Friday, March 18, 2005

A Third Post about Timing Holds

In two recent posts I've been dissecting hold timing. The goal is to fly a racetrack pattern such that, despite winds, it takes exactly one minute to fly the inbound leg. Clearly, if the first inbound leg takes you less than a minute, you need to fly longer outbound, and if the first inbound leg takes more than one minute, you need to fly for a shorter time outbound. The challenge lies in how much more or less.

As I mentioned before, I subtract two thirds of any excess inbound time from my outbound leg, and add one and a third times any shortfall. A colleague just hacks off half the excess and adds all the shortfall. He's good at math, and conscientious, so I pulled out Excel to find out what he knew that I didn't.

As I made up the table below I realized that if the first calculation gives me an inbound time of less than 20 seconds inbound, I fly for 20 seconds, which helps me not overshoot. I also don't make a correction based on the very first inbound leg, because that's part of the entry, and not to be trusted. I probably have more arbitrary rules like these that I haven't even noticed.

The values in the table below use the inbound leg distance calculations from my last hold post and the formula distance equals speed times time, with speed being the true airspeed of the airplane plus a tailwind or minus a headwind.

IB #1 time OB #2 my way OB #2 his way IB #2 my way IB #2 his way OB #3 my way OB #3 his way IB #3 my way IB #3 his way
10 126 110 43 35 149 134 54 47
20 113 100 52 44 125 116 58 53
30 100 90 55 49 106 101 59 56
35 93 85 61 54 93 90 60 59
40 87 80 61 56 86 84 61 59
50 73 70 60 57 73 72 60 60
60 60 60 60 60 60 60 60 60
75 50 53 64 66 48 50 59 57
83 45 49 65 69 42 44 59 57
91 39 45 65 71 36 39 59 56
113 24 33 62 75 23 26 60 56
150 20 15 80 71 7 9 64 59

The above data leads me to conclude that with light to medium winds, both methods work fine, mine a bit better, but not worth bragging about. For strong headwinds on the outbound, giving an initially very short times on the inbound, adding one and a third the shortfall works better than just adding the shortfall. For strong tailwinds on the outbound, subtracting the whole excess works faster than subtracting two thirds of it. Which makes perfect sense.

I initially set up my spreadsheet looking at aircraft of different speeds in different winds, but discovered that it makes no difference what combination of wind speed and airspeed produce a given inbound time, any initial inbound time corresponds to a particular outbound time that will give a one minute inbound. One could make a computerized hold timer that contained a database, such that when you hit stop on the inbound leg it automatically looked up the value and set up a countdown timer for the appropriate time on the outbound leg. Or you could have a printed table stuffed in with your approach plates, giving the same effect.

No one needs to, because as long as you are correcting in the appropriate direction, a few iterations will converge on the appropriate outbound time. I even tried setting up my spreadsheet so that the correction was a random number and if I limited the number so that the values didn't overshoot, after four turns in the hold, the timing was within a few seconds.

Why do we do this? Why don't they protect airspace based on a one minute outbound, and tracking inbound, so no one would have to calculate anything? I think they even do that in Europe. Perhaps it's a bit like the practice of putting mirrors in elevator lobbies. It gives you something to do while you are waiting, thus making the wait seem shorter.

I've had enough of holds for now, even though I haven't touched correction angles or hold entries.


Anonymous said...

They usually do things better in Europe, except for the UK, where we seem inexplicably bound to American practice in most things

Anonymous said...

These 'Holding Timing' posts make my head hurt! How happy am I for FMS that figures these things out and just leaves me alone!

Anonymous said...

I was taught to not worry about the actual time of the inbound leg and just add whatever the headwind component is to 60 to give me the time in seconds to track outbound.

However if you wanted to be precisely correct you could do what I have detailed below (with explanations). Trying to work it out in the air while maintaining straight and level on the inbound would be a challenge though. (The working and notes were more for my benefit as opposed to yours btw.)

GS(I) = inbound groundspeed
GS(O) = outbound groundspeed
w = wind component in the direction of the beacon parallel to the inbound leg of the hold

GS(I) = TAS + w
GS(O) = TAS - w

You said you want the inbound leg to be exactly one minute from when you roll wings level at the end of the turn inbound.

Therefore let's call the distance inbound d(I) and we can say that:~
d(I) = st
= GS(I) × 1/60
= GS(I)/60

Assuming that there is a tailwind on the inbound part of the hold you will have already rolled wings level before getting abeam of the beacon to head outbound due to the wind pushing you further inbound in the turn. So the timing will start from when you are abeam the beacon. We know that the distance travelled has to be the same as on the inbound plus a little extra.

Let us break this down into two parts. One part is you travelling to a point abeam where you would roll out of your inbound turn to do your one minute inbound. I am going to call this distance outbound one or d(O1). However if you start turning here, due to the wind, we know that you will end up too close to the beacon and your inbound leg will take less than one minute. So you need to fly a little further and I will call this distance required distance outbound two or d(O2).

We know that d(O1) = d(I) so we can work out the time taken to fly this portion of the outbound. Let's call the time required t(O1):~
t(O1) = d/s
= d(O1)/GS(O)
= d(I)/GS(O)
= GS(I)/60 / GS(O) (since d(I) = GS(I)/60)
= GS(I) / 60GS(O) (rearranging)

The little extra distance travelled needs to be just enough to allow for the wind pushing us towards the beacon as we turn inbound. We know that the turn takes one minute so we can figure out how much you will get pushed by the wind.

d(O2) = st
= w × 1/60
= w/60

Now we can work out the extra time needed to fly outbound, which we will call t(O2):~
t(O2) = d/s
= w/60 / GS(O)
= w / 60GS(O) (rearranging)

So the total outbound time, t(O):~
= t(O1) + t(O2)
= GS(I) / 60GS(O) + w / 60GS(O)
= (GS(I) + w) / 60GS(O)

But currently our speeds are in distances per hour so our time will be in hours. Therefore the total time in minutes which is what we want is 60 times this and:~
= (GS(I) + w) / GS(O)

Or in English your time outbound is the ratio of your inbound groundspeed plus the component of wind acting inbound all over your outbound groundspeed.

Alternatively we could substitute back in the equations with our TASs in which case:~
t(O) = (TAS + w + w) / (TAS - w)
= (TAS + 2w) / (TAS - w)

This will only work if you have a tailwind on the inbound leg. If you have a headwind inbound you have the complication of already being past the abeam point when you start your outbound timing which will screw things up. I'm sure it would be possible to figure out but my brain hurts right now so maybe some other time.

Interestingly enough though, as I mentioned to start with, the rule of thumb I was taught for your outbound timing is that you should travel outbound for 60 seconds plus one second per knot of headwind. I am learning to check rules of thumb though as they have been wrong in the past. This one is no exception as I have just done the maths (which is quite long but relatively easy in this case) and this rule only gives you a one minute inbound if your TAS happens to be 180 knots plus the wind (or the wind happens to be your TAS minus 180 knots ). The plane I am training in has a holding speed of 120 knots so the rule we have been taught will not work precisely, although I am sure the difference is probably so slight as to be inconsequential.

As an aside though, is the inbound supposed to take one minute or is one lap of the hold supposed to take four minutes?

Aviatrix said...

The inbound is supposed to take one minute.