What does your intution tell you...?

It looks to me like the "whole story" is that you would rather persist in disagreeing with the opinions of the people at Black Diamond, and Metolius, who actually engineered these cams, and who have offered their opinion (contrary to yours) on this topic--and also with other climbers who know far more than you do, than to admit you are wrong.

Curt

"P" represents the load applied along the stem of the cam with "b" representing the angle that the stem makes (off horizontal). It can be just the weight of the runner, biner and associated hardware, rope drag or a fall.

So what does that mean for keeping your cam from walking or other stabilty issues, well, we need more algebra.

In the dynamic world, there are so many variables that changes from one second to the next, using just one set of diagrams is not enough to see all outcomes. For example, consider this: cam walking is generally caused by rope drag and slinging the cams too short. The climber's UPWARD progress places an upward and side to side movement to the cam's stem. The angle of b in this case could be 30 degrees ABOVE horizontal at the instant when the climber is experiencing rope drag. With this assumption, we follow all your calculations and get F2 is 33% greater than F1, and we can surmise that by placing the outer lobes UP in a horizontal crack, it will make the cam more resistant to walking caused by rope drag.

In reality, I think the system is too dynamic and too complex to say with certainty which way is better. My intuition tells me the difference is probably too small to consider when I'm trying to plug in a cam in a horizontal while worrying rope drag, direction of route, my next rest stance, and everything else that's on my mind at the moment.

Many of us were taught that all things being equal (it does not fit better either way), to place it with the lobes down. Others never heard this, or have other reasons to believe that it really doesn't matter.

in the first drawing, when summing all forces for newtown's 2nd law, you did not include the force of friction on the lobes or the gravitational force

Not only is this true, but with a stiff cam stem (as found on forged friends, for example) the force applied to the upper cams in a fall can easily exceed that put on the lower cams because the stem itself acts as a sort of "lever" to transfer the force. This can be true even if the cam is properly tied off short--through one of the lightening holes.

As far as the rigid friend loading the upper lobes more, I can't say I follow you. True, the forged friends use a bar that will act as a beam instead of a cable, so the assumption that the load acts in the dirrection of the stem no longer holds. Are you saying that the stem pushing on the rock will transmit a shear into the bar, which the upper lobes will have to offset? I would be interested to see that problem worked out. I wouldn't trust my intution on that one. It would take a moment balance to make it statically determinate.

As far as the rigid friend loading the upper lobes more, I can't say I follow you. True, the forged friends use a bar that will act as a beam instead of a cable, so the assumption that the load acts in the dirrection of the stem no longer holds. Are you saying that the stem pushing on the rock will transmit a shear into the bar, which the upper lobes will have to offset? I would be interested to see that problem worked out. I wouldn't trust my intution on that one. It would take a moment balance to make it statically determinate.

As far as the rigid friend loading the upper lobes more, I can't say I follow you. True, the forged friends use a bar that will act as a beam instead of a cable, so the assumption that the load acts in the dirrection of the stem no longer holds. Are you saying that the stem pushing on the rock will transmit a shear into the bar, which the upper lobes will have to offset? I would be interested to see that problem worked out. I wouldn't trust my intution on that one. It would take a moment balance to make it statically determinate.

This merely requires that the placement be deep enough so that the fulcrum formed by the crack's edge is somewhere between the cam tie in point and the cams themselves. Here's a pic that may help.

http://www.rockclimbing.com/...p.cgi?Detailed=33042

first lemme say that i bow to all the engineers on this site and am usually entertained reading the physics/etc breakdowns of how gear works...
the extent of my knowledge regarding levers forces, etc.. is severely limited.

with that out of the way,

curt, wouldn't the fulcrum always be between the tie in point and the cam lobes? unless you tied into the cam somewhere that was closer to the lobes than the edge?

If a cam is placed in a horizontal crack, does it matter if the outer lobes are up or down?

Many of us were taught that all things being equal (it does not fit better either way), to place it with the lobes down. Others never heard this, or have other reasons to believe that it really doesn't matter.

Well, this thread is to hopefully help resolve this question.

I will start with an analysis of the forces involved. Bear with me as it appears complicated (I show all the algebra), but in principle it is really very simple.

The following is just adding up all the forces. If the cam stays put (as we hope it will), the forces add up to zero.
I think you can click on the image to see the detail.
http://img203.exs.cx/...talcam20011zy.th.jpg
"P" represents the load applied along the stem of the cam with "b" representing the angle that the stem makes (off horizontal). It can be just the weight of the runner, biner and associated hardware, rope drag or a fall.
"F1" and "F2" are the forces of the rock against the cam, "F1" being the lower force. The cam angle is "a."
For the sake of making the math mean something, I used a cam angle of 13.75 degrees and assumed the stem to be 30 degrees off verticle.
With those angles, the force into the rock on the lower cam is about 33% greater than the upper cam lobe.

