Mixing
Rotate points from one plane to another
$begingroup$
I'm trying to create a algorithm that will rotate points given on plane 1 to plane 2.
I have found two different ways of doing this.
My question is ...
Why are the transformation matrices different for solution 1 solution 2?
What am i Missing?
Solution 1
Plane 1
X Y Z
Pt A' 0.000 0.000 0.000
Pt B' 0.000 1.000 0.000
Pt C' 1.000 0.000 0.000
Va 1.000 0.000 0.000
Vb 0.000 1.000 0.000
X Axis = YxZ
Y Axis = Vb
Z Axis = Plane Normal
X Axis = 1.000 0.000 0.000
Y Axis = 0.000 1.000 0.000
Z Axis = 0.000 0.000 1.000
Plane 2
X' Y' Z'
Pt A' 814925.0 818261.8 20.8
Pt B' 814926.9 818258.6 20.8
Pt C' 814920.9 818259.3 20.4
Va -4.077 -2.439 -0.395
Vb 1.907 -3.194 0.032
X' Axis = YxZ
Y' Axis = Vb
Z' Axis = Plane Normal
X' Axis = -56.417 -33.753 -5.461
Y' Axis = 1.907 -3.194 0.032
Z' Axis = -1.339 -0.621 17.672
Transformation Matrix
Fx1 Fy1 Fz1
Fx'1 X.X' Y.X' Y.X'
Fy'1 X.Y' Y.Y' Y.Y'
Fz'1 X.Z' Y.Z' Y.Z'
Fx1 Fy1 Fz1
Fx'1 -0.855 -0.512 -0.083
Fy'1 0.513 -0.858 0.009
Fz'1 -0.075 -0.035 0.997
Final Transformation Matrix (Transpose of the one above)
Fx1 Fy1 Fz1
Fx'1 -0.855 0.513 -0.075
Fy'1 -0.512 -0.858 -0.035
Fz'1 -0.083 0.009 0.997
Solution 2
Plane 1 Normal = 0x + 0y + 1z
Plane 2 Normal = -1.339x + -0.621y + 17.672z
Rotation Axis = |N1| x |N2|
Rotation Axis = -0.421x + 0.907y + 0.000z
Cos Theta = A.B/|A|*|B|
Cos Theta = 0.99653
c = costheta = 0.9965
s = sqrt(1-c*c) = 0.0832
T = 1-c = 0.0035
Rotation Matrix (Right Handed)
xxT+c xyT-zs xzT+ys
yxT+zs yyT+c yzT-xs
zxT-ys zyT+xs zzT+c
0.997 -0.001 0.075
-0.001 0.999 0.035
-0.075 -0.035 0.997
Rotation Matrix (Left Handed)
xxT+c xyT+zs xzT-ys
yxT-zs yyT+c yzT+xs
zxT+ys zyT-xs zzT+c
0.997 -0.001 -0.075
-0.001 0.999 -0.035
0.075 0.035 0.997
Point Check (Graphical)
Original
X (ft) Y (ft) Z (ft)
0.00 0.00 1.00
0.00 1.00 1.00
1.00 1.00 1.00
1.00 0.00 1.00
Solution 1
X (ft) Y (ft) Z (ft)
-0.075 -0.035 0.997
0.437 -0.894 1.005
-0.418 -1.405 0.922
-0.931 -0.547 0.914
Solution 2 Left
X (ft) Y (ft) Z (ft)
-0.08 -0.04 1.00
-0.08 0.96 1.03
0.92 0.96 1.11
0.92 -0.04 1.07
transformation 3d rotations plane-geometry
asked Nov 30 '15 at 16:07
T ChaT Cha
1
$endgroup$
add a comment |
$begingroup$
I'm trying to create a algorithm that will rotate points given on plane 1 to plane 2.
I have found two different ways of doing this.
My question is ...
Why are the transformation matrices different for solution 1 solution 2?
What am i Missing?
