Absurdly good gear since 1983

Re: Fall factors and forces involved 
Author: Tony Bubb
Email:   Bubb@xxxxxxxxxxxx
Date: 1997/05/01
Forums: rec.climbing

Mike Garrison wrote:
>
> Mike Kubicka wrote:
> >
> > I'm doing a physics project for school involving fall factors, forces in
> > a fall on protection and the climber, and how dynamic ropes affect these
> > forces. I've already read the Petzl catalog, but I was looking for some
> > more in depth information. Thanks in advance for any help.
>
> It's a pretty simple problem. Why not solve it for yourself? Physics is more
> fun that way.

Yeah, really simple, sure. Just do the math, 100's of ignorant
rec.climbers have done it before. Oh... as a side note, nobody
has been very close yet.

Fall factor is not the dominant system in short falls? Know why?
Hint: Model the human spine and hips into the equasion. Simple
math really!

-T.

Re: Fall Forces when belayer runs downhill  
Author: Clyde Soles
Email:   Clyde_Soles@xxxxxxxxxxxx
Date: 1995/04/15
Forums: rec.climbing

In message ID <503-QB7.jwadsworth@xxxxxxxxxxx> on 4/15/95,
jwadsworth@delphi.com wrote:
> Apparently no physicists read this, and no one who replied has studied the
> puysics of climbing.

Maybe, but what is your excuse?

> Remember this simple formula: Impact Force = Distance
> fallen divided by rope paid out. More simply, this means if you take a 20 foot
> fall on ten feet of rope, you get an IF of 2.
> If you take a 20 footer and your belayer reels it in to 5 feet of rope, you get
> an IF of 2.5.

Non sequitor. First off, impact force and fall factor (what is described above)
are not at all the same thing. It is impossible to generate a fall factor
greater than 2, no matter what you do. (You can't even achieve 2 in the real
world but we won't get into that) Think about it. If you have 10 feet of rope
out and the belayer takes in 5, you cannot fall 20 feet without cutting the
rope. Impact force is determined by numerous factors including the amount of
rope out, rope characteristics, mass of climber, and quality of belay.

Re: Fall Factor Math - Was Need Help Phys/Engr 
Author: BillFoote
Email:   billfoote@xxxxxxxxxxxxx
Date: 1995/04/19
Forums: rec.climbing

Anthony Bubb replied directly to author and wrote (paraphrased)

|> No... the rope will exceed the elastic limit and go into plastic
deformation...

I would suggest that if a rope is designed to withstand a factor-2 fall,
then it will not reach it's elastic limit (yield point) when loaded below
that factor-2 force, so the analysis should be fairly correct in that load
range.

If the rope DOES go into plastic deformation, it is in the process of
failing. OK, it might in a really bad fall (factor 2 perhaps), but as
illustrated below
the max forces calculated should still be conservative numbers.

As shown below, (QUALITATIVELY), when a material reaches it's elastic
limit, or yield point, Y, the stress-strain curve levels off. The slope
of the curve shown below is the material's Modulus of Elasticity, or the
reciprocal
of the "spring constant" k used in my analysis. So, as the material
is loaded beyond Y, the change in the slope (modulus) will result in the
actual forces being less than those calculated using the original
equation.
(The equation assumes a straight line stress-strain relationship).

As originally stated, this is a CONSERVATIVE analysis, and ACTUAL forces
should not exceed the values produced by this equation. This is not to
say
that the effect on the rope is less severe. Exceeding the elastic limit
means severe damage to the rope, and it should be retired!

| *<-results from the provided equation
| *
| *
| * + U = ultimate strenght
(breaks here)
| * +
| * + ^
| * + |__ more realistic stress-strain curve
| * Y
| *+
Force | +
(stress) | +
| +
| +
| +
|+
+----------------------------------------------------

Stretch
(strain)

BillFoote@xxxxxxxxxxxx

P.S. A interesting thing to note: According to this analysis, there is no
difference (impact force - wise) between a 2-foot factor-2 fall and a
160-foot factor-2 fall, or a 2-inch factor-2 fall!

Should a rope be retired if one, for example, falls 4 inches with only 4
inches of rope paid out? THIS IS A FACTOR-2 FALL!!!

