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expanding flakes: nut vs. cam?
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By jcurl
May 23, 2011

Assuming the appropriate idealizations, if you put a tapered nut into a constricting crack, which expands with an associated spring constant, the nut would expand the crack until the (vertical) components of normal force and proportional frictional force combined were sufficient to offset the load. But this is also true for a cam in a constricting crack. The only difference would be that a nut would need to slide deeper into the crack to achieve this equalization whereas a cam could stay in place and expand with the crack, provided the cam's range had not been exceeded.


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By brenta
From Boulder, CO
May 23, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

A little nit picking. The outward force in a vertical, parallel-sided crack, when the cam angle is arctan(1/4) is twice the load, not four times. There are two frictional forces, one on each side, that together must balance the load. Each frictional force is related to the normal force by the tangent of the cam angle. Mix, shake well, and the result follows.


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By brenta
From Boulder, CO
May 23, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

Just to add a little clarity or confusion--pick one--when a nut slides down the constriction, part of the work done by the load goes into bending the flake and part is lost to friction. So, if one wants to analyze an actual fall, one has to take into account that the fixed quantity (OK, almost fixed) is the energy to be dissipated, while the load may peak at different values depending on the details of the anchor.


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By DannyUncanny
From Vancouver
May 23, 2011

brenta wrote:
A little nit picking. The outward force in a vertical, parallel-sided crack, when the cam angle is arctan(1/4) is twice the load, not four times. There are two frictional forces, one on each side, that together must balance the load. Each frictional force is related to the normal force by the tangent of the cam angle. Mix, shake well, and the result follows.


Your nit picking is incorrect. Friction force adds up to load force, normal force adds up to 4 x friction force, you can add it any way you want, it's still 4 times.


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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

DannyUncanny wrote:
Your nit picking is incorrect.

Wrong. Let's see why.
DannyUncanny wrote:
Friction force adds up to load force,

True.
DannyUncanny wrote:
normal force adds up to 4 x friction force,

Normal force equals four times the friction force on each side. The normal forces are equal in magnitude and opposite in direction and cancel each other. Draw a free body diagram if you are still incredulous.
DannyUncanny wrote:
you can add it any way you want, it's still 4 times.

Wrong.


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By Aric Datesman
May 24, 2011

Oh my.



Anyone have a single end beam load cell handy? That'd be the easy way to settle this. All of mine are S-beam, which would take an annoying amount of fixturing to get to work for this.


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By Optimistic
From New Paltz
May 24, 2011

Aric Datesman wrote:
Oh my. Anyone have a single end beam load cell handy? That'd be the easy way to settle this. All of mine are S-beam, which would take an annoying amount of fixturing to get to work for this.


I (the OP) was hoping that Aric would show up on this thread eventually!

The reason that quite a lot more science got done in the last 350-ish years than in the preceding 300,000 was that we decided to go into the lab and MEASURE WHAT IS ACTUALLY HAPPENING as opposed to sitting around in our togas talking about what "makes sense".

But I digress...

I took a shot at measuring the angle between the wide faces of my HB Offset. It appears to be 7.5 degrees.

So if it is in fact true (I don't know if it is) that
a) both a cam and a nut can be viewed as governed by the equation describing an inclined plane and
b) outward force increases in inverse proportion to the cam/nut angle

Then: nuts, or at least my HB Offset, will put more outward force on a given placement, for a given load, than cams will, making the cam less likely to move the flake.

But I still think I won't really know until someone measures it.


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By slim
Administrator
May 24, 2011
tomato, tomotto, kill mike amato.

not sure if brenta and dannyuncanny are maybe arguing the same thing, exept brenta is talking as a function of the friction force at one side and danny is talking as a function of total downward force at the stem(?), but each side (pair of lobes) will horizontally exert twice the vertical load applied at the stem. so, for a 4 lobe unit, each individual lobve would exert the equivalent of the vertical load at the stem.


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By Optimistic
From New Paltz
May 24, 2011

slim wrote:
not sure if brenta and dannyuncanny are maybe arguing the same thing, exept brenta is talking as a function of the friction force at one side and danny is talking as a function of total downward force at the stem(?), but each side (pair of lobes) will horizontally exert twice the vertical load applied at the stem. so, for a 4 lobe unit, each individual lobve would exert the equivalent of the vertical load at the stem.


Again, I don't really know, but that's not where the 4x comes from in the Kodas paper cited at the beginning of the thread. It comes from the load being divided by the cam angle coefficient, which is 0.25. 1/0.25=4.

But you do raise an interesting point: is the whole cam unit equal to one inclined plane, as some are asserting here, or two (two sides to the crack) or four? As far as the nut goes, should it be treated as one inclined plane, or two planes opposing the midline? Does that make a difference? I don't know.

