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• To: <seaint(--nospam--at)seaint.org>
• From: "John H. Cato, jr, R.A." <jcato(--nospam--at)accessatc.net>
• Date: Wed, 8 Oct 2003 12:35:47 -0400

```Joe,

I've read the other responses (as of this morning).  One possible other
'approach'...

Background:  One of my hobbies is rocketry (from model up to bigger stuff)
and one of my main interests therein is recovery.  The hobby had a history
of failed recoveries and I set out to try to find out why.  Did some

One thing that surfaced from that research effort was the concept of
"Equivalent Static Load" (ESL) to quantify the energies of moving objects
'hitting the end' - what describes parachutes inflating and imposing shock
loads to the 'connectors' (textile here) -- but that also seems to describe
a log 'hitting the end' (your 'stop').

That generally reduced to a 'Impact Factor' (I.F.) that was nothing more
elaborate than a 'multiplier' of the static load (tied to the static
deflection).  There were two ways it was expressed - either related to
distance fallen or velocity of impact.  This second seems to fit here.

The equation was:

I.F. = SQRT (v^2 / (g * d(st)))

SQRT= square root
v = velocity (INCHES per second)
g = gravitational acceleration (INCHES per second)
d(st) = STATIC deflection (inches)

(don't forget to convert all terms to INCHES)

then...

ESL = SL * I.F.

At that point, you would simply design the 'stop' to a larger load than the
actual static load of the object under consideration.

=-=-=-=-=

Now, I haven't given much thought to how this 'vertical' situation may or
may not translate into the 'horizontal' situation you are facing (and it
would seem the 'g' term gets curious in this regard).

Nevertheless, it seems the concept here (ESL) would have applicability - and
maybe give you a point of departure for further research (or spur discussion
here).  It would seem, at the very least, that a log hitting a stop
horizontal (at 200 fpm) is little different than a log falling vertically
and hitting that same stop (at 200 fpm) - at least from the standpoint of
assessing the minimum strength needed of that stop.

HTH,

-- john.

----- Original Message -----
From: "Joseph Grill" <j.grill(--nospam--at)comcast.net>
To: <seaint(--nospam--at)seaint.org>
Sent: Wednesday, October 08, 2003 8:24 AM

> Thanks Scott and Steve,
>
> I've gotten that far with this and with other problems of a similar
nature.
> I was just hoping that from that last guard rail type problem someone has
> joined the list with a great "snap your fingers" easy way to go about it.
> Guess I'm back to armwave the movement to get an answer.
>
> Thanks
> Joe
> ----- Original Message -----
> From: "Scott Maxwell" <smaxwell(--nospam--at)engin.umich.edu>
> To: <seaint(--nospam--at)seaint.org>
> Sent: Tuesday, October 07, 2003 10:35 PM
>
>
> > Joseph,
> >
> > The problem is easy, but not really.  <grin>
> >
> > If you know how quickly the log stops from 200 ft/s to zero and over
what
> > distance, then the problem is relatively simple.  You would be able to
> > determine the decleration and multiply that by the mass of the log to
get
> >
> > But, the problem is that determining how quickly it stops and over what
> > distance is not easy.  Thus, the problem becomes difficult.  As Michael
> > pointed out, it will be come a function of the stiffness of the stop
(i.e.
> > is it really stiff such that it does not compress much and stops the log
> > rather quickly or somewhat soft so that the log takes a little longer to
> > stop over a longer distance).
> >
> > Beyond the above, I am not much help.  I could only suggest that you
take
> > a "reasonable" guess.  If you take a guess as to the distance that it
> > takes to stop (i.e. how much the stop compresses), you should be able to
> > get an approximate decleration value.  But, to really be accurate, you
are
> > talking about dealing with the mechanics and dynamics of non-rigid
> > bodies...stuff that make my head hurt (and not because my non-rigid head
> > would rather be bounced on the more rigid desk than think of such
things).
> >
> > HTH,
> >
> > Scott
> > Ypsilanti, MI
> >
> >
> > On Tue, 7 Oct 2003, Joseph Grill wrote:
> >
> > > Gentlemen and ladies of this fine list,
> > >
> > >
> > > I know I am taking you California engineers away from the election
this
> evening, but I need to ask a question that I am not too familiar with.  I
> have a situation where a 6200# log is moving along a support driven by
> chains.  The log will contact a stop mechanism which is supposed to stop
the
> log.  The log will be moving at 200 ft/min.  I need to calculate the force
> on the stop.  Can you give me any help or send me to a reference that I
hope
> I have that may explain what I need to do.  I have been looking at a 25
year
> old dynamics book from college, but to be honest it isn't helping me very
> much.  It is a little late this evening, but I need to get going on it in
> the morning.  I can scratch out some numbers which I think will be pretty
> conservative, but would like some other input if I can get it.  The stop
> mechanism needs to be designed after the force is determined.
> > >
> > > Thanks,
> > > Joseph R. Grill

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