I think you need to say if you are allowed to cut the rope in order to
insert a transducer (load cell etc).
Also the question appears a bit homework-like! So if it isn't, perhaps
you should tell us a bit more about the application so that people are
not put off replying.
No, this is not homework, and I'm not a student.
Imagine a parachute. How can you measure stress/tension on a parachute line?
You can not access the either end of the string/line. (One end connectected
to parachute, the other is connected to carabiner.)
Can we contact the string/line at all? Do we know what it is made of?
If so we could potentially measure stretch over a small sample if we can
access it before the load is applied and can previously know the stretch
characteristics.
Would it be acceptable to fit a sensor to the harness and assume the
load is equally distributed amongst the risers?
Otherwise, is this what you are looking for?
Collecting Parachute Test Drop Data:
http://www.industrologic.com/cptdd.htm
-- Franc Zabkar
Please remove one 'i' from my address when replying by email.
You probably could make a simple gizmo to measure the strain
using inexpensive force sensors like below (search
www.digikey.com for force sensors). You would just slip the
parachute line in the gizmo sideways and when the string is
pulled tight the gizmo flexes pinching the sensor in preportion
to the pull on the line.
http://tinyurl.com/96srw
You might try calling the U.S. Navy NAWS, China Lake, Ca.
at 760-939-9011 and ask for the parachute department. I
vaguely recall seeing something about the group being
disbanded but it may still be in business and I would
imagine they've instrumented a parachute at some time
or another.
Without access to the string, it can't be done in any way that could
justifiably be called "electronically". You're talking about doing
some serious physics here. Like: shoot a lot of x-ray intensity at it
and have an expert interpret the diffraction pattern for you to
determine the lattice length of the string, from that (assuming you at
least know the material) the deformation and from that, in turn, the
tension. Or shoot acoustic energy at it over a wide spectrum and try
to find its resonance frequency.
--
Hans-Bernhard Broeker ( snipped-for-privacy@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.
You have described a "pendulum" : oscillation frequency is
proportional to pendulum length and weight on the end.
If you know the time for one oscillation, the gravitational
acceleration, length of the rope, you should be able to
solve for the mass of the pendulum (weight + rope).
Once you have the mass at the end and the velocity of it you
should be able to calculate the force exerted on the rope
through centrifugal force.
Sounds like a lot of mucking around.
Unless the rope is infinitely rigid (wonderful first year physics assumption),
the period and the swing itself will vary with the mass, because the rope will
stretch with the angular acceleration. I don't want to do the math, even if I
could.
Not that this applies to this case, but just to be complete.
Just to be pedantic, that's the oscillation period, not frequency.
And "proportional" implies a linear relationship, not sqrt.
And the formula strictly only holds for a "simple" pendulum, ie one
where the oscillation angles are small.
See http://hyperphysics.phy-astr.gsu.edu/hbase/pendl.html
As for your interpretation of Hooke's Law, how long is a string when
no force is applied to it? ;-)
Hint #1: F = k.dL, not F = k.L.
See http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/SHM/HookesLaw.html
Hint #2: Would a 100kg mass swing with 10 times the period of a 1kg
mass, if released at the same angle?
-- Franc Zabkar
Please remove one 'i' from my address when replying by email.
I haven't seen all of this post, but the answer seems obvious.
Deflect the rope with a roller, bewtween two other rollers. Measure the force on
the
deflecting roller perpendicular to the rope. This could be fitted to a rope
without
cutting it.
Andrew M
I didn't bother to read all of this post, but if the answer seems obvious,
it's likely that someone has already proposed what you did. I guess we'll
never know.
Mitch
Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here.
All logos and trade names are the property of their respective owners.