So what does that mean for keeping your cam from walking or other stabilty issues, well, we need more algebra.
This time, instead of summing forces, we sum what engineers call moments. Think of them as torques or twisting forces you get from pushing on a lever. The Bending Moment "M" will be the force times the length of the arm pushing on it, or "moment arm." They will add up to zero if the cam is to stay put. For this analysis, the moments are summed around the point where the cable attaches to the body of the cam.
http://img219.exs.cx/...zontalcam13nw.th.jpg"P" will now be a load pulling sideways on the cable with "moment arm" of length "l" (think levers) and rock the cam in a circle. Only 2 of the 4 lobes are in a position to stop it, one on each side. The fricitional forces acting on the 2 lobes are "R1" and "R2." The distance between the middle lobes and the center, and also the middle lobes and the outer lobes is represented as length "a." I assumed the distance to be equal to simplify the math and is not far off from a typical cam's geometry.
The maximum friction force is related to the force into the rock by the "coeficient of friction," represented by the greek letter mu (resembles a "u" with a tail or a sharp "m").

Did you follow any of that?

In the end you can see that putting the lobes downward does help a little (at least according to my calculations). Not enough to makea hard and fast rule about which way to orient the cam, but enough to give the outer lobes down an edge.

The lower picture is just another orientaion where stabilty could be an issue, although it would take some flukey loading to twist a cam out of the rock in a horizontal crack. The math will work out the same, but intution is stronger with the lower picture. Proabable because it resembels an SUV vs sports car wheel arrangement.

What this really means:
The lower lobes carry more downward force than the upper lobes. The outer lobes resist twisting and walking better. if you put the lower lobes down, you make the cam a little more stable and likely to stay put.

Okay, so what are the implications of the model?

Well, the lower lobes will be loaded more than the upper lobes in a way proportional to sin b. So, if the stem is horizontal, the lobes will all carry the same load. This would also be the case for a cam in a vertical crack. Therefore, if the crack is deep or the cam small and the stem is coming out straight, there really is not much difference.

But if the cam is shallow, making the angle large (remember 30 degrees was used in my calculations, which is not a very large angle), the advantages you get from having the wider lobes down will be greater.

However, note the terms on the left of the column of the second page. If you can lower "l" (lenght of the stem), or make "mu" (coeficient of friction) or "a" (space between lobes) higher, it will do more for you. Of course stiffer springs will help (which will just increase the force into the rock). None of these can really be changed, except maybe when you are buying the gear.

I could go on, but I don't want to spoil your fun.

What are some of the finer points of placing cams horizontaly?

Do you see anything wrong with the model (perhaps assumptions that aren't quite right). Algebraeic or conceptual mistakes are possible. Feel free to call me on them if you find any.

What does your intution tell you and how does it comapre with the math?

Thanks curt for the picture establishing clearly the interest of tieing short rigid stem friends.

It also allow us to put a plausible maximum angle of misalignment on simple "everything equal" setup:

picture a horizontal plank above your kitchen board, as if we had a "letterbox" slot in which to put the friend,
the friend is inserted horizontally, each lobe goes the same depth,
if we where to pull on the stem horizonally, each side up and down would get the same force (neglecting weight of the friend)
if we were to pull on the stem as it is resting naturally against the bottom but without a lever effect thanks to the tie-in
the angle of misalignment will be determined by the triangle
axle, stem contact point with bottom, stem contact projection on the horizontal plane (the point where the stem should have been).

The closer the tie-in to the axle, the less chance of leverage, the greater the misalignment.
The deeper the friend, the less misalignment, the less unequal force.

OK, so does it matter ??
I do not have the dimensions of a cam with me
but let's take length of stem till lip of letterbox = 10cm
and 1/2 height of properly opened cam = 1 cm
tan(a) = 1/10 a= 5 degrees => no need to worry
1/2 heigth of cam =2.5 cm , a~= 15 degrees
1/2 heigth of cam =5 cm , a~= 30 degrees, in that case the lower lobes get 33% more force than the upper lobes.

in "proper english", it means that if the axle is as deep as the cam is opened in a horizontal slot, despite tie-off, one risk unequal loading of upper vs. lower cam lobes .

I would have to get a picture to make this clearer I am afraid