Torque
Here's a reply on the force from another newsgroup: (it suggests that the spreader tip out of line with the mast head and chainplate can be worth dealing with. )
How to calculate the torque? Well, if you know what the tension in the
shroud is going to be (and given that the spreader is hinged, I guess we
can assume the tension in the upper and lower parts will be the same),
call it T, the shroud will exert a force F at the spreader tip equal to
2*T*cos(A/2), where A is the angle between the two parts of the shroud
(probably around 140 degrees). The direction of the force will be in
the plane defined by the two shroud parts, and will be in a direction
bisecting the angle between them.
Now identify the plane defined by this force direction and the axis of
the spreader, and note the angle B between them (this will be about
3 degrees). The force will be resisted by the spreader in two
components, one compressive along the spreader, equal to F*cos(B),
and one seeking to wrench the spreader sideways, equal to F*sin(B),
at right angles to the spreader and in the plane just identified (which
will be horizontal (and the wrenching force will therefore be directed
forward) unless the spreader is at the incorrect vertical angle (in
which case the wrenching force will also have a vertical component
(probably down))). The torque will then be 41 inches times that but
the action of the torque on the socket, if its mountings are perhaps
4 inches apart, will be to amplify the wrenching force at the spreader
tip by a factor of 10. Your rivets (if that's what holds the socket
to the mast) will be pulled out with a force equal to about 1/3 the
shroud tension (1/3 being roughly 10*sin(3)*2*cos(70)), so I reckon
it's definitely worth correcting the problem before you pile on the
pressure. But you could be lucky. Having pulled out the aft rivets,
the forward ones could still hold, as the system would by then have
re-balanced itself. There's a good chance the mast will stay up. :-)
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