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Following on from this question on the descent rate formula.

It appears from the answers and comments that a 3 degree slope is pretty standard across the board.

I would like to know why a 3 degree descent rate is used? I thought it would vary based upon the scenario.

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Basically, it's a combination of historical significance, passenger comfort, and ease of mental calculation.

Historically, before aircraft were pressurized a rapid descent would be uncomfortable for passengers. At the speeds typically flown, a 300ft/min descent was deemed to be comfortable enough for the typical passenger.

Many aircraft had a cruising speed between 100–120 miles per hour (160–190 km/h). Three miles would be traveled in about 1.5–1.8 minutes, resulting in a rate of descent of about 550–660 feet per minute. That was about as fast as passengers could comfortably adapt to the changing pressure on their eardrums. However, many pilots used a 300-feet-per-minute descent rate because doing so is almost unnoticed by passengers. A pilot cruising at 10,500 feet would calculate that in order to be at 1,000 feet at his destination, he would have to descend 9,500 feet. Dividing 9,500 feet by 300 feet per minute, that descent would require about 32 minutes. If his ground speed was 120 miles per hour (190 km/h), he would begin his descent about 64 miles (103 km) from his destination (traffic permitting).

As you can see, there is a reasonably simple mental calculation that can be done there. This is commonly referred to as the rule of three which is in part quoted above.

That covers descent from cruising altitude, however you are right that flight in IFR will depend on procedures, and scenario. An example is London City Airport which uses a much steeper descent profile of 5.5 degrees on approach to land.

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    $\begingroup$ Three nanometers per 1000ft?! Surely you mean n.m. ;) $\endgroup$ – J... Sep 9 '16 at 14:04
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    $\begingroup$ @J... oh jokes based on SI. You dont get more pedantic than that ;) So as there is no SI unit for Nautical mile ive gone with the one used by ICAO according to en.wikipedia.org/wiki/Nautical_mile $\endgroup$ – Jamiec Sep 9 '16 at 14:14
  • $\begingroup$ I was stuck at NM also but nano is n not N and meter is m not M. After coming away with Newton Mega which does not scan I realized it was not an SI unit. Since it was not expanded in full or linked I figured it must be well known to this community. (Thanks to the popular network questions not having fikters you get a fair number of readers who are not among the community.) $\endgroup$ – JDługosz Sep 9 '16 at 15:57
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    $\begingroup$ I thought these were Newton-meters. Amittably not a very useful unit either. $\endgroup$ – John Dvorak Sep 9 '16 at 17:28
  • $\begingroup$ I'm more stuck on the "at typical speeds" part. Is there any positive speed at which the ratio of vertical to horizontal distance traveled is not equal for a given angle? As far as nautical miles are concerned, I commonly see "nmi" used for it in order to help prevent confusion. $\endgroup$ – reirab Sep 9 '16 at 19:43
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Descent is a matter of managing the aircraft's energy budget.

To descend, the aircraft must shed all its potential, and big part of its kinetic energy. Pilots reduce engine power to minimum to stop adding energy, but the energy the plane already has is still only dissipated by drag. The angle at which the aircraft can descend without accelerating is equal to its lift-to-drag ratio. Descend any steeper and the aircraft will accelerate.

Now higher lift-to-drag ratio means lower fuel burn, so aircraft are designed to have that ratio high. A typical modern airliner has lift-to-drag ratio, at optimal speed, around 18. The older generation a bit less, 15-16, the newest aircraft a bit more, 20-21. And $18\times 1,000\ \mathrm{ft}\ \dot=\ 2.96\ \mathrm{NM}$.

The aircraft can descend less steeply by using some engine power, but the engines are less efficient at low altitude, so it is not desirable to descend slower.

The aircraft can also to some extent increase its drag, usually by using the speed brakes. However descending faster this way means that the engines were running longer before the descent started and thus burnt more fuel, the energy of which is now being wasted to speed brakes.

So pilots try to descent at around the optimal glide angle. And with the aerodynamics of current airliners it happens to be around 3 miles per 1,000 ft. Rounded to that value if you need to do the calculation mentally instead of having the FMS do it for you.

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  • $\begingroup$ That would had been my answer, too. +1. $\endgroup$ – Peter Kämpf Sep 10 '16 at 18:18
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    $\begingroup$ Jan, I think that the 3 deg was arrived at before FMS were standard fare. (Heh, when I got my wings, my "FMS" was my brain and a wiz wheel). $\endgroup$ – KorvinStarmast Oct 3 '16 at 12:52
  • $\begingroup$ @KorvinStarmast, of course it was. The lift-to-drag was a bit lower, but not that much, so that was already reasonable rounding down of the actual glide angle. $\endgroup$ – Jan Hudec Oct 3 '16 at 15:46
  • $\begingroup$ You already got my up vote, it's a nice concise answer. $\endgroup$ – KorvinStarmast Oct 3 '16 at 15:53

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