I asked the following a number of posts ago:
In the interests of discussion, any interested parties care to share their thought processes in endeavouring to complete the above visual in minimum track miles?
Thats what I wanted to know...
I'm quite aware of all the formulas but the point was, in the absence of an applicable aide memoire to calculate the turn radius, bearing in mind I'd rather be looking out the window on a visual than punching numbers into my calculator (as I hazard a guess you still did to evaluate the formula in your previous post? If not, you obviously have some aide memoire for calculating \tan(\phi)...), I realised I wasn't sure if my lateral displacement from the centreline was enough to ensure I didn't travel through it whilst making the turn...
Of course, in reality the differential equations
do need to be solved
in situ because I chose to decelerate and change the configuration in the turn. That was the whole point of the exercise in terms of reducing the track miles of the visual to the minimum by using all the appropriate aides available.
The exercise then became one of approximation...
If I started my turn at 330kts and was endeavouring to complete it at 180kts, is it geometrically possible to do so within 5 miles? (I chose to approximate such a turn by using an average of 250kts, whereupon, your/my calculations show that one only needs a lateral displacement from the centreline of 4 miles to turn through 180 degrees....even less with a crosswind in the favourable direction. Whereupon, one may relax the bank angle towards the end of the turn to roll out on the correct inbound track.)
If so, is it also possible to loose 1000' and be in the slot?
If not, what are my options?
Shall I extend further downwind and decelerate before making the turn?
So, in short, I wanted to know quite a lot of things...
