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Inferring depth cues from a changing viewing angleSarah
Constantin
December 13, 2011
We consider the problem of two skew lines in 3-d space. Using
orthographic projectionfrom some viewing angle gives a 2d
projection of the scene, in which the two lines appear tointersect.
The goal is to determine, from the rates of change in the location
of the apparentintersection point as the viewing angle varies, the
actual depth difference between the twolines.
1 Formulating the Problem
We write the two lines as(axt + cx, ayt + cy, azt + cz)
(bxt + dx, byt + dy, bzt + dz)
In our case, we are concerned with two skew lines at different
depths, which we canrewrite as
A = (axt + cx, ayt + cy, 0)
andB = (bxt + dx, byt + dy, dz)
We want to deduce dz.The viewing angle is a matrix derived from
the Euler angles x, y, z:
V =
1 0 00 cos(x) sin(x)0 sin(x) cos(x)
cos(y) 0 sin(y)0 1 0
sin(y) 0 cos(y)
cos(z) sin(z) 0sin(z) cos(z) 0
0 0 1
For any given (x, y, z), the corresponding viewing matrix V(x,
y, z) applied to thepair of lines gives a two-dimensional image in
which the lines appear to cross. Call the
point of apparent intersection P(A, B, V). We attempt to solve
for dz, the distance betweenthe lines (the depth difference) in
terms of the derivatives of the point of intersection withrespect
to the viewing angle:
xP(A,B,V) and so on.
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First we solve for the location of the apparent intersection,
given V and the coefficientsof A and B; we want to know the value
of t.
V(axt + cx, ayt + cy, 0) = V(bxt + dx, byt + dy, dz)
t =V(dx cx, dy cy, dz)
V(ax bx, ay by, 0)
This gives us
t =cos(z)cos(y)(dx cx) cos(y)sin(x)(dy cy) + sin()dz
cos(z)cos(y)(ax bx) cos(x) sin(x)(ay by)
Now we computedV(P)
dx,dV(P)
dy,dV(P)
dz,
Each of these is a pair of expressions, in fact: (
dV(P)
dx x,
dV(P)
dx y), for instance.Computing this explicitly, in the case of
differentiation with respect to x, we get that
the first coordinate is
cos(x) cos(y)ax(dy cy)
cos(y)cos(z)(ax bx) cos x(ay by)sin x
ax(cos2 x(ay by) + (ay by)sin
2 x)(cos y cos z(dx cx) cosy sin x(dy cy) + dz sin y)
(cos y cos z(ax bx) cos x sin x(ay by))2
Denote = cos y cos z(dx cx) cos y sin x(dy cy) + dz sin y
and = cos y cos z(ax bx) cos x sin x(ay by)
So the first coordinate is
ax[cos x cos y(dy cy)
(ay by)(1 2cos
2 x)/2]
Since were solving for dz, and the only component of this
expression containing dz is ,we have to solve for . Denote by Dx
the (scalar) expression
cos x cos y(dy cy)
(ay by)(1 2cos
2 x)/2
which is the first component of the derivative divided by ax, or
equivalently the second
component divided by ay. Likewise,
Dz = cos y sin z(dx cx)
+ cos y sin z(ax bx)/
2
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= Dx2
(ay by)(1 2cos2 x)
cos x cos y(dy cy)
(ay by)(1 2cos2 x)
dz = 1/ sin y cos y cos z(dx cx) cos y sin x(dy cy)
= 1/ sin y(Dx2
(ay by)(1 2cos2 x)+ cos
x cos y(dy cy)(ay by)(1 2cos2 x)
)cos y cos z(dxcx)cos y sin x(d
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