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Model Forum / General / Rockets / July 2005



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Physics 101 was a long time back...

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Kevin OClassen - 28 Jul 2005 10:42 GMT
I'm trying to determine the force required to break a piece of cord.

Setup: A fixed boom at the top, with tensionless attachment suspends a piece
of cord of the type being tested. A weight of 8 pounds is secured to the
lower end, and the weight dropped from successively greater height until the
cord breaks. A new piece of cord is used for each drop to avoid cumulative
damage to the sample being tested.

Now, my (probably erroneous) thinking says that the velocity of the weight
(derived from v=at) in feet/second * mass of the weight (8 pounds) yields a
force in foot/pounds.

Example: The 8 pound weight does not break the cord when dropped 18". A drop
of 22" breaks the cord. My method suggests a breaking strength of about 80
foot/pounds, which was significantly higher than I expected.  The rated
strength of the line is 22 lbs, so my numbers are not out of the ballpark,
but are perhaps way out in left field... May be apples and oranges as well,
i.e. dynamic vs static loads.

This is all about engineering parachute shroud lines, and managing the
dynamic loads of deacceleration at opening. TIA for setting me straight on
the proper method of figuring the forces involved. (yeah, yeah, I know...
the Knacke manual)

Kevin OClassen
Dan Major - 28 Jul 2005 16:29 GMT
In a word, No.  You need to hook a load cell or other force transducer to
the string so you can measure the force in real time.  your calculations
are 8way* too simplistic, and do not take into consideration things like
the accelleration and jerk that are accutaly applied to the string.  The
forces are not instantanious due to the stretch of the string.  A simple
real-time data logging system is inexpensive and easy to put together, even
if you boild it yourself.
Kevin OClassen - 28 Jul 2005 16:43 GMT
> In a word, No.  You need to hook a load cell or other force transducer to
> the string so you can measure the force in real time.  your calculations
[quoted text clipped - 4 lines]
> even
> if you boild it yourself.

Agreed. Am I closer to say that what I described essentially calculates the
total force, which is then dissipated over a (relatively short) period of
time?

KO
Dan Major - 28 Jul 2005 18:23 GMT
> Agreed. Am I closer to say that what I described essentially
> calculates the total force, which is then dissipated over a
> (relatively short) period of time?

Well, if you're looking for a simple numeric formula to do something with,
I'd recomend using a formula that puts more emphasis on the velocity - mv^2
than a simple linear formula.
Gary - 28 Jul 2005 20:28 GMT
> I'm trying to determine the force required to break a piece of cord.
>
[quoted text clipped - 21 lines]
>
> Kevin OClassen

Your calculations and units are off.  Velocity times mass is momentum,
not a force.  In English units, a measured pound weight in not a mass,
it is a force.  If you convert to SI units, you can eliminate the
units issue,
since most SI scales are calibrated in mass units (kg), not weight
(Newtons).
A pure calculation of dynamic forces will be difficult without knowing
precise times and distances (cord stretch), or making some assumptions
about them.  (And I assume you do not have tech info on the cord you are
testing.)

I would imagine a static pull would tell you the ultimate yield
strength of the cord to suit your purposes.  You could try a spring scale
or tie a bottle on it and add weight (water?) till the cord breaks.
If you
can get cord stretch info (zero to max weight), you could estimate
the dynamic performance of the cord.

Signature

Gary Bolles

summum jus, summa injuria est

To contact me; bollesg at comcast dot net
http://home.comcast.net/~bollesg/rockets/rockets.html
Help make r.m.r. a No Flame Zone

Rick Dickinson - 28 Jul 2005 20:55 GMT
>I'm trying to determine the force required to break a piece of cord.

>This is all about engineering parachute shroud lines, and managing the
>dynamic loads of deacceleration at opening. TIA for setting me straight on
>the proper method of figuring the forces involved. (yeah, yeah, I know...
>the Knacke manual)

Here's the simple way:

Send a sample of your cord to "Doc" Damerau, and he'll run tests on
honest-to-goodness materials testing equipment, and post the results
online at http://www.rocketmaterials.org for the rest of us to use.

It's what I've done, and what lots of others of us have done, too.
Check out the site -- lots of great info there!

- Rick "Pay it forward" Dickinson
Signature

Rick Dickinson - rtd@notesguy.com
National Association of Rocketry (NAR) # 73975 Level 2
Tripoli Rocketry Association (TRA) #10498 Level 2
            http://www.crayonrocketry.com

Jerry Irvine - 29 Jul 2005 02:04 GMT
> >I'm trying to determine the force required to break a piece of cord.
>
[quoted text clipped - 6 lines]
>
> Send a sample of your cord

3+ samples.

> to "Doc" Damerau, and he'll run tests on
> honest-to-goodness materials testing equipment, and post the results
[quoted text clipped - 4 lines]
>
>  - Rick "Pay it forward" Dickinson

Signature

Jerry Irvine, Box 1242, Claremont, California 91711 USA
Opinion, the whole thing. <mail to:01rocket@gte.net>
Please bring common sense back to rocketry administration. (too late)
Produce then publish.  http://www.usrockets.com
Ebay. http://tinyurl.com/6wlp8

Rick Dickinson - 29 Jul 2005 17:36 GMT
>> Send a sample of your cord
>
>3+ samples.

Well, one really long sample is what I sent, and Doc seemed to do ok
with that....

- Rick "How long is a cord?" Dickinson

Signature

Rick Dickinson - rtd@notesguy.com
National Association of Rocketry (NAR) # 73975 Level 2
Tripoli Rocketry Association (TRA) #10498 Level 2
            http://www.crayonrocketry.com

Dave Grayvis - 29 Jul 2005 17:38 GMT
>>>Send a sample of your cord
>>
[quoted text clipped - 4 lines]
>
>  - Rick "How long is a cord?" Dickinson

I'm pretty sure Doc has some scissors.
Doc - 30 Jul 2005 11:48 GMT
>>>>Send a sample of your cord
>>>
[quoted text clipped - 6 lines]
>
> I'm pretty sure Doc has some scissors.

Actually... I don't have a pair at work. I use a blueprint shear. :-)

Doc

Signature

Drake "Doc" Damerau
President, NEPRA
NAR Section 614
L3CC member
TRA 9934

www.nepra.com
www.rocketmaterials.org
http://home.sprynet.com/~monel/home.htm

David Erbas-White - 29 Jul 2005 19:07 GMT
> - Rick "How long is a cord?" Dickinson
>
>  

I believe it's 4 ft. by 4 ft. by 8 ft...

OOOHHHH....

You weren't talking about wood...

David Erbas-White
Kevin OClassen - 29 Jul 2005 09:53 GMT
> Here's the simple way:
>
[quoted text clipped - 7 lines]
>  - Rick "Pay it forward" Dickinson
> --

Thanks for the link. That is one useful site.

Kevin OClassen
Doc - 30 Jul 2005 11:40 GMT
>> Here's the simple way:
>>
[quoted text clipped - 11 lines]
>
> Kevin OClassen

Send me samples and I'll be happy to test them for you. I have a back log of
rocket materials to test. But I'm going to be doing some Kevlar cord in a
week or so, so I can test it then.

The Doctor is IN!

Doc

Signature

Drake "Doc" Damerau
President, NEPRA
NAR Section 614
L3CC member
TRA 9934

www.nepra.com
www.rocketmaterials.org
http://home.sprynet.com/~monel/home.htm

 
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