So try it with a 12 spoke wheel.

Just because the load's shared so much you can't see an increase in tension doesn't mean you should throw out the only explanation that makes sense, IMO...

Imagine a three-spoked wheel for a minute; one of the spokes pointing down. To say the lower spoke bears any weight applied to the hub sounds pretty wrong, doesn't it? Surely you'd notice an increase in tension in the hanging spokes then.

What would happen to this loaded wheel if you backed off the tension in the lower spoke until it was out of the picture? The rim would become slightly egg-shaped and the hub would move up. So how that spoke could possibly be holding the hub up completely eludes me.

spokes in bike wheels (at least standard wheels and not carbon) act in tension. They offer no support when they are loaded in compression.

think of a rope and rock climber. The rope only works under tension, that is, being pulled apart at the ends. If you compress the rope, it offers no support until it has folded in on itself.

check this out: emergency repair kit for a broken spoke.
http://www.peterwhitecycles.com/fiberfix.htm

Nobody has suggested that spokes in a normal bicycle wheel act in compression.

yes, you did.
the spoke would have to act in compression in this situation.

No, that is wrong.

no, you're wrong. :rolleyes:

desconhecido is 100% correct.

I have no idea why this is so difficult to understand. Someone looks at a bicycle wheel and thinks, "when a load is applied to the hub, it must be the spokes above which bear this load." As a first thought, this seems reasonable. Perhaps at first it seems to be the only description which makes sense. But, when the forces acting on the spokes and rim are actually analyzed and tested, when changes in spoke length, rim shape, and the tensions in the spokes are actually analyzed and evaluated empirically, it becomes obvious that the upper spokes do not change significantly in length and do not see a significant increase in tension. What can actually be observed is in contradiction to what seemed to be the only explanation that made sense. How can what is observed be reconciled with the only explanation that makes sense when there is a contradiction? I suppose the answer is "turtles all the way down."

Rofl.

Since we're having so much fun discussing whether losing pre-tension is comparable to acting in compression - why don't we continue with an equally invigorating and rewarding discussion about the difference between deceleration and negative acceleration while we're at it?

well, all I know for sure is that the rim acts in compression and the spokes act in tension. Neither functions properly in reverse, unless they are carbon.

to say that the spokes next to the contact patch on the ground support the wheel through compression is silly.

So what's your beef? is it with the spokes between hub and ground seeing the biggest change in load, or with the admittedly clumsy turn of language stating that the hub is standing on the spokes?

If it's the former there's nothing I can do, and if it's the latter why don't you try to come up with a more fitting phrase? I suppose you could say "the hub is hanging from all spokes simultaneously , but it hangs the least from the bottom spoke(s)". Doesn't exactly roll of the tongue either.

Just got back from work in time to propose the correct language:

"The hub is suspended by the sum of the vectored forces in the spokes. This sum is always zero in the non-collapsed wheel."

------

As to the "what's your beef" question posted by Dabac above:

1. My beef is not with the fact that the bottom spokes experience the biggest change in load. This is true. I have no idea why anybody would have a vendetta against this perfectly normal result of physics.

2. I have a very big problem with the incorrect use of language in the context of structural analysis, in particular with the use of the phrase "standing on the bottom spokes". Not only is "standing" not a regularly used word in this field, it's connotation is the opposite of what it is intended to mean. This word cannot be used to describe a tensile member. When I stand I am supported by my rigid legs. If I'm hanging off a tree branch from my hands, and someone is trying to pull me down to the ground with a rope tied to my ankles, I am not "standing" on that rope. Even if a monkey suddenly jumps onto my head, causing a decrease in tension in the rope, and a smaller (since I have two arms and the force is divided between them) increase in tension in each of my arms, I am still not standing on the rope. That's absurd. Substitution of the correct language with "whatever language makes sense to me" is not trivial, in the same way that you cannot invent your own random mathematical symbols or chemical syntax. Judging from everyone's posts, the misuse of the word "standing" has clearly created a great deal of confusion.

Yep.

But this example is a bit misleading: if you look at the various measurements and FEAs you find that the increase in tension on top does not, even when summed across all spokes, come close to equaling the decrease in tension at the bottom.

