Drift estimation

[Cron]

Would be most grateful for a rule of thumb/estimation technique for drift.

Sure I saw one on here some time ago but can't find it.

Thank you.

[Leo45]

First of all, calculate the Max Drift, knowing the windspeed and the aircraft TAS.
Max Driftº=windspeed X 60/TAS
Then assess the angle Aº between the desired track and the wind in order to work out the Drift along the required track.
Drift= Max Drift X Sin Aº
Example
Wind is 360º/30kts ,TAS= 180 kts, DTk= 330º
Max Driftº=30 X 60/180= 10º
Angle Aº is 30º, Sin30º=0.5 so Drift= 10ºx .5= 5º
Therefore, Hdg should be 335º to maintain a track of 330º
For mental calculations use the following figures: Sin30º= .5, Sin45º= .7, Sin60º= 1, and for 60/TAS: 60/100=.6 , 60/120 =.5 , 60/180= 1/3 , 60/240= 1/4 , 60/300= 1/2

[Gillespie]

Leo45 is bang on. I use the exact same method, but to help you remember think of the clock face..

eg, if the wind is 30* off the nose, 30 minutes on the clock is half it's face therefore use half the max drift. 45* = 3/4 of the face so use 0.75ish..so on and so forth...

you can also use the clock face method to mentally calculate ground speed. For wind angles of 0-30* use all the wind, for 30 t0 45 use 3/4, for 45-60 use 1/2, for 60 to 70ish use 1/4 and for any angle beyond ignore it.

[Cron]

Thanks L and G - that's just what I wanted. That clock business is the business.

Cron

[High Wing Drifter]

First of all, calculate the Max Drift, knowing the windspeed and the aircraft TAS.
Max Driftº=windspeed X 60/TAS
Just a personal minor addition that I find useful that doesn't necessarily work for others. I work with the same equation but in this form: MD = WindKts/(TAS/60). This is because for mental calcs I personally find working in NMs/Min much easier than NMs/Hour (100 = 1.5, 120 =2, 150=2.5, etc). The TAS/60 plugs straight into lots of other mental calcs so if I am already thinking in NMs/Min then its a doddle. For example, for me 30/2.5 (or 60/5 as I would do it in my head) is much easier than 30*60/150

[Leo45]

There isn't much mental arithmetics involved.

Actually, the term (60/TAS) should be learned by heart for a set of speeds.

TAS=60 kts ---> 60/TAS=1

TAS=80 kts ---> 60/TAS=3/4

TAs=90 kts ---> 60/TAS=2/3

TAS=100 kts --> 60/TAS=.6

TAS=120 kts --> 60/TAS= 1/2

TAS=150 kts --> 60/TAS= .4

TAS=180 kts --> 60/TAS=1/3

TAS=240 kts --> 60/TAS=1/4

TAS=300 kts --> 60/TAS=.2

It is very useful not only for calculating Max Drift (i.e Max Driftº= Windspeed X (60/TAS)) but also for calculating ETAs/EETs!

Time (min) = Distance (NM) X (60/Ground Speed(kts))

Example: 45 NN to run , GS= 100 kts, therefore Time to go= 45 X .6 = 27min

If the ground speed was 110 kts it's only an extra 10% speed increase therefore, you would take 27min - 2 min= 25 min to cover 45 NM.

How do we assess ground speed? Well, the ground speed is roughly:

GS= TAS + TWC or TAS - HWC

To calculate the tail/head wind component, use: Tail/Head Wind Component = Wind Speed X Cos Aº

We can assume that for : Aº= 30º, Sin Aº= 0.5 Cos Aº= 1
Aº= 45º, Sin Aº= 3/4 Cos Aº=3/4
Aº= 60º, Sin Aº= 1 Cos Aº= 0.5

Example: You fly a PA-31 Chieftain. TAS= 175kts call it 180 kts! So 60/TAS = 1/3. Wind is 240º/35kts. Max Drift is therefore 35 X 1/3 = 12º

Ok so far? If you're Desired Track is, say, 090º, the angle between the wind and the track is 30º (just form a mental picture of what's going on looking at your DI/ Compass Rose, you'll find the wind has a TWC and it's also blowing from the right). Drift= 12º x Sin 30º= 12º X .5 = 6º

You can easily deduct then that you need to steer 096º to maintain a track of 090º. There is a tail wind component which is: 35kts X Cos 30º = 35kts X 1 = 35 kts. Therefore, Ground Speed is 175kts + 35kts = 210 kts

You have 52 NM to run. How long does it take? Well 210 kts still close to 180 kts ( it's roughly an extra 15%) so use 60/180= 1/3.
Therefore time = 52 x 1/3= 17min, however you fly faster by 15% (= 10% + 5% time reduction) so time to run is in fact: 17 - 1.7 - 8= 14.5 min

Bear in mind, if you have a relatively slow aircraft and a strong X wind component, the drift will have an effect on the ground speed. Why?

Well simply because, in fact: GS= TAS X Cos (Drift Angle) + TWC or - HWC

Example: wind = 60 kts TAS= 90 kts, Max Drift is 60X2/3= 40º

If the wind is blowing perpendicular to the track, the drift angle is the Max Drift angle (you have no TWC or HWC), so GS= TAS X Cos( Max Drift)= 90 x Cos 40º= 90 X .77= 70kts. Roughly a 20% reduction in speed, purely because the Hdg is 40º off the desired track.