Lots of answers might appear sensible. One answer would trump all of them: measuring it in the lab.


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By slim
Administrator
May 24, 2011
tomato, tomotto, kill mike amato.

i think that you are forgetting that the load on each side is F/2 though, which is the basis of my point. if you are talking relative to the load on one side (ie saying that F/2 is your F), or as a function of the total vertical load on the cam.


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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

slim wrote:
so, for a 4 lobe unit, each individual lobe would exert the equivalent of the vertical load at the stem.

That's exactly it. The point is that the outward force is not the sum of the forces at the two sides. And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve.


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By Aric Datesman
May 24, 2011

Thinking on this a bit more it's not to hard to measure indirectly.... simply bolt the ends of a pair of plates together (in a U shape) a fixed distance apart (with a bit of a constriction cut into them so the nut can hold), insert the cam/nut, load to a specific force and then measure the amount the plates spread. From there you could back into the outward force provided you know the modulus of elasticity of the material, but since you're only really looking for a comparison between the two it doesn't really matter.

I really don't have time to spend on this at the moment, unfortunately.


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By slim
Administrator
May 24, 2011
tomato, tomotto, kill mike amato.

even after 15 to 20 years of doing this, i always have an intuitive hangup on that one brenta (ie not summing the 2 forces such that the outward force on the flake is 4 times). i always pretend that the force is reacted by a vertical beam reaching down to a floor (instead of reacted by the other cam lobe). then it feels like it makes more sense. not sure if you know what i am talking/babbling about?


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By Optimistic
From New Paltz
May 24, 2011

brenta wrote:
That's exactly it. The point is that the outward force is not the sum of the forces at the two sides. And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve.


That's what the Greeks said about all matter being composed of four elements.

But anyway, my original question was about making a comparison between nuts and cams, and it doesn't seem like much has been said here to make an actual, numerical comparison between the two types of gear. Some folks have said very authoritatively that cams exert more force, some have said nuts. Which is it?


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By Optimistic
From New Paltz
May 24, 2011

David Horgan wrote:
That's what the Greeks said about all matter being composed of four elements. But anyway, my original question was about making a comparison between nuts and cams, and it doesn't seem like much has been said here to make an actual, numerical comparison between the two types of gear. Some folks have said very authoritatively that cams exert more force, some have said nuts [Edit: because nuts have a narrower angle than cams]. Which is it?


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By doligo
May 24, 2011
Jose Cuervo Fruitcups dirtbag style

dumb question - i feel like the best gear for expanding flakes are sliding nuts, am I wrong?


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By DannyUncanny
From Vancouver
May 24, 2011

brenta wrote:
That's exactly it. The point is that the outward force is not the sum of the forces at the two sides. And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve.


I don't think you get how forces work yet. If the coefficient of friction is 0.25, and you only have twice the load in compression, your cam would be slipping.

If you look at the contact point of your cam, you need to have four times the load being exerted in order to resist pull, do you agree with that? Where you are going wrong is that you add 2 times on one side, and then add 2 times on the other side and say that adds to four times. This is the completely wrong part.

The forces are a force pair that create compression on the cam. You don't treat them individually. If you took one away, the cam would go zooming off into space.

If instead of a cam, suppose you had just a block of aluminium against a flat rock face. How hard would you have to push it against the wall to keep it from slipping? You would need to apply a force equal to four times it's weight correct? But by your logic, you are actually pushing it with eight times it's weight, because you apply four times on one side, and the wall applies four times against it in the other direction. Do you see where your logic has gone askew?

As for the original question, it depends on the taper angle. A very high taper (like you get on a hex) will exert less force than a cam. A very small taper less than something around 8 deg will in theory exert more outwards force than a cam, but this will depend on the coefficient of friction.


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By Aric Datesman
May 24, 2011

brenta wrote:
And there's no more reason to go to the lab to test this than there is to write a proposal to get some time on the supercomputers at Oak Ridge National Labs to settle the dispute as to whether three plus three equals six or twelve.


Then again, a couple minutes in the lab would have easily prevented several pages of arguing about it. Seems worthwhile to me, but YMMV.


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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

David Horgan wrote:
That's what the Greeks said about all matter being composed of four elements.

No, they didn't.


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By Aric Datesman
May 24, 2011

While Sicily is currently part of Italy, in 490BC was actually part of Greece. The significance of this is that Sicily is the birthplace of Empedocles, who was the first to posit the 4 element theory that was held as true for the next thousand or so years in the Western world.

(/tangent)


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By jcurl
May 24, 2011

DannyUncanny wrote:
The forces are a force pair that create compression on the cam. You don't treat them individually. If you took one away, the cam would go zooming off into space.

Danny, as you know it's the frictional force that keeps the cams from slipping. Although you are correct that the normal force is the same on both sides and is not additive, the two frictional forces do add. This is why the normal force is half of what you are claiming.


