What is the in-flight trick?

I know how to calculate it in theory, but how can I calculate it fast? For example, I know I can use this formula for calculating the rate of descent:

[ground speed + add 0]\2

I'm trying to find a formula like that for TOD.


Any approximate method is going to vary for different types of plane, depending on their speed, altitude and descent rate, but here's how I figure it for my Cessna 172.

I know I'm traveling around 2NM a minute (actually a little slower), and I like to descend around 500 fpm.

So if I have to drop 5,000 feet to pattern altitude, the TOD is 10 minutes out (5,000 feet / 500 fpm). At 2NM a minute, that is 20 miles out.

So if I have my GPS / DME set to my destination airport, I am looking for both ETE: 10 min, and also Dist: 20 miles, then I start my descent.

Its approximate, but its pretty close for planning purposes. Its always just:

[Altitude Drop] / 500 = Minutes
Minutes * 2 = Miles
  • $\begingroup$ what is your Resource for this formula? $\endgroup$
    – Arian
    Jan 30 '18 at 20:48
  • $\begingroup$ Altitude drop mean decent rate? $\endgroup$
    – Arian
    Jan 30 '18 at 20:49
  • $\begingroup$ could you explain the formula? $\endgroup$
    – Arian
    Jan 30 '18 at 21:00
  • $\begingroup$ No, altitude drop and descent rate are not the same thing. If I'm at 6,500 feet, and I need to descend to 1,500 feet to enter the pattern, then I need to go down 5,000 feet. That is an altitude drop of 5,000 feet. Descent rate is is the Feet-Per-Minute rate at which you lose altitude. $\endgroup$
    – abelenky
    Jan 30 '18 at 21:01
  • $\begingroup$ The formula comes from basic math and physics. This should be covered by high school. See math.com/school/subject1/lessons/S1U2L3DP.html $\endgroup$
    – abelenky
    Jan 30 '18 at 21:02

On a large aircraft, an often cited trick is to use 3 x altitude to airfield, divide it by 1000, and then that is still air distance from airfield to top of descent in nautical miles. For example, from 30000ft, TOD would be 90NM out. This works to roughly achieve an idle descent (if the aircraft’s glide ratio is around 18 - which, give or take, it is for modern airliners in clean configuration at moderate speeds).

  • $\begingroup$ This rule of thumb is based on a 3° descent path. $\endgroup$
    – J Walters
    Jan 30 '18 at 22:10
  • $\begingroup$ @JWalters Yes, same difference - it works in clean configuration from cruise to approach, and then again down the glide path in approach/landing configuration. $\endgroup$ Jan 30 '18 at 22:13
  • 1
    $\begingroup$ You can refine that by considering if you're approaching from a downwind, base, or straight-in direction. Also, one can refine it further by considering the distance required to slow from descent speed (between 280-320 knots, typically) to 250 knots before descending below 10,000', and then slowing from 250 knots to approach speed. The "close-as-duplicate" question/answers go into these computations. But 3x is a pretty good general ballpark. $\endgroup$
    – Ralph J
    Jan 31 '18 at 2:08
  • $\begingroup$ So you say this formula based on 3dgree ? $\endgroup$
    – Arian
    Jan 31 '18 at 6:05
  • $\begingroup$ It’s the same - 3° translates into a descent gradient of 0.052 which translates (for idle thrust) into a glide ratio of roughly 19. The three are the same physical relationship, so it can be based on either. In flying context, I choose to think of the enroute part based on glide ratio (as I‘m ideally descending with idle thrust, accepting as descent angle whatever the aircraft achieves) and the final descent based on 3° (as I‘m following a fixed glide path with thrust set to match whatever required). For modern jets, the values are roughly the same, but they are based on different assumptions. $\endgroup$ Jan 31 '18 at 8:01

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