The following figure is extracted from Boeing FCOM that I downloaded from the Internet:


I noticed that Mach and KIAS values are inconsistent. An example is in red box. Let me show what I found:

$$ \text{IAS} = \sqrt{\sigma} \times \text{TAS} = \sqrt{\sigma} M a $$

For ISA @35,000 ft: $ \text{IAS} = \sqrt{0.3099} \times 0.797 \times 576.4 = 256 \text{kts} $

Speed of Mach 0.797 is calculated to be 256 KIAS but the red box shows 271 KIAS. Does anyone know why IAS and Mach are inconsistent?


What you calculated is Equivalent Airspeed (EAS), not Indicated Airspeed (IAS) or Calibrated Airspeed (CAS). The formulas to calculate these speeds from the True Airspeed (TAS $ = Ma $) are:

$$ \text{EAS} = \text{TAS} \sqrt{\sigma} = M a \sqrt{\sigma} $$ $$ \text{CAS} \approx \text{EAS} \left( 1 + \frac{1}{8} (1 - \delta) M^2 \right) $$

(formulas from Wikipedia, valid below Mach 0.85)

Here, $a$ is the speed of sound, $ \sigma = \rho / \rho_0 $ is the density ratio and $ \delta = P / P_0 $ is the pressure ratio. You can find the values for ISA e.g. in the table in this answer. At FL350, they are $ a \approx 573 \, \text{kt} $, $ \sigma \approx 0.3099 $ and $ \delta \approx 0.2353 $.

Plugging in the given Mach number of 0.797 then gives:

$$ \text{EAS} \approx 254 \, \text{kt} \qquad \text{and} \qquad \text{CAS} \approx 270 \, \text{kt} $$

The difference between CAS and IAS is usually small (only a few knots). It is possible that Boeing took that into account in the table, or they use the terms interchangeably and my result of 270kt is just a rounding error.


The speeds in Boeing's table are consistent.

To demonstrate that, we will have to know how to convert between Mach and CAS. There is no need to convert to true airspeed, the relation between CAS and Mach is well defined.

Calibrated airspeed is purely based on impact pressure, which can be measured by a pitot-tube.

For subsonic speeds:


  • $V_{CAS}$ is calibrated airspeed
  • $a_{0}$ is the speed of sound under standard sea level conditions (340.3 m/s)
  • $P_0$ is the static air pressure under standard sea level conditions (101325 Pa)
  • ${q_c}$ is the impact pressure

The impact pressure for compressible flow is:

$\;q_c = P\left[\left(1+0.2 M^2 \right)^\tfrac{7}{2}-1\right]$

  • $P$ the static pressure
  • $M$ the Mach number

In standard atmospheric conditions at FL350:

  • the pressure is 23842 Pa.
  • the temperature -54.35° C or 218.8 K.

Filling in the numbers gives:

$\;q_c = 23842\left[\left(1+0.2\cdot 0.797^2 \right)^\tfrac{7}{2}-1\right] = 12393.4 $Pa

$V_{CAS}=340.297\sqrt{5\left[\left(\frac{12393.38}{101325}+1\right)^\frac{2}{7}-1\right]} = 139.3 \textrm{m/s} = 270.8 \textrm{KCAS}$

Close enough!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.