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# Determining Whether Two Algebraic Primitives Intersect in a Box

IP.com Disclosure Number: IPCOM000120372D
Original Publication Date: 1991-Apr-01
Included in the Prior Art Database: 2005-Apr-02
Document File: 3 page(s) / 104K

IBM

## Related People

Woodwark, JR: AUTHOR

## Abstract

Disclosed is a test, based on interval arithmetic, that determines approximately but conservatively whether two arbitrary algebraic primitives intersect in a given box. This offers the opportunity of identifying near-parallel primitives (e.g., concentric cylinders) at an early stage in the division of space during shape modelling processes of all kinds with potential for a significant speed increase by reducing the amount of work required deciding whether primitive intersection occurs.

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Determining Whether Two Algebraic Primitives Intersect in a Box

Disclosed is a test, based on interval arithmetic, that
determines approximately but conservatively whether two arbitrary
algebraic primitives intersect in a given box. This offers the
opportunity of identifying near-parallel primitives (e.g., concentric
cylinders) at an early stage in the division of space during shape
modelling processes of all kinds with potential for a significant
speed increase by reducing the amount of work required deciding
whether primitive intersection occurs.

In the VOLE (*) precursor to WINSOM (The Winchester Solid
Modeller System) an increase in efficiency was obtained by not always
allowing recursion to descend to a pixel level, but stopping it where
it could be shown that there was a primitive that occluded all the
others in a sub-space;  VOLE had planar primitives only.
Description

The technique is as follows.  Given two functions (i.e.,
surfaces in three dimensions) A = 0 and B = 0, the surface A +
mB = 0 meets A and B where they cross.  Suppose that the surfaces A
= 0 and B = 0 pass through a given box defined by intervals in
x, y and z.  If a value of m is found so that the surface
A + mB = 0 does not pass through that box, then A does not
intersect B within the box.

Put another way, where A and B meet, they are both 0, therefore
A +  mB must also be 0.  By proving that A + mB is never
0 in the box, then A = 0 and B = 0 do not cross
in the box.

It is unknown whether a surface A + mB which does not cut the
box can be found for all pairs of surfaces A and B which do not
intersect in the same box.  (There is the further possibility of
replacing m by an arbitrary function:  an unnecessary complication at
present.)  In any case, this test is certain to be approximate, as
the only way we know to determine whether A + mB cuts the box is
by interval arithmetic. Therefore, there will be times when A does
not intersect B, but the test tells us the reverse.  In this case,
the box is subdivided and the test repeated.  The test will not,
however, separate surfaces falsely;  this is what is meant by a
conservative test.

In searching for a value of m that passes the test, it is
clearly inadequate to determine intervals on A and B, and then look
for a value of m.  Since the intervals on A and B must contain zero
(because A and B pass through the box), the interval on A + mB will
also contain zero.  It is necessary to rearrange the polynomial A +
mB by collecting terms together and adding their coefficients.  The
interval computed from this will then in gene...