a way to simplified look at it is to consider the chain/rode/line to have
zero
weight pulled between two points (say 100 feet apart), then hang a 1# weight
in
the center point and check how much strain it put on the end points when the
weight hangs 20 feet, then 10 feet, then 5 feet, then 1 foot, then 1 inch,
then
1/10th inch. Just use trig to figure the forces.

So what would the forces be using your example?

I don't have a trig calc handy, but do this. divide 50 feet by 20 feet, then
10 feet, then 5 feet, then 1 foot, then 1 inch, then 1/10th inch. that will
give you the tangent of each angle.

look up each tangent, then divide each number into 1#. that will give you the
#'s force on the end points of the line.

a catenary is worse and much, much, much more difficult to calculate, but the
above will give you an idea of the HUGE forces involved once the chain starts
to pull tight

Later,

Tom

On 18 Sep 2004 23:14:45 GMT, (JAXAshby) wrote:

a way to simplified look at it is to consider the chain/rode/line to have
zero
weight pulled between two points (say 100 feet apart), then hang a 1# weight
in
the center point and check how much strain it put on the end points when the
weight hangs 20 feet, then 10 feet, then 5 feet, then 1 foot, then 1 inch,
then
1/10th inch. Just use trig to figure the forces.

So what would the forces be using your example?

I don't have a trig calc handy, but do this. divide 50 feet by 20 feet, then
10 feet, then 5 feet, then 1 foot, then 1 inch, then 1/10th inch. that will
give you the tangent of each angle.

look up each tangent, then divide each number into 1#. that will give you the
#'s force on the end points of the line.

a catenary is worse and much, much, much more difficult to calculate, but the
above will give you an idea of the HUGE forces involved once the chain starts
to pull tight

Thanks - that's pretty much what I figured.

Later,

Tom