The physics of climbing & falling

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Hi,

I'm a climber and I'm trying to understand the physics involved in climbing and falling. In particular I'm trying to understand the factors that are used within the climbing community to describe it and to unite this with my (high school) physics understanding. I also would like to be able to calculate the forces involved in a couple of simple examples.

The first and foremost term used in climbing to determine the severity of a fall is the fall factor which can be seen from this diagam:

www[dot]getoutofglasgow .com/mikey/fallfactor.png

where the fall factor = length of fall / length of rope

It is understood that the fall factor is a number <2 where the closer to 2 it is the more force is applied to the protection as a result of the fall. This happens where the climber falls from above the belayer before putting the rope through a piece of protection.

Now From what I remember of high school physics

The force applied by the rope is F = kx where k is the spring constant of the rope (I'm not sure how to get this) and x is the stretch on the rope.

No this has to absorb the kinetic energy of the falling climber who's KE can be calculated from.

v^2 = v0^2 + 2ad

where v0 = 0 for a slipping climber and a = g or 9.81 m/s^2 for a earthly climber

so v^2 = 2gd

and KE = 1/2 * m v^2

so KE = m g d

Now this is where things get confused for me.

Can I say KE = Fd

where d is the stretch distance x and

F = kx to give

KE = k x^2 and therefore

k = mgd / x^2

and for force

F = mgd / x

?

Other questions I have are about the friction on the protection and how this can be accounted for. The force applied on the top protection (which acts as a pulley) and on the belayer. And how to go about calculating k for ropes from manufacturers information - and how k varies as the length of the rope extends as there is more rope to stretch.
 

Answers and Replies

  • #2
Redbelly98
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That's pretty close, but for a spring or rope you missed a factor of (1/2) for the energy:

Energy = (1/2) k x^2

So that

m g d = (1/2) k x^2

The maximum force F = kx (x is the maximum stretch here):

F = k x = 2 mgd / x

Also of possible use is the maximum acceleration when the rope is at max stretch x:

a = F/m = 2 gd / x

Note, d is the fall distance here.

Other questions I have are about the Friction on the protection and how this can be accounted for. The force applied on the top protection (which acts as a pulley) and on the belayer. And how to go about calculating k for ropes from manufacturers information - and how k varies as the length of the rope extends as there is more rope to stretch.
Not sure if calculating k is possible, it may require measurement. k should be inversely proportional to rope length.
 
  • #3
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Does a = F/m = 2 gd / x not show g the gravitational pull which will result in a maximum stretch x? If so I'm not so interested in the whole space man climbing thing :).

F = k x = 2 mgd / x thanks but I'm still not sure how to use this because I don't know what x will be.

k should be inversely proportional to rope length.
So k could be replaced in the formula by rope stretch per unit meter / length of rope?
 
  • #4
Redbelly98
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Does a = F/m = 2 gd / x not show g the gravitational pull which will result in a maximum stretch x? If so I'm not so interested in the whole space man climbing thing :).
Okay, I'll stick to force in that case :-)

F = k x = 2 mgd / x thanks but I'm still not sure how to use this because I don't know what x will be.
Good point. Mainly I'm correcting your original formula here. But yes, we will want an equation for F in terms of known or knowable quantities.

We'll have to know either k or the maximum x to get this. Let's rule out measuring x during a fall, so that means we need to know k. (Which I think you have asked about already).

From before,

mgd = (1/2) k x^2 = (1/2) k (F/k)^2 = (1/2) F^2 / k

so

F^2 = 2 mgdk

or

[tex]F = \sqrt{2mgdk}[/tex]


So k could be replaced in the formula by rope stretch per unit meter / length of rope?
Since k is inversely proportional to rope length, we can say

k = A/L

where A is a constant for a given rope, no matter what length we cut it to, and
L is the rope length.

So

F = k x = A x / L

I think you'll need to do a measurement to get k. Hang a known weight (perhaps yourself, or even several people if possible) and see if you can measure x, the change in length of the rope. Note, you needn't actually measure the complete length while people hang from the rope, you just need the distance that end of the rope moves or stretches, before and after the people get on it.

If x is a very small amount, this will be a difficult or impractical measurement. I'm not familiar with how much "stretch" is typical for a climber's rope. The longer the rope length, the more it will stretch under a given force and the easier it will be to measure the amount of stretch.
 
  • #5
Redbelly98
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Playing with the force result some more, I substituted A/L for k and get

[tex]
F = \sqrt{2mgAd/L}
[/tex]

Interestingly, if the ratio d/L is maintained, the maximum force exerted by the rope does not change.

