorder but I am mostly on top of it now thanks to perseverence and the great way it IS explained in the manual if you take the time to stop, read and THINK about it.

Might be a couple of easier ways forward in a couple of areas in the Nav manual, but no urgency, just on next revision. Its the hardest subject I have encountered so far to self study, I think I was going great till one in 60, even the first part of that was simple, it was the additional angles and concepts in 1 in 60 I found gave me a good workout. This might be another one to put online sooner than later if you haven't already considered it.

If anything, I reckon on next revision it might just pay to have a few more of the actual exam question workings elaborated on, that's it. Sometimes I "reverse engineer an answer" until I actually get it if you know what I mean. Once I know it back to front like I could teach it, I remember it for life most of the time!

Thanks

Brook

Hi Brook,

Yep, your working is correct. In the wind component grid, find the angle line corresponding to the situation (60 degrees in this case). Follow the relevant wind angle line inwards towards the origin of the graph until it intersects the wind speed arc. You can read the value of each arc off the scale on the headwind or crosswind axes. Our example is 20 kts.

Drop the line down to read off the crosswind or go horizontally across to read off the head wind component (17 and 10 kt respectively) Please see the attached image for the working.

The 10 kts you read off the wind side of the E6B (image #1 in your post) is the headwind component not the crosswind. Also, as this example shows, headwind + crosswind do not equal the total speed. For example, let's say you have 20 kts of wind and you find there's 5 kts of headwind. This does not mean there's going to be 15kt of crosswind.

Just as your example problem has shown, we have 20kt wind but 10 kt headwind and 17kt crosswind. You might think "But that's 27kt total wind!! Where's the extra 7kt coming from?"

Well, the reason is, because of the direction of the wind, it's actually headwind and crosswind simultaneously. Don't worry too much about it, just remember to use your E6B, the wind component table in ERSA (or even the sin/cos function of your calculator ) to get the components you need.

Cheers,

Rich