On 1/20/2009 10:08:53 AM, lpoulo wrote:
>Lea: "However I failed to
>convey that I want the
>physical path taken from the
>top of the cube to the bottom
>to be a straight line. This is
>an important part of what I am
>trying to do."
>
>
>I'll take two conditions to
>start, and I'll skirt the
>issue of coordinate systems
>per se, since that is not the
>core of the problem.
>
>1. You have repeated the
>statement that the path is a
>straight line along the
>unfolded cube, per your
>diagram, so I'll assume no
>ambiguity here.
>
>2.In your file, you have
>another condition: "Z =
>starting from the top - the
>decent is a constant rate
>downwards to the bottom of the
>cube..."
>
>These two statements constrain
>the motion along the path to
>be at constant speed, since a
>constant downward velocity
>implies a constant horizontal
>component of velocity (look at
>the triangle of the unfolded
>cube).
This is not true. If the angular velocity is constant and the vertical (dz/dt) velocity is constant, then the horizontal velocity will vary with the distance from the axis of rotation. You can keep a "straight line" and solve for the horizontal velocity required to keep it.
In this case, the
>length of the path travelled
>is simply proportional to
>time. The path length is that
>of the hypotenuse =
>sqrt(6^2+24^2) = 24.74. If the
>total time to traverse the
>four sides is T, then the
>length of the path at an
>intermediate time t is simply
>given by 24.74*(t/T).
>
>Now let's look at the constant
>angular velocity with respect
>to the center axis of the cube
>you have also stated as a fact
>which may be used in the
>analysis. As Tom has pointed
>out twice, the "decent is a
>constant rate downwards" and
>"the angular velocity is also
>constant @ 1 complete rotation
>per 1 unit of time" are
>inconsistent with each other.
>The solution depends on which
>condition you choose to
>impose. I suspect it is the
>"constant rate downwards,"
>since I think you
>inferred(incorrectly) that
>this would also imply constant
>angular velocity - a secondary
>condition. However, you will
>need to tell us which
>condition you really want.
>
>In your plan view, pick a
>point P on the path along the
>left side of the cube (say
>between the horizontal radius
>and the diagonal radius). Call
>the center of the cube O, and
>let C be thepoint at the left
>end of the horizontal radius
>(on the cube face). In
>triangle OCP, let Y be the
>distance CP (actually the
>horizontal distance component
>of the path along the face.
>Let A be center angle COP. the
>distance OC is half the cube
>side = 3. The relation between
>Y and A is Y = 3*tan(A).
>Taking time derivatives, we
>get dY/dt =
>3*[sec(A)^2]*dA/dt.
>
>dY/dt is the horizontal
>velocity of a point on the
>path. dA/dt is the angular
>velocity. Since sec(A) is not
>a constant with time (a
>varies), the two velocities
>cannot both be constant with
>time.
>
>If the path is a straight line
>as you have reconfirmed, then
>the horizontal(dY/dt) and
>vertical(dZ/dt) velocity
>components are either both
>constant or both changing.
>Therefore, constant dZ/dt is
>not consistent with constant
>dA/dt(angular velocity). You
>need to pick one or the other.
>
>Lou
Fred Kohlhepp
fkohlhepp@sikorsky.com