And not surprisingly, force is more for heavier (mg factor) people, or longer falls (d) that use a fixed-length rope.

Edit added:
Rereading your 1st post above, d/L is the fall factor that you mentioned before. This equation shows its relevance.
 
Last edited:
  • #6
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Rereading your 1st post above, d/L is the fall factor that you mentioned before. This equation shows its relevance.
Thanks that's a big help. I might try contacting the manufacturers rather than trying to measure A myself. Hopefully this will allow me to get started with some calculations.
 
  • #7
Redbelly98
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Sounds good, best of luck to you.
 
  • #8
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or do it the old fashioned way. Take a length of rope, tie it up, and put some mass on the end of it. The amount it stretches should tell you:
Fg = Fk
mg = k x
k = mg/x where x is the displacement from no-mass. (stretch)
bam you have a first order approximation of k.

Or am I thinking about this the wrong way?
 
  • #9
Redbelly98
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No, that's the correct way.

I just wonder, if the rope didn't stretch very much then it's not an accurate measurement. But probably these ropes have a fair amount of stretch, since their purpose is to stop a person who has already fallen some distance. So that method should work to find k fairly accurately.

Then mikeym could determine A = k L, and do the force calculations for any length of that type of rope.
 
  • #10
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Hi,

I'm a climber and I'm trying to understand the physics involved in climbing and falling.
Its great to see a fellow climber interested in physics! I have often pondered on the physics of climbing and really things like dyno's and crimping seem to be much more interesting. I recall one of my friends from Kennesaw State University did his dissertation on the dynamics of dynamic climbing techniques using high speed cameras and such. Let me know if you have any interest in other applications of physics in climbing. Oh also climbing shoe rubber is a fascinating science. The company Evolve Sports has come up with an amazing formula for their shoe rubber. Best of luck mikeym!
 
  • #11
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Yes, I've been a fan of evolv shoes since I bought a pair in a sale. :)

First off I've got a question about what we've got so far. How did you get the 1/2 in the energy equation

[tex]Energy = \frac{1}{2} k x^2[/tex] ?

I can't see where I went wrong. Is it something to do with the force being applied 2 ways?

You can see from the manufacturer's website that the elongation of the rope with an 80KG weight on it is 9%.

www[dot]mammut[dot]ch/mammut/katalog[dot]asp?view=detail&did=9&dart=4&tid=54298

So using

[tex]
F = \frac{A x}{L}
[/tex]

[tex]
A = \frac{F L}{x} = \frac{80 * 9.81 * L}{0.09*L} = \frac{784.8}{0.09} = 8720 N
[/tex]

Which is orders of magnitudes different than my lame attempt to measure it on my rope.

So to calculate the largest force I could create in a fall (where the fall factor d/L is 2)

[tex]
Fmax = \sqrt{4 m g A} = \sqrt{4 * 76.2 * 9.81 * 8720} = 5106 N = 5.1 kN
[/tex]

Now to calculate the force on the protection that's holding the fall (this is is a different fall than the fall factor 2 fall as a FF2 can only happen when no protection has been placed yet). So for a fall where there is 4m of rope out and the protection is 1m below the climber:

[tex]
F = \sqrt{2 m g A \frac{d}{L}} = \sqrt{2 * 76.2 * 9.81 * 8720 * \frac{2}{4} } = 2553 N = 2.6 kN
[/tex]

So the force on the carabiner is

[tex]
F = Fclimb + Fbelay
[/tex] where [tex]
Fbelay = Fclimb * J
[/tex]

where J is the friction constant on the carabiner (according to petzl this value is ~ 66%).

www[dot]southeastclimbing[dot]com/faq/faq_pulley[dot]htm

So

[tex]
F = Fclimb + Fbelay = Fclimb + 0.66 Fclimb = 1.66 Fclimb = 1.66 * 2553 = 4237 N = 4.2 kN
[/tex]

Which would be held on a nut of size 3+ which are rated for 5kN-10kN but wouldn't be held by a 1 or 2 size which hold 2 kN.

evosport[dot]com[dot]au/product[dot]php?productid=16844

Alright does all this make sense?
 
  • #12
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OK the potential problems I'm seeing with this way of calculation are:

  • It takes into account the friction on the carabiner for calculating the force on the pulley but not how the stretch propagates along the rope.
  • It ignores the protection + friction + rope drag on the protection that isn't catching the fall.
  • It ignores sideways force (as experience tells you that when you fall you're coming into the wall at a fair speed.
  • The spring constant per unit length is calculated from a static weight and the manufacturers cite stretches for up to 33% for first drops but don't give weights used.
 