Solution 1
Plane 1
X Y Z
Pt A' 0.000 0.000 0.000
Pt B' 0.000 1.000 0.000
Pt C' 1.000 0.000 0.000
Va 1.000 0.000 0.000
Vb 0.000 1.000 0.000
X Axis = YxZ
Y Axis = Vb
Z Axis = Plane Normal
X Axis = 1.000 0.000 0.000
Y Axis = 0.000 1.000 0.000
Z Axis = 0.000 0.000 1.000
Plane 2
X' Y' Z'
Pt A' 814925.0 818261.8 20.8
Pt B' 814926.9 818258.6 20.8
Pt C' 814920.9 818259.3 20.4
Va -4.077 -2.439 -0.395
Vb 1.907 -3.194 0.032
X' Axis = YxZ
Y' Axis = Vb
Z' Axis = Plane Normal
X' Axis = -56.417 -33.753 -5.461
Y' Axis = 1.907 -3.194 0.032
Z' Axis = -1.339 -0.621 17.672
Transformation Matrix
Fx1 Fy1 Fz1
Fx'1 X.X' Y.X' Y.X'
Fy'1 X.Y' Y.Y' Y.Y'
Fz'1 X.Z' Y.Z' Y.Z'
Fx1 Fy1 Fz1
Fx'1 -0.855 -0.512 -0.083
Fy'1 0.513 -0.858 0.009
Fz'1 -0.075 -0.035 0.997
Final Transformation Matrix (Transpose of the one above)
Fx1 Fy1 Fz1
Fx'1 -0.855 0.513 -0.075
Fy'1 -0.512 -0.858 -0.035
Fz'1 -0.083 0.009 0.997
Solution 2
Plane 1 Normal = 0x + 0y + 1z
Plane 2 Normal = -1.339x + -0.621y + 17.672z
Rotation Axis = |N1| x |N2|
Rotation Axis = -0.421x + 0.907y + 0.000z
Cos Theta = A.B/|A|*|B|
Cos Theta = 0.99653
c = costheta = 0.9965
s = sqrt(1-c*c) = 0.0832
T = 1-c = 0.0035
Rotation Matrix (Right Handed)
xxT+c xyT-zs xzT+ys
yxT+zs yyT+c yzT-xs
zxT-ys zyT+xs zzT+c
0.997 -0.001 0.075
-0.001 0.999 0.035
-0.075 -0.035 0.997
Rotation Matrix (Left Handed)
xxT+c xyT+zs xzT-ys
yxT-zs yyT+c yzT+xs
zxT+ys zyT-xs zzT+c
0.997 -0.001 -0.075
-0.001 0.999 -0.035
0.075 0.035 0.997
Point Check (Graphical)
Original
X (ft) Y (ft) Z (ft)
0.00 0.00 1.00
0.00 1.00 1.00
1.00 1.00 1.00
1.00 0.00 1.00
Solution 1
X (ft) Y (ft) Z (ft)
-0.075 -0.035 0.997
0.437 -0.894 1.005
-0.418 -1.405 0.922
-0.931 -0.547 0.914
Solution 2 Left
X (ft) Y (ft) Z (ft)
-0.08 -0.04 1.00
-0.08 0.96 1.03
0.92 0.96 1.11
0.92 -0.04 1.07
transformation 3d rotations plane-geometry
asked Nov 30 '15 at 16:07
T ChaT Cha
1
$endgroup$
add a comment |
$begingroup$
I'm trying to create a algorithm that will rotate points given on plane 1 to plane 2.
I have found two different ways of doing this.
My question is ...
Why are the transformation matrices different for solution 1 solution 2?
What am i Missing?
