# Is there good information available on maximum operating altitudes whilst maintaining a sea level cabin?

I am in the process of reviewing the aircraft we regularly commission for air ambulance transfers. Although the majority of transfers are undertaken with a standard cabin altitude ( 4 - 8,000 ft) some are conducted with a sea level cabin (or cabin altitude restriction). The reasons for this are not really necessary for the purposes of the question, but the consequences of requesting one are that the operating maximum altitude of the aircraft is reduced (to maintain a safe cabin wall differential pressure) and the range / endurance of the aircraft is reduced.

Although this has low operational impact (and would normally not affect our choice of providers) I wondered if anyone had stumbled upon a list of airframes and the corresponding max altitudes that they would be limited to while maintaining a sea level cabin environment.

Any help would be much appreciated.

The definitive answer would come from the manufacturer, but you can calculate some estimates using the barometric formula in combination with known pressure differential limits for specific aircraft. Below are some examples for several aircraft types that are used as air ambulances.

Type psid Maximum Sea Level Altitude
King Air 90 (A90) 4.6 psi 10,000 ft
King Air 100 4.7 psi 10,200 ft
King Air 90 (C90) 5.0 psi 11,000 ft
Pilatus PC12 5.8 psi 13,200 ft
King Air 200/300 6.5 psi 15,200 ft
Learjet 35 9.4 psi 25,500 ft

Some of the psid values that I used came from the AOPA website and BeechTalk Forums.

If you are interested in how I calculated the estimates for these planes so that you can calculate estimates for other aircraft, I have provided an explanation in the next section.

Some private jets can go even higher, for example the Bombardier Global Express has a maximum pressure differential of 10.3 psi which allows a cabin pressure of 5,680 feet at its ceiling of 51,000 feet (Wikipedia). So it should be able to maintain sea level cabin pressure up to 30,000 feet. The leader currently seems to be the Syberjet SJ30i which has a maximum pressure differential of 12 psi which the company website says allows a sea level cabin pressure up to 41,000 feet. And an honorable mention goes to Concorde which had a differential of 10.7 psi which allowed a cabin pressure of 6,000 ft at its service ceiling of 60,000 ft.

## Calculation for Other Aircraft

The above estimates were calculated using the barometric formula:

Using a simplified version of the barometric formula along with a few additional steps you can calculate what should be a pretty good estimate of sea level cabin pressure maximum altitude for any aircraft for which you know the maximum psid (psi differential). By simplified version I mean with all of the several variables already filled in.

Step 1

The first step is to subtract the known maximum psid for a particular aircraft from standard sea level pressure to determine the psi at the maximum sea level altitude. For example if calculating for the King Air 100 you would subtract 4.7 psi from 14.7 psi, which tells us that the air pressure at the maximum altitude for sea level cabin pressure is 10 psi.

Step 2

So now we just need to determine what altitude has a pressure of 10 psi. There are tables online that have this information, but for more accuracy we can run the barometric formula for different altitudes until we find one that has the exact psi that we are looking for.

The simplified equation with all variables except altitude filled in looks like this:

14.6953 ((1 – (0.000006875586  h))5.25588 ) 

h is the altitude in feet that you are calculating the air pressure for. In this case your first guess might be 15,000 feet. You can use a scientific calculator, including the built-in Windows calculator (switched to scientific mode) and do the calculation using the following keystrokes:

MC 15000 X 0.000006875586 M+ 1 – MR = xy 5.25588 X 14.6953

which gives a result of 8.3 psi.

Even easier is to enter the formula into Excel. A1 represents the cell where you enter altitude h.

=14.6953*POWER(1-(0.000006875586*A1),5.25588)

We are looking for an altitude that has a psi of 10.0 and 8.3 is too low. So next we might try 10,000 which is 10.1 psi, then after a couple more tries arrive at 10,200 which is 10.0 psi.