  • #13
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Testing the rope results in disposable rope, since anytime a rope has been stressed in a major fall it should be discarded. Stressing to its test strength would qualify. I'm sure the manufacturers can supply al the data you would ever need!
Edmund
 
  • #14
Redbelly98
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First, as a public service, I'll post clickable versions of the links from mikeym's message #11:

http://www.mammut.ch/mammut/katalog.asp?view=detail&did=9&dart=4&tid=54298 [Broken]
http://www.southeastclimbing.com/faq/faq_pulley.htm
http://evosport.com.au/product.php?productid=16844

First off I've got a question about what we've got so far. How did you get the 1/2 in the energy equation

[tex]Energy = \frac{1}{2} k x^2[/tex] ?

I can't see where I went wrong. Is it something to do with the force being applied 2 ways?
Okay, I think I can explain this. As you stretch a rope or spring's elongation from 0 to x, the force increases from 0 to kx during the elongation.

If the force were equal to kx during the entire elongation from 0 to x, then the potential energy would be

PE = force times distance = kx times x = kx^2.

But the force is actually less than kx, up until the elongation reaches x. So the potential energy is definitely less than kx^2.

The average force during elongation is kx/2, and so the potential energy is

PE = (1/2) k x^2

This result can also be found from integral calculus, but I wanted to use a non-calculus explanation if possible. Do you still have your high school physics text? You can look up where it gives the potential energy of a spring, and find this same formula there.

So to calculate the largest force I could create in a fall (where the fall factor d/L is 2)

[tex]
Fmax = \sqrt{4 m g A} = \sqrt{4 * 76.2 * 9.81 * 8720} = 5106 N = 5.1 kN
[/tex]
Yup, looks good so far.

Now to calculate the force on the protection that's holding the fall (this is a different fall than the fall factor 2 fall as a FF2 can only happen when no protection has been placed yet). So for a fall where there is 4m of rope out and the protection is 1m below the climber:

[tex]
F = \sqrt{2 m g A \frac{d}{L}} = \sqrt{2 * 76.2 * 9.81 * 8720 * \frac{2}{4} } = 2553 N = 2.6 kN
[/tex]
Doesn't the climber fall 5m, so the fall factor is 5/4 or 1.25?

I'll try to follow the rest of your post later this weekend. I'm a little too tired right now to follow the pulley discussion at the link you provided.

Later,

Mark
 
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  • #15
stewartcs
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Hi,

I'm a climber and I'm trying to understand the physics involved in climbing and falling. In particular I'm trying to understand the factors that are used within the climbing community to describe it and to unite this with my (high school) physics understanding. I also would like to be able to calculate the forces involved in a couple of simple examples.

The first and foremost term used in climbing to determine the severity of a fall is the fall factor which can be seen from this diagam:

www[dot]getoutofglasgow .com/mikey/fallfactor.png

where the fall factor = length of fall / length of rope

It is understood that the fall factor is a number <2 where the closer to 2 it is the more force is applied to the protection as a result of the fall. This happens where the climber falls from above the belayer before putting the rope through a piece of protection.

Now From what I remember of high school physics

The force applied by the rope is F = kx where k is the spring constant of the rope (I'm not sure how to get this) and x is the stretch on the rope.

No this has to absorb the kinetic energy of the falling climber who's KE can be calculated from.

v^2 = v0^2 + 2ad

where v0 = 0 for a slipping climber and a = g or 9.81 m/s^2 for a earthly climber

so v^2 = 2gd

and KE = 1/2 * m v^2

so KE = m g d

Now this is where things get confused for me.

Can I say KE = Fd

where d is the stretch distance x and

F = kx to give

KE = k x^2 and therefore

k = mgd / x^2

and for force

F = mgd / x

?

Other questions I have are about the friction on the protection and how this can be accounted for. The force applied on the top protection (which acts as a pulley) and on the belayer. And how to go about calculating k for ropes from manufacturers information - and how k varies as the length of the rope extends as there is more rope to stretch.
Using the principle of conservation of mechanical energy we should have,

[tex] \Delta K + \Delta U_g + \Delta U_{rope} = 0 [/tex]

[tex] \Delta K = 0 [/tex] since the object is stationary at the initial and final positions.

Now let H be the distance from the anchor point to the object at the initial position, d be the distance the object falls, and L be the total length of the rope.

[tex] \Delta U_g = mg \Delta y = mg(-2H - d) [/tex]

and

[tex] \Delta U_{rope} = \frac{1}{2} kd^2 [/tex]

Now substituting these into the first equation we get,

[tex] \frac{1}{2} kd^2 - mgd - 2mgH = 0[/tex]

which, when solved using the quadradic equation, gives...

