Søgning i bogen
Den bedste måde at søge i bogen er ved at downloade PDF'en og søge i den.
Derved får du fremhævet ordene visuelt direkte på billedet af siden.
Digitaliseret bog
Bogens tekst er maskinlæst, så der kan være en del fejl og mangler.
PIERHEADS, QUAYS, AND LANDING-STAGES. 209
the water-line passes through
quadrangular shape.
The second phase of the
problem, therefore, is to deal
with values of A between C
and C-D (fig. 182), where C
is the full depth of the pon-
toon and D the depth of im-
mersion in the initial position.
First, we must détermine
the length of base-line on HN
corresponding to any assigned
value of A, say TS.
Let 6 (fig. 182) be the
length of the base required.
Then, since the area of the
triangle of immersion must
equal the area of the rectangle
RHNT, we have
^(D+ A)=LD;
A+D
P, from which point onward it resumes the
or, we can find & geometrically thus:—In fig. 182 \ 'I
produce NH to U, so that UH=HN. Join SU. \ 1
Through T draw TV parallel to SU, cutting NH in \ I
V. Join SV. Then SV is the desired water-line, \ 1
and VN=6. \ 1
The proof of the construction is simple. The \ I
triangles SVU and STU are equal, being on the \ I
same base and between the same parallels. Deduct \ 1
the common portion SYU, and add to each, in place \ I
of it, the trapezium VYTN. Then the triangle SVN \I
is equal to the triangle UTN, which is also equal \I
to the rectangle RHNT, being on double the base,\
between the same parallels.\
Reverting to the algebraical value of & = -?^-\ \I
X A + D/
and taking x and y as the co-ordinates of the centre
of gravity of the triangle SNV, referred to the same \
axes as before intersecting at B, then
„,L l/2LD\
2' s(a+d)
L 3A-D
6’ A+D
Fig 182.
14