Assume a spherical horse moving through a vacuum... ;-)
Regards,
-=Dave
Assume a spherical horse moving through a vacuum... ;-)
Regards,
-=Dave
-- Change is inevitable, progress is not.
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.
Is this really a measurement on a parachute?
If so, Franc Zabker's link to our web site is a good start:
Collecting Parachute Test Drop Data:
But as some others have pointed out, there are some non-invasive tension measuring devices available also.
-- Gary Peek mailto:mylastname@mycompanyname.com Industrologic, Inc. http://www.industrologic.com Phone: (636) 723-4000 Fax: (636) 724-2288
a
string?
Piece of cake. Build a simulator, that is, an exact replica of the size or wire or string, the length, and fix one end of the string to an anchor, and the other end over a pulley with a weight on it. Now take a spring scale such as those used by fishermen and hook your stretched string in the middle. Have a piece of paper behind the line and mark the paper along the straight line. Now hook onto the middle of the string and apply pressure. Measure the distance in inches or centimeters, and write down the pull in pounds or grams at different distances. Now change the weight on the string and repeat the experiment. With this information in hand you can do a simple proportional formula extrapolation for the unknown in your actual case.
Let us know what happens since this seems to be a very interesting thread.
Wayne
On Fri, 09 Sep 2005 18:14:21 +1000, Mark Harriss put finger to keyboard and composed:
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
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
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
The tighter the rope the higher resonant frequency becomes like stretching an elastic band. It would require something to give it a twang and a microhone to pick up the resulting ringing.
----------------------------------------------------------------------- Ashley Clarke
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Here's something that the OP may be interested in.
I dont know the maths (too long ago when I did this sort of thing) but surely applying a force to the object that moves it a distance by a transducer then you can calculate the weight of the object - eg m=force x distance or something - then by measuring the period of the swing you can calculate the length of the string.rope and by using a lazer 'device' measure the diam of the rope/wire etc you can calculate it own weight (or close to it) - then the stress etc becomes the sum of the object plus the weight of the string
Or am I just rambling?
David - who loved physics but was h> T> > (a) Imagine an 5-100kg (we do not know exact weight) object is hanged with a
Exactly - The 3 sheave dynomometer. This method has been used in cranes for years to measure the weight on the end of the hook.
Andy
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