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Topic: How do I create a circular lamina?
Replies: 8   Last Post: Aug 4, 2012 6:03 AM

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David Park

Posts: 1,560
Registered: 5/19/07
Re: How do I create a circular lamina?
Posted: Aug 1, 2012 5:00 AM
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The Presentations Application has the following routines that might be
useful in this regard:

Circle3D[position, normal, radius, anglerange:{0,2\[Pi]}, plotoptions] will
draw a circle with the specified position and radius. The orientation of the
circle is given by the normal vector.

Disk3D[position, normal, radius, anglerange:{0,2\[Pi]}, plotoptions] will
draw a disk with the specified position and radius. The orientation of the
disk is given by the normal vector.

DrawArrow3DAxes[location, size, headsize:0.35, colors:{Blue,Green,Orange}]
will draw an orthogonal triad of 3D arrows at location, each arrow being of
length size. The arrows will point in the x, y and z directions.

DrawLabeled3DAxes[{location, axessize, outlinedirective, fillcolor},
xspecifications, yspecifications, zspecifications] will draw labeled 3D axes
centered at location. The xyz-specifications take the form {label,
labelsize, position, angle, alignment}. The labelsize is the vertical height
of the label expression, position is in terms of the axessize, angle is the
rotation of the reading direction from the x axis, and alignment is the rhs
of an Alignment option.

AngleDisk3D[center, {vector1, vector2}, radius, opts] will draw a disk
segment of the given radius between vector1 and vector2. Options suitable
for Disk3D may be passed.

AngleSquare3D[center, {vector1, vector2}, size, sidedirective:EdgeForm[]]
will draw an outlined square of the given size between vector1 and vector2,
which are assumed to be at right angles.

EulerAngles[matrix, seqstring, opts] will return the Euler angles
corresponding to a sequence of axes rotations specified by seqstring. An
axes sequence of "XYZ" means rotation around the X axis by \[Psi], followed
by rotation about the Y axis by \[Theta], followed by rotation about the Z
axis by \[Phi]. Two sets of rotation angles {\[Psi],\[Theta],\[Phi]} are
returned. The first, canonical set, has -\[Pi]/2 <= \[Theta] <= \[Pi]/2 for
ABC sequences and 0 <= \[Theta] <= \[Pi] for ABA sequences. The second
solution corresponds to the other value of \[Theta] in the range -\[Pi] <=
\[Theta] <= \[Pi]. The answers are in terms of the standard alibi matrices,
unless alias is specified in the EAMode option. If the option EAMatrixTest
is set to True, the routine will check if the matrix is a proper rotation
matrix. If the matrix is degenerate for a given sequence, only \[Psi]
\[PlusMinus] \[Phi] can be determined. In this case, the \[Phi] = 0 solution
is returned, with the corresponding second solution. The option
EADegeneracyCriterion gives the criterion for determining degeneracy.

RotationAngleAndAxis[rotationmatrix] will generate the axis of rotation and
the associated rotation angle in radians for a 3 x 3 rotation matrix. They
are returned as {angle, axis}. An equally valid answer is obtained by
reversing the signs of both the angle and the axis vector.


David Park
djmpark@comcast.net
http://home.comcast.net/~djmpark/index.html



From: drmoose94@gmail.com [mailto:drmoose94@gmail.com]

Hi,

I want to be able to visualize Euler's Rotation Theorem (I don't think
there's an existing visualization/demonstration of that anywhere?), but in
order to do so I need to draw two intersecting great circles of spheres.

I can't see any primitive that allows a circular lamina or disk to be drawn
in 3D. There's Disk[] and Circle[] in 2D, but Mathematica won't allow them
to be converted to 3D. I can make one via revolution plotting a constant
but that seems a really ugly hack.

Is there a good way of getting a circular lamina in 3D/






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