Brian Ratliff

The rim does not deform much at the top when you push down on the hub. Rather, the wheel collapses down slightly at the bottom. The rim will not assume an ovalized shape; as seen in the pdf I posted earlier, the wheel takes on an "egg" shape whereas the bottom flattens and bulges out and the top doesn't do much at all.

Your thought experiment is invalid because we are talking about a tensioned wheel, not a weight suspended between two vertical springs. Totally different problems from a stress analysis standpoint. Apples and oranges. A more apt analogy for a thought experiment is to imagine a soft beachball resting on the ground. The top remains round; the bottom flattens out. Or a flat (not fully inflated) basketball. Drop it on the ground, the contact point will form a flat spot; the top will remain round.

Finally, the bending stiffness in the rim will play a large roll in this. The more stress that is taken up by the rim, the less the spoke stress will vary. For an infinitely stiff rim, you'll see a minimum spoke stress at the bottom, with spoke stress gradually increasing to a maximum at the very top. For a more typical bicycle rim which is relatively soft in bending and really only accepts stress in pure compression, the bottom-most spokes will be extremely deloaded, the spokes next to those disproportionately loaded, and the rest of the spokes equal and with slightly more loading.

rpenmanparker 

I finally worked out the model. Assume you have an infinitely stiff rim. Two spokes of any length, upper and lower. 100 kgF tension in each to start. Starting tensioned spoke length of 300 mm or 600 mm sum total. You apply 50 kgF weight to the hub axle. The final upper spoke tension is x, the lower is y. In order for equilibrium to occur, the upper and lower forces must be equal, so x = 50 kgF + y. Since the sum of the spoke lengths remains the same, and the stretch factor (modulus) for spokes is a constant over in the region they are used, the new tension sum in the spokes must equal the old tension sum. Otherwise the spokes wouldn't "meet" at the hub like before the outside load is applied. So x + y = 200 kgF. Combining the two equations, we get 2y + 50 = 200, 2y = 150, y = 75 kgF. Therefore x = 125 kgF.

The upper spoke increases in tension from 100 kgF to 125 kgF while the lower spoke decreases in tension to 75 kgF. In a more general case, I think you can easily show that 1/2 the applied load amount is always removed from the lower spoke and added to the upper spoke in tension.

Now I'm tired. Someone tell me how this affects total energy for the perfectly incompressable rim.

Bike Gremlin

Brian Ratliff has explained it well, IMO. +1

@rpenmanparker
I think you are considering the spokes to be springs, which they are not. The load does not necessarily stretch the spokes, nor decrease their length. Just like a concrete beam does not always compress when loaded. But it still takes load and preload.

That is why I think your calculation is not correct. The sum of upper spoke tension change is close to zero, and the lower spoke tension change is close to the previous tension - minus extra load on the hub.

rpenmanparker 

I disagree...at least until proven otherwise. The beach ball is not a good example as the top and bottom are not connected by rigid spokes. Same with a tire. True you only get a flat patch at the bottom, but the air inside providing the shape is fluid. The more rigid the rim, the more similar will be the flattening at top and bottom. Of course the magnitudes of flattening will be less as the rim rigidity increases, but the shape will be more symmetrical. More oval and less egg-like.

rpenmanparker 

You are totally incorrect. The spokes are nearly perfect springs in the range of tensions that they operate. They increase and decrease in length nearly perfectly in response to the tension. A 265 mm long, 2.0 mm diameter spoke elongates about 1 mm/120 kgF. A Laser spoke at about 1.5 mm diameter elongates about twice that much (proportionally to the square of the radius) under the same tension. Of course when building a wheel, you take up the slack inside the nipple, but once the wheel is built, forces applied to the wheel change the spoke lengths instantaneously according to the spring constant of the spokes. In the lingo of material tensile properties, the spring constant is known as modulus. Outside of a limited region the modulus is not constant, but that doesn't apply to the tension range in which spokes are used.

rpenmanparker 

Adding more spokes to the perfectly rigid rim model just requires you to invoke the cosine modifier for the angle of each spoke relative to the vertical. The principal is the same as with just the two vertical spokes.

