-
It seems to me that including friction makes things very complicated, even in 2D, I think we have to make some simplifying assumptions.
Sliding friction would be very difficult to model, but if we are modeling a collision in outer space with no gravity or constraining force, then I guess we can ignore this case.
But even in a pure impulse situation I am not sure what simplifying...
2009-11-04 10:55:34 UTC in mjbWorld
-
Hi Ian,
It seems to me that we can draw a conceptual box around any subset of components, provided that we are clear about what is internal to the box, what is external and that we account for all the forces going across the boundary and we account for the inertia then we will have valid equations.
For instance if we just concentrate on the front/back forces, when we are braking and the...
2009-10-19 15:09:59 UTC in mjbWorld
-
Robert,
Thank you very much for this. Looking back on it I am now reasonably sure that I reversed the order of multiplication, that is it should have the rotation on the left hand of the multiplication and the initial orientation on the right hand of the multiplication. When I do this the results are consistent and the rotation is anticlockwise as expected. I have updated the text and digram...
2009-10-14 10:33:40 UTC in mjbWorld
-
Thanks very much, even though the original idea didn't work out you have given me some interesting things to think about,
Cheers - Martin.
2009-10-04 17:10:23 UTC in mjbWorld
-
we want to simplify:
-sq*(vp•vq) + vq•(-vp x vq + sq * vp)
expanding out the second part gives:
-sq*(vp•vq) - vq•(vp x vq) + vq•(sq * vp)
the scalar part can be moved outside the brackets: vq•(sq * vp)=sq*(vq•vp)
unlike the cross product the dot product commutes so: vq•vp=vp•vq
so we get:
-sq*(vp•vq) - vq•(vp x vq) + sq *(vp•vq)
canceling out the...
2009-10-03 08:22:43 UTC in mjbWorld
-
Thanks for this. I don't understand as much of this as I would like, so I appreciate any help.
I am not quite clear why we can set:
O=C*M
where:
+ O is orthogonal: O^t = O^-1
+ C is symmetric: C^t = C
+ M is almost orthogonal but not quite.
As you say, eigenvalue decomposition looks similar to SVD:
+ eigenvalue decomposition: Q*D*Q^T
+ SVD: [U][D][V]^t
The difference...
2009-10-02 15:18:22 UTC in mjbWorld
-
teamterradactyl,
Thanks very much I have made these corrections to the webpage.
I think "veracities" must have happened when I was using the spelling checker, I must have allowed a substitution in error, I intended to say vertices!
I appreciate the corrections, so please feel free to use whatever form is easiest, a single thread would be fine.
I assume the lack of an...
2009-09-26 17:24:55 UTC in mjbWorld
-
Thank you very much, I have made the changes that you suggested. I also removed the: width="100%" from the table (this means that the right justification has less effect) but I think its best to do this as display screens are getting wider and so the table might get more stretched.
Cheers,
Martin.
2009-09-20 15:03:55 UTC in mjbWorld
-
Claus,
Thank you very much for letting me know about this. I have corrected the page and while I was doing this I noticed some other issues so I have made some other changes as well. I think the page is a lot better now.
When I get the chance I will check similar pages to make sure I have not made similar errors elsewhere.
Thanks again,
Martin.
2009-08-02 17:33:23 UTC in mjbWorld
-
Hi Jay,
I think you are right because we are looking for the largest of tr1, tr2 or tr3. The first test finds the case where tr1 is the largest so in the second test we only need to test if tr2 or tr3 is the largest.
In the end I didn't put this version on the web page for the reasons that Ethan explained, however this method may show more explicitly what is happening? I would like to find...
2009-07-25 07:10:08 UTC in mjbWorld
