Is bigger truly better.........cams that is, you sickos!

[johnny]

OK, I finally get the motivation to learn to aid climb. My partner and I find a cool, right leaning hand crack (actually a super fun 5.8 lead)and I flop my way up it in the aiders I tied up the night before with much huffing and puffing, learning lots about weight distribution, balance and how to keep the whole magilla from becoming a single large knot at my waist. I lower off and then climb it free to clean my gear.

Alan does the same as I next, including placing an old #3 camalot behind a block about 40 feet up just like I had done. Difference is that when he weighted it the block blew out sending him back to the stopper below. The 50 lb block nicked his (thankfully helmeted) head, put a gash in his chin and nearly took me out at the deck. Funny thing happened after that, Alan asked to be lowered after that.

OK now to the point. We discussed what had happened and had a few questions. The biggest one involved the physics behind cams. I realize that they work by transfering a downward pull into an opposing lateral force but to what degree? Is it multiplied in any way by being distributed between two directions? How about the size of the cam. Does a larger cam create more outward force than a smaller one?

I am wondering if anyone more versed in physics out there can help???

[johnny]

Jeez, for once I have an honest question and what do I get?? Huh?? Nuttin', absolutely nuttin'. Fine then, guess I'll just go and lever off a few more hunks of rock with my oversize cams.

C'mon people, I know at least Will and Dru have an opinion on this one???????

[allthumbs]

quote:

---

Originally posted by johnny:

OK now to the point. We discussed what had happened and had a few questions. The biggest one involved the physics behind cams. I realize that they work by transfering a downward pull into an opposing lateral force but to what degree? Is it multiplied in any way by being distributed between two directions? How about the size of the cam. Does a larger cam create more outward force than a smaller one?

I am wondering if anyone more versed in physics out there can help???

---

...WTF? Johnny, you're a blowboater aren't ya? That and an engineer. Take yur vitamins!

[johnny]

huh???

[johnny]

and what's a blowboater?? should I be annoyed??

Perhaps I should try another thread title or maybe an entirely new bb????

[chucK]

...WTF Johnny yer a blowboater aren't ya!? Noone answers your question between 6:30 and 7:30 am and you get all crazy. You got a paper due this morning?

I think cams do multiply the force due to something kind of lever and fulcrum business. Don't know how much. I think it varies dependent on the logarithmic taper of the cam. The tapers vary from brand to brand so bigger or smaller is not really the determining factor of how much the force is multiplied. A bigger cam will continue to push over a greater distance though (more work) and would thus probably have a better chance and pushing a block far enough out to topple.

There how was that? Might be correct? Probably get you a better grade than a blank piece of paper.

You're welcome,

Chuck

[Bronco]

Any weight difference between you and Mr. Alan?

maybe he moved onto the peice slightly harder than thou with juuuuuuust enough force to cause the blowout.

[ 01-24-2002: Message edited by: Bronco ]

[willstrickland]

Ok, I can beat Dru to the punch I guess since it's early.

The size of the cam lobes themselves only affect the distribution of the lateral force, for example those "fat cams" metolius makes simply spread the lateral force over a greater area.

The degree of lateral force that is transmitted is related to the cam angle (basically related to the radius of curvature of the lobe). The new metolius "power cams" or some such moniker claim a better holding power. They achieve this by changing the cam angle to transmit more force laterally. Old school cams tended to have a strait circular radius curvature, some now incorporate a spiral/logarithmic curvature to a small degree although they claim a constant cam angle will perform better.

The multiplier effect is actually very small in the case of cams. It's more a function of the axle to contact point length acting in a similar way as a lever than of the physics of the cam lobe multiplying force during the transmission to lateral force.

Any ME out there feel free to make this analysis more accurate. I'm much better with water/hydraulics/pollutants etc and statics/dynamics was a long time ago.

[johnny]

No, I don't have a paper due, I got my BS in bs about 8 years ago..... but I do tend to forget that it is 11:30 from where I sit so I have been up for quite a while now (cleaned the kitchen, took the kids to school, went through yesterdays gear, took a shit, what have you done so far )

Interesting about the cam angles, suppose it would make a difference, especially the distance from the axle or fulcrum, whatever you call it........How come I have to be a pointdexter to think about such shit? would it make you feel better that I once used a bent bong stem (with bowl of course) to protect a sketchy boulder problem??????

