I need to find a way to do two things. Find the time it will take for
a 3d point( lets this point ball) traveling at a constant velocity to
intercept the surface of a sphere. I also need to find the time it
will take for ball traveling at a constant velocity to intercept a
Frustum of Right Circular Cone. Remember this is in 3d space.

I figured if I can some how find the closest point (on the sphere of
Frustum of Right Circular Cone) that the velocity vector of ball will
intercept. I would have to find the distance to that intercept point
from ball?s position. Once that is done all I would have to do is
(time till intercept) = MagnitudeOf(ball?s velocity
vector)/DistanceBetween(ball and intercept point).

The only part I don?t know how to do is come up wit the equation to
shoot the vector through the 3d objects and find the intercept
point(s).

If there is another way to do this I am also willing to accept it.

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Request for Question Clarification by mathtalk-ga on 05 Apr 2004 14:54 PDT

Two quick things:

A constant velocity means the point is moving in a straight line, right?

And by an analysis of units:

time til intercept = distance / speed

which is the reciprocal of what you have.

To find the point of (closest) intersection of a line (point's
trajectory) and a quadratic surface (either the sphere or the frustum
of a right circular cone), use a parametric equation for the point's
position in time:

(x(t),y(t),z(t))

where each coordinate x,y,z is a linear function of time (because of
the constant velocity).  Plug these expressions for x,y,z in terms of
t into the equation of the surface and you'll get a quadratic to solve
for t.  If "now" is time zero, then the first intercept is the
smallest positive root t.  Note that a trajectory may have no
intercept (no real roots), or it may have one or two roots "in the
past" (negative t) that you should ignore.

You may also want to test whether the point is "now" inside the
surface, in case your point is suppose to have an effect only when
penetrating the surface from the outside.

You then introduce another degree of difficulty, I believe, when you
refer to a "ball" intercepting a frustum of a cone.  Just to be clear,
is the "ball" a sphere of known radius?  If so, you can largely treat
the ball as point and the cone as a larger cone (based on the
observation that when the original ball is tangent to the orginal
cone, its center lies on the surface of a larger cone).  The only
catch is dealing with boundaries where the cone is truncated, where
it's possible for the "ball" to reach these boundaries later than when
its center would pass the surface of the enlarged cone.

regards, mathtalk-ga

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Clarification of Question by focusedny-ga on 05 Apr 2004 15:24 PDT

Yes you?re right time till intercept = distance / speed and the ball
is a point is just named it ball to differential it from the intercept
point. Yes it is moving in a straight line.

Just to clarify
Known
1-Balls position and velocity
2-Sphere position and radius
3-Frustum of a cone?s position of both points and radii.

Unknown
1-Interception point on object?s surface

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Clarification of Question by focusedny-ga on 05 Apr 2004 15:25 PDT

I meant differentiate.

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Clarification of Question by focusedny-ga on 11 Apr 2004 22:33 PDT

thanks for the response

Perhaps I would organize the solution in this way.

Let (x0,y0,z0) be the current position of the ball/point.

Let (x',y',z') be the its constant velocity, with time t measured in
consistent units.

Then the parametric form of the point's trajectory is:

(x0 + t*x', y0 + t*y', z0 + t*z')

It remains only to provide an equation for the surface to be
intersected.  This is quite easy for the sphere, whose equation, given
radius r and center:

(xc, yc, zc)

is of course:

(x - xc)^2 + (y - yc)^2 + (z - zc)^2 = r^2

Upon substituting the parametric expressions:

x = x0 + t*x'
y = y0 + t*y'
z = z0 + t*z'

you will obtain a quadratic equation for t, which may have one, two,
or no real roots.  As discussed before, you are interested in the
smaller of any positive real roots (except that you may wish to check
for one positive and one negative root and interpret that a the point
emerging from inside the sphere).  A root at time t = 0 would indicate
the point is one the sphere's surface "now", and this too may have
some special significance for your application.

Treatment of the frustum of the cone would be similar.  However the
frustrum of a cone is necessarily less symmetrical than the sphere,
which manifests itself in awkward expressions for the general equation
for the cones surface.

I would apply a rigid transformation (rotation and translation of
coordinates, which would need to be applied equally to the
ball/point's trajectory) so that the axis of the cone becomes simply
the z-axis (say).  Then your data for the cone would consist of two
points for the top and bottom circles positioned some equal distance
apart on the z-axis as they were originally, together with their
respective radii (for the top and bottom circles).  That is:

(0,0,zA) with radius rA

(0,0,zB) with radius rB

Now the surface of the full (dual) cone consists of the rotation of a
line, e.g. the one in the xz-plane given by:

x = m (z - zB) + rB

where m = (rA - rB)/(zA - zB), all the way around the z-axis.

The equation of the full (dual cone) is then:

x^2 + y^2 = ( m(z - zB) + rB )^2

One substitutes the parametric expressions for x,y,z in terms of t as
before, and solves the resulting quadratic equation for any (positive)
real roots t.  However there is an additional check to perform, given
such real roots t, to verify that the points of contact lie in the
frustum region and not beyond.  This check is easily performed,
however, by evaluating the z-coordinate from t and seeing if this
z-value is between zA and zB.

Again there are circumstances such as a point emerging through the
frustum from within the cone whose significance is best determined by
your intended application.

Hopefully this account is sufficiently detailed for you to be able to
apply its principles to the specifics of your own notations (which I
assume will involve some programming!).

best wishes, mathtalk-ga