Once you have done this a couple of times it goes pretty quickly, especially if using a spreadsheet. You might even use an online table to provide the first guess, and then fine tune it using the formula. Of course in the end these are just estimates, it’s up to you how accurate the estimate needs to be.

I am making an assumption that the maximum psid for flying at high altitudes can be used to establish sea level cabin pressure limits at lower altitudes. Someone may point out that the Boeing 737, which has a maximum psid of 8.3 at its service ceiling, has lower psid’s for lower altitudes. But that is for maximizing the lifetime cycles of the airframe by using a higher cabin altitude at lower altitudes than the 8.3 psi limit would otherwise allow (b737.org.uk). For example at 26,000 feet the 8.3 psi limit would allow a cabin altitude of 2,300 feet. But the maximum psid for the 737 under 28,000 feet is 7.45 psi, which requires a cabin altitude of 4,000 feet, not from a safety standpoint, but economic in terms of airframe life.

As further evidence of this, the Boeing 737 has an estimated lifespan of 75,000 cycles. The BBJ (Boeing Business Jet) however is certified with a psid of 9.0, similar to smaller business jets. This is apparently not because of any structural modifications to the BBJ compared to the airline version of the 737, but simply because they expect that in VIP use the lifetime cycles will be lower. The result of the higher psid limit is that it reduces the BBJ lifespan to 60,000 cycles instead of 75,000. This makes it pretty evident that the 8.3 psid for the 737 is not a strict structural limit, otherwise the BBJ would not have been certified to 9.0 psid.

Another example is that private jets often cruise with a cabin altitude of around 4,500 feet, however reportedly when only the pilots are onboard they often fly at a cabin altitude of 8,000 feet to increase airframe life. So it’s possible that a manufacturer might certify an air ambulance for a higher psid than the normally published limit, with the condition that the number of lifetime cycles will be reduced.

## Other Altitudes

Besides calculating sea level cabin pressure maximum altitude this method can also be used to calculate other cabin altitudes that may be of interest. I read your comments in the other question about the medical reasons for the 8,000 ft cabin pressure limit for airliners which was quite interesting. As you may know there are different rules for general aviation. Even though judgment can be affected above 10,000 feet, FAA regulations for general aviation don’t begin until 12,500 feet, and even then only if pilots spend more than 30 minutes above that altitude. Above 14,000 feet pilots must use oxygen at all times, and above 15,000 feet passengers must be provided with oxygen. It’s interesting (and useful) to calculate at what altitude a particular pressurized airplane will have a cabin altitude of for example 10,000 feet. Using the King Air 100 again as an example,

Step 1

Determine the pressure at 10,000 feet. Using the formula this is determined to be 10.1 psi.

Step 2

Subtract the King Air 100 psid of 4.7 from 10.1, the result is 5.4 psi.

Step 3

Use the formula to determine what altitude has a psi of 5.4. The result is 25,250 ft. So that would be the estimated altitude where the King Air 100 would have a cabin altitude of 10,000 feet. Of course in actual flying pilots will be using their pressure instruments and/or tables to make this determination, but still it’s interesting to do some estimates.

Along the lines of your question, air ambulances seem to sometimes operate at higher than sea level cabin pressure, but they seem to be limited to a 5,000 foot cabin altitude or similar. Here are some estimates for our example aircraft where I have included 5,000 and 8,000 foot cabin altitudes, along with the service ceilings that I could find for these airplanes.

Type psid Cabin 0 Cabin 5000 Cabin 8000 Service Ceiling
King Air 90 (A90) 4.6 psi 10,000 ft 17,000 ft 21,500 ft 30,200 ft
King Air 100 4.7 psi 10,200 ft 17,500 ft 22,000 ft 24,850 ft
King Air 90 (C90) 5.0 psi 11,000 ft 18,500 ft 23,000 ft 29,800 ft
Pilatus PC12 5.8 psi 13,200 ft 21,000 ft 26,500 ft 30,000 ft
King Air 200/300 6.5 psi 15,200 ft 24,000 ft 29,500 ft 35,000 ft
Learjet 35 9.4 psi 25,500 ft 39,000 ft 52,000 ft 45,000 ft

## The Equations

Optional reading in case you are interested in how the barometric formula works. I am using the equations (plural) shown in Wikipedia.