Re: Fall Factor Math - Was Need Help Phys/Engr 
Author: Anthony R Bubb
Email:   bubb@xxxxxxxxxxxxxxx
Date: 1995/04/19
Forums: rec.climbing

billfoote@aol.com (BillFoote) writes:
>Anthony Bubb replied directly to author and wrote (paraphrased)
>
>|> No... the rope will exceed the elastic limit and go into plastic
>deformation...
>
>I would suggest that if a rope is designed to withstand a factor-2 fall,
>then it will not reach it's elastic limit (yield point) when loaded below
>that factor-2 force, so the analysis should be fairly correct in that load
>range.

Why do you think it only survives a certain number of them???

>If the rope DOES go into plastic deformation, it is in the process of
>failing. OK, it might in a really bad fall (factor 2 perhaps), but as
>illustrated below
>the max forces calculated should still be conservative numbers.

No, if you say teh spring can extend infinitely without limit,
then you may continue elastic deformation linearly until the fall
energy is absorbed. If you consider an elastic limit, the force
may curve upward exponetually until you are in a purely plastic
region. The force may well exceed the calculations you've done,
and the result is that your estimates were NOT conservative.

>As shown below, (QUALITATIVELY), when a material reaches it's elastic
>limit, or yield point, Y, the stress-strain curve levels off. The slope
>of the curve shown below is the material's Modulus of Elasticity, or the
>reciprocal
>of the "spring constant" k used in my analysis. So, as the material
>is loaded beyond Y, the change in the slope (modulus) will result in the
>actual forces being less than those calculated using the original
>equation.

Just before that point, you may get quite a swell though,
Like a function that approaches some huge value (parabolically)

The problem is that much of the "springiness" of the rope comes
NOT from the material, but from the WEAVE. This WEAVE can only
re-align so far before it's at it's max.

An example. A BW 10.5mm enduro stretches about 7% under an 80kg
mass load ( aprox 170lbf) Do you think that under 1700lbf it will
stretch 70%?!?!?!?! Now, what about the 2200 lbf you predicted?
Do you think the rope will stretch 91%?!?!? No, it won't.
The "spring" gets stiffer as you go, and all linear approximations are
irrelivant and wrong.

>As originally stated, this is a CONSERVATIVE analysis, and ACTUAL forces
>should not exceed the values produced by this equation. This is not to say
>that the effect on the rope is less severe. Exceeding the elastic limit
>means severe damage to the rope, and it should be retired!

Yup. Not that on tests by the UIAA, after the first or second fall,
that the impact forces exceed safe limits.

> (ascii art nice, but not relivant)
>
>P.S. A interesting thing to note: According to this analysis, there is no
>difference (impact force - wise) between a 2-foot factor-2 fall and a
>160-foot factor-2 fall, or a 2-inch factor-2 fall!

We get into system dominant effects. A knot and your body
can absorb most of a 2ft fall, but are insugnificant in a 200' fall.

>Should a rope be retired if one, for example, falls 4 inches with only 4
>inches of rope paid out? THIS IS A FACTOR-2 FALL!!!

No, system domination achieved by spinal compression.
I did some rough calculations on this 2 years ago, and again last fall.
They were posted.

-T.

Fall Factor Math - Was Need Help Phys/Engr 
Author: BillFoote
Email:   billfoote@xxxxxxxxxx
Date: 1995/04/18
Forums: rec.climbing

ASpencerS (aspencers@aol.com) wrote:
|> Calling all rock climbing physicists
|> Impact Force
|> In Cimamondís book ìClimbing Rock and Iceî
|> He describes Impact force as
|> I = mg + mg sqroot(1+ 2fM/mg )
|> Were
|> I = impact force in ft-lb/sec2
|> M= rope modulus in Ft-lb/sec2
|> f= fall factor (distance fallen /amount of rope paid out )
|> g= acceleration constant of gravity ( 32 ft/sec2)
|> m = mass

|> He goes onto say that a 180 pound climber falls onto a top rope with no
|> slack he generates 360 pound of force (this must refer to the top anchor
|> on a social belaye or lead ) he go on to state a fall factor of 0.4, 0.8,
|> 1.0, 1.4, 1.6, and 2.0 would generate forces of 1137; 1521, 1676; 1947;
|> 2067; 2288; pounds of force respectively.
|> for compete text refer to cinamons Climbing Rock and Ice pg. 157 -159.

|> I attempted recreate his calculations with no avail
|> I was unable to factor in M-Rope Modulus

|> What I was attempting to create was a spread sheet that would have length
|> of fall on one axes and amount of rope played out (say in 5 ft incments)
|> on the other axes. Given rope stretch of x percent say 7% you could
|> conceivably form a matrix of impact forces

|> Can any one give me some help with this excercis on how to use this
|> formula in a spread sheet. It may help others understand impact forces
|> and how they differ as rope is played out.

|> Thanks Spencer

Reading so much talk on fall factor, whether to run uphill, downhill,
sideways, "what if..." , etc., I was interested to know if the maximum
force in a fall could REALLY be correlated to fall factor. So, here is
my attempt to answer my own question... just thought some of you may also
be interested.

