edit: Considering that the recent (23-Nov-2016) ESA update suggests that a saturation of the rotational sensors in the IMU may have set off a series of miscalculations leading to mission loss, the quote below continues to be relevant. I'm still looking for a posted answer to What are transonic oscillations (wrist mode) and would Schiaparelli have been transonic at 2 to 4 km altitude?

In the jan woerner's blog (director-general of ESA) post titled Spacecraft are Tricky ... and Engineering is an Art Form there is a comment that I don't understand. Could someone explain what is being discussed here more clearly to a non-rocket-scientist? And add a bit about how it applies to Schiaparelli's entry, descent, and landing?

What about transonic oscillation ? .

MSL descent scenario planned that wrist mode oscillations be mitigated by using little thrusters of its reaction control system to counteract these oscillations.

Contact was lost at 2 to 4 km altitude in the crtical area of transonic speed inducing resonant wave with possible parachute twist,.disturbing attitude control system and leading to a crtically reduced rocket burst.

I've used the NASA JPL site for the Mars Science Laboratory - Entry Descent and Landing and found that at 4km to 2km altitude the velocity of the MSL drops from about 95 to 80 meters per second. I know that the atmosphere is cold, but that still sounds safely sub-sonic.

Was Schiaparelli in a transonic state in that altitude range?

edit: some items incorporated from comments...

Wrist mode:



1 Answer 1


Wrist mode:

These large changes in parachute loading can couple with the “wrist mode” of the (MSL) entry body where, as the two bodies descend, the entry body rotates about its center of mass.

At 4 km, Schiaparelli was planned to be subsonic, at ~300 km/h:

Schiaparelli descent sequence

Note that speed decreases rapidly from 1700 km/h at 11 km altitude to 320 km/h at 7 km altitude, the transonic region is between these altitudes.

(at 4 km, Mach 1 is about 800 km/h)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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