Pirkaus

I'm no engineer, but wouldnt all the spoke carry some portion of the load? In a 20 spoke wheel wouldn't all the upper spokes increase in tension, and all the lower spokes decrease, save maye those perpendicular to the load? Can one of you engineer people do the maths for that one and see what happens?

rpenmanparker 

No math necessary. Yes, you are right. As I said above, however, the simple case of just two spokes provides a good analysis.

Bike Gremlin

But it has been proven otherwise - by experiments - measuring spoke tension change.

2 such docs have been quoted in this thread only - Jobst Brandt's book, and an independent research for a engineering degree.

waterlaz

They are springs. Unless you load them to the point they start to irreversibly deform, spokes behave like perfect springs, i.e. the elongation is proportional to the force applied.

rpenmanparker 

I don't dispute the egg shape as the real life situation. I am just saying that oval is the limiting case. Certain factors shift the deformation toward oval.

waterlaz

You are right. But this works only for a perfectly rigid rim.

wphamilton

Why only a perfectly rigid rim? You're visualizing tension from the hub simply hanging from the rigid hoop, right? and of course all of those spokes bear the weight, the more vertical the more weight.

But in a less rigid rim something similar will apply. Depending on how rigid the rim is, the rim's shape will change and you will get less tension increase in the vertical spokes, which I'm guessing is your model. But in order for that to happen the rim has to squash which will mean more increase in tension in the more horizontal spokes. The distribution changes but his idea remains intact.

joejack951

Two spokes might make the math easy but it wouldn't make a rideable bicycle wheel. I think at a minimum any model of a bicycle wheel needs to consider at least four spokes spaced 90 degrees from each other which changes the math considerably.

rpenmanparker 

You are right, but simplified models have their place.

joejack951

But when using a simplified model you need to be careful not to over-simplify and change the problem. if you tried to solve your simplified model rotated 90 degrees you'd get an answer that makes no sense when compared to a real bicycle wheel.

rpenmanparker 

Why would I rotate it 90 degrees? That wouldn't be an appropriate model, would it? Point is rhe two vertical spoke model can answer the original question: do the spokes take on higher or lower energy when weight load is applied to the wheel?

joejack951

You have highlighted my point. If your simplified model doesn't hold up when analyzed in anything but one scenario, then it is not an accurate model. A bicycle wheel rotates. So should your simplified model.

rpenmanparker 

Thst isn't so. The rotating wheel just repeats the physical state endlessly with a different point in contact with the ground. Nothing changes that relates to the problem that was presented on page 4. You don't set up a wheel model to make a functional bicycle. You do it to answer the question that was posed. If the model does that, it is successful. KISS

ThermionicScott 

I learned a couple of good jokes since it got bumped. So, thanks.

joejack951

It repeats a physical state which is vastly different than the one you have presented, one where the hub is attached to the rim by both vertical and non-vertical spokes resulting in a structure than can support loads 360 degrees around the circumference.

You are trying to get an answer about how a bicycle wheel reacts to loads. You need to model a bicycle wheel to do that. It doesn't need 36 spokes, or even 20, but it needs more than 2 to represent a bicycle wheel constructed of a rim, hub, and tensioned spokes.

rpenmanparker 

I disagree, but won't pursue it further. In any case good discussion. Let's do it again sometime.

big chainring 

Mmm,huh,hum,huh...load.

MikeyBoyAz

Is it my turn to typ(o)?

Ahem, here I go.

Two spoke simulation arguments:
Valid: Perfectly rigid, non-rotating model. In this scenario we claim that the sum of the force must be balanced, so any force applied to the axle must be supported through the spokes and ultimately by the ground. The sum of the forces are equal, so as you push on the axle, you push on the tire.