There are two equal and opposite forces on the wheel: the force on the hub downwards, and the force from the ground on the rim upwards.

A wheel stays on the ground; it does not accelerate vertically. If you only think about the force applied to the hub, you've only analyzed a wheel that's under constant acceleration, as if you attached a rocket engine to the hub of wheel floating in space. You have to consider both the load on the hub and the load on the rim.

In your example you would have not just a monkey jumping on your head (load on the hub) but another monkey on the ground pushing upwards on the bottom dude's hands (the one who's pulling down on your legs) with force equal to the weight of monkey 1 (load on the rim). Draw that one out on paper, and you will see that the bottom rope loses a lot of tension and the top two ropes don't gain any tension!

The answer to your question is actually indeterminate until you say something about the rim. There are a lot of ways three spokes can change tension to support a load at the hub. Say each spoke has a tension of 200 kgf, and you put a load of 100kgf on the hub (which is matched by a load of 100kgf from the ground on the rim.) This could result in:

1. Top two spoke tensions do not change, and bottom spoke tension drops to 100 kgf.
2. Top two spoke tensions increase to 250 kgf and bottom spoke tension drops to 150 kgf.
3. Something in between.

In all cases the sum of forces on the hub is zero, (in the first case it is obvious that the detensioning of the bottom spoke matches the load on the hub; in the second case two angled spokes increase by 50kgf and are 30 degrees above horizontal so you have 2*sin(30 degrees)*50 kgf + 50 kgf = 100 kgf vertical change in force from spokes, which matches the load on the hub), so each case is plausible.

As it turns out,

(1) is the same as Yan's monkey example, once you correct it for equal and opposite forces; i.e. the rim is doing nothing to spread the load around.
(2) is what happens when the rim is infinitely stiff and rigid, which no rims are.
(3) All actual wheels are somewhere in between.

Many people's intuition about traditional wheels with shallow rims and lots of spokes is something like (2) but the truth when you measure it is closer to (1) than intuition would suggest.

The top members (my arms) have to gain tension. The rope on my ankle does nothing to hold me up. There is an additional monkey on my head. If my arms don't gain tension, what's holding up the increased weight of the monkey?

Anyway:

The "top" and "bottom" categories are misleading as well. the hub is not suspended from just the top spokes. The hub is suspended from all the spokes, whose net force on the hub equals zero. In tension, the direction of the force is colinear to the member. Therefore the direction of these forces causes them to all cancel out to zero at the hub, no matter what the load. If they do not cancel out, the hub would experience a net force in a direction. This means the wheel has collapsed.

For the purposes of structural analysis, it's perfectly reasonable to ignore the rim and focus on only the hub. This sounds like the hub is floating in space, but it is not. It is still tied to a system of external forces (from the spokes). You only have to consider the forces immediately in contact with the hub. Afterall, the hub doesn't actually know that the spokes are connected to a rim. It only sees the spokes. It does not see the rim. This is called free body analysis. A stable free body is one whose forces all cancel out to zero (this relates to the previous paragraph).

You can draw a free body diagram for the hub, each individual spoke, and the rim. Assuming the 36 spoke wheel. The hub has 37 forces acting on it. One for each spoke and one for the load. The rim also has 37 forces acting on it. One from each spoke and one up from the ground. The spokes each have two forces acting on them. The sum of the forces in each item, be it a spoke (sum of two opposing forces, i.e. the tension), the hub, or the rim, is always zero (forces in opposite directions cancel out, force in the same direction add, forces in random directions in relation to eachother can be added or subtracted trigonometrically).

The forces on the spokes are matched to the corresponding forces on the hub and rim, according to their location. For example, let's say the tension in a spoke is 900 Newtons. The force on the rim, at that location, is also 900N; and likewise for the hub. When we consider this same type of relationship for all the individual components, we will see that the net force of all the components, that is to say the wheel, is zero. When we consider the wheel in the context of the load on the axle and the support provided by the ground, we can see that the force pushing up from the ground is equal to the force pushing down on the axle (but in opposing directions). Therefore the net force of the entire system is still zero.

