I actually knew it but I consciously decided to remove it from the equation with the idea of make it easer for users. Let me explain.Originally Posted by jdarling
Imagine a flight simulator that allows to add new planes just adding a 3D model description and a file that describe their "physics". If this plane actually exists an user may dive in Internet or a library to find the characteristics, but may be he/she wants to add a non-real plane (i.e. a ship from his/her favorite sci-fi world, or the legendary F-19 ). Providing the "lift vectors" will make it hard to "calibrate" in any case but specially if plane "doesn't exists". My idea was that using "rotation factors" instead of actual forces it would be easer.
BTW I have no idea how to rotate a matrix applying a vector (but I know how to apply a transformation matrix to a vector ).
[edit] I'm reading the IOCCC code you linked... Impressive!
On the contrary, I find your comment very useful. If I have time I'll try to update the document this night.Originally Posted by grudzio
But I don't understand your suggestion about "divide the left hand side by mass since on the right you have acceleration and on the left forces". Isn't mass in the M vector enough? And isn't force the same than (or equivalent to) acceleration? (You see I have very limited knowledge about physics. ).
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