QUOTE (Siman @ Nov 20 2014, 06:40 AM)
On location A, only 86 seconds after touchdown, horizontal component of Philae speed is 0.314 m/s (it had no time to change), and on location B average horizontal component of speed up to that point is 0.335 m/s. There is another unused parameter. We have been told that bounce velocity was 0.38 m/s (Philae acceleration sensors). Because we have horizontal component of that speed, it now possible to calculate bounce angle. It is arccos(31/38)=34.28 (40 degrees in pictures is from previous, not so precise measurement, sorry). ...
That's about consistent with my first estimates. But we need to be careful with the extrapolation, since due to the height of the jump the rotation of the comet under Philae cannot be neglected. Philae's angular velocity relative to the center of mass of 67P slows down with height. This applies to the velocity over ground, too. For a mean height (integrated over time) of 1 km (as an upper bound estimate) this could be up to about 0.9 km (within 1h50m for a 12.7h rotation period, 1km * 2 pi * 1h50m / 12.7h), which may have to be subtracted, not ruling out, that Malmer's simulations may be roughly correct.
With 0.335 m/s the jump would be 2211 m in 1h50m, minus 900m would be 1311 m over ground.
The value depends mainly on the angle (vertically and horizontally) between the tangent of 67P's rotation and Philae immediately after first td. For a rough estimate needed for the integration a parabolic trajectory should do the job, a better approximation would be a Kepler ellipse. A detailed simulation may return even better results, provided the initial conditions would be known better.
Edit: If the sensors measured the vertical velocity component, we need the arctan instead of arccos, arctan (38/31) = 50.8 degrees.
Edit 2: For the (very much simplified) parabolic trajectory with 50.8° vertical at 1st td, I get a mean height of 452 m ( = 2/3 * (2211m / 2) * 1/2 * 38/31, after some calculus and geometry: mean height = 2/3 max height by subtracting 1/(2a) times integral over x² from -a to a, from a², ascent time = descent time, (mx²)' = 2mx), hence about 405 m to be considered (vectorially) for the horizontal jump distance near the equator. This estimate is most likely too low, since it assumes a homogenious field of gravity.