Mixing
3d transformation two triangles
$begingroup$
I have two triangles in 3d. I need to calculate transformation matrix(3X3) between two triangles in 3D.
1)How can I calculate the transformation matrix(rigid) while fixing one of the points to the origin(to get rid of the translation part)?
2)How does it affect if the deformation is non rigid?
transformation 3d
asked Jun 15 '12 at 5:52
nimmienimmie
412
$endgroup$
add a comment |
$begingroup$
I have two triangles in 3d. I need to calculate transformation matrix(3X3) between two triangles in 3D.
1)How can I calculate the transformation matrix(rigid) while fixing one of the points to the origin(to get rid of the translation part)?
2)How does it affect if the deformation is non rigid?
transformation 3d
asked Jun 15 '12 at 5:52
nimmienimmie
412
$endgroup$
add a comment |
$begingroup$
I have two triangles in 3d. I need to calculate transformation matrix(3X3) between two triangles in 3D.
1)How can I calculate the transformation matrix(rigid) while fixing one of the points to the origin(to get rid of the translation part)?
2)How does it affect if the deformation is non rigid?
transformation 3d
asked Jun 15 '12 at 5:52
nimmienimmie
412
$endgroup$
I have two triangles in 3d. I need to calculate transformation matrix(3X3) between two triangles in 3D.
1)How can I calculate the transformation matrix(rigid) while fixing one of the points to the origin(to get rid of the translation part)?
2)How does it affect if the deformation is non rigid?
transformation 3d
transformation 3d
asked Jun 15 '12 at 5:52
nimmienimmie
412
asked Jun 15 '12 at 5:52
nimmienimmie
412
asked Jun 15 '12 at 5:52
nimmienimmie
412
asked Jun 15 '12 at 5:52
nimmienimmie
412
asked Jun 15 '12 at 5:52
nimmienimmie
412
412
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
1) The transformation matrix is in general 4x4:
begin{equation}
begin{bmatrix}
R&t\
0&1\
end{bmatrix}
end{equation}
with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3inmathbb{R}^3$ be the first triangle and $bar{x}_1,bar{x}_2,bar{x}_3inmathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
begin{equation}
c = frac{1}{3}(x_1 + x_2 + x_3), quad bar{c} = frac{1}{3}(bar{x}_1 + bar{x}_2 + bar{x}_3), quadtext{(calculate centroids)}
end{equation}
begin{equation}
y_i = x_i-c, quad bar{y}_i = bar{x}_i-bar{c}, quadtext{(subtract centroids from points)}
end{equation}
begin{equation}
H = y_1bar{y}_1^top + y_2bar{y}_2^top +
y_3bar{y}_3^top, quad
UcdotSigmacdot V^top = H quadtext{(singular value decomposition)}
end{equation}
begin{equation}
R = V U^top, quad t = bar c - Rc
end{equation}
2) The solution above is a least-squares fit, so if there is a tiny deformation
this should not affect the result too much. It is hard to say more about it,
if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).
edited Jan 22 at 21:22
LWChris
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
$endgroup$
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
add a comment |
$begingroup$
Let the two triangles be $(O,A,B)$ and $(O,A', B')$, put ${bf a}:=OA$, $,{bf b}:=OB$, and similarly for $A'$, $B'$. Then you are looking for a linear transformation $T: {mathbb R}^3to{mathbb R}^3$ such that $T{bf a}={bf a}'$,
$ T{bf b}={bf b}'$. In the first place such a transformation is not uniquely determined. In ${mathbb R}^3$ you can prescribe the images ${bf a}_i'$ of three linearly independent vectors ${bf a_i}$ in order to determine a linear map $T$ uniquely.
Maybe there are extra circumstances present in your situation. For instance, the two triangles might be euclidean congruent to begin with, and $T$ should be a rotation. In this case the conditions
$$T{bf a}={bf a}', quad T{bf b}={bf b}', quad T({bf a}times{bf b})={bf a}'times {bf b}'$$
determine a certain rotation $T$ (whence the matrix of $T$ with respect to the standard basis) uniquely. To prove this, consider the three vectors ${bf a}$, ${bf b}$, ${bf a}times{bf b}$ as a new basis of ${mathbb R}^3$.
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
1) The transformation matrix is in general 4x4:
begin{equation}
begin{bmatrix}
R&t\
0&1\
end{bmatrix}
end{equation}
with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3inmathbb{R}^3$ be the first triangle and $bar{x}_1,bar{x}_2,bar{x}_3inmathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
begin{equation}
c = frac{1}{3}(x_1 + x_2 + x_3), quad bar{c} = frac{1}{3}(bar{x}_1 + bar{x}_2 + bar{x}_3), quadtext{(calculate centroids)}
end{equation}
begin{equation}
y_i = x_i-c, quad bar{y}_i = bar{x}_i-bar{c}, quadtext{(subtract centroids from points)}
end{equation}
begin{equation}
H = y_1bar{y}_1^top + y_2bar{y}_2^top +
y_3bar{y}_3^top, quad
UcdotSigmacdot V^top = H quadtext{(singular value decomposition)}
end{equation}
begin{equation}
R = V U^top, quad t = bar c - Rc
end{equation}
2) The solution above is a least-squares fit, so if there is a tiny deformation
this should not affect the result too much. It is hard to say more about it,
if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).