Mathematics of the Fall Factor

This analysis is a simple derivation from first principles of the maximum
force |> generated in a fall as a function of Fall Factor and other
parameters. Several simplifying assumptions are made which include:

* The rope is treated as a simple spring, and no attempt is made to
account for energy absorption in the rope, or dampening, which really
only results in this analysis yielding LARGER than actual maximum forces.

* Does not account for energy absorbed by either the belayer or fallen
climber, both of which will also reduce the actual maximum forces
generated.

Both of the above assumptions make this a CONSERVATIVE analysis, meaning
that actual forces ON THE ROPE will be less than those
calculated here. However, these are just the forces on THE ROPE, not on
the pro, and not concentrated by a knot.

The following notations will be used:

L = rope length (rope paid out before stretching)
l = rope stretch

S = %stretch = l / L

d = distance of fall, NOT INCLUDING stretch

F = fall factor = d/L

m = mass (of fallen climber)

g = gravitational constant

W = weight of climber = m * g

K = spring constant, force per unit length

k = K / L = spring constant per unit length or force per length per length

Z = max force

The analysis basically uses the concept of conservation of energy,
assuming that the potential energy difference between two different
heights will be equal to the energy stored in the rope (assumed to behave
as a simple spring).

Potential energy = Energy stored in spring

m * g * h = 1/2 * K * x * x

W * (d + l) = 1/2 * (k / L) * l * l (fall + stretch)

W * (d/L + l/L) = 1/2 * k * (l/L) * (l/L)

W * (F + S) = 1/2 * k * S * S

The max force on the rope is Z = K * l = (k / L) * l = k * S

W * (F + Z/k) = 1/2 * Z * S = 1/2 * Z * Z / k

or

2 * W * F * k + 2 * W * Z = Z * Z

Z * Z - 2 * W * Z - 2 * W * F * k = 0

using binomial equation to solve,

Z = (2 * W +/- sqrt(2W * 2W - 4 * (-2WFk))) / 2

Z = W + sqrt(WW + 2WFk)

So, we have all the necessary information to correllate maximum force Z to
fall factor F, except the spring constant k.

Looking at rope specs in a gear catalog, I found that climbing ropes
stretch anywhere between 4% and 8% under an 80kg load.

therefore, by my notation, k = 80 kg / s, where here lower case s is %
stretch for 80kg.

Z = W + sqrt(WW + 2WF(80 / s))

Z = W + sqrt(WW + 160WF / s)

For an 80kg climber (about 176 pounds) here are some results from that
calculation:

F s Z(kg) Z(lb)

1 8% 488 1073
1 4% 651 1433
2 8% 651 1433
2 4% 884 1945
2 2% 1214 2671
2 1% 1682 3700
2 0.5% 2344 5157

Now a couple of extra comments... notes.

1) As stated, for most dynamic climbing ropes, s=4% to 8%.

2) I have no idea what s is for webbing or static line, but probably down
in the range 1% or less.

3) Once again, this is the force on the rope (or more precisely, the
fallen climber). Knots will concentrate the force.

4) If the climber is belayed through a piece of pro, and assuming the
biner acts as a perfect frictionless pulley (it's really not), the force
on the climber must be opposed by the belayer. Therefore the total force
on the PRO is 2X!. As someone suggested on this thread,
ROPES DON'T BREAK! People and placements DO!

Just food for thought

Flames welcome. I'd like to know if I've made any horrible mistakes.
But, I think you'll find that it's equivalent to the formula provided by
ASpencer.

