Yes the Skype SDK has been discontinued and is no longer supported. To be honest I have finished this project and don't have the Skype kit myself. It is, however, still possible to download it here http://developer.skype.com/account/tools You have to first create the Skype developer account in order to access the SDK. The SDK itself is not enough to run the Skype, you also need development and distribution key pairs. These keys have limited lifetime (the dev is valid for 60 days) and can be generated from the Skype project page Unfortunately Skype will be issuing only the development keys for the next 10 months or so. After that the SDK won't be usable at all. The distribution keys are not available anymore. Skype officially announced that on the forum:

SkypeKit will be supported until July 2013, after that time we will stop issuing keypairs and no longer provide an official support channel.

I'm sorry I cannot help you with this 🙁

All the best.

I would love to give this a try on my Cubieboard V2 but when I went to the Skype developer portal I am told they are not taking on new developers?

Could you make available the updated developer tools you used from Skype, so that I can try and follow your tutorial?

I have tried to email you but i am not sure if you will get it so I thought i should post the question.

Thanks

]]>Sorry for this late reply and BIG thanks for your comment and spotting my mistake. The (v⋅n)n is indeed nnTv and not nTnv.

Regarding the outer product identity I have update the article with the proof. The identity holds true only for normalized vector n which is probably why it didn't work for you. In the Matlab/Octave you can try this.

a=[1;2;3];

b=a/norm(a);

b*b'

b_skew = [ 0 -b(3) b(2) ; b(3) 0 -b(1) ; -b(2) b(1) 0];

b_skew^2 + eye(3)

I understand almost all of it except for 2 things in the simplification part .

1.) For this part, vp=(v⋅n)n=(nTv)n=nTnv (converting the dot and cross products to multiplication)

How were you able to swap the order of (nTv)n to nTnv?

I thought matrix multiplication was non-commutative? (Since vectors are just matrices with one of its dimensions being 1)

It also appears differently in the next step:

u=vcosα+nnTv(1−cosα)+[n]×vsinα (nnTv != nTnv)

2.) For this part, nnT=[n]2×+I (Taking the identity of the outer product)

I'm not too clear on how this identity holds true.

I tried giving concrete values to n1=1, n2=2, and n3=3 of the vector n and I'm getting different results on the 2 sides of the equation.

Once again, thanks for this awesome explanation!

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