I'm doing a lot of talking and am not really making my point very clear. This topic is somewhat complicated for a discussion forum. At the very least I hope the above explains the reasoning behind the "free floating components" type of analysis.

The second monkey who is pushing up and lessening the pull of the rope on your ankles down! You have an additional weight of 1 monkey on your head and an additional weight of -1 monkey on your feet (added to whatever tension is already on your feet). Your arms don't change tension at all.

As to the rest of your point, I suggest actually attempting the math with the free body diagrams. Try with three or four spokes. It works out differently than you suppose -- if you actually try to work it out using only free body diagrams you will fail, because you will come face to face with the problem of static indeterminacy, which you did not touch on in your post, and is essential to understand here. In my answer to Kimmo I described two scenarios that you would find are both completely compatible with your type of analysis (if you bothered to write it down instead of internet arguing about what you think it would be if you wrote it down someday). That's the problem of indeterminacy.

Moving on from post 92. Edit: note this is not a response to post 91. See post 93 for response to post 91.

It is not just the top spokes which hold the hub up. When load is applied to the hub, the forces in each of the spokes change. These forces are carried to the rim. The rim is a compression member. Since we want to analyze the wheel as a simple system, with no bending forces, we simplify the rim into a 36 sided polygon, with each side a line between adjacent spokes. In reality, the curvature of each of these "segments" apply bending force to the rim. This is why deep rims are stronger than shallow rims. The deeper section has a higher moment of inertia (the ability to resist bending).

The rim, with all the forces acting on it, wants to change shape (likely into a D, with the flat side against the ground). However, in order for the rim to become a D, the distance between the "sides of the D" has to become longer than the diameter of the of rim (try it with some string). The spokes, all pulling in, will not allow this. Therefore, the spokes at the 3 and 9 o'clock positions also increase in tension. In effect the tension is spread across all the spokes whose contact points on the rim are not within the flat section of the D. The flat spot on a failed rim is usually limited to a tiny area (one, two spokes at the most?). All the other spokes have increased in tension to allow this. The increase in each spoke is small, because there are so many of them. This is why tension increase is hard to measure.

Ofcourse I read the quote. Remember the system does not end at the guy pulling me. He's standing on the ground, which is connect to the tree, which is connected to me from above, all compression members (analogous to the rim). We don't need the monkey pushing his hand up because the ground does that. The ground IS the monkey. Realize though that the force of the ground (your second monkey) pushing up is not going to decrease the tension on the rope, because that upwards force is sucked up and used to support me via the tree (in a roundabout way)! You can't use the same force twice.

Agree on the indeterminancy. However that's for the universe to figure out, not us. :D

Hey you edited your post after I replied to it. I can't keep track of everything like that!

Responding to your added bit on indeterminate systems: You are correct. I can't actually draw a free body diagram of the system and figure out all the numbers. We're in (way, very) much too high of an order or determinacy to do that. You'll admit that the principle stands though, no?

Hey, since I've got your attention here: I bought one of those fork mounted cantilever brake hangers, and as you know it doesn't fit on the Cross Check fork. Did you ever try that "power hanger" thing? How did that work out?

If the ground is the monkey, then your analysis is incomplete until you've accounted for the ground as a fully fledged member of the system. There's that, and there's also a bit of a fudge you made, in that the guy who "pulls down with constant force" can't actually DO that in static analysis. He can only pull down as much as the difference between his body weight and the force with which the ground is supporting him. In static analysis he's not allowed to adjust the length of his arm muscles in order to respond to the forces on his body changing, either. If the tree branches you're hanging from bend down at all, he's going to be holding a slack rope and not pulling down any more. He'll have to be another static body too.

The principle stands, it's fine, but it doesn't (by itself) allow you to say which spokes change tension.

Re: indeterminacy, you could let the universe figure it out, or you could consider each member as a deformable body (including bending of the rim sections, alas) and each member that connects to another member has to deform compatibly. That gives you enough extra constraints to pursue a solution.

The power hanger is terrible. Impossible to keep the brake centered, wears the pads unevenly. I came up with something a lot better using a screen door roller from the hardware store: http://www.flickr.com/photos/1325451...n/photostream/

In the analogy the ground acts as both a portion of the rim as well as the surface supporting the entire system. Otherwise I agree with you on all points. Thanks for the rewarding discussion.

