PE&RS July 2019 - page 499

r
D
p
r
m
max
sel
image
cos
=
− ( ) 



( )










*
*sin
π θ α
π
θ α
2
2 2
2
where Dm is the mean diameter of the target tree i three lowest
sections matched (see full description of the matching process
in Sánchez-González et al. 2016) and the θ(α) is the zenith
angle of the viewing direction at each azimuthal direction for
the the third lowest section estimated from all the matched
trees using cubic splines (Sánchez-González et al. 2016).
The effect of occlusions should be corrected using the
information contained in the sample of apparent (i.e. not
occluded) trees. To estimate the actual nonshaded sampling
area, the theoretical sampling area Ai (Equation 17) is multi-
plied by the probability of detection Pi of each apparent tree i.
N is then computed as:
N A
i
n
i i = ( )
= ∑1
10 000 P * (18)
If S0 is the nonshaded area, the probability of detecting tree
0 in a sampling plot of radius rmax , considering the potential
shadowing of all other apparent trees is determined as:
P max 0 0
2 = S r/ π (19)
The term shadowing in this case refers to nondetection
during the matching process. Matching fails to detect or iden-
tify a tree when the stem is partially or completely missed in
any of the two stereo-images. In contrast, the method devel-
oped for application to TLS data by Seidel and Ammer (2014)
estimates the shadowed area for a single scan.
In order to calculate S0, let d01 and d02 be the horizontal
distances from the left and right cameras to tree 0 (Figure 5a)
and analogously di1 and di2 the horizontal distances from the
left and right cameras to tree i, (i = 1 … n, n being the number
of apparent trees). S0 can be computed as the integral of the
circumference length, from d01 = 0 to the maximum sampling
radius rmax (remember rmax depends on the tree size) multiplied
by the probability of no occlusion at d01:
S d d d
rmax
0
0
01 01 01 2 = ( ) ∫ * *π P d (20)
For each distance d01, the probability of no occlusion P(d01)
is the product of no occlusion probability in the left image
P(S01) and no occlusion probability in the right image, condi-
tional on being visible in the left image P(S02|S01):
P P P( | d S S S
01 01 02 01 ( ) = ( )⋅ )
P S S01 01 2 ( ) = π (21)
P S S S S02 01 02 01 | ( ) =
where S01 is the angle where there is no occlusion in the
left image and S02 is the angle where there is no occlusion in
the right image conditional on being visible in the left image.
In order to calculate S01 and S02, let ε01 and ε02 be the angle
span covered by tree 0 in the left and right images respective-
ly (Figure 5b)—and analogously εi1 and εi2 for tree i. S 01 equals
a complete sampling round 2π minus the sum of occlusion
angles in the left image οi01 produced by all shading trees i
closer to the device (Equation 22, where I(d01 < di1) equals 1 if
d01 < di1 and 0 in any other case).
S d d
i i 01 01 01 1 2 = − ⋅ < π Σ( ( )) ο I (22)
Likewise, S02 can be computed as the difference of the
visible angle in the left image S01 minus the sum of occlusion
angles ο i02 of trees located closer to the device:
S S d d
i i 02 01 02 02 2 = − ⋅ < Σ ( ( )) ο I (23)
Tree 0 occlusion angle produced by tree i (ο i01) is the angle
at which tree 0is totally or partially shaded by tree i, calcu-
lated as a sum of covering angles:
Figure 5. (a) Angular displacement δ0 between tree 0 projection in the left and right images and (b) tree i (with covering angle
in the left image εi1) shading angle over tree 0 (with covering angle ε01).
PHOTOGRAMMETRIC ENGINEERING & REMOTE SENSING July 2019 499