Invalid: Imperfectly rigid, or rotating model. If a force is applied to the axle of a deflecting rim surface, the force is transmitted to the rim hoop, which applies the force to the ground. The ground pushes back (it doesn't move down) and the structure of the rim absorbs the energy, rather than transmitting it into the tire. This causes the points furthest from the two spokes to equalize the stress on the rim in the only directions not constrained by either a force, or a spoke... causing an elliptical shape.

Infinite or scaled measure-ability argument:
Perfectly Rigid: For each measuring location on the rim for which tension exists, assume at rest that the opposite spoke to that point (or in the case of 2x and 3x the system of spokes) has equal tension on the rim. For each spoke apply the point force to the nipple and sum the forces, one for gravity, one for tension, and a third to add the new axle force. Notice that once the calculations are complete, the closer to the bottom of the wheel you go, the higher the tension on the spoke will be because the angular force, performing work, increases as the spoke aligns with the force vector (the spokes which are horizontal simply rotate around the nipple virtual origin). The spokes at the bottom will decrease equally to the spokes on their 180 degree opposing sides. In this model the tension on the spokes will graduate up (potentially sinusoidal) up to the apex top spokes.

Limited Elastic: In this final (real wheel) we assume a few important components. First, the wheel hoop does NOT in-elastically deform, permanently changing shape. Second, the spokes continue to apply a positive tension (outward radial force) from the hub; no lose or hanging spokes. Third, the spokes are strained within their elastic limits and do not change tension coefficients. Fourth, gravity is constant (haha). Fifth, the hub does not deform (not a Zipp hub apparently).

In this scenario we take an unloaded wheel and tension all spokes so that the wheel is true and dished completely center; apparently a front wheel. Next, we apply no rotational torque, so the wheel is stationary also.

We start the experiment by applying a load which increases the top spoke (measured individually) by some non zero amount. This deflects the hub downward a finite distance and increases the force on the bottom of the wheel, deflecting the tire some distance, which applies the force radially around the wheel by increasing the PSI by some measurable amount. The force from the deflection of the tire is transmitted, in a distribution up to all spokes within the deflection area (the contact patch). The forces are equalized with the threshold of tire deflection stopping when the pressure is balancing the energy necessary to reshape the tire. Each spoke which is affected by the deflection is at some finite measurement pushed up, which decreases the work it performed on the hub to create tension on the hoop. This work is balanced because the hoop is elastic, so although the circumference cannot change, the radius at each spoke can.

This finite, but measurable change in radius at the spoke points (within the contact patch) must be balanced and the former radius shape the contact patch section of the rim once held now changes in the direction of a triangle. (for measurement sake, we would argue the two points are the ends of the contact patch... and we are arguing a 2D wheel, because 3D would be ugly at this point). The distance from the two corners of the triangle approaching shape are closer in the pie shape and the radius must go somewhere once flattened. This increases the diameter of the wheel some finite amount, and with that change the work on opposing spokes around the entire wheel, which are not included in the contact patch, are equally affected because of opposing force conservation.

The points where the hoop is actually deformed infinitesimally also receive work changes as the corners of the triangle section of the contact patch inclusive are no greater in diameter than the rest of the wheel, but as you approach the point where the force vector from the axle is parallel, the radius is the smallest, and therefore the work applied by the spoke is the smallest.

The only spokes which can be different are those which have work directly applied to them. The remaining spokes around the wheel will simply rotate around the hub spoke hole and balance their cross side spokes forces. This means that the spoke tensions around the whole wheel will be almost the same (one to the next), and the only spokes which will change in tension from the rest of the wheel are being directly acted upon by the tire at the contact patch.

TL;DR: Foo.

tomato coupe

Regardless of whether or not your two-spring system is an appropriate model for a bicycle wheel, your conclusion is wrong -- loading the springs increases the energy stored in the springs. It's very easy to calculate the energy increase for your two-spring model. Do the math and you'll see this is the case.