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Old 15th April 2026 | 05:17
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Noknoipobin
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Joined: Mar 2012
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From: Somewhere
Originally Posted by 400guy2
I think you mean 5.03. No?

Thank you ,I mean it’s actually 5.3, not 5.03.

tan 3° = height / distance
0.0524 = X / 6076 (1 NM)
→ X = 6076 × 0.0524 ≈ 318.4 ft

So, over 1 NM the height is about 318 ft.

Make 1 NM = 1 minute then→ GS = 60 kt
if flying 60 kt → ROD ≈ 318 ft/min
If flying at Y kt: ROD = (318 / 60) × Y ≈ Y × 5.3

When we truly understand the actual values, it helps improve our situational awareness and caution.
In real operations, it may be difficult to fly exactly at an ROD of 530 (GS × 5.3)(in V/S mode). However, if we are flying at around 500 (GS × 5), we can still anticipate how the aircraft profile will develop. This helps avoid confusion or surprise about why, even with a “correct” ROD (GSx5), the aircraft may appear high.

Try looking at an example AIP chart for EGLL.




Last edited by Noknoipobin; 15th April 2026 at 05:37.
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