Let's say an aircraft has an abs ceiling of some value at the start of its cruise and then it consumes some percentage of its fuel during cruise. How will this change the new abs ceiling at the end of its cruise? Is there any mathematical way of finding the new abs ceiling?

For example: At the start of its cruise a jet aircraft has an absolute ceiling of 35 000 ft. During the cruise it consumes fuel equal to 20% of the aircraft’s starting weight, what will the aircraft’s absolute ceiling at the end of the cruise segment be?

  • $\begingroup$ It might be helpful to mention a specific aircraft, as Mach and the ability to generate thrust at a higher altitude will be important along with weight. $\endgroup$ Commented May 8, 2022 at 3:14
  • $\begingroup$ The question is applicable for any given commercial airliner. I thought the info provided in the question was enough to find the new abs ceiling. I'm just not sure about the maths behind finding it $\endgroup$ Commented May 8, 2022 at 9:21
  • $\begingroup$ Probably Mach limited, and with an airliner, a max safety altitude. My guess is, from a thrust point of view, airliners do not come anywhere near their absolute (or service) ceiling. So, it becomes a matter of "how much does my weight savings allow me to slow down (indicated) at my Mach limit (TAS) without wasting fuel (maintaining optimal AoA). Climbing allows you to fly at a lower IAS for the same TAS, but lift is proportional to V$^2$. Within your Mach limit, your allowable IAS reduction will be $square root$ weight reduction. $\endgroup$ Commented May 8, 2022 at 10:47
  • $\begingroup$ Cheers. I'm trying to get a mathematical view of this scenario. Since less weight means greater altitude, I'm assuming at the end of its cruise it will be in the stratosphere. I'm also assuming weight is proportional to the density ratio (sigma, σ). From digitaldutch.com/atmoscalc , density at 35000 ft is 0.3796 giving me a sigma value of 0.3099. After 20% weight gone, sigma should decrease by 20% as well (0.2479). Interpolating from the ISA table, I get the new ceiling to be around 39880 ft. This method looks kind of broken nevertheless. Lmk if this sounds alright or not $\endgroup$ Commented May 8, 2022 at 13:16
  • $\begingroup$ yes, if your engines can produce adequate thrust for the matching IAS at a higher altitude and your Mach is ok. Check your TAS. $\endgroup$ Commented May 8, 2022 at 17:26

1 Answer 1


Short answer: The new aerodynamic ceiling is 40,000 ft density altitude.

Constant Temperature Condition

It is instructive to imagine the following scenario: The aircraft is flying at the coffin corner at 35k feet (density altitude).

$L = W = ½ \cdot ρ \cdot V² \cdot C_lmax \cdot S$

The 'W' represents weight (in Newtons), which we are assuming to be equal to lift. In this case, the 'V' (TAS) is at it's maximum limit, which cannot be exceeded without having a shock stall (and a subsequent collapse in $C_l$). 'V' is also at it's minimum limit, which cannot be subceeded since that requires an increase in $C_l$, which is not possible since $C_l$ already equals $C_lmax$. It can be seen that any change in 'V' would ultimately result in a stall.

Now reduce the weight by 20% (W' = 0.8W), this means that the $C_l$ will be 80% $C_lmax$. If the aircraft is slowed down now, it won't stall, since there is a headroom available in $C_l$. This will also alter the shock stall speed (shock stall speed changes with $C_l$), but for the purposes of this question, it does not make any difference, and so we will assume that it doesn't change.

Now, we will start a climb at a constant Mach number - the same Mach number that we had earlier in the coffin corner. Doing so ensures that we never exceed the Mach number for shock stall (we're basically climbing alongside the buffet boundary). At some point in the climb, the aircraft will again attain the $C_lmax$ (stall boundary intersects buffet boundary), and will again find itself in the coffin corner - the altitude at which this happens is the new aerodynamic ceiling.

Since in the question we started at 35k feet, the aircraft was basically at tropopause, and so we can assume that the temperature is constant throughout the climb (how convenient is that!). This means that the Mach number of the aircraft varies ONLY with TAS - this also means that a climb at a constant Mach number is same as climb at a constant 'V' (TAS). And with that, this particular case is solved:

$W = ½ \cdot ρ \cdot V² \cdot C_lmax \cdot S$

removing all the constants from this equation (including 'V'), we get the following relationship:

$W ∝ ρ$

And so reducing the weight by 20% reduces the minimum necessary density $(ρ)$ by 20%. Which altitude has 20% less density than that at 35k feet? According to ISA charts, that altitude is 40k feet.

ISA chart

ISA Chart

Variable Temperature Condition

In the previous case, we conveniently assumed that the temperature was constant, and so a climb at constant Mach number was the same as that at a constant TAS. This is not always true, so let's look at a more general case.

Consider the same previous case of climb at constant Mach number, except that 'V' is no longer constant; as the altitude is increasing, the temperature and thus the speed of sound (a) are decreasing. For a constant Mach number, a reduction in 'a' requires a reduction in 'V'. From the previous equation, if we again remove all the constants, we get:

$W ∝ ρ \cdot V²$

'V' was not removed this time since it is no longer a constant - it's a variable, the value of which is not obvious (again, V is reducing as altitude is increasing). However, it is possible to substitute it with something known... We are climbing at a constant Mach number. What is Mach number?

Mach number $(M) = V / a$

$or, V = M \cdot a$

What is 'a' (speed of sound)?

$a = √(γ \cdot R \cdot T)$

This implies that:

$V = M \cdot √(γ \cdot R \cdot T)$

Again, removing all the "unnecessary" constants from this equation, we get the following relationship:

$V ∝ √T$

or $V² ∝ T$

Simply substituting this into the previous relation, we get:

$W ∝ ρ \cdot T$

Where 'T' is the absolute temperature (in Kelvin). Now as we can see, there are two variables to worry about: $ρ$ and T. But that's not a huge problem, since each altitude generally corresponds to a particular density and a particular temperature. By looking up these two values in the chart, the new aerodynamic ceiling can be determined for a given change in weight.


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