Projection of arbitrary rotated circle on planeCenter of circle on a stereographic projectioncalculate out-of-plane and in-plane rotation from virtual camera position.Calculating rotated relative positions on 2D planeCalculating angle on ellipseGet the camera transformation matrix (Camera pose, not view matrix)Width of rotated planeComputing the properties of the 3D-projection of an ellipse.Section of cone through the rotated plane (with and without offset)What is equation of moved and rotated ellipse, parabola and hyperbola in XY plane?what is the equation of a rotated ellipsoid?
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Projection of arbitrary rotated circle on plane
Center of circle on a stereographic projectioncalculate out-of-plane and in-plane rotation from virtual camera position.Calculating rotated relative positions on 2D planeCalculating angle on ellipseGet the camera transformation matrix (Camera pose, not view matrix)Width of rotated planeComputing the properties of the 3D-projection of an ellipse.Section of cone through the rotated plane (with and without offset)What is equation of moved and rotated ellipse, parabola and hyperbola in XY plane?what is the equation of a rotated ellipsoid?
$begingroup$
I have a camera that is trying to work out the angle that a disc is rotated at.
Assuming this was an orthographic projection, how would I work out the angle that the circle is at from the ellipse that I can see?
The circle will have some 3D rotation applied to it and then it will be projected on to the plane that is the camera.
Below is an example of such a transform:
geometry rotations
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
$endgroup$
add a comment |
$begingroup$
I have a camera that is trying to work out the angle that a disc is rotated at.
Assuming this was an orthographic projection, how would I work out the angle that the circle is at from the ellipse that I can see?
The circle will have some 3D rotation applied to it and then it will be projected on to the plane that is the camera.
Below is an example of such a transform:
geometry rotations
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
$endgroup$
add a comment |
$begingroup$
I have a camera that is trying to work out the angle that a disc is rotated at.
Assuming this was an orthographic projection, how would I work out the angle that the circle is at from the ellipse that I can see?
The circle will have some 3D rotation applied to it and then it will be projected on to the plane that is the camera.
Below is an example of such a transform:
geometry rotations
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
$endgroup$
I have a camera that is trying to work out the angle that a disc is rotated at.
Assuming this was an orthographic projection, how would I work out the angle that the circle is at from the ellipse that I can see?
The circle will have some 3D rotation applied to it and then it will be projected on to the plane that is the camera.
Below is an example of such a transform:
geometry rotations
geometry rotations
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
asked Oct 20 '15 at 10:49
MorgothMorgoth
1088
1088
add a comment |
add a comment |
1 Answer
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$begingroup$
Orthogonal projection has the property that a line segment which is parallel to the plane of projection gets mapped onto a segment of the same length. All other segments will become shorter. The more so the farther they are from being parallel.
So the original diameter of the circle corresponds to the length of the major axis of the ellipse. Along that axis, the angle between original and projected segment is zero. Perpendicular to that is the minor axis of the ellipse. There the angle between original and projection is maximal, i.e. equal to the angle between original and projected plane.
Imagine cutting the scene along a plane which is perpendicular to the plane of projection, and passes through the minor axis. The projection there appears as a right triangle: one leg is the direction of the projection, one leg is the projected segment (i.e. the minor axis of the ellipse) and the hypothenuse is the original diameter of the circle (i.e. the major axis of the ellipse). Do you know how to work out the angles of a right triangle if you know the hypothenuse and one of the legs?
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
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add a comment |
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1 Answer
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$begingroup$
Orthogonal projection has the property that a line segment which is parallel to the plane of projection gets mapped onto a segment of the same length. All other segments will become shorter. The more so the farther they are from being parallel.
So the original diameter of the circle corresponds to the length of the major axis of the ellipse. Along that axis, the angle between original and projected segment is zero. Perpendicular to that is the minor axis of the ellipse. There the angle between original and projection is maximal, i.e. equal to the angle between original and projected plane.
Imagine cutting the scene along a plane which is perpendicular to the plane of projection, and passes through the minor axis. The projection there appears as a right triangle: one leg is the direction of the projection, one leg is the projected segment (i.e. the minor axis of the ellipse) and the hypothenuse is the original diameter of the circle (i.e. the major axis of the ellipse). Do you know how to work out the angles of a right triangle if you know the hypothenuse and one of the legs?
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
$endgroup$
add a comment |
$begingroup$
Orthogonal projection has the property that a line segment which is parallel to the plane of projection gets mapped onto a segment of the same length. All other segments will become shorter. The more so the farther they are from being parallel.
So the original diameter of the circle corresponds to the length of the major axis of the ellipse. Along that axis, the angle between original and projected segment is zero. Perpendicular to that is the minor axis of the ellipse. There the angle between original and projection is maximal, i.e. equal to the angle between original and projected plane.
Imagine cutting the scene along a plane which is perpendicular to the plane of projection, and passes through the minor axis. The projection there appears as a right triangle: one leg is the direction of the projection, one leg is the projected segment (i.e. the minor axis of the ellipse) and the hypothenuse is the original diameter of the circle (i.e. the major axis of the ellipse). Do you know how to work out the angles of a right triangle if you know the hypothenuse and one of the legs?
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
$endgroup$
add a comment |
$begingroup$
Orthogonal projection has the property that a line segment which is parallel to the plane of projection gets mapped onto a segment of the same length. All other segments will become shorter. The more so the farther they are from being parallel.
So the original diameter of the circle corresponds to the length of the major axis of the ellipse. Along that axis, the angle between original and projected segment is zero. Perpendicular to that is the minor axis of the ellipse. There the angle between original and projection is maximal, i.e. equal to the angle between original and projected plane.
Imagine cutting the scene along a plane which is perpendicular to the plane of projection, and passes through the minor axis. The projection there appears as a right triangle: one leg is the direction of the projection, one leg is the projected segment (i.e. the minor axis of the ellipse) and the hypothenuse is the original diameter of the circle (i.e. the major axis of the ellipse). Do you know how to work out the angles of a right triangle if you know the hypothenuse and one of the legs?
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
$endgroup$
Orthogonal projection has the property that a line segment which is parallel to the plane of projection gets mapped onto a segment of the same length. All other segments will become shorter. The more so the farther they are from being parallel.
So the original diameter of the circle corresponds to the length of the major axis of the ellipse. Along that axis, the angle between original and projected segment is zero. Perpendicular to that is the minor axis of the ellipse. There the angle between original and projection is maximal, i.e. equal to the angle between original and projected plane.
Imagine cutting the scene along a plane which is perpendicular to the plane of projection, and passes through the minor axis. The projection there appears as a right triangle: one leg is the direction of the projection, one leg is the projected segment (i.e. the minor axis of the ellipse) and the hypothenuse is the original diameter of the circle (i.e. the major axis of the ellipse). Do you know how to work out the angles of a right triangle if you know the hypothenuse and one of the legs?
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
answered Oct 20 '15 at 12:45
MvGMvG
31k450105
31k450105
add a comment |
add a comment |
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