Base turn outbound timing
 

Hey guys,

I recently had an interview where I was asked about the outbound timing of a base turn.

The question was: when flying a base turn, let's take a 30 degree angle between outbound and inbound, no outbound timing published, what time should I fly outbound in order to roll out on the inbound track using a 25 degree bank angle (ofc dependable on the speed) in zero wind conditions.

For a class A airplane I know a formula

Time = 36/angle between inbound and outbound

So in this case it would be 1,2 minutes but this is for a rate one turn only.

Does anyone know the exact math behind my problem or a rule of thumb?
If so, please help me out :)

Thanks in advance!

I have never come across a procedure which doesn't dictate when you turn, so surely the question is hypothetical?

Whats the TAS

Variable, must be in the formulla.

I have no clue of what they expected me to answer..

But lets take 180 knots :)

I very much doubt there is any formula or rule of thumb for exactly this problem. It seems the question was merely brought up to see if you can apply the basic rules of thumb that do exist.

By approximation, your turn radius will be 1/200 of your ground speed. So, at 180 kt you'll need roughly 1.8 NM for the turn. I would round this up and continue with 2 NM. Using the 1:60 rule, you can estimate that, for a 30 degrees' base turn angle, you'll have these 2 miles once you're 4 miles out. Now, how long does it take to fly 4 miles at a speed of 180 knots? Answer: 1:20 minutes.

Sounds easy, doesn't it. But I must admit I'm glad it was you who was asked this in an interview and not me.

"Base turn"?

That's in the circuit, right? How does timing come into that?

That's a procedure turn approach or what is sometimes called a no delay approach?
The turn is at 30 degrees to the outbound leg into a left or right hand procedure turn to then commence a constant rate one turn onto the final approach intercept, NDB, VOR or ILS?
If that's what they wanted to have answered then in still air it's 60 seconds for a Category A aircraft from commencing the procedure turn to commencing the base turn onto finals. This gives a still air time of 45 seconds from the completion of the rate one procedure turn until the commencement of the base turn.
The answer or the confirmation of the above probably lies in the pages of Jeppesen, Air Traffic Control, Flight Procedures.
That type of question at an interview is not likely to relate to a visual circuit.

The official definition of a base turn in ICAO Doc 8168, PANS Ops Volume 1 is, "A turn executed by the aircraft during the initial approach between the end of the outbound track and the beginning of the intermediate or final approach track". In the turn construction parameters the angle of bank is assumed to be 25 degrees and the timing to be one minute from the start of the turn for Cat A and B aircraft and 1 minute 15 seconds for Cat C,D and E aircraft.

According to PANS-OPS, Volume II, Part I, Section 4, Chapter 3, 3.5.4.1 c), the 1 minute, respectively 1 minute and 15 seconds, are the times prescribed for procedure turns. Times for the construction of base turns are given in 3.5.5 of the same chapter.

Circuit? No! @ Agor Bis
 

No! Base turn is in a hold to final or a procedure towards the inbound, but is actually outbound - its part of an instrument procedure.

You are talking about base leg in a visual circuit - not an instrument procedure.

Never heard it called a base turn before, always a procedure turn.

A let-down plate that didn't give a distance outbound or a time outbound, modified for ground speed would be a liability lawyers delight.

Been called a base turn in Aus AIP and Jepp for years. You have some reading to do PBUM! :)