To determine this location we
draw the line _B c_ from _B_ (the pallet center) through the
intersection of the arc _h_ with the pitch circle _a_.
Again, it follows as a self-evident fact, if the pallet we are dealing
with was locked, that is, engaged with the tooth _D''_, the inner angle
_n_ of the exit pallet would be one and a half degrees inside the pitch
circle _a_. With the dividers set at 5", we sweep the short arc _b b_,
and from the intersection of this arc with the line _B c_ we lay off ten
degrees, and through the point so established, from _B_, we draw the
line _B d_. Below the point of intersection of the line _B d_ with the
short arc _b b_ we lay off one and a half degrees, and through the point
thus established we draw the line _B e_.
LOCATING THE INNER ANGLE OF THE EXIT PALLET.
The intersection of the line _B e_ with the arc _h_, which we will term
the point _n_, represents the location of the inner angle of the exit
pallet. We have already explained how we located the position of the
outer angle at _o_.
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