Last things first,
Pack2, the formula offered by
square leg is absolutely correct for practical purposes, but if you’re doing the maths, forget the percentage sign,
(Rule of thumb: 3.5%x120kts=420' per minute)
Just multiply the Groundspeed in Knots by the percentage gradient, i.e. multiply 120 by 3.5 in the example, DON’T take 3.5% of the G/S. 3.5% of the G/S gives you the Vertical Speed in Knots, and as the conversion from Knots to Feet per minute is as near as dammit to 100 (101.27 to split hairs), the ‘built in’ factor of 100 eliminates the need to work percentages.
With regard to my earlier discussions with
flybubba and
FullWings, my few days ‘think’ time is up, and I should have listened to my own words in a post which I made on another thread (difference between Vy and Vx), when I said –
Essentially all aircraft performance depends upon the 4 basic vectors, Weight, Drag, Lift, and Thrust. These may all be resolved geometrically, which, when there is an excess of thrust, yield a climb angle. Following on from this, all climb performance, when related to the 4 basic vectors indicate that all climb performance is at a particular angle, thus, the amount of excess thrust determines the climb angle.
Similarly, all descent performance, when related to the 4 basic vectors indicate that descent performance is at a particular angle, with the amount of excess DRAG Vs Thrust (Idle) determining the descent angle. The descent rate will then be determined as a function of descent angle and TAS. Variations in descent angle will then depend directly upon drag for the speed schedule flown (ignoring variations in Net Idle thrust during descent). To examine total drag, we must consider the Low speed Polar, due to Equivalent Airspeed (EAS), and the High speed Polar due to wave drag when flight is above Mcrit.
Take, as an example, an aircraft with a Mcrit of M0.73, and a descent profile speed schedule of M0.78 / 300 KIAS (These are ‘ball-park speeds for B737, A320, DC9, Learjet aircraft). Mach/CAS changeover Pressure Height for this speed schedule would be 29323 feet.
(1) For the descent to 29323 ft, High speed (wave) drag would be constant, due to the constant Mach number, but Low speed (dynamic) drag will be increasing due to increasing CAS and EAS. Thus, total drag and TAS would be increasing during descent, resulting in increasing descent angle and rate.
(2) At Mach/CAS changeover Pressure Height (29323 ft) the EAS is 286 knots whilst the Mach number is M0.78. Further descent at a constant CAS of 300 knots will be occasioned by decreasing wave drag as Mach number now reduces, until 25943 feet, when Mach = 0.73 (Mcrit) and wave drag is now zero. In this same phase of descent, EAS increases slightly to 288 knots. Thus, in this phase, wave drag decreases significantly, whilst dynamic drag increases slightly. The net result is that, in this phase, descent angle, TAS, and descent rate all reduce.
(3) Below the level where Mcrit was passed (25943 ft), at a constant CAS of 300 knots, EAS increases from 288 knots towards equality with CAS at Sea level. At 10000 feet EAS is 297 knots, and dynamic drag has increased by 6.3% between 25943 ft and 10000 ft. This leads to a steepening descent angle, but, as TAS decreases at a faster rate, Rate of Descent steadily decreases.
The summary is that initially, at the ‘Mach stage’, descent angle and rate constantly increase. At the CAS phase, descent angle and rate initially decrease, but then descent angle increases to a small degree, whilst rate of descent steadily decreases. In the ‘win a little, lose a little’ CAS descent phase, the average is fairly close to a constant angle, and may be considered as thus for PRACTICAL purposes.
I think (I hope) that I got it right this time.
Regards,
Old Smokey