Solution 1
Plane 1
X Y Z
Pt A' 0.000 0.000 0.000
Pt B' 0.000 1.000 0.000
Pt C' 1.000 0.000 0.000
Va 1.000 0.000 0.000
Vb 0.000 1.000 0.000
X Axis = YxZ
Y Axis = Vb
Z Axis = Plane Normal
X Axis = 1.000 0.000 0.000
Y Axis = 0.000 1.000 0.000
Z Axis = 0.000 0.000 1.000
Plane 2
X' Y' Z'
Pt A' 814925.0 818261.8 20.8
Pt B' 814926.9 818258.6 20.8
Pt C' 814920.9 818259.3 20.4
Va -4.077 -2.439 -0.395
Vb 1.907 -3.194 0.032
X' Axis = YxZ
Y' Axis = Vb
Z' Axis = Plane Normal
X' Axis = -56.417 -33.753 -5.461
Y' Axis = 1.907 -3.194 0.032
Z' Axis = -1.339 -0.621 17.672
Transformation Matrix
Fx1 Fy1 Fz1
Fx'1 X.X' Y.X' Y.X'
Fy'1 X.Y' Y.Y' Y.Y'
Fz'1 X.Z' Y.Z' Y.Z'
Fx1 Fy1 Fz1
Fx'1 -0.855 -0.512 -0.083
Fy'1 0.513 -0.858 0.009
Fz'1 -0.075 -0.035 0.997
Final Transformation Matrix (Transpose of the one above)
Fx1 Fy1 Fz1
Fx'1 -0.855 0.513 -0.075
Fy'1 -0.512 -0.858 -0.035
Fz'1 -0.083 0.009 0.997
Solution 2
Plane 1 Normal = 0x + 0y + 1z
Plane 2 Normal = -1.339x + -0.621y + 17.672z
Rotation Axis = |N1| x |N2|
Rotation Axis = -0.421x + 0.907y + 0.000z
Cos Theta = A.B/|A|*|B|
Cos Theta = 0.99653
c = costheta = 0.9965
s = sqrt(1-c*c) = 0.0832
T = 1-c = 0.0035
Rotation Matrix (Right Handed)
xxT+c xyT-zs xzT+ys
yxT+zs yyT+c yzT-xs
zxT-ys zyT+xs zzT+c
0.997 -0.001 0.075
-0.001 0.999 0.035
-0.075 -0.035 0.997
Rotation Matrix (Left Handed)
xxT+c xyT+zs xzT-ys
yxT-zs yyT+c yzT+xs
zxT+ys zyT-xs zzT+c
0.997 -0.001 -0.075
-0.001 0.999 -0.035
0.075 0.035 0.997
Point Check (Graphical)
Original
X (ft) Y (ft) Z (ft)
0.00 0.00 1.00
0.00 1.00 1.00
1.00 1.00 1.00
1.00 0.00 1.00
Solution 1
X (ft) Y (ft) Z (ft)
-0.075 -0.035 0.997
0.437 -0.894 1.005
-0.418 -1.405 0.922
-0.931 -0.547 0.914
Solution 2 Left
X (ft) Y (ft) Z (ft)
-0.08 -0.04 1.00
-0.08 0.96 1.03
0.92 0.96 1.11
0.92 -0.04 1.07
transformation 3d rotations plane-geometry
asked Nov 30 '15 at 16:07
T ChaT Cha
1
$endgroup$
I'm trying to create a algorithm that will rotate points given on plane 1 to plane 2.
I have found two different ways of doing this.
My question is ...
Why are the transformation matrices different for solution 1 solution 2?
What am i Missing?
Solution 1
Plane 1
X Y Z
Pt A' 0.000 0.000 0.000
Pt B' 0.000 1.000 0.000
Pt C' 1.000 0.000 0.000
Va 1.000 0.000 0.000
Vb 0.000 1.000 0.000
X Axis = YxZ
Y Axis = Vb
Z Axis = Plane Normal
X Axis = 1.000 0.000 0.000
Y Axis = 0.000 1.000 0.000
Z Axis = 0.000 0.000 1.000
Plane 2
X' Y' Z'
Pt A' 814925.0 818261.8 20.8
Pt B' 814926.9 818258.6 20.8
Pt C' 814920.9 818259.3 20.4
Va -4.077 -2.439 -0.395
Vb 1.907 -3.194 0.032
X' Axis = YxZ
Y' Axis = Vb
Z' Axis = Plane Normal
X' Axis = -56.417 -33.753 -5.461
Y' Axis = 1.907 -3.194 0.032
Z' Axis = -1.339 -0.621 17.672
Transformation Matrix
Fx1 Fy1 Fz1
Fx'1 X.X' Y.X' Y.X'
Fy'1 X.Y' Y.Y' Y.Y'
Fz'1 X.Z' Y.Z' Y.Z'
Fx1 Fy1 Fz1
Fx'1 -0.855 -0.512 -0.083
Fy'1 0.513 -0.858 0.009
Fz'1 -0.075 -0.035 0.997
Final Transformation Matrix (Transpose of the one above)
Fx1 Fy1 Fz1
Fx'1 -0.855 0.513 -0.075
Fy'1 -0.512 -0.858 -0.035
Fz'1 -0.083 0.009 0.997
Solution 2
Plane 1 Normal = 0x + 0y + 1z
Plane 2 Normal = -1.339x + -0.621y + 17.672z
Rotation Axis = |N1| x |N2|
Rotation Axis = -0.421x + 0.907y + 0.000z
Cos Theta = A.B/|A|*|B|
Cos Theta = 0.99653
c = costheta = 0.9965
s = sqrt(1-c*c) = 0.0832
T = 1-c = 0.0035
Rotation Matrix (Right Handed)
xxT+c xyT-zs xzT+ys