Me and Alan; within 5 pounds.

Still don't know what a blowboater is!!!!!

[willstrickland]

Let me correct myself a little:

quote:

---

Originally posted by willstrickland:

The size of the cam lobes themselves only affect the distribution of the lateral force, for example those "fat cams" metolius makes simply spread the lateral force over a greater area.

---

This isn't entirely correct, since a larger cam (i.e. a #3 vs a #2) will have a great axle to contact point length and have a great "lever" factor for lack of a better term. This point was implied in my post before, just want to be clear.

[pope]

The force that cams put on the rock can be divided into two component vectors, one orthogonal to the rock (straight in), one tangential to the rock (down). Each of the four cams has these two force components, and the tangential force that each cam applies to the rock is about 1/4 of your body weight. The orthogonal force each cam applies to the wall is always some multiple of the tangential force, and it depends on the cam angle (which supposedly stays constant throughout the range of the unit).

Sound complicated? Here's a simple thing to remember: when a heavier climber steps on a cam, both the tangential force and the orthogonal force increase proportionally. The frictional force the rock puts back on the cam (upward, keeping the climber from winging off) is a function of that orthogonal force, and so a cam should hold equally well for a heavier climber.

A cam behind a loose flake....well, the heavier climber puts more orthogonal force on the rock, which might result in a greater chance of prying the flake out.

[johnny]

Thanks Will , I knew you had nothing better to do! I get the wider lobes/force distribution part, that makes sense. Also we (me and Alan) pretty much came to the same conclusion about point of contact to axle distance. (both of us have played on see-saws recently!)

Whats up with the curvature thing? Intuitively, round would seem better as it would appear to apply more consistant force at different levels of expansion. What about radius affecting the curve? Is it directly related to contact/axle distance? Shit, this is getting intense, maybe I should go back to school...

Anyone know of any sites that have relatively company-line free info on such topics?

Oh, BTW, sorry for getting so technical so early, maybe y'all should just switch to EST so we could be in sync?????

[johnny]

Throughout all my last comments I forgot to add

TRASK IS A WANKER!!!!!!!

[specialed]

Sounds like a question for Chongo Chuck.

[allthumbs]

A blowboater is a sailboater you whacko...

Sailboaters always ask very technical questions, and stare menacingly with eyes that look like black holes in the sky. They tend to drive most of us "Normal" people nuts.

[willstrickland]

Here ya go, courtsey of Middendorf:

CAMS-A Technical Review (Article circa 1985)by John Middendorf

Without a doubt, cams have revolutionized climbing. Ever since Jardine invented the first fully functional unit (with the proper camming angle) to the climbing community with his introduction of Friends, climbing achievement standards have advanced considerably through the use of superior technology. The way camming devices work is describable by basic engineering concepts. The principles of cams (more specifically, logarithmic spirals) described below are prerequisite to a functional camming unit design and use.

Logarithmic Spirals

The modern camming unit utilizes the logarithmic spiral (also known as an equiangular spiral). The logarithmic spiral is a mathematical curve which has the unique property of maintaining a constant angle between the radius and the tangent to the curve at any point on the curve (figure 1).

A logarithmic spiral cam (a "constant angle cam") ensures that the line between the axle and the point of contact (the "line of force") is at a constant angle to the abutting surface, independent of how the cam is oriented. Thus the force diagram for a given camming unit will be identical no matter how it is positioned in a crack, i.e. whether it be compressed or expanded (figure 2).

Friction

Camming units completely depend on the friction created between the camming unit and the rock. The force created by friction is best analyzed by understanding the engineering basic law of friction: the Frictional force is equal to the coefficient of friction (u) times the Normal (perpendicular) force (F=uN). For example, a 100 pound object with a coefficient of friction of 0.50 between the object and the floor requires 50 pounds of force to move it in a horizontal direction (0.50 times 100 lbs. equals 50 lbs.). The value of the coefficient of friction is determined by the materials and shapes (on a macroscopic level) of the adjoining surfaces.

