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Wind-corrected timings in the hold

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Wind-corrected timings in the hold

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Old 14th Mar 2004, 01:26
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Wind-corrected timings in the hold

Wind 210/20
Airspeed 120kt
Max Drift 10 degrees

Inbound track 270

Outbound track 090
Outbound heading 120 degrees (3 x max drift applied into wind)

Outbound timing = ?????

The wind-to-track angle is 60 degrees
The wind-to-heading angle is 90 degrees

I don't know which one to use.

Cam anyone help?

Thanks in anticipation!
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Old 14th Mar 2004, 13:24
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Thanks Notso!

However, still a bit confused. Thought the timing correction would be nil (wind 90 degrees off) or 20 seconds (20 knots at 60 degrees off, therefore giving the full force as a tailwind).

You're absolutely right about theoretical accuracy and real life.................and that calmed me down a bit!
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Old 16th Mar 2004, 19:52
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It's worth remembering that for the IR you only have to make the adjustments in the correct sense. The adjustments themselves do not have to necessarily be correct.

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Old 16th Mar 2004, 20:23
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There's not much to it. Even if you didn't know where the wind was coming from, you can easily make a nice 4minute'ish pattern.

Once tracking inbound for the first time, start your time. Also note the wind correction required, since you'll need that for the outbound correction (as you already stated). Now, once hitting the fix for your turn outbound, look at the time (before turning). Was your inbound leg longer than 1 min? Simply subtract 1 minute from it, half the rest, and subtract it from the outbound leg. On the other hand if your inbound leg was short of 1 min, simply note the difference, half it, and add it to your outbound leg. You might need 2 turns in the pattern to get the first timing estimate, but it ain't a big deal. If it takes you more turns, then thats not a big deal either!

Anyway, here's an example:

Lets say you tracked inbound for 1min 26sec (Quite a nice headwind). Lets take the excess of 26 seconds and half it = 13 sec. Lets subtract that from the standard outbound leg (1 min), and we get 47 seconds.

Just remember that any correction you make has to be made on the outbound leg, so that your inbound leg equals to 1 min.
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Old 16th Mar 2004, 22:11
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This seems quite an interesting thread. I'd like to ask a question about mattpilot's post though. I've never been in a hold in my life (well not flying it anyway) so I'm not saying anyone is wrong, I'd just like something explained please. Where would you start the timing for the inbound leg on your entry to the pattern from a direct entry? I followed the example he gave but I'm not sure where you'd start the timing, as I'm under the impression that the hold would be defined by the holding fix, the inbound track, the direction of turns and the leg times, or in special cases a distance, but usually 1 minute legs. If there wasn't a specified distance for the inbound leg then how do you know where to start timing? And even if there is a distance then surely timing is somewhat irrelevant now since it's the distance that counts?? I'd be grateful if someone could point out what I'm missing thanks. Capt Wannabe, I didn't know that about the IR either so thanks for that as well.

Thanks very much for your time.
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Old 16th Mar 2004, 23:23
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Sorry for being vague

A hold's length is usually detirmined by the time, unless its otherwise specified to be a DME hold (either depicted or instructed by ATC), where you would then use the DME distance.

So an aircraft cruising at 100kts will make a much smaller circle than an aircraft cruising at 200kts. Now if the controller specifies a DME hold of 5miles, then it will take the aircraft going 100kts a little bit longer to accomplish that hold than it would take the 200kts aircraft.



Where would you start the timing for the inbound leg on your entry to the pattern from a direct entry?
You generaly always start your timing when the wings are level on your desired heading. Now this isn't supposed to be rocket science, so even if you are off course (due to wind) and have to make a correction to get back on course, you would still start your time when you completed your 180 degree turn. Though as i said, you don't have to be very exact - a lot of estimating won't harm you.


You specifically asked about the direct entry - here's an example:

Say your already on the holding course and cruising towards the fix. Start your time when entering the outbound turn. After established outbound, time 1 minute and go for it. At this point you don't really have a wind correction so you just have to guess based on the information you have (preflight weather briefing), or don't do anything. though this can blow you across your course (happened to me in actual) - so just watch out for it. After you've flown outbound for 1 minute, reset the time and make your inbound turn (there is no point IMO for setting time in a turn, other than if you have an instrument failure to give you a heads up once the time is up). Once established inbound and wings level, start your time again. Now this is the time i refered to in my original post. Say you arrive at the fix in 40 seconds. thats 20 seconds early - half it, and add it to your outbound leg. So the next time you level your wings for the outbound turn, fly outbound for 1min10sec. This should get you closer to your 1 minute inbound mark you should ultimately obtain.
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Old 17th Mar 2004, 13:56
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There's one issue that hasn't really been addressed so far, which is what the timing in the hold is intended to achieve. The US way, as mattpilot describes, is to adjust the outbound leg so that the inbound leg is 1 min long. The ICAO PANS-OPS way is simply a 1 min outbound "adjusted for wind". I've always interpreted that as half the correction that the US uses. In reality, as others have suggested, nobody cares.

To answer Hansard's original question, your groundspeed along the outbound leg is airspeed * cos(30 deg) or airspeed/1.15. Therefore to get a still-air minute of distance, add 15% or about 10 seconds.
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Old 17th Mar 2004, 18:49
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Adjust outbound by 1 sec / 1 min of outbound timing / guestimated knot of head or tailwind component & stuff trying to time the inbound.

Alternatively use 0, 10, or 20 sec as standard adjustments if the wind is light, medium or strong. Works accurately enough.
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Old 18th Mar 2004, 09:47
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Thanks for the replies. I understand your explaination mattpilot, thanks very much. From your previous post I got the impression that you had some method of estimating the timing corrections from your initial entry into the hold on the inbound leg. I see what you're actually saying is take on board what you learned from the 1st time round the pattern and apply it to the outbound leg on the 2nd time. Guess you're not psychic after all

Thanks again for the replies.
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