- #1

edmundfo

- 5

- 0

I assume quadratic damping. Furthermore I assume that this quadratic damping is substantially smaller along its body-centric x-axis than along its body-centric z-axis. The modelling can be found in the two attached Matlab files. It is (with gross simplifications and some changes of convention) based on Fossen & Fjellstad: Position and Attitude tracking of AUV's: A Quaternion Feedback Approach, IEEE-JOE vol 19. No. 4 Oct. 1994. The chosen values for damping parameters (xUU, zWW and mQQ) are based on Timothy Prestero's modelling of an AUV in water (master's thesis at WHOI), but I cannot see any reason why his approach should not be valid for an object moving in air as well. All units are standard SI units (kg, m, s).

The weird result is that the torpedo after something like 60 seconds of free fall reaches a vertical velocity close to zero. This only happens when x-damping (xUU) is much smaller than z-damping (zWW). If you run the script with xUU closer to zWW, you will see the torpedo reach a terminal velocity around 50 m/s downwards, which I find much more reasonable.

So my questions would be:

1. Is it plausible that a torpedo falling in air can behave like this?

2. If not, then why?

a) Any bugs in the code that I have overlooked?

b) Because I have (deliberately) ignored lift-forces or any other forces?

c) Because quadratic damping isn't applicable for high velocities?

d) Because damping coefficients for water cannot be transfered to damping coefficients for air just like that?

e) Because of numerical inaccuracies due to transformations between body and world frame etc, possibly related to the Coriolis terms which I find a bit confusing?

f) Any other reason?

Any thoughts on this would be greatly appreciated.