
kachoong
Mar 4, 2009, 9:28 AM
Post #1 of 13
(2871 views)
Shortcut
Registered: Jan 22, 2004
Posts: 15304

I figured you techy lab types might know the in's and out's of this. I read somewhere that smaller camming ranges increase the power of the camming action. Is this true and what's the theory behind it? Also, how does this affect the security of the placement in softer rock? Which would compromise the placement more in soft rock.... a larger or smaller camming range? Thanks!!





hafilax
Mar 4, 2009, 9:55 AM
Post #2 of 13
(2850 views)
Shortcut
Registered: Dec 11, 2007
Posts: 3025

The cam angle is more important than the range in regards to the outward force. The double axle design of the C4 gives more range for a given cam angle for example. In general the smaller the cam angle, the greater the outward for on the rock for a given pull. As for blowing out soft rock? That also depends on the pressure exerted by the cam which is a function of the contact area. Metolius made some fat cams that had a small cam angle but a large surface area. I would ask around and see what cams people find hold better in soft rock since it is a complex system that can't really be assessed by looking at the specs.





kennoyce
Mar 4, 2009, 10:07 AM
Post #3 of 13
(2834 views)
Shortcut
Registered: Mar 6, 2001
Posts: 1320

In reply to: I read somewhere that smaller camming ranges increase the power of the camming action. Is this true and what's the theory behind it? so the math behind it is this. If you draw a line horizontally from the cam's axel to the rock, then draw a line from the cam's axel to where the lobe contacts the rock, this is the cam angle. Because the forces in both the horizontal and vertical dirrections have to be balanced for the cam to hold, the frictional forces have to equal the forces exerted on the cam by the climber. The frictional forces are proportional to the normal force exerted by the cam on the rock which if you do the math turns out to be the 1/2 the force of the climber divided by the tangent of the cam angle (for each side of the cam thereby making the horizontal forces equal). Basically what I am saying is that as the cam angle decreases, the normal force on the rock increases which in turn increases the friction between the cam and the rock, but also has the possibility of more easily breaking the rock.





adatesman
Mar 5, 2009, 8:05 AM
Post #4 of 13
(2758 views)
Shortcut
Registered: Jul 13, 2005
Posts: 3479






kachoong
Mar 5, 2009, 8:57 AM
Post #5 of 13
(2735 views)
Shortcut
Registered: Jan 22, 2004
Posts: 15304

Thanks you guys! That's awesome and your explanations helped me out heaps! I appreciate your thoughts!





basilisk
Mar 8, 2009, 8:18 PM
Post #6 of 13
(2662 views)
Shortcut
Registered: Oct 1, 2005
Posts: 636

We had a cool discussion related to this a while back: http://www.rockclimbing.com/...d;page=unread#unread It sorta gets into the math of it. I tried my damnedest to get a formula to figure out how much force a cam generates with a given angle, but I either suck at math (which is true), or none of the ones provided actually work. It's a good read nonetheless
(This post was edited by basilisk on Mar 8, 2009, 8:20 PM)





adatesman
Mar 8, 2009, 8:40 PM
Post #7 of 13
(2645 views)
Shortcut
Registered: Jul 13, 2005
Posts: 3479






rgold
Mar 8, 2009, 9:32 PM
Post #8 of 13
(2632 views)
Shortcut
Registered: Dec 3, 2002
Posts: 1801

Hoho, it's tan theta. But tan theta and arctan theta are nearly equal for small values of theta, as you can infer (using l'Hopital's rule) from the fact that the limit as x approaches zero of tan(tan x)/x = 1.





brenta
Mar 8, 2009, 10:06 PM
Post #9 of 13
(2625 views)
Shortcut
Registered: Aug 24, 2006
Posts: 50

As a mnemonic, remember that the force multiplier comes out of a proportion: the ratio of the forces equals the ratio of the sides of a right triangle. That ratio is a tangent, not an arctangent. Since we are on the subject, it's worth noting that the formula 1/2tan(theta) gives the horizontal component of the force. The reaction at the rocklobe interface, however, has a vertical component. Hence, the total force applied to the rock is obtained by multiplying the load by 1/2sin(theta) and is directed along the line that connects the axle to the contact point. In practice, the difference is very small for the angles we are interested in.





basilisk
Mar 9, 2009, 6:20 AM
Post #10 of 13
(2590 views)
Shortcut
Registered: Oct 1, 2005
Posts: 636

adatesman wrote: Must be the former, since the two ways of calculating it are listed in the thread.... [..] Fx=Cos[theta]*(.5*Sqrt[1 + 1/(Tan[theta]^2)]) Fx=1/(2*Tan[theta]) I expected as much. poop.





adatesman
Mar 9, 2009, 7:35 AM
Post #12 of 13
(2553 views)
Shortcut
Registered: Jul 13, 2005
Posts: 3479






brenta
Mar 9, 2009, 8:18 AM
Post #13 of 13
(2538 views)
Shortcut
Registered: Aug 24, 2006
Posts: 50

I prefer to set up things differently from Kodas. Theta is the spiral's pitch and is an angle. Then the equation of the logarithmic spiral is r = r_0 exp(tan(theta)(phiphi_0)). This reduces confusion, also considering that mu often denotes the coefficient of friction.





