Page 329 - Plastics Engineering

312 Processing of Plastics
be estimated by assuming that the volume remains constant.

2nR2( 1 - cos a)h = 2n(R + dR)2( 1 - cos a)(h + dh) + 2xrh dS sin a

Substituting for r(= R sina) and for R from (4.29) this equation may be reduced
to the form
sin2 a tan a sin ads
h 1 - cosa (H - S sina) (4.31)

This equation may be integrated with the boundary condition that h = hl at
S = 0. As a result the thickness, h, at a distance, S, along the side of the conical
mould is given by
)
H - s sina seca-l
h=h1( (4.32)

Now consider again the boundary condition referred to above. At the point
when the softened sheet first enters the mould it forms part of a spherical
bubble which does not touch the sides of the cone. The volume balance is
therefore

2(0/2)2( 1 - cos a)h1
sin2 a
sin2 a
so. hi = *h
2( 1 - cos a)
Making the substitution for hl in (4.32)

h=
2( 1 - cos a)
1 +cosa H --L
or (4.33)
hlh=( 2 )[XI
This equation may also be used to calculate the wall thickness distribution in
deep truncated cone shapes but note that its derivation is only valid up to the
point when the spherical bubble touches the centre of the base. Thereafter the
analysis involves a volume balance with freezing-off on the base and sides of
the cone.

Example 4.8 A small flower pot as shown in Fig. 4.56 is to be thermoformed
using negative forming from a flat plastic sheet 2.5 mm thick. If the diameter
of the top of the pot is 70 mm, the diameter of the base is 45 mm and the
depth is 67 mm estimate the wall thickness of the pot at a point 40 mm from
the top. Calculate also the draw ratio for this moulding.