Force diagrams

A force diagram is an analytical tool used by engineers to determine the individual forces exerted on an object or mechanical device. Vectors depict the forces and represent the inclination and magnitude of each force. The sum of all forces exerted on an object in static equilibrium equals zero, allowing us to calculate unknown forces. Figure 3 is a force diagram for an opposed 2-cam unit (here we define Y' as the constant camming angle). Here, the vector sum of the two side forces and the pull force equals zero. Each side force vector can be represented by two perpendicular vectors, the horizontal component, N (the outward force) and the vertical component, F (the frictional force). The horizontal and vertical components of the side force are related both geometrically (F = N times the tangent of Y'), and physically. The physical relation stems from a basic law of friction explained above. Combining these two relations allows us to pinpoint the factors involved in the holding ability of a camming unit.

Force diagram analysis (figure 3)

S1=S2 (sum of horizontal forces)

P= 2F (sum of vertical forces)

Note that 2F must be greater or equal to P, the pulling force, for holding to occur. Each side force contributes one half to the overall upward component (the holding force).

Geometric relation: F=N tanY'

Law of friction: F=uN

Combining above equations and simplifying, we arrive at the relation u>tanY' for holding to occur.

Restated in plain English, this relation tells us that the coefficient of friction (u) must be greater than the tangent of the camming angle (Y') for holding to occur. Thus, if the trigonometric tangent of the camming angle is greater than the coefficient of friction, the camming unit will pull out of the rock under load.

Coefficients of friction

The coefficient of friction (µ) can be determined experimentally between two materials. A good general design figure determined empirically for the coefficient of friction between aluminum and rock is 0.30. Some types of rock have a greater coefficient than this with aluminum; the sandstone coefficient,however, is sometimes effectually less due to thin shear planes of the large grained, loosely cemented sandstone crystals, which explains the occasional failure of camming units in sandstone cracks. A good deal of experimental research could be done on the coefficients of friction between cams of various materials and surface characteristics and various rock-types (including icy cracks) under load, allowing cam designers to optimize even further for varying conditions and rock types.

Range and Camming Angles

The camming angle utilized for the shape of a camming device determines the cam's range (that is, the cam's maximum/minimum size ratio), and the cam's holding power. A larger camming angle results in a more elongated cam (use CAM program below to see this visually). Range and holding power are inversely proportional. Increasing the camming angle of a single-axle camming unit increases the range but decreases the holding power. If the tangent of the camming angle (Y') exceeds the coefficient of friction (u), the cam will pull. Note visually that a camming unit placement in a downward flare effectively decreases the camming angle and thus decreases the holding power. The larger camming angle results in a cam with more range, yet requires a higher coefficient of friction to hold the same force. Modern camming units optimize range and holding power with cam angles near 14.5 degrees.

Draw your own...

The mathematical equation for a logarithmic spiral is R=beaØ. To draw a logarithmic spiral curve, plot points along the curve, and connect the dots. All you need is a calculator with trigonometric functions, and a sheet of graph paper.

In the equation above, R equals the radius of the curve at a given point; b is a scaling factor and can be chosen arbitrarily; e is the natural log (=2.718); a is the sine of the camming angle; and Ø is the angle of the radius (in radians). R and Ø determine points on a curve plotted in polar coordinates; for a given Ø , R can be calculated and the point (R, Ø) can be plotted. Alternatively, we can use (x,y) coordinates (allowing us to use normal graph paper) to plot points along the logarithmic spiral curve. To do this, first create a chart as shown in figure 5. The conversion factor of Ø from degrees to radians is: Ø(radians) = 2/360 x Ø(degrees). Pi () is equal to 3.1416.

The x and y values can be plotted and connected creating a curve as in figure 6. Once the curve is plotted, the axle hole must be drilled at the origin (x,y = 0,0); material must be added around the hole. To create smoother curves, create a chart with more values of Ø and plot more points. To create curves with differing cam angles, substitute the desired camming angle in the equation a=sin(Y).

Determining camming angles of an actual cam.

It is possible to calculate the camming angle of an actual cam with two measurements of the length of the cam at two points 90 degrees apart (Figure 7). Plug the two values of R1 and R2 into the equation on figure 7 to arrive at the camming angle.

