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Given a ViewMatrix setting of the form {t, p}, how can one determine the ViewVertical?

Let's say we have 

vm = {
  {
    {0.10402567469787839`, 0.05634724046135078`, 0., 0.06671318616834762`}, 
    {-0.033304268882567746`, 0.061484804090894324`, 0.09542953968274223`, -0.10079092470343654`}, 
    {0.045451481623891156`, -0.08391042761333754`, 0.06992535634444795`, -0.030220722382910785`}, 
    {0.`, 0.`, 0.`, 1.`}
  }, 
  {{1.`, 0.`, 0.`, 0.5`}, {0.`, 1.`, 0.`, 0.5`}, {0.`, 0.`, -1, -0.5`}, {0.`, 0.`, 0.`, 1.`}}
};

The (normalized) ViewPoint can be found with

qr = QRDecomposition[vm[[1]]];

vpn = Sign[Diagonal[qr[[2, 1 ;; 3, 1 ;; 3]]]] qr[[1, 1 ;; 3, 3]]

{0.384185, -0.709265, 0.591054}

Norm[{1.3, -2.4, 2}] * vpn

{1.3, -2.4, 2.}

Can the ViewVertical be found in a similar fashion?

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  • 1
    $\begingroup$ From Heike's formulae from here, one could use EulerAngles[] to recover rotation angles, but retrieving ViewVertical from those does not look immediately obvious to me. $\endgroup$
    – J. M.'s eventual burnout ♦
    May 11, 2020 at 1:33
  • 1
    $\begingroup$ Thanks, I went down that rabbit hole too. I should have added that in the question. $\endgroup$
    – Greg Hurst
    May 11, 2020 at 1:54
  • 1
    $\begingroup$ I'm not sure how to check, and given the time since you asked, bounty and all, I'm assuming the obvious doesn't work, right? Inverse[vm[[1]]].{0, 1, 0, 0} // Most ? $\endgroup$
    – Rojo
    May 11, 2020 at 11:12
  • $\begingroup$ No, I don't think it does. $\endgroup$
    – Greg Hurst
    May 11, 2020 at 11:42

1 Answer 1

Reset to default
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May be this code could be supportive 

vm = {{{0.10402567469787839`, 0.05634724046135078`, 0., 
     0.06671318616834762`}, {-0.033304268882567746`, 
     0.061484804090894324`, 
     0.09542953968274223`, -0.10079092470343654`}, \
{0.045451481623891156`, -0.08391042761333754`, 
     0.06992535634444795`, -0.030220722382910785`}, {0.`, 0.`, 0.`, 
     1.`}}, {{1.`, 0.`, 0.`, 0.5`}, {0.`, 1.`, 0.`, 0.5`}, {0.`, 
     0.`, -1, -0.5`}, {0.`, 0.`, 0.`, 1.`}}};

Let define some function

gvm = Graphics3D[{Opacity[.3], Sphere[]}, ViewMatrix -> vm]

To get point we simply use 

AbsoluteOptions[gvm, ViewPoint]

Out[]= {ViewPoint -> {1.3, -2.4, 2.}}

We see that result exactly the same what you calculated. Now we can get vertical as follows 

AbsoluteOptions[gvm, ViewVertical]

Out[]= {ViewVertical -> {0., 0., 1.}}
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    $\begingroup$ It looks like AbsoluteOptions is just returning the default values. The same results are returned for the examples in the ViewMatrix reference page. $\endgroup$
    – Greg Hurst
    May 13, 2020 at 15:26
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    $\begingroup$ @ChipHurst So you think it is not ViewVertical as it suggested in vm? $\endgroup$
    – Alex Trounev
    May 13, 2020 at 16:45
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    $\begingroup$ In this case they agree, but for any vm this will always return {0., 0., 1.}. $\endgroup$
    – Greg Hurst
    May 13, 2020 at 16:53
  • $\begingroup$ @ChipHurst It looks like option ViewVertical -> {0., 0., 1.} is default option for any ViewMatrix->{t, p}. $\endgroup$
    – Alex Trounev
    May 13, 2020 at 19:01

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