Pacenti SL23 Spoke tensions

[R2D2]

Originally Posted by Mark McM

Maybe I'm showing my age, but when I started cycling, standard everyday wheels used 36 spokes, wheels meant for racing used 32 spokes, and "special event" wheels (only expected to last for short mileage) used 28 spokes.

Of course, those old Torrington and Robergel spokes weren't too reliable in those days ...

Used to be the case for standard 3x wheels. But many factory botique wheels are now 24 rear and 20 front with the front and non drive side radially spoked. A lot of bladed /oval spokes too.

Those with more knowledge than me can fill in all the details, but I believe that rims are stronger and have welded seems versus a joint with an insert.

[Mark McM]

Originally Posted by R2D2

Those with more knowledge than me can fill in all the details, but I believe that rims are stronger and have welded seems versus a joint with an insert.

Rim alloys are definitely stronger now, but welded seams are more about streamlining manufacturing processing and/or making a rim air-tight (for tubeless tires) than about strength or durability. Even without a weld, the seam is usually not the weakest point on the rim. The weakest point is usually the point with the smallest cross-section, due to the biggest hole drilled through it (i.e. at the valve hole).

[oldpotatoe]

Originally Posted by Kirk Pacenti

110kgf really is all you need, and returns on higher tension diminish rather quickly once you get above 120kgf. imo, ymmv, etc, etc...

His name is on the rim, I'd listen to Kirk.

[oldpotatoe]

Originally Posted by bluesea

Semi O.T.

Tension for 32h 6700 Ultegra hubs, DT Comp all around, and DT 465 rims?

100-110 KGF right side rear and front..same for just about any wheelbuild.

[ColonelJLloyd]

Originally Posted by oldpotatoe

His name is on the rim, I'd listen to Kirk.

+1 Why someone questions the guy who designed and built the rim is beyond me.

[cachagua]

Why someone questions the guy who designed and built the rim is beyond me.

Need we mention the Pinto? Not that I'm asserting these rims would burst into flames if rear-ended -- but the principle that someone who does something must know what they're doing has many, many counterexamples.

Meanwhile: measuring tension first with no tire on the rim, and then inflated to 120 psi, I'm getting in the neighborhood of 5% reduction in spoke tension. My sample is small and my methods aren't as tight as a testing laboratory, but if there's any validity to that number, it's far smaller than the range of recommended tensions in other posts (100-145kg). It's in the range of variation necessitated by irregularities in rims.

I'll admit I'm surprised -- although perhaps I shouldn't be; the author Hjertberg mentions (J. E. Gordon) makes an example of the deformation of solid stone columns under the load of the roof above them. So even though compressing the rim loads it in its strongest dimension, it shortens it a little after all. Let's hear it for elasticity!

Hjertberg doesn't say anything about the translation of radial load into circumferential load, and he also doesn't mention the force the spokes themselves exert on the rim, which I calculate at twice what tire inflation causes. (28 spokes at 125 kg = 7700 pounds of compressive force, against his figure of 3850 from a tire at 100psi.) Counting the effects of both the tire and the spokes, the circumferential stress is large enough that, again, it shouldn't be surprising to see the rim give up a little length (circumference).

Learn something new every day. I don't think it'll change the way I build wheels, but this has been interesting to think about.

[Mark McM]

Originally Posted by cachagua

Hjertberg doesn't say anything about the translation of radial load into circumferential load, and he also doesn't mention the force the spokes themselves exert on the rim, which I calculate at twice what tire inflation causes. (28 spokes at 125 kg = 7700 pounds of compressive force, against his figure of 3850 from a tire at 100psi.) Counting the effects of both the tire and the spokes, the circumferential stress is large enough that, again, it shouldn't be surprising to see the rim give up a little length (circumference).

I think the above analysis is based on several incorrect assumptions.

In Hjertberg's example, the 3850 lb. of radial load on the tire from the innertube pressing against the rim (100 psi over an area of 38.5 in^2) is correct - but not complete. He neglects to consider that the innertube also presses with the same force radially outward against the tire casing - and that the rim constrains the tire at the bead, pulling the rim radially outward by the same force. So the pneumatic pressure in the innertube presses inward on the rim tire bed at the same time as the rim beads are pulled outward by the tire casing under the same pneumatic force - these two force cancel out so the net radial force on the rim directly due to air pressure is thus zero. See my previous post on the actual mechanism that causes the constrictive force on a rim due to air pneumatic pressure.