yxT+zs yyT+c yzT-xs
zxT-ys zyT+xs zzT+c
0.997 -0.001 0.075
-0.001 0.999 0.035
-0.075 -0.035 0.997
Rotation Matrix (Left Handed)
xxT+c xyT+zs xzT-ys
yxT-zs yyT+c yzT+xs
zxT+ys zyT-xs zzT+c
0.997 -0.001 -0.075
-0.001 0.999 -0.035
0.075 0.035 0.997
Point Check (Graphical)
Original
X (ft) Y (ft) Z (ft)
0.00 0.00 1.00
0.00 1.00 1.00
1.00 1.00 1.00
1.00 0.00 1.00
Solution 1
X (ft) Y (ft) Z (ft)
-0.075 -0.035 0.997
0.437 -0.894 1.005
-0.418 -1.405 0.922
-0.931 -0.547 0.914
Solution 2 Left
X (ft) Y (ft) Z (ft)
-0.08 -0.04 1.00
-0.08 0.96 1.03
0.92 0.96 1.11
0.92 -0.04 1.07
transformation 3d rotations plane-geometry
transformation 3d rotations plane-geometry
asked Nov 30 '15 at 16:07
T ChaT Cha
1
asked Nov 30 '15 at 16:07
T ChaT Cha
1
asked Nov 30 '15 at 16:07
T ChaT Cha
1
asked Nov 30 '15 at 16:07
T ChaT Cha
1
asked Nov 30 '15 at 16:07
T ChaT Cha
1
1
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Essentially, you’re looking for a $3times3$ matrix $A$ with $det A=1$ that maps a given line to another given line (the normals to the source and destination plane). There are many such matrices: given such a matrix, composing it with a rotation about the destination plane’s normal gives another. If you want a unique solution, you’ll have to add other conditions to the problem.
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Essentially, you’re looking for a $3times3$ matrix $A$ with $det A=1$ that maps a given line to another given line (the normals to the source and destination plane). There are many such matrices: given such a matrix, composing it with a rotation about the destination plane’s normal gives another. If you want a unique solution, you’ll have to add other conditions to the problem.
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
$endgroup$
add a comment |
$begingroup$
Essentially, you’re looking for a $3times3$ matrix $A$ with $det A=1$ that maps a given line to another given line (the normals to the source and destination plane). There are many such matrices: given such a matrix, composing it with a rotation about the destination plane’s normal gives another. If you want a unique solution, you’ll have to add other conditions to the problem.
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
$endgroup$
add a comment |
$begingroup$
Essentially, you’re looking for a $3times3$ matrix $A$ with $det A=1$ that maps a given line to another given line (the normals to the source and destination plane). There are many such matrices: given such a matrix, composing it with a rotation about the destination plane’s normal gives another. If you want a unique solution, you’ll have to add other conditions to the problem.
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
$endgroup$
Essentially, you’re looking for a $3times3$ matrix $A$ with $det A=1$ that maps a given line to another given line (the normals to the source and destination plane). There are many such matrices: given such a matrix, composing it with a rotation about the destination plane’s normal gives another. If you want a unique solution, you’ll have to add other conditions to the problem.
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
answered Nov 30 '15 at 18:16
amdamd
31.7k21052
31.7k21052
add a comment |
add a comment |
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