edited Jan 22 at 21:22
LWChris
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
$endgroup$
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
add a comment |
$begingroup$
1) The transformation matrix is in general 4x4:
begin{equation}
begin{bmatrix}
R&t\
0&1\
end{bmatrix}
end{equation}
with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3inmathbb{R}^3$ be the first triangle and $bar{x}_1,bar{x}_2,bar{x}_3inmathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
begin{equation}
c = frac{1}{3}(x_1 + x_2 + x_3), quad bar{c} = frac{1}{3}(bar{x}_1 + bar{x}_2 + bar{x}_3), quadtext{(calculate centroids)}
end{equation}
begin{equation}
y_i = x_i-c, quad bar{y}_i = bar{x}_i-bar{c}, quadtext{(subtract centroids from points)}
end{equation}
begin{equation}
H = y_1bar{y}_1^top + y_2bar{y}_2^top +
y_3bar{y}_3^top, quad
UcdotSigmacdot V^top = H quadtext{(singular value decomposition)}
end{equation}
begin{equation}
R = V U^top, quad t = bar c - Rc
end{equation}
2) The solution above is a least-squares fit, so if there is a tiny deformation
this should not affect the result too much. It is hard to say more about it,
if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).
edited Jan 22 at 21:22
LWChris
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
$endgroup$
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
add a comment |
$begingroup$
1) The transformation matrix is in general 4x4:
begin{equation}
begin{bmatrix}
R&t\
0&1\
end{bmatrix}
end{equation}
with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3inmathbb{R}^3$ be the first triangle and $bar{x}_1,bar{x}_2,bar{x}_3inmathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
begin{equation}
c = frac{1}{3}(x_1 + x_2 + x_3), quad bar{c} = frac{1}{3}(bar{x}_1 + bar{x}_2 + bar{x}_3), quadtext{(calculate centroids)}
end{equation}
begin{equation}
y_i = x_i-c, quad bar{y}_i = bar{x}_i-bar{c}, quadtext{(subtract centroids from points)}
end{equation}
begin{equation}
H = y_1bar{y}_1^top + y_2bar{y}_2^top +
y_3bar{y}_3^top, quad
UcdotSigmacdot V^top = H quadtext{(singular value decomposition)}
end{equation}
begin{equation}
R = V U^top, quad t = bar c - Rc
end{equation}
2) The solution above is a least-squares fit, so if there is a tiny deformation
this should not affect the result too much. It is hard to say more about it,
if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).
edited Jan 22 at 21:22
LWChris
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
$endgroup$
1) The transformation matrix is in general 4x4:
begin{equation}
begin{bmatrix}
R&t\
0&1\
end{bmatrix}
end{equation}
with $R$ being a 3x3 rotation matrix and $t$ being a translation vector.
Let $x_1,x_2,x_3inmathbb{R}^3$ be the first triangle and $bar{x}_1,bar{x}_2,bar{x}_3inmathbb{R}^3$ the second one.
Then following Arun et al. 1987, we get:
begin{equation}
c = frac{1}{3}(x_1 + x_2 + x_3), quad bar{c} = frac{1}{3}(bar{x}_1 + bar{x}_2 + bar{x}_3), quadtext{(calculate centroids)}
end{equation}
begin{equation}
y_i = x_i-c, quad bar{y}_i = bar{x}_i-bar{c}, quadtext{(subtract centroids from points)}
end{equation}
begin{equation}
H = y_1bar{y}_1^top + y_2bar{y}_2^top +
y_3bar{y}_3^top, quad
UcdotSigmacdot V^top = H quadtext{(singular value decomposition)}
end{equation}
begin{equation}
R = V U^top, quad t = bar c - Rc
end{equation}
2) The solution above is a least-squares fit, so if there is a tiny deformation
this should not affect the result too much. It is hard to say more about it,
if we don't know what kind of deformation we are expecting (scale change only, affine deformation, or something else).
edited Jan 22 at 21:22
LWChris
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
edited Jan 22 at 21:22
LWChris
1055
edited Jan 22 at 21:22
LWChris
1055
edited Jan 22 at 21:22
LWChris
1055
1055
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
answered Jun 15 '12 at 17:05
B0rk4B0rk4
1,3861126
1,3861126
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
add a comment |
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
1
1
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
$begingroup$
Can you clarify how $H$ is actually formed? It's difficult to see how you can write a sum of the multiplication of row/column vectors as a 3 x 3 matrix. I also think your expression for $t$ is wrong. referring to Arun et al 1987, it is suggested that $R = V U^top, quad t = bar{c} - Rc$
$endgroup$
– Matt Esch
Dec 21 '12 at 17:19
1
1
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Just to confirm. Arun et al consider, their vectors to be column vectors, so you're actually computing $H = Y cdot bar{Y}^top$ where $Y$ is $[y_1, y_2, y_3]$ as augmented column vectors and likewise for $bar{Y}$. If the real transformation is not rigid this will give you the least squares solution, i.e. it minimises the square of the distance between the points.
