My last post on this (I know I said that before) because pissing contests are not my thing. Exception being if someone questions my maths below!
I did some maths homework to see where our theoretical maximum 5 degree bank would lead me and the results are here:
In very simple terms if one was stuck with 5 degrees bank when engine out then ATZs will have to be considerably larger to accommodate MEP and IR training and tests else EFATO asymmetric-circuit-to-land will all be way outside the airport's ATZ.
A practical example: In the case of Blackbushe we'd have to co-ordinate with Farnborough because we'd be overhead them when downwind [Blackbushe]
Maths below including knowns:
Blackbushe and Farnborough runways are within 10 degree of parallel, approx. 4nm (centre to centre) separates the two.
Turn Radius = TAS^2 /(Tan Bank Angle X G)
Where TAS is in Ft/sec, Bank Angle is in degrees, and G is in Ft/sec^s
6076 ft/nm
This example Vyse (blueline) is 107 KIAS [I have for this example assumed TAS=IAS as at these low heights the difference is minimal]
107 KIAS = 180.596 FPS
Bank angle = 5 degrees
Gravity = 32.174 ft/s/s
[for completeness, a climbing rate 1 turn (practised engine-out for any ME-IR candidate) at that speed would require about 17 degrees of bank]
t/f
radius of turn is
(180.596 * 180.596)
-------------------
Tan(5) * 32.174
=
32614.9
-------
2.815
=
11586.1 (feet)
=
1.9 NM RADIUS
NB diameter of turn (to where you would be start of ‘downwind’ leg) is min 3.8NM in nil wind; more if slight northerly component (in this example)