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$begingroup$

As I'm studying differential geometry of surfaces, I encountered the possibility of the first fundamental form $F_I$ being an identity matrix making the shape operator $S=F_I^-1F_II$ equal to the second fundamental form $F_II$ hence, the principle curvatures $kappa_1$ and $kappa_2$ would be the eigenvalues of $F_II$.

If $F_I$ is given as
$F_I = beginbmatrixf_xf_x&f_xf_y\f_xf_y&f_yf_yendbmatrix$

then for it to be an identity matrix, $f_x$ and $f_y$ should be orthogonal unit vectors.

While reading this article Spek et al. (2017) - A Fast Method for Computing Principal Curvatures From Range Images, at Section $(3.1)$ Theory, they started with $hatu$ and $hatv$ two orthogonal vectors in the tangent plane to the surface at point $p$.

My first question is, where does this orthogonality come into account and does it offer any advantages over using standard Cartesian coordinates system with the origin shifted to the point $p$ and rotating the surface unit normal vector to become parallel to the $Z-(0,0,1)$ axis?

Say I'm going with a (Monge patch parametrized) local surface around point $p$.

A surface of the form $F(x,y)=[x, y, z(x,y)]$ where the local surface normal is rotated to a position where it is parallel to Z-axis at $(0, 0, 1)$ and the origin is shifted to point $p$.
$z(x, y)$ is approximated by fitting a bi-quadratic surface of the form $z(x,y) = p_0x^2 + p_1y^2 + p_2xy$.

Now what should be changed to ensure that $F_I$ is an identity matrix thus, along with the rotation of the surface normal to $(0,0,1)$, $F_II$ becomes
$F_II = beginbmatrixz_xx&z_xy\z_xy&z_yyendbmatrix$

and its eigenvalues would be the principal curvatures.

Apologize if the post is not coherent (or theoretically correct), as I seem to not be able to connect the dots in this topic yet and some parts are still unclear to me.

differential-geometry curvature

share|cite|improve this question

edited Mar 21 at 23:03

Elia

asked Mar 21 at 15:14

EliaElia

1125

$endgroup$

add a comment | 

0

$begingroup$

As I'm studying differential geometry of surfaces, I encountered the possibility of the first fundamental form $F_I$ being an identity matrix making the shape operator $S=F_I^-1F_II$ equal to the second fundamental form $F_II$ hence, the principle curvatures $kappa_1$ and $kappa_2$ would be the eigenvalues of $F_II$.

If $F_I$ is given as
$F_I = beginbmatrixf_xf_x&f_xf_y\f_xf_y&f_yf_yendbmatrix$

then for it to be an identity matrix, $f_x$ and $f_y$ should be orthogonal unit vectors.

While reading this article Spek et al. (2017) - A Fast Method for Computing Principal Curvatures From Range Images, at Section $(3.1)$ Theory, they started with $hatu$ and $hatv$ two orthogonal vectors in the tangent plane to the surface at point $p$.

My first question is, where does this orthogonality come into account and does it offer any advantages over using standard Cartesian coordinates system with the origin shifted to the point $p$ and rotating the surface unit normal vector to become parallel to the $Z-(0,0,1)$ axis?

Say I'm going with a (Monge patch parametrized) local surface around point $p$.

A surface of the form $F(x,y)=[x, y, z(x,y)]$ where the local surface normal is rotated to a position where it is parallel to Z-axis at $(0, 0, 1)$ and the origin is shifted to point $p$.
$z(x, y)$ is approximated by fitting a bi-quadratic surface of the form $z(x,y) = p_0x^2 + p_1y^2 + p_2xy$.

Now what should be changed to ensure that $F_I$ is an identity matrix thus, along with the rotation of the surface normal to $(0,0,1)$, $F_II$ becomes
$F_II = beginbmatrixz_xx&z_xy\z_xy&z_yyendbmatrix$

and its eigenvalues would be the principal curvatures.

Apologize if the post is not coherent (or theoretically correct), as I seem to not be able to connect the dots in this topic yet and some parts are still unclear to me.

differential-geometry curvature

share|cite|improve this question

edited Mar 21 at 23:03

Elia

asked Mar 21 at 15:14

EliaElia

1125

$endgroup$

add a comment | 

$begingroup$

As I'm studying differential geometry of surfaces, I encountered the possibility of the first fundamental form $F_I$ being an identity matrix making the shape operator $S=F_I^-1F_II$ equal to the second fundamental form $F_II$ hence, the principle curvatures $kappa_1$ and $kappa_2$ would be the eigenvalues of $F_II$.

If $F_I$ is given as
$F_I = beginbmatrixf_xf_x&f_xf_y\f_xf_y&f_yf_yendbmatrix$

then for it to be an identity matrix, $f_x$ and $f_y$ should be orthogonal unit vectors.

While reading this article Spek et al. (2017) - A Fast Method for Computing Principal Curvatures From Range Images, at Section $(3.1)$ Theory, they started with $hatu$ and $hatv$ two orthogonal vectors in the tangent plane to the surface at point $p$.

My first question is, where does this orthogonality come into account and does it offer any advantages over using standard Cartesian coordinates system with the origin shifted to the point $p$ and rotating the surface unit normal vector to become parallel to the $Z-(0,0,1)$ axis?

Say I'm going with a (Monge patch parametrized) local surface around point $p$.

A surface of the form $F(x,y)=[x, y, z(x,y)]$ where the local surface normal is rotated to a position where it is parallel to Z-axis at $(0, 0, 1)$ and the origin is shifted to point $p$.
$z(x, y)$ is approximated by fitting a bi-quadratic surface of the form $z(x,y) = p_0x^2 + p_1y^2 + p_2xy$.

Now what should be changed to ensure that $F_I$ is an identity matrix thus, along with the rotation of the surface normal to $(0,0,1)$, $F_II$ becomes
$F_II = beginbmatrixz_xx&z_xy\z_xy&z_yyendbmatrix$

and its eigenvalues would be the principal curvatures.

Apologize if the post is not coherent (or theoretically correct), as I seem to not be able to connect the dots in this topic yet and some parts are still unclear to me.

differential-geometry curvature

share|cite|improve this question

edited Mar 21 at 23:03

Elia

asked Mar 21 at 15:14

EliaElia

1125

$endgroup$

share|cite|improve this question

edited Mar 21 at 23:03

Elia

asked Mar 21 at 15:14

EliaElia

1125

share|cite|improve this question

edited Mar 21 at 23:03

Elia

asked Mar 21 at 15:14

EliaElia

1125

asked Mar 21 at 15:14

EliaElia

1125

asked Mar 21 at 15:14

EliaElia

1125