Is this coming?

With all this presumption I thought I would run some real calculations in an attempt to provide a realistic prediction of the release time frame:

ΠN(x1,…,xn,xn+

Therefore, the F-14 Module will be released when it is released.

:thumbup:

:unsure:

Whatever the release schedule is, there is a lot in the pipeline... it will be enough to keep us busy for a very long time to come. Waiting is tough, but it will be worth the wait!

Do u have some insider info about how far along the F14 artwork is? :O All we have been show is the placeholder textures (apart from a few pics of specific parts)

Blacklion / Nick is one of our testers and advisors.

Blacklion / Nick is one of our testers and advisors.

Lets hope that means before the end of the year then.

Bring on the TOMCAT !!!!

Blacklion / Nick is one of our testers and advisors.

aren't you also an insider? :) tell us something

aren't you also an insider? :) tell us something

I think "insider" is understating it...

With all this presumption I thought I would run some real calculations in an attempt to provide a realistic prediction of the release time frame:

 

ΠN(x1,…,xn,xn+1)Π−1N(t1,…,tn)=11−xn+1(x1,…,xn),=(2t1,…,2tn,∥t∥2−1)∥t∥2+1.

ΠN(x1,…,xn,xn+1)=11−xn+1(x1,…,xn),ΠN−1(t1,…,tn)=(2t1,…,2tn,‖t‖2−1)‖t‖2+1.

 

In these coordinates, the induced (round) metric on the unit sphere is well-known (and easily checked) to be conformally-Euclidean:

 

g(t)=4(dt21+⋯+dt2n)(∥t∥2+1)2.

g(t)=4(dt12+⋯+dtn2)(‖t‖2+1)2.

 

Stereographic projection from the north pole (0,…,0,r)(0,…,0,r) of Sn®Sn® is given by the scaled mapping x↦t=rΠN(x/r)x↦t=rΠN(x/r), whose inverse is t↦x=rΠ−1N(t/r)t↦x=rΠN−1(t/r), i.e.,

 

rΠN(x1/r,…,xn/r,xn+1/r)rΠ−1N(t1/r,…,tn/r)=1r−xn+1(x1,…,xn),=(2t1,…,2tn,r(∥t/r∥2−1))∥t/r∥2+1.

rΠN(x1/r,…,xn/r,xn+1/r)=1r−xn+1(x1,…,xn),rΠN−1(t1/r,…,tn/r)=(2t1,…,2tn,r(‖t/r‖2−1))‖t/r‖2+1.

 

The induced metric in these coordinates is consequently

 

r2g(t/r)=4(dt21+⋯+dt2n)(∥t/r∥2+1)2=4r4(dt21+⋯+dt2n)(∥t∥2+r2)2.

 

Therefore, the F-14 Module will be released when it is released.

 

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If i calculate this right.... the outcome = in 2 weeks

You've almost got it wright. You need to apply the Banach fixed-point theorem to arrive to a fixed release date. Once you do it, the result is: 2 weeks from now + sqrt( (it's released when it's released)^2) - sqrt(x^2) just to avoid a miscalculation when it had been already released ;)

And here I was thinking that it was just an English Comp multiple choice test question. The answer naturally being:

C: "It will be released."

Just sit tight.

April 2018 :smilewink:

Remember autumn 2015 that Leatherneck want to show us 2 new modules.

No offense:)

With all this presumption I thought I would run some real calculations in an attempt to provide a realistic prediction of the release time frame:

 

ΠN(x1,…,xn,xn+1)Π−1N(t1,…,tn)=11−xn+1(x1,…,xn),=(2t1,…,2tn,∥t∥2−1)∥t∥2+1.

ΠN(x1,…,xn,xn+1)=11−xn+1(x1,…,xn),ΠN−1(t1,…,tn)=(2t1,…,2tn,‖t‖2−1)‖t‖2+1.

 

In these coordinates, the induced (round) metric on the unit sphere is well-known (and easily checked) to be conformally-Euclidean:

 

g(t)=4(dt21+⋯+dt2n)(∥t∥2+1)2.

g(t)=4(dt12+⋯+dtn2)(‖t‖2+1)2.

 

Stereographic projection from the north pole (0,…,0,r)(0,…,0,r) of Sn®Sn® is given by the scaled mapping x↦t=rΠN(x/r)x↦t=rΠN(x/r), whose inverse is t↦x=rΠ−1N(t/r)t↦x=rΠN−1(t/r), i.e.,

 

rΠN(x1/r,…,xn/r,xn+1/r)rΠ−1N(t1/r,…,tn/r)=1r−xn+1(x1,…,xn),=(2t1,…,2tn,r(∥t/r∥2−1))∥t/r∥2+1.

rΠN(x1/r,…,xn/r,xn+1/r)=1r−xn+1(x1,…,xn),rΠN−1(t1/r,…,tn/r)=(2t1,…,2tn,r(‖t/r‖2−1))‖t/r‖2+1.

 

The induced metric in these coordinates is consequently

 

r2g(t/r)=4(dt21+⋯+dt2n)(∥t/r∥2+1)2=4r4(dt21+⋯+dt2n)(∥t∥2+r2)2.

 

Therefore, the F-14 Module will be released when it is released.

 

:thumbup:

Exactly - that's precisely the same result I got !