Oh yes, you posted that brake setup in another thread. Ingenious. One of these days I'll try that.

To completely dismiss the unanimous results of laboratory tests (one you can even try at home!) and finite element models because it sounds silly is, well, silly.

The guy with the monkey on his head is pretty silly, too. He must be a manager. Perhaps it's a team-building exercise.

Nobody said that any spoke supports any load while in compression. Perhaps it would be silly if someone had made that claim, but nobody has. There is nothing inherent in the term "standing" which implies a member in compression, no matter what somebody's free body diagram shows. I would much rather use the "standing" metaphor and say that the wheel stands on the bottom spokes than the "hanging" metaphor and say that the load hangs negatively from the bottom spokes. Or, as implied above, that the load hangs from all the spokes equally but some more equally than others.

To state that the bottom spokes do nothing to support the load on the hub is what is silly. Notice that the person who originally made that claim is no longer making it. Now we are treated to inept arguments about monkeys on heads and how nowhere in the engineering literature do we find anyone describing a load being supported by members in tension as "standing" on those members.


Now, the problem with analyzing this with a free body diagram describing just the 37 non-zero forces on the hub is that there are infinitely many arrangements of those 37 non-zero forces which will result in no acceleration of the hub. That is what zzyzx_xyzzy said above (quite correctly) when addressing the 3 spoke wheel problem. You can't say anything about how a load is supported by the spokes in a three spoke wheel until you describe how the rim behaves when the spokes change tension. It's a very complex problem which is why people resort to numerical methods/FEA to figure out what's going on.

I wish people would look at the site I linked to before:http://www.astounding.org.uk/ian/wheel/

The guy has gone through the trouble of explaining the problem, the terminology, and his analysis including what is meant by statically indeterminate.

Perhaps considering a very simple example that can be solved discretely will help in understanding this.

Let's look at a 2 spoke wheel with one spoke going straight up and one spoke going straight down. Let's say that the modulus of elasticity is 1(force unit per unit of area), the area of the cross section of each spoke is 1(unit of area), prior to applying any load the length of each spoke is 100(unit of length) and the tension applied to each spoke is 100(force unit). Hub is weightless, as are the spokes, and the diameter of the hub is zero.

Now, apply a load of 10 (straight down) to the hub. If you draw a free body diagram of the loaded hub with the two spokes you will see that you can't solve for the tensions in the spokes because you haven't specified how the rim constrains the problem. That is, it is indeterminate. So, we can assign any characteristics to the rim we desire. Let's say that the rim is infinitely stiff so that the addition of the load of 10 to the hub does not change the combined length of the spokes (200) at all. This constraint results in four equations and four unknowns. The unknowns are: tension in the top spoke (T1), tension in the bottom spoke (T2), length of the top spoke (l1) length of the bottom spoke (l2). The four equations are:

l1 + l2 = 200
T2 = l2
T1 = l1
T2 + 10 = T1

note that these equations have been simplified because of the choices of modulus and area and represent the magnitudes.

Solve these equations and you find that the tension in the top spoke after the load is applied is now 105 and in the bottom spoke it is 95. The lengths match the tensions (because of the choice of modulus and cross section area). Is it proper in this case to say "[t]he bottom spokes do nothing to resist vertical load on the hub?"

So, now change the behavior of the rim so that when the hub is loaded that the bottom spoke shortens but the top spoke does not change in length. This is a little simpler to solve. Because the top spoke does not change in length at all, the tension of the top spoke must remain the same as before the load was applied -- it is 100. In order for the forces to balance, the tension in the bottom spoke is reduced to 90 and the length also decreases to 90 (modulus of elasticity is 1. cross section area is 1). This is sort of what happens with an actual normally constructed bicycle wheel. I think it's reasonable to say that, in this case, the applied load stands on the bottom spoke though the bottom spoke remains in tension.

What happens to the bottom spoke in this example if tension in the bottom spoke is reduced to zero, you can decide. Perhaps it tears a hole in the space-time continuum. Plugh!