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By Optimistic
From New Paltz
May 24, 2011

brenta wrote:
No, they didn't.


I'm sure you're correct, I'm no historian, and was just kind of fooling around. And I certainly don't know how to say "Oak Ridge supercomputer" in Ancient Greek.

But the point behind the fooling around, which I will definitely stand by, is that there are a whole lot of examples in which very sensible applications of idealized scientific principles turned out to be incorrect when measured experimentally.


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By kennoyce
From Layton, UT
May 24, 2011
Climbing at the Gallery in Red Rocks

DannyUncanny wrote:
Your nit picking is incorrect. Friction force adds up to load force, normal force adds up to 4 x friction force, you can add it any way you want, it's still 4 times.


Not correct. Draw yourself a free body diagram (or check out the awesome MS Paint one I drew below), you have the downward force "F" on the cam, and you have 2 forces "f" from the rock onto the cam lobes. Now the vertical components of the 2 "f" forces have to add up to the total downward force "F", and the horizontal component of the force "f" is the force causing the flake to expand. Now we can create equations for the sum of the forces equal to zero in both the horizontal and vertical dirrections, but lets just worry about the vertical since there are only the two horizontal components of the two "f" forces that will just cancel each other out. So in the vertical dirrection we have:

0 = f sin (theta) + f sin (theta) - F

Now we want to solve for "f" since the horizontal component is the expansion force:

f = F / (2 sin (theta))

Now we pick a random cam angle for thata, lets say 15 degrees to make it simple and we get:

f = 1.932 F

Now the expansion force is the horizontal component of "f", fx, which is found as "f" times the cos of theta:

fx = 1.866 F

so for a cam angle of 15 degrees and neglecting the friction between the cam lobe and the axel, the expansion force on the flake is 1.87 times the force of the fall. Of course as the cam angle decreases the expansion force will increase, but even a 13 degree cam angle only gives an expanding force of 2.17 F, nowhere near 4 times the downward force.

FBD of a cam
FBD of a cam


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By brenta
From Boulder, CO
May 24, 2011
Cima Margherita and Cima Tosa in the Dolomiti di Brenta.  October 1977.

DannyUncanny wrote:
I don't think you get how forces work yet.

Philosophus mansisses si tacuisses.

DannyUncanny wrote:
If the coefficient of friction is 0.25, and you only have twice the load in compression, your cam would be slipping.

No. The relation between tangent of the cam angle and friction coefficient is easily derived as follows:

1. From the moment equation that says that the lobe is not rotating around the axle, we get that the ratio between friction force on one side and the normal force equals the tangent of the cam angle.

2. It is this ratio that must be less than the coefficient of friction lest the cam should start slipping. We are applying Coulomb's Law here. It follows that as long as the tangent of the cam angle is less than the coefficient of friction, enough friction will be available to avoid slipping.

DannyUncanny wrote:
If you look at the contact point of your cam, you need to have four times the load being exerted in order to resist pull, do you agree with that?

No, for the above reason.

DannyUncanny wrote:
Where you are going wrong is that you add 2 times on one side, and then add 2 times on the other side and say that adds to four times. This is the completely wrong part.

Whatever it is you're trying to describe, it's not what I'm doing. Let's write the translational and rotational balance equations and solve them: it's easy-peasy. Let F be one frictional force, L the load, N the normal force and beta the cam angle:

1. 2F = L
2. F cos(beta) = N sin(beta)

whence N = F/tan(beta) = L/(2 tan(beta)).

DannyUncanny wrote:
The forces are a force pair that create compression on the cam. You don't treat them individually. If you took one away, the cam would go zooming off into space.

Obviously.

DannyUncanny wrote:
If instead of a cam, suppose you had just a block of aluminium against a flat rock face. How hard would you have to push it against the wall to keep it from slipping? You would need to apply a force equal to four times it's weight correct?

You are ignoring the friction between the block and whatever you use to push the block against the wall. If you used a frictionless device, like a roller, to push the block, then you'd be right, but the lobes of a cam don't work like that.

DannyUncanny wrote:
But by your logic, you are actually pushing it with eight times it's weight, because you apply four times on one side, and the wall applies four times against it in the other direction. Do you see where your logic has gone askew?

I hope by now you see how nonsensical this is.


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By Aric Datesman
May 24, 2011

David Horgan wrote:
That's what the Greeks said about all matter being composed of four elements.

.
Brenta wrote:
No, they didn't.

.
David Horgan wrote:
I'm sure you're correct, I'm no historian, and was just kind of fooling around.


Actually, he's not; see post above re: Empedocles. As for the cam/nut thing, I'd rather see someone put it to rest by measuring something rather than continue watching this turn into RC.


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