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Maybe if possible you could help from the below?

I'm at 140 knots following 354° inbound (CGS VOR) at Nice Airport LFMN. From my heading 354°, I need to join 43°inbound for RWY 04L ( Not allowed ti make any turn prior to 5NM from CGS VOR). I would need to make a right turn of 49° to join the inboung 43°.

If I calculate my time for turning, would be (IAS speed/3° standard turn) = 140knots/3=16.3 seconds.

Then using the thumb rule, my bank angle should be (140/10)+7= 21°.

From these assumptions and the fact that I cannot turn before 5 NM from VOR, how do I know then when to turn to 43° inbound?

If I use the other formula IAS (approx to GS as close to RWY) squared / 10 , I get a radius of (140/60)2/10= 0.9 NM. But this assumes a standard bank angle of 25-30° right?

Then in first assumption, my bank angle should be 21°. I'm confused there.

If I use a third formula, 1% of GS, I get a radius of  0.7 NM (1.4/2) correct?

Should I calculate the circumference of the circle and start turning from that circumference?

It would be great if you could help me understand this!

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  • $\begingroup$ Are you asking this for some kind of simulation or theoretical exercise, or are you asking how the pilot actually does it? Because the pilot has no time to do these calculations in flight, and does it in a much simpler (but less precise) way. $\endgroup$
    – Ben
    Commented May 7, 2023 at 12:40
  • $\begingroup$ By the way, CGS is no longer a VOR/DME (it used to be), it's just a DME now, so your aircraft needs to be RNP capable for flying this approach. $\endgroup$
    – Bianfable
    Commented May 7, 2023 at 13:17

1 Answer 1

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What you are describing is the VPT A (Visual Maneuvering with Prescribed Tracks) for Rwy 04L or 04R into Nice Côte d'Azur (LFMN) following the RNP A approach (which ends at MN04A):

LFMN VPT A Rwy 04L/R

Note that MN04A is the MAP (missed approach point) for the RNP approach with an MDA (minimum descent altitude) of 2000 ft. This means that you must be visual from that point on. The entire procedure shown above is flown entirely with visual references (you may use assistance to initially stay on the 354° track towards CGS, as the chart says, if equipment permits).

The point at which you would turn towards the runway is therefore determined by looking out of the window. There is no need to calculate anything, you just visually fly towards the runway. The only restriction is that you shouldn't deviate to the west on the initial segment towards CGS (to reduce noise in the city of Antibes). If you turn early, you can still correct to the left; if you turn late, you can still correct to the right. If you get the turn just right, you can intercept the PAPI for a visual descent towards the runway just after the turn.

If you misjudge the turn so bad that you can no longer safely correct your flight path to land, you would follow the Visual Balked Landing procedure (dashed lines) and then fly to NERAS (end of the missed approach procedure of the RNP A).

If the visibility is too poor to see the runway at MN04A or when there are clouds below 2000 ft, you would be cleared for the ILS or RNP Z approaches instead, which are both straight-in approaches overflying Antibes.


If you really want to calculate the point to start turning, consider the following sketch:

Sketch

The first thing we need is the intersection point of the track 354° from MN04A and the extended runway centerline (I chose runway 04L here because that is the default runway for landing in Nice). I looked up the coordinates for the relevant points in my navigation database:

Point Latitude Longitude
LFMN 04L 43° 39' 9" N 7° 12' 17" E
LFMN 04R 43° 38' 48" N 7° 12' 9" E
MN04A 43° 33' 43" N 7° 9' 14" E
CGS 43° 38' 43" N 7° 8' 45" E

You can use the calculator on movable-type.co.uk to figure out the coordinates for the intersection point: 43° 36′ 28″ N, 007° 08′ 50″ E.

The next thing we need is the distance $D$ from the intersection point to CGS. The same page also has a calculator for that and gives $ D \approx 4.168 \, \text{km} \approx 2.251 \, \text{NM} $.

Now we calculate the turn radius $R$ using the formulas from this answer:

$$ R [\text{ft}] = \frac{V [\text{kt}]^2}{11.26 \tan \theta} \; , \qquad \omega [^\circ/\text{s}] = \frac{1 \, 091 \tan \theta}{V [\text{kt}]} $$

When flying a standard turn at 3 degrees per second, we can solve the second equation for $\tan \theta$ and insert it into the first one:

$$ R [\text{ft}] = \frac{V [\text{kt}]^2}{11.26 \, \omega [^\circ/\text{s}] \, V [\text{kt}] / 1 \, 091} \approx 96.89 \frac{V [\text{kt}]}{\omega [^\circ/\text{s}]} $$

Assuming 140 kt TAS, this gives a turn radius of $ R \approx 4522 \, \mathrm{ft} \approx 0.7442 \, \mathrm{NM} $. The corresponding bank angle is then given by (source):

$$ b = 57.3^\circ \arctan \left( \frac{V [\text{kt}]}{362.1 \, \text{kt}} \right) \approx 21.14^\circ $$

The last thing missing is the distance $d$ before the intersection point, where the turn should be started. With a bit of geometry, one finds that this is given by

$$ d = R \tan \left( \frac{\alpha}{2} \right) $$

with our known $ \alpha = 49^\circ $. This results in $ d \approx 0.3392 \, \text{NM} $.

This gives the final answer for the distance from CGS where the turn should be started:

$$ D + d \approx 2.59 \, \text{NM} $$

Note that this entire calculation does not take wind into account!

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  • $\begingroup$ thank you for your response. It's for simulation purposes. I eeally would want to understand the how to calculate if there's no help from FMS or no chart. Would anyone be able to help me calculate that time to initiate the turn to 04L? $\endgroup$
    – R O
    Commented May 8, 2023 at 10:05
  • $\begingroup$ I cannit find words to thank you again! This is absolutely perfect. For the radius I think that the thumb rule of 1% of your IAS giving you the diameter works perfectlt as it gives you a turn radius of 0.7 NM, would you agree. Oh 1 more thing is the time for the turn, would you agree 49°/3= arpund 13 seconds? $\endgroup$
    – R O
    Commented May 8, 2023 at 12:21
  • $\begingroup$ Sorry forgot about the bank angle, do you agree it's (140/10)+7=21°? $\endgroup$
    – R O
    Commented May 8, 2023 at 12:32
  • $\begingroup$ Sorry not able to edit the comment so adding it separately. What about the bank angle? Do you agree it's 140/10 +7=21°? $\endgroup$
    – R O
    Commented May 8, 2023 at 12:35
  • $\begingroup$ And the time for the turn is also correct using 49°/3= 13 seconds? $\endgroup$
    – R O
    Commented May 8, 2023 at 12:43

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