[tex] d = \frac{mg + \sqrt{(mg)^2 + 4kmgH}}{k} [/tex]

Note that only the principal root matters here.

Now since the system will follow Hooke's Law we have,

[tex] F_{max} = kd [/tex]

Plugging in d to that gives,

[tex] F_{max} = mg + \sqrt{(mg)^2 + 4kmgH} [/tex]

The "springiness" factor can be described as [tex]k = \frac{ e_{rope}}{L} [/tex]

[tex] e_{rope} [/tex] is the elasticity of the rope. I looked up a book value for it and a typical one is ~20 kN.

Hence...

[tex] F_{max} = mg + \sqrt{(mg)^2 + 2e_{rope}mg \frac{2H}{L}} [/tex]

(2H is the distance the object falls before the rope begins to stretch)

Hope that helps.

CS
 
  • #16
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Stewartcs, I'm not sure what it is you're doing here.

[tex]
\Delta U_g = mg \Delta y = mg(-2H - d)
[/tex]?

From the way I'm interpreting what you've written for your definitions of H and d

[tex]
2H = d
[/tex]

So mg(-2H - d) doesn't make sense?

Redbelly98, don't bother with the pulley page I just posted it because it's the only page I could find that gives a value for the friction lost on the carabiner. (I checked this against the force calculator on petzl's website and it seems to be what they use.)

Thanks for the explanation on the work done. I can see that I wasn't taking into account that the work done was the sum of the force's over x which needs to be integrated when F depends on x.

[tex]
W = \int^{x}_{0}F.dx
[/tex]

And since F = kx

[tex]
W = \int^{x}_{0}kx.dx = \frac{1}{2} kx^2
[/tex]

Doesn't the climber fall 5m, so the fall factor is 5/4 or 1.25?
Nope, if the climber is one meter above the protection then they fall twice that distance (2m).

To clear things up as to the set up I'm talking about I've posted a couple of more diagrams.

www[dot]getoutofglasgow[dot]com/physics/fallfactor2[dot]png
www[dot]getoutofglasgow[dot]com/physics/fallfactor3[dot]png

Thanks for all the help again.
 
  • #17
Redbelly98
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Stewartcs, I'm not sure what it is you're doing here.

[tex]
\Delta U_g = mg \Delta y = mg(-2H - d)
[/tex]?

From the way I'm interpreting what you've written for your definitions of H and d

[tex]
2H = d
[/tex]

So mg(-2H - d) doesn't make sense?
I'm also puzzled by that. If d is the fall distance than change in grav. potential is just mgd.

Nope, if the climber is one meter above the protection then they fall twice that distance (2m).

To clear things up as to the set up I'm talking about I've posted a couple of more diagrams.

www[dot]getoutofglasgow[dot]com/physics/fallfactor2[dot]png
www[dot]getoutofglasgow[dot]com/physics/fallfactor3[dot]png
I see now, I wasn't understanding the role of the belayer.

For other readers, you may simply click here:
http://www.getoutofglasgow.com/physics/fallfactor2.png [Broken]
http://www.getoutofglasgow.com/physics/fallfactor3.png [Broken]
 
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  • #18
stewartcs
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Stewartcs, I'm not sure what it is you're doing here.

[tex]
\Delta U_g = mg \Delta y = mg(-2H - d)
[/tex]?

From the way I'm interpreting what you've written for your definitions of H and d

[tex]
2H = d
[/tex]

So mg(-2H - d) doesn't make sense?
Sorry if I wasn't clear.

H is the distance from the point of the object (the person climbing) to the anchor (runner). d is the distance during the stretching of the rope. So the leader will fall the total change in distance or H + H + d which equals -2H - d since he is falling downward.

Hope that helps.

CS
 
  • #19
Redbelly98
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I had neglected including the stretch distance as part of the change in gravitational potential. If that is valid, our results should agree very closely. Looks like you were more careful in your calculation, so I'll have a closer look at your result.
 
  • #20
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That's quite a significant difference to the dynamic of the factors involved.

If I rewrite it into my terms d for distance and x for stretch

[tex]
x = \frac{mg + \sqrt{(mg)^2 + 2mgkd}}{k}
[/tex]

and

[tex]
F_{max} = mg + \sqrt{(mg)^2 + 2mgA \frac{d}{L}}
[/tex]

That puts weight back as the dominant factor in a fall, above fall factor. Perhaps the climbing adage that fall factor is the most important factor is not entirely true.

Something else I'd like to throw into the mix is from the idea of the top piece of protection acting as a pulley which made me think about the belayer and what happens to them in a large fall. They get lifted off the ground. This must effect the severity of the fall. Although in the situation where you are getting a fall factor between 1 and 2 they won't be getting lifted in this way as the fall will be onto them from below.
 