BillFoote@xxxxxxxxxxx

Fall Factor: Summary and corrections 
Author: Anthony R Bubb
Email:   bubb@xxxxxxxxxxxxx
Date: 1995/04/17
Forums: rec.climbing

>Clyde Said:
>It is impossible to generate a fall factor
>greater than 2, no matter what you do. (You can't even achieve 2 in the
>real world but we won't get into that)

Casey Boyce Replied:
WRONG!!! Say you're at a belay, and your partner starts up a climb.
He's just about to clip into the first piece of pro, when, slip--he's off.
He falls back to the belay, and as far below the belay as he was above
it. Do the math & it works out to fall factor 2. (say 10 ft. above the
belay. this means a 20 footer, and 20/10=2) It is also possible to
generate falls above factor 2, but I forget how (possibly by taking long
falls on static rope?). See the Petzl ad for more info.
Casey Boyce
--------------------------------------------------------------------
Casey,
Why is Clyde wrong?
OK, so how about 2.0?!?! Well, was there NO
rope below the anchor? Were you belaying the leader
dircetly off of it!?!?!? SHeesh... I'll stick with
my regular climbing partners!
-----------------------------------------------------------------------

: SOmeone said:
: Shortening rope is a Bad Idea, period. You should be placing enough pro to
: prevent groundfall in the firstplace, & by shortening rope you increase the
: liklyhood of ripping out you gear.

Chuck Replied:
Taking in rope during a fall will reduce the amount of rope available to
absorb the fall but it will also reduce the length of the fall.
If you are expecting a fall of less then factor 1.0, taking in rope will
reduce the fall factor.
If you are expecting a fall of factor 1.0, taking in rope will have no
effect on the fall factor.
If you are expecting a fall of more then factor 1.0, taking in rope will
increase the fall factor.
---------------------------------------------------------------------
Essentailly the thread we had 2 months ago. Genearlly
correct, but again, TAKING IN rope is NOT the same as
RUNNING with it. (Which was the issue at hand.) You
aren't setting a large inertia in motoin the other way...
------------------------------------------------------------------

In article <503-QB7.jwadsworth@xxxxxxxxxxxx> jwadsworth@delphi.com writes:
>Apparently no physicists read this, and no one who replied has studied the
>puysics of climbing.
---------------------------------------------------------------------------
Not you either eh? Crap, my little sister (12) could tell
you what was wrong with your analysis. I'll give you her
Email address if you want to discuss it with her.
---------------------------------------------------------------------------
>Remember this simple formula: Impact Force =
>Distance fallen divided by rope paid out.
--------------------------------------------------------------------------
Forget that. The only thing simple here was the poster.
--------------------------------------------------------------------------
>More simply, this means if
>you take a 20 foot fall on ten feet of rope, you get an IF of 2.
>If you take a 20 footer and your belayer reels it in to 5 feet of rope,
>you get an IF of 2.5.
-------------------------------------------------------------------------
This analysis is _SOOOO_ wrong, I am not even going to
bother with it point by point. Just ignore it. It is
flawed in every fundimental sense possible.
-------------------------------------------------------------------------
>Any impact force of more than 1.5 can cause whiplash or break a few bones of
>the climber, as well a rip gear right out.
------------------------------------------------------------------------
First of, all you're talking FALL FACTOR, not impact force.
Not! If the rope has not lost dynamic charactoristics, it
should not be as severe as implied. UIAA dictates that a
rope must survive the FF= 1.78 without a max impact force
over what a human can survive. That's not to say injuries
can't happen, but really...
-------------------------------------------------------------------------
>Shortening rope is a Bad Idea, period. You should be placing enough pro to
>prevent groundfall in the firstplace, & by shortening rope you increase the
>liklyhood of ripping out you gear.

SOmeone else replied:
a common mistake in several of the responses in this thread is the thinking
that running downhill shortens the rope length. It doesn't! Only taking in
rope through your belay device can shorten the rope. Think about it a little.
----------------------------------------------------------------------------
Amen. Don't confuse the two... inertial diffs. again.
----------------------------------------------------------------------------
>The old "dynamic hip belay" of pre-belay device times was safer in terms of
>impact force generated. This involved feeding rope out, not shortening it.
>Summary: DON'T shorten rope, run downhill, etc. You will probably cause more
>harm by doing this.
------------------------------------------------------------------------------
There is a second issue being ignored. The attached file at the
end of this post discusses "Short-Roping" in detail.
-----------------------------------------------------------------------------
>Read the two page ad on safety that petzl has recently printed in Climbing
>Magazine if you want a better expanation of the physics involved. John Forrest
>Gregorys Rock Sport also explains it well.
-------------------------------------------------------------------------------
Hopefully far better!
------------------------------------------------------------------------------

DTWS46B@xxxxxxxxxxxx (Casey Boyce) writes:
>> Clyde Wrote:
>> It is impossible to generate a fall factor
>>greater than 2, no matter what you do. (You can't even achieve 2 in the
>>real world but we won't get into that)

