I am hard to understand to calculate service ceiling and absolute ceiling in PA-44 Seminole. Is there anyone know how to calculate it in the condition(T/O weight: 3599.24 L/D weight: 3354.44 PA:2329 Temp:20)

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You can use the given chart to define (single engine) ceilings. There should be separate graph for both engines running case.

Note: it is important to understand that the definition absolute ceiling is absolute: zero climb rate while service ceiling is defined by local regulations.

In Europe, in commercial operations for class B aircraft (which Seminole is) service ceiling is the altitude where both engines running aircraft is able to climb at 300 ft/min at present gross weight and local OAT. A single engine service ceiling is defined as "positive gradient" so it is the first usable altitude below single engine absolute ceiling.

To define required altitude enter the graph from the right from the desired vertical speed, go upwards until you meet your current (or estimated) gross weight. From there draw a horizontal line all the way across the left side of the graph. The point where to local OAT (at an altitude) and your horizontal lines cross is the ceiling according to the selected climb requirement.

When interpolating between the altitude lines make sure to go perpendicular between altitude lines - not up and down along the grid.


With the chart you uploaded you don't calculate the ceiling, but the climb rate when you have only one engine operating. The TO weight and LD weight don't matter, what you need is the GW (which should lie somewhere between the TO weight and the LD weight).

I have added red lines to the chart to show you how to use it:

  • Select your OAT (20°C in your case)
  • Move up until you meet your PA (2329 in your case)
  • Move right until you meet your weight (let's say 3599 at take off)
  • Move down and you get your climb rate (in this case 120ft/min)*

*At your LD weight it is 190ft/min

NOTE: at the bottom of the chart you see dashed lines with arrows. Those are used to understand the order in which you have to read the chart.

Climb performance

  • $\begingroup$ Thank you so much! $\endgroup$ – Mun Park Mar 3 '19 at 2:24

Service ceiling I believe is the altitude at which an aircraft will no longer climb at a rate of 100 ft/minute. It should be published in the POH for the plane.

For example, in a 1977 Cessna 150

SERVICE CEILING ..... 14, 000 FT

That is a single engine, 2 seat propeller plane, with no turbocharging/turbonormalizing.

The Seminole POH should have a similar number. Absolute ceiling can be higher but it will take a while to get higher. Use of O2 is required above certain altitutes, and after certain time periods above a lower altitude as well. If one departed a hot California airport at 45C (110F, definitely possible out in the desert areas), say Rosamond Skypark (L00) at 2415ft, then 14000-2400 = 11,600 feet x 2C/1000 ft = 23.2C, and 45C - 23.2C = 21.8C!

Will feel nice & cool while airborne, but really hot when that door is opened.

We were flying the US southwest a few years ago, July 4th weekend. Flying southeast away from Albquerque, New Mexico, to Carlsbad, Arizona, we couldn't climb above 9,000 ft, my 180 HP normally aspirated engine just couldn't make any more power to get higher. So the rated Service Ceiling for my plane,

SERVICE CEILING ------------ 14,600 FT

made no difference. 107F on the ground. Hot!

  • $\begingroup$ Thank you so much! $\endgroup$ – Mun Park Mar 3 '19 at 2:25
  • $\begingroup$ to acheive a density altitude of 14,600 at a pressure altitude of 9000 you would need an OAT of 50c(122f); the ground in Carlsbad is about 3200 which implies the ground air temp would be about 61.6c(143f) (9000-3200=5800, 5.8*2c=11.6c) Allowing for a rather low(storm low) altimeter setting 29.00 this is a pressure altitude of 9862ft and would need an OAT of 39.5c for a density alt. of 14600, and an implied surface temp of 51c(124f). In Pheonix (yes slightly different, just data I have), the average low altimeter setting is 29.6 and the average high in summer 30.0. (add 0.6 to winter highs) $\endgroup$ – Max Power Mar 9 '19 at 22:59

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