  • #21
Redbelly98
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That's quite a significant difference to the dynamic of the factors involved.

If I rewrite it into my terms d for distance and x for stretch

[tex]
x = \frac{mg + \sqrt{(mg)^2 + 2mgkd}}{k}
[/tex]

and

[tex]
F_{max} = mg + \sqrt{(mg)^2 + 2mgA \frac{d}{L}}
[/tex]

That puts weight back as the dominant factor in a fall, above fall factor. Perhaps the climbing adage that fall factor is the most important factor is not entirely true.
I think the equation shows that fall factor is significant. Maybe not the most important factor, but important nonetheless.

The largest term in that expression is [itex]2mgA \frac{d}{L}[/itex]. Since A is significantly larger than mg, several thousand N vs. 500-1000 N, that makes mgA much larger than the (mg)^2 term. The fall factor is part of that term, so it still plays a significant role.

(Of course, if the fall factor is small enough the above won't be true, but you're probably not too concerned about that case.)

p.s. Put another way, you can't control what mg is, or A (once you've chosen a rope). Fall factor becomes the only variable in the equation that you can change.
 
Last edited:
  • #22
stewartcs
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That's quite a significant difference to the dynamic of the factors involved.

If I rewrite it into my terms d for distance and x for stretch

[tex]
x = \frac{mg + \sqrt{(mg)^2 + 2mgkd}}{k}
[/tex]

and

[tex]
F_{max} = mg + \sqrt{(mg)^2 + 2mgA \frac{d}{L}}
[/tex]

That puts weight back as the dominant factor in a fall, above fall factor. Perhaps the climbing adage that fall factor is the most important factor is not entirely true.

Something else I'd like to throw into the mix is from the idea of the top piece of protection acting as a pulley which made me think about the belayer and what happens to them in a large fall. They get lifted off the ground. This must effect the severity of the fall. Although in the situation where you are getting a fall factor between 1 and 2 they won't be getting lifted in this way as the fall will be onto them from below.
The weight for the climber is constant. The only factor that is controllable is the distance between the climber and the anchor point. That is why the fall factor (2H/L) is the most important consideration when climbing. The higher the fall factor, the more likely it is for the rope to snap for a given climber.

CS
 
  • #23
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0
OK I messed up that conversion of terms across to my values.

[tex]
d \not= 2H
[/tex]

[tex]
d = 2H + x
[/tex]

where H is the height above the protection that the fall occurs and x is the distance the stretches and d is the distance fallen.

So I can show stewartcs' equation

[tex]
F_{max} = mg + \sqrt{(mg)^2 + 4mgA \frac{H}{L}}
[/tex]

is equivalent to

[tex]
F = \sqrt{2mgA \frac{d}{L}}
[/tex] when [tex]
d = 2H + x
[/tex]

[tex]
F = \sqrt{2mgA \frac{(2H+x)}{L}}
[/tex]

[tex]
F^2 = 2mgA (\frac{2H}{L}+\frac{F}{A})} = \frac{4mgAH}{L}+2mgF
[/tex]

[tex]
F^2 - 2mgF - \frac{4mgAH}{L} = 0
[/tex]

Completing the square

[tex]
(F-mg)^2 - (mg)^2 - \frac{4mgAH}{L} = 0
[/tex]

[tex]
(F-mg)^2 = (mg)^2 + \frac{4mgAH}{L}
[/tex]

[tex]
F-mg = \sqrt{(mg)^2 + \frac{4mgAH}{L}}
[/tex]

[tex]
\underline{F = mg + \sqrt{(mg)^2 + \frac{4mgAH}{L}}}
[/tex]
 
  • #24
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0
Which gives the conclusion that the maximum fall factor is not 2 as is accepted in the climbing community but

[tex]
f = \frac{d}{L} = \frac{2H+x}{L} = \frac{2H}{L}+\frac{x}{L}
[/tex]

And

[tex]
x = \%x L
[/tex]

[tex]
f = \frac{2H}{L}+\frac{\%x L}{L}
[/tex]

Where the height H = the length of rope out L

[tex]
f_{max} = \frac{2L}{L}+\%x = \underline{2+\%x}
[/tex]

or using the weight of the climber rather than the %x stretch which we don't know

[tex]
f_{max} = \underline{2+\frac{mg}{A}}
[/tex]
 
Last edited:
  • #25
There is a chief problem..

First of all, How do you know that the rope obeys Hookes law(ie F is proportional to x)? ot all ropes obey the F=kx relation.. Infact only very few of them do..
 

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