[willstrickland]

Here's the url if you wanna look at the figures:

Middendorf cam analysis

[johnny]

I have on several occasions stuck my shell's hood into my paddle-end and held onto the bottom edges to grab a free ride while kayaking...and my eyes are capable of incredibly malevelant stares. Does that count Trask?????

[Chongo]

Cams are not allowed when learning to aid climb. start off with nuts and hexes, johnny. only move to cams when you can do so without hurting yourself. perhaps you could try some SMC Cam locks as an intermediate step?

[pope]

This is difficult without a diagram. Imagine looking into a crack at cam. See the cam axle? See the shaft? Imagine a plane perpendiculat to the shaft, running through the cam axle. See where the plane intersects the rock? What do you notice? Right, the cams actually contact the rock at a point lower than where this imaginary plane contacts the rock.

Now make a cross sectional picture. Label: "C" is the point at which the cam touches the rock, "A"is the center of the cam axle, and "P" is the point at which the previously described imaginary plane contacts the rock. Now consider the angle CAP: when this angle is small, the cams contact the rock just barely below where our imaginary plane does. In this case, the orthogonal force applied by the cams to the wall is ENORMOUSLY bigger than your body weight, and the cam generates superb holding power. Such a design would be ideal, but I'm told that using a small cam angle results in a unit with a smaller application range (only fits a small range of cracks).

A larger cam angle generates lower orthogonal forces, and so the unit can be expected to generate less friction in a given setting, resulting in lower holding power. The advantage is that choosing a design with a large cam angle can result in greater application range.

Finally, fat cams really don't hold any better. Friction is a function of materials (rock type, cam metal) and orthogonal force. As mentioned, bigger cams spread our the force, but they don't create more friction. Bigger cams might be stronger, perhaps less likely to deform...but I'm speculating now.

I was under the impression that Lowe or Jardine had worked out the cam shape that would allow for a constant camming angle.

[johnny]

Will, You ROCK! Good karma and scentless farts to you and yours.......

[allthumbs]

quote:

---

Originally posted by johnny:

I have on several occasions stuck my shell's hood into my paddle-end and held onto the bottom edges to grab a free ride while kayaking...and my eyes are capable of incredibly malevelant stares. Does that count Trask?????

---

Count as a blowboater, you mean? Why, Johnny, is this important? I'm just givin' ya some morning love, Gumby. Cam on bra!

[willstrickland]

quote:

---

Originally posted by pope:

Finally, fat cams really don't hold any better. Friction is a function of materials (rock type, cam metal) and orthogonal force. As mentioned, bigger cams spread our the force, but they don't create more friction. Bigger cams might be stronger, perhaps less likely to deform...but I'm speculating now.

---

I'd add this:

The rock type (sandstone is what we're getting at here) is the factor where the fat cams MAY hold better. By spreading the lateral/orthognoal force over a greater surface area you are not increasing the friction per se, but the cam is less likely to track out for the following reasons:

The distribution of force is less likely to crush/deform the rock which can create a slip plane (think of a bunch of tiny ball bearings)Depending on the angularity of the individual grains (and this will be a function of the particular layer of sandstone) this slip plane could provide very little in the way of friction. The fat cams have very little space between the lobes and therefore contain the crushed material (as well as crushing less in the first place) under the lobes more than a skinny lobe where the material can displace to the area between the lobes. I've blown decent placements out in dakota sandstone on aid and tracked a tcu out of wingate twice...leaving pretty impressive grooves in the rock. Believe what you want, but I had a set of fat cams when I lived in Utah and used them, maybe in reality they don't hold any better, but the extra confidence (real or imagined) was helful.

[pope]

Good point, Brother Will, and believe me, I believe you. In a related cam story, I once climbed with a lunatic named James who, one rainy Seattle day, accompanied me to the jam cracks under I5 down below Capitol Hill, back when Cam-O-Lots (or whatever they're so cleverly called) first appeared. To find out whether they'd hold, he climbed up about 15 feet, smacked in a cam, climbed another five feet, then cautioned, "Look out!" and jumped. I couldn't believe it. You talked about slip planes....these cracks are grimy with urban filth, almost greasy! Anyway, I think he's still alive, although ten years ago I never would have predicted it.