0
$\begingroup$

A traditional bicycle wheel looks like the left wheel in the image. The spokes are not concurrent at the center. They connect with the hub at a near tangent.

Given their length-to-diameter ratio, it is clear that all spokes must be under tension at all time. Even a minuscule compression would result in buckling.

wheelset, with both concurrent and traditional spokes

I always thought that the idea behind spokes arriving at the hub at a near tangrent is that some spokes take increased tension (and "pull" the rim) when the cyclist is torquing (for the rear wheel), and some spokes are under increased tension when the cyclist is braking using disc brakes (for both front and rear wheels). Even though they are not connected to the drivetrain, front wheels still need to have alternative spokes to maintain the wheel's symmetrical balance.

But it is getting increasingly common to see bike wheels with concurrent spokes (wheel on the right of the image, plus the non-drive side in the wheel on the left). Front wheels have them on both side, and rear wheels have them on the non-drive side (with the traditional arrangement on the drive side).

Applying even a small torque on the front wheel (with disc brakes) would seem to require that the hub rotate a little before activating the tension on the spokes necessary to induce rotation at the rims. In other words, the rim and the hub do not function as a solid object.

Of course a bicycle wheel is so light precisely because it is not a solid object, but the rim and hub have no relative "wobble".

Even with enormous spoke tension, there would need to be a slight wobble in the hub when turned against the rim.

How can wheels with spokes concurrent at the center be solid?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ The right wheel is spoked radially. $\endgroup$
    – fred_dot_u
    Mar 21, 2022 at 14:00
  • $\begingroup$ Hmmmm at what point did "concentric" start applying to anything other than circles? Seems like a bastardization to me. $\endgroup$
    – DKNguyen
    Mar 21, 2022 at 14:07
  • $\begingroup$ @CarlWitthoft I updated the wording 28 minutes before your comment. We're good already, no? $\endgroup$
    – Sam
    Mar 21, 2022 at 16:29
  • $\begingroup$ @Sam ahh, the famous "comment lag" . I'll delete $\endgroup$
    – Carl Witthoft
    Mar 21, 2022 at 16:30
  • $\begingroup$ @Sam If you ask over at the bicycling Stack Exchange, I think you'll get a lot more info on the materials and parameters used in wheel construction. Suffice it to say that there's always some compression and some torsional distortion, but that's figured into the materials requirements. $\endgroup$
    – Carl Witthoft
    Mar 21, 2022 at 16:31

2 Answers 2

Reset to default
-1
$\begingroup$

Basically because the mass of the load ie frame and rider, is applied in tension and the spokes are much stronger in tension than compression.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes they're stronger in tension, but any hub-originating load simultaneously attempts to apply both. The spoke in tension 'refuses' to lengthen, thus limiting the compressive force on the other spoke. (I'm not the downvoter BTW) $\endgroup$
    – Carl Witthoft
    Mar 21, 2022 at 16:29
  • $\begingroup$ If I take a wheel by itself and bounce it off the ground, even if mounted and loaded, it makes little to no difference how the spokes are arranged. But if I apply torque at the hub, I expect the traditional arrangement to be much stronger. $\endgroup$
    – Sam
    Mar 21, 2022 at 16:30
-1
$\begingroup$

I agree with you. Unless the breaks on the front are old-fashioned type and not the newer disk type.

Any disk brake would impart a huge torque to the front hub and has no reaction from the spoke (they are octagonal the tangential force of the disk and radial of the spoke).

So the whole wheel will jerk. and the spokes can break due to fatigue.

Edit

To respond to the comment about brake torqe on the wheel has the same effect as the disk:

The braek on the wheel has no torque, because it directly stops the wheel but the brake on the disk stops the hub and then a huge torque stops the wheel.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Even assuming you mean "brakes," the torque from rim brakes is on the order of the torque from the drive hub out along the spokes to the rim. $\endgroup$
    – Carl Witthoft
    Mar 21, 2022 at 16:27
  • 1
    $\begingroup$ @kamran There is no torque applied on the front wheel at the hub with rim brakes. But you're right; it's important to make the distinction. Fixing. $\endgroup$
    – Sam
    Mar 21, 2022 at 16:32
  • $\begingroup$ Torque from rim brakes would be due to the decelerating mass of the bike and rider. No matter how the braking occurs, this torque must pass through the spokes. $\endgroup$
    – Tiger Guy
    Mar 21, 2022 at 17:40
  • $\begingroup$ @TigerGuy, the declaration will cause the back set of spokes to tension, with no torque to the hub. Even if the deceleration is going to cause the bike to tumble! $\endgroup$
    – kamran
    Mar 21, 2022 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.