The reason that there are two equations is because at some altitude ranges the temperature does not change much with altitude, so they have a different formula for those altitude ranges. Fortunately the first altitude range is from 0 to 11,000 meters (36,089 feet), which covers a high percentage of airline flights. The second range, using a different formula, is from 11,000 to 20,000 meters (36,089 to 65,617 feet). However for the first several thousand feet of that range the calculation results are very similar. So the 0-FL36 formula is generally adequate for all commercially flown altitudes. That’s why at the beginning of this answer I only mentioned one formula.

Using 41,000 ft as an example, using the correct FL36-FL65 formula for that altitude gives a result of 2.59 psi. Using the 0-FL36 formula gives a result of 2.58 psi. You do start to see some differences however when you get up to Concorde height at 60,000 feet. Using the correct FL36-FL65 formula for that altitude gives a result of 1.04 psi. Using the 0-FL36 formula gives a result of 0.90 psi which is a 13% difference.

I will explain both formulas here, but I will only go into detail on the 0-FL36 formula, since the main points apply to both.

As mentioned before the 0-FL36 equation looks like this:

Some of these are a fixed constant, for others the Wikipedia page includes a table for the variables for the different altitude ranges. The formula works with either metric or imperial units, and the tables in Wikipedia have values for both. For this explanation I am using imperial units.

Hb – the reference altitude. For the 0-FL36 formula that would be sea level or 0 feet. For the FL36-FL65 formula the reference altitude is 11,000 meters, or 36,089 feet.

h – this is the altitude that we are calculating the pressure for. After simplifying this will be the only variable left in the formula.

Tb – the standard temperature at the reference altitude in kelvin, for the 0-FL36 formula this is 288.15 K (59 °F)

G0 – gravitational acceleration, this is a fixed constant at 32.17 ft/s2

M - molar mass of the atmosphere, a fixed constant 28.96 lb/lbmol

R – universal gas constant, a fixed constant 8.949×104 lbft2/(lbmol•K•s2)

Pb – the standard pressure at the reference altitude and temperature, for the 0-FL36 formula this is 29.92 inHg (inches of mercury).

Lb – temperature lapse rate due to altitude change, for the 0-FL36 formula this is 0.001981 K/ft (yes that unit works)

P – the result in inHg, which we will then convert to psi.

With the constants and variables inserted the equation now looks like this:

The result in inHg is then multiplied by 0.4912 to get psi. I won’t go through the steps of simplifying but this is the resulting simplified formula that was mentioned earlier:

14.6953 ((1 – (0.000006875586  h))5.25588 ) 

And as mentioned the Excel version of the formula would be:

=14.6953*POWER(1-(0.000006875586*A1),5.25588)

Running the equation for 0 altitude we get 14.7 psi as expected. Looking at some other altitudes

 5,000 ft = 12.2 psi
6,000 ft = 11.8 psi
8,000 ft = 10.9 psi
10,000 ft = 10.1 psi
20,000 ft =  6.8 psi
35,000 ft =  3.5 psi


For altitudes between 36,089 and 65,617 feet there is an exponential function used which assumes that the temperature does not drop with altitude:

And there are different values for some of the variables.

Here is the simplified version of the FL36 - FL65 formula:

3.28281 e(-0.0000480634 h) + 1.73456)

Using a scientific calculator, using 41,000 feet for example this would be:

MC 41000 X 0.0000480634 +/– + 1.73456 = M+ e xy MR = X 3.28281

with a result of 2.6 psi.

The Excel version is:

=3.28281*EXP((-0.0000480634*A1+1.73456))