> WRONG!!! Say you're at a belay, and your partner starts up a climb.
>He's just about to clip into the first piece of pro, when, slip--he's off.
>He falls back to the belay, and as far below the belay as he was above
>it. Do the math & it works out to fall factor 2. (say 10 ft. above the
>belay. this means a 20 footer, and 20/10=2) It is also possible to
>generate falls above factor 2, but I forget how (possibly by taking long
>falls on static rope?). See the Petzl ad for more info.

malley@xxxxxxxxxxx replies:
I think your right. Suppose you're 10 feet up from the belay and
fall. The belayer is on top of things and decides to take in as much rope
as possible before you weight the rope. Suppose he takes in 3 feet. The
leader will fall the 10 feet to the belay, and then 7 more feet to the end
of the rope. He has taken a 17 foot fall and there is only 7 feet of rope
out when the fall stops. Fall factor = 17/7 = 2.43. Is this right?

----------------------------------------------------------------------------
Possible, yes. assuming that the FF= 2.0 was possible to
start with. How were you belaying in the first place though???
See prior comment. FF = 2.0 shouldn't be happening!

------------------------------Attatched File--------------------------------

Short-Roping the leader (a blue-print for injury).
This applies to sport-climbing, primarily, and more to overhangs than
anything else, though the same will be true of dead-vertical climbs.
Slightly slabby stuff DOES NOT APPLY TO THIS REASONING.

Simply put:

Given a lead fall of a certain lenght, there are several
possibilities. These amounts of "slack" are not the distance between the
climber and last piece of pro. They are the EXTRA rope payed out, or
"looseness" of the system.

#1) No slack in the rope (rope is almost tight before fall).
#2) A bit of slack in the rope (say 2').
#3) A LOT of slack (say, 5' or more) in the rope.
#4) SHORT_ROPING The belayer takes rope or puts on tension before
or durring a fall. THIS IS BAD!

In case #2, what you will have under normal circumstances, the leader falls
a bit, and is yanked to a stop below his pro. Normally, this is not a problem,
but may be less than ideal at times, with the leader being swung into the
cliff somewhat.

Case #1 Is not quite short-roping , but is almost as bad. In this situation,
the leader has progressed up what we'll assume to be a constant incline
(overhang, I guess) on the cliff. (s)he is now a certain vertical
distance and horizontal distance above the pro. WHen the leader falls,
they will come onto tension just slightly further out than they started
(due to "popping" off of the cliff), then proceed to be swung/catapoulted
into the wall at about the same distance on the other side of the pro...
I.E., right into the cliff. Pull back a swing a certian distance, it will
swing about the same amount the other way. Applies (though not ideal,
probably better than a swing) to climbers as well.

Case #4 Is SHORT_ROPING. The climber ends up falling less PAST the pro than
(s)he started above it. IE, climber is 4' above the pro, and 2' of rope
is taken in. In a high-factor fall, the leader might stop 3' below
the pro after rope stretch (2' if you don't count it). Well, let's say it was
a 45 degree angle cliff. Now you're 4' OUT from one side of your pro, plus
a foot or two (we'll say 2) because you "popped"; this totals a horizontal
distance of 6 feet out. The cliff is only 3' on the other side of the bolt.
Guess who gets slammed into the cliff.

It could be worse yet. Say, for example that your belayer DID put tension
on the line as you cruised past the last piece of pro... No you have an
added inward accelleration. DUMB! Smack, evolution isn't always working
like it should. The belayer makes the error, and the leader pays.

SO, the ideal situation could be either #2 or #3. The main point is that
the leader is further than twice the distance from the cliff when the rope
comes tight than (s)he started out horizontally from the pro.

Someone also mentioned jumping out from teh cliff when they fall. On
overhangs and verticals, I'd say it's a raw deal. Don't do it. You'll
get slammed in with an equal amount of energy added. Fall vertically.
On slabs, it will "depend" upon the fall. If you think you can avoid
impact that way, go for it, if not, slide!

Obviously, you don't give or allow slack if there is a ledge coming up.
The math was greatly simplified for the sake of brevity. Flame me if you
really want to, but all it's gonna get you is ridicule and possibly a more
elaborate model. Yes, I could do an energy analysis, but it's gonna yield
about the same #'s. Different amounts of overhangs produce the same effect
to varrying degrees, and I'll just let you guys figure out vertical walls
by yourselves (which you are capable of). Swing a pendulum of given mass
and initial horizontal displacement on a long string then a short string
to see what collides more violently with a wall, and you have a perfect
example of what happens on vertical-climbs if short-roped.