$endgroup$
– Matt Esch
Dec 24 '12 at 10:03
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
$begingroup$
Yes, you are right, I meant the outer product. This was a typo/error (fixed now).
$endgroup$
– B0rk4
Jan 2 '13 at 22:12
add a comment |
$begingroup$
Let the two triangles be $(O,A,B)$ and $(O,A', B')$, put ${bf a}:=OA$, $,{bf b}:=OB$, and similarly for $A'$, $B'$. Then you are looking for a linear transformation $T: {mathbb R}^3to{mathbb R}^3$ such that $T{bf a}={bf a}'$,
$ T{bf b}={bf b}'$. In the first place such a transformation is not uniquely determined. In ${mathbb R}^3$ you can prescribe the images ${bf a}_i'$ of three linearly independent vectors ${bf a_i}$ in order to determine a linear map $T$ uniquely.
Maybe there are extra circumstances present in your situation. For instance, the two triangles might be euclidean congruent to begin with, and $T$ should be a rotation. In this case the conditions
$$T{bf a}={bf a}', quad T{bf b}={bf b}', quad T({bf a}times{bf b})={bf a}'times {bf b}'$$
determine a certain rotation $T$ (whence the matrix of $T$ with respect to the standard basis) uniquely. To prove this, consider the three vectors ${bf a}$, ${bf b}$, ${bf a}times{bf b}$ as a new basis of ${mathbb R}^3$.
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Let the two triangles be $(O,A,B)$ and $(O,A', B')$, put ${bf a}:=OA$, $,{bf b}:=OB$, and similarly for $A'$, $B'$. Then you are looking for a linear transformation $T: {mathbb R}^3to{mathbb R}^3$ such that $T{bf a}={bf a}'$,
$ T{bf b}={bf b}'$. In the first place such a transformation is not uniquely determined. In ${mathbb R}^3$ you can prescribe the images ${bf a}_i'$ of three linearly independent vectors ${bf a_i}$ in order to determine a linear map $T$ uniquely.
Maybe there are extra circumstances present in your situation. For instance, the two triangles might be euclidean congruent to begin with, and $T$ should be a rotation. In this case the conditions
$$T{bf a}={bf a}', quad T{bf b}={bf b}', quad T({bf a}times{bf b})={bf a}'times {bf b}'$$
determine a certain rotation $T$ (whence the matrix of $T$ with respect to the standard basis) uniquely. To prove this, consider the three vectors ${bf a}$, ${bf b}$, ${bf a}times{bf b}$ as a new basis of ${mathbb R}^3$.
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
$endgroup$
add a comment |
$begingroup$
Let the two triangles be $(O,A,B)$ and $(O,A', B')$, put ${bf a}:=OA$, $,{bf b}:=OB$, and similarly for $A'$, $B'$. Then you are looking for a linear transformation $T: {mathbb R}^3to{mathbb R}^3$ such that $T{bf a}={bf a}'$,
$ T{bf b}={bf b}'$. In the first place such a transformation is not uniquely determined. In ${mathbb R}^3$ you can prescribe the images ${bf a}_i'$ of three linearly independent vectors ${bf a_i}$ in order to determine a linear map $T$ uniquely.
Maybe there are extra circumstances present in your situation. For instance, the two triangles might be euclidean congruent to begin with, and $T$ should be a rotation. In this case the conditions
$$T{bf a}={bf a}', quad T{bf b}={bf b}', quad T({bf a}times{bf b})={bf a}'times {bf b}'$$
determine a certain rotation $T$ (whence the matrix of $T$ with respect to the standard basis) uniquely. To prove this, consider the three vectors ${bf a}$, ${bf b}$, ${bf a}times{bf b}$ as a new basis of ${mathbb R}^3$.
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
$endgroup$
Let the two triangles be $(O,A,B)$ and $(O,A', B')$, put ${bf a}:=OA$, $,{bf b}:=OB$, and similarly for $A'$, $B'$. Then you are looking for a linear transformation $T: {mathbb R}^3to{mathbb R}^3$ such that $T{bf a}={bf a}'$,
$ T{bf b}={bf b}'$. In the first place such a transformation is not uniquely determined. In ${mathbb R}^3$ you can prescribe the images ${bf a}_i'$ of three linearly independent vectors ${bf a_i}$ in order to determine a linear map $T$ uniquely.
Maybe there are extra circumstances present in your situation. For instance, the two triangles might be euclidean congruent to begin with, and $T$ should be a rotation. In this case the conditions
$$T{bf a}={bf a}', quad T{bf b}={bf b}', quad T({bf a}times{bf b})={bf a}'times {bf b}'$$
determine a certain rotation $T$ (whence the matrix of $T$ with respect to the standard basis) uniquely. To prove this, consider the three vectors ${bf a}$, ${bf b}$, ${bf a}times{bf b}$ as a new basis of ${mathbb R}^3$.
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
answered Aug 3 '12 at 15:13
Christian BlatterChristian Blatter
175k8115327
175k8115327
add a comment |
add a comment |
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StackExchange.helpers.onClickDraftSave('#login-link');
});
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
