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Re: How do I create a circular lamina?

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  • Subject: [mg127528] Re: How do I create a circular lamina?
  • From: Mark Green <drmoose94 at>
  • Date: Wed, 1 Aug 2012 04:58:11 -0400 (EDT)
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  • References: <24471452.12823.1343701025853.JavaMail.root@m06> <000101cd6ec6$4dd495d0$e97dc170$>


Thanks.  I was aware of Presentations but the problem is that I need to be able to make the file into a CDF that I can share with others taking classes.
  As I understand it, if I used Presentations I would be unable to do this because I would have to include Presentations in the CDF file which I cannot do because it is commercial software. 


On 31 Jul 2012, at 03:43, "djmpark" <djmpark at> wrote:

> 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 at
> From: drmoose94 at [mailto:drmoose94 at]
> 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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