Instructions

I believe that you should be able to follow the pulley program without comments now
that you have calculated the tangent points.
Our next problem is linking to arcs together, and in this case the arcs are going in
opposite directions. What I want you to notice in this particular program is the fact that
you enter cutting the second 0.300 radius from the opposite direction as you did in the
same radius on the opposite side. This changes the i and j values going in and coming
out.
Figure 17
When you have to solve a problem like this, the first thing you have to do is determine
what you have to work with. In this case, we know the two radii of the arc and the width
of the straight section leading to it. Studying the drawing you will find that with the
information known there is only one position where we can begin to solve for the missing
dimensions to write our program.
The radii will give us the hypotenuse of the triangle we need to form, and we can
determine the second side of the triangle by adding together the radius of the first arc and
the distance the second arc’s center is from the edge. Now we can calculate not only the
starting point of the first arc (b=√c² – a² = 0.7437), but also the angle of that triangle (sin
A° = a ÷ c = 36.4837°). With the angle now known we can calculate the x, y ending
points of the arc, and we have already calculated the i and j values for that radius.
The ending point of our first arc is also the starting point of our second arc; therefore,
because the part is symmetrical, the ending point of the second arc is located the same
distance from the center point of that arc. To start the third arc new i and j values must be
determined, because we are approaching this arc from the opposite direction even though
this arc is identical to arc one.
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