Hope this helped, but I know I didn't do the best job explaining it all.
I'll welcome net.discussion on this subject matter (won't fire the
flame thrower up except to retaliate), as I believe it's appropriate
material for this news group.

So... Any questions?

-T.

Re: Force of a fall? 
Author: Kenneth Cline
Email:   cline+@xxxxxxxxxxx
Date: 1995/08/17
Forums: rec.climbing

In article <40vt2b$1s3m@rover.ucs.ualberta.ca>, cscobie@gpu.srv.ualberta.ca (Corey Scobie) writes:
|> I was wondering if anyone could supply me with the correct procedure to
|> determine the force of a fall (math equations). I am going to trivialize the
|> question a little with the following stipulations:

Oh, no.

Everybody duck! We're in for another round of bad physics...

Please: If you feel compelled to respond to this question be
absolutely sure that any technical discussion is correct. I hereby
offer to proofread articles for technical content before posting.
I also threaten to savagely flame anyone who gets it wrong.

The simple answer to the questions posed is that despite numerous
discussions of the physics of falling on rec.climbing, we don't
know. I can assure you that a rope is more complex than a spring.
It is also clear that for a given rope, fall force is positively
correlated with fall factor, the ratio of distance fallen to length
of rope absorbing the fall. Similarly with climber's mass.

Note that ropes absorb energy by several different means, including
stretching of individual fibers, tightening of yarns and braids,
friction between fibers, and probably others that I have
overlooked. Manufacturers design their ropes to act differently at
different loads.

Ken

Fall factor spec. 
Author: David T. Wilson
Email:   dtw@xxxxxxxxxxxx
Date: 1997/03/30
Forums: rec.climbing

I had requested some information on what a UIAA FALL is and I had one
reply then founf it in a book I had over looked so I thought I would
pass it on .

The drop Test.

In a climber's eyes the drop test is the most critical test that could
be made. This test is designed to simulate a leader fall. The UIAA
measures both the ropes impact force and the number of falles a rope
can with stand.

To conduct the test, The uiaa anchors one end of a 2.8 meter length of
some manufacturers rope, which they run through a carabiner-like
device. Next they load its free end with either an 80kg. load (if
they are test a single rope), or a 55kg. loAD (FOR ONE DOUBLE OR TWIN
ROPE). They also test two double and twin ropes at a time, dropping
the 80kg. weight.

The anchored end is 0.3m away from the carabiner-like device, so the
actual lenght of the slack rope is 2.5m. The loadis dropped from this
hight, and ends up falling a bit more than 5 meters, due to the rope's
stretch. The result is a fall factor of 1.78.

In order to pass the test, a single rope must have an impact force of
less than 12 kilo newtons, and with stand at least 5 falls befor it
breaks. A double or twin rope must rigister an impact force of less
than 8 kn., and with stand at least 5 falls.

Taken from Rock Gear by: Layne Gerrard

Re: Fall factor spec. 
Author: Clyde Soles
Email:   csoles@xxxxxxxxxxxx
Date: 1997/04/01
Forums: rec.climbing

David T. Wilson <dtw@xxxxxxxxxxx> wrote:

Just to set the record straight, your source is a bit misinformed.

> The drop Test.
> In a climber's eyes the drop test is the most critical test that could
> be made.

Should read, "In the uneducated climber's eyes..." Fall ratings are
about the least valuable bits of info when deciding on a rope. Great
marketing tool though.

> They also test two double and twin ropes at a time, dropping
> the 80kg. weight.

Only the twin, not the half rope.

> The result is a fall factor of 1.78.

Close enough. It's really 1.73. Still very severe.

Re: Fall Factors  
Author: Clyde Soles
Email:   csoles@xxxxxxxxxx
Date: 1997/04/01
Forums: rec.climbing

D B FRAZ <dbfraz@xxxxxxxxxx> wrote:

> Any rope manufacturer will tell you that
> there ropes will take hundreds if not thousnads of factor .166 falls.

No, they won't. Bluewater has done tests which showed repeated
low-factor falls results in significantly lower tensile strength and
higher impact forces. This is something which is not practical to create
a standard for due to the very high costs of testing. There is also the
problem of the sheath wearing out roughly fifteen feet from the end on
ropes used for taking lots of wingers -- hence the reason behind
"Program" and gym ropes. Ropes also deteriorate with age somewhat, even
when sitting in a wrapper.

You might educate yourself a bit more before calling everyone else
idiots. There is no easy answer on when to retire your ropes from lead
duty.