The 7700 lb. of radial compressive force on the rim due to the spokes (28 spokes at 125 kg each) is true as far as gross force, but is not the same as the rim circumferential load. You can't simply add the force of each spoke, because each spoke pulls at a different angle - you have to do a bit of vector arithmetic to find the actual circumferential load on the rim. For a large number of spokes (more than 16 or so), the circumferential load on the rim works out to be about:

C ~= ( N x T ) / (2 x pi )

where:

C = circumferential load
N = number of spokes
T = tension in each spoke

For the 28 spokes at 125 kg, the circumferential load on the rim is about 557 kgf, or about 1225 lb.

[cachagua]

So, no net circumferential force from air pressure in a clincher, some from the Chinese-finger-trap effect (how much?), and circumferential force from the spokes = negative(2pi)(sum of spoke tension)? Am I with you on all that?

I pictured the circumferential force as similar to undoing the rim joint, straightening out the rim, holding it vertically, and pushing down with x amount of weight on the top end. Not that you could really do this, for several reasons, but it's the way the circumferential force acts in the rim. So using your figure, the load from the spokes is like putting 1225 pounds on top of the rim's straightened length.

I figured, for simplicity's sake, that the spokes' vectors sum to radii. Is that really way off? I began to explore a conversion from gross compressive force to circumferential force, but didn't get all the way into it so I listed the sum of tensions. Presumably we'd use a similar conversion for Hjertberg's 3850 pounds of compressive force (if we accept his account), and/or to the compressive force from the Chinese-finger-trap effect, and add the results to the 1225 lbs. to get the total circumferential force?

And am I understanding your earlier post about the Chinese-finger-trap effect to say that tubulars and clinchers apply about the same compressive force to the rim, since direct pressure isn't a factor?

Just at first blush, it's a little hard to picture an awful lot of (radial) force making its way through the tire bead/rim hook interface. . . is that the way it'd work, in a clincher? The air pressure would try to fatten the tire, and thereby shorten the tire's circumference, and the bead (being the only contact between the tire and the rim) would press radially inward, toward the center of the rim, against. . . the inner bed of the rim, the bottom of the slot the bead sits in? (Grasping for terminology here.) What if a rim had an awful lot of vertical wall underneath the hook at the top?

Or is the comparatively sharp outward curve where the tire's sidewall emerges and begins to balloon out wider than the rim, is that the contact where the force would be applied? And if so, would different combinations of tire and rim width yield different amounts of compressive force? And let's remember that in the case of clinchers, the bead/rim interface must also resist the circumferential force *inside* the tire/rim combination, so there's a considerable force opposing any compression the bead would apply to the rim.

Anyway, clincher beads are made of about as inelastic a material as we can find, as Hjertberg notes. H'mmm. . . still pondering this. Yes, if you could go further into the mechanism of the Chinese-finger-trap effect in clinchers, that'd help me. Thanks!

I should mention that inbetween this post and the last, I got a couple of hours in on the bike, and I'm very pleased to note that regardless of all our theorizing, the system works.

[cachagua]

Sorry, gross error in my interpretation of your formula. Let me try again:

C ~= ( N x T ) / (2 x pi )

where:

C = circumferential load
N = number of spokes
T = tension in each spoke

So circumferential force is roughly gross radial (compressive) force divided by twice pi. Thus a nominal 7700 pounds of spoke tension / 6.282 = 1225 pounds of circumferential force, as you say, and that would convert Hjertberg's 3850 pounds from air pressure (again, assuming he's right about that) to 613 pounds.

I'm liking this so far -- how can we estimate the Chinese-finger-trap effect?

[soulspinner]

Originally Posted by ColonelJLloyd

+1 Why someone questions the guy who designed and built the rim is beyond me.

Ya id bet some testing went into the rim, Im getting a pair of these. Following the 110 recommend...........

[10-4]

I ended up building my set to 110 kgf WITH tires mounted. When I stopped at 110 kgf before mounting the tires, then mounted tires, tension dropped to about 98-102 kgf.

Maybe I should pull the tires and remeasure without them.

Also, Kirk mentions "110 is all you need" in this thread, but 125 in another.

[Kirk Pacenti]

Originally Posted by 10-4

I ended up building my set to 110 kgf WITH tires mounted. When I stopped at 110 kgf before mounting the tires, then mounted tires, tension dropped to about 98-102 kgf.

Maybe I should pull the tires and remeasure without them.

Also, Kirk mentions "110 is all you need" in this thread, but 125 in another.

125kgf is the maximum recommended tension. I typically run less than that (110), but many wheelbuilders seem to prefer higher tensions...