Agreed. If I didn't make that part clear then I said it wrong.


It was all great up to the bolded part. That bit just doesn't wash. For a given torque in the hub radial spokes will see a much higher torque related rise in tension compared to tangential spokes due to the geometry of how the tension in the hub is applied to the spokes. Due to the radial nature the hub has a lot more mechanical advantage due to the angle the radial of the hub meets the spoke at. That mechanical advantage means the hub torque is going to be able to pry on the spokes with a better leverage ratio so the tension buildup in the spokes from the torque will be a lot higher than for the tangential setup.

A great way to show this is with a vector analysis of the forces. But it's late and my brain is mush at the moment. I'll see if I can remember enough of my Grade 19 geometry to figure it out tomorrow. But in the mean time if you've ever used an over center cam lock you'll remember how easy it is to move the cam when the rod is moving up closer to the center and then over center point. That's the same principle I'm trying to describe that will result in torque in the hub being able to generate a lot more force in the spokes with radial lacing vs tangential.

For the same reason a full radial drive wheel will suffer from a lot more wind up. When the force stretches the spokes that little bit more that little bit of stretch translates to a lot more angular travel in the hub compared to the same amount if the spoke is tangetial. So the radial lacing will feel sort of rubbery at the pedals to the rider vs the tangential lacing. That part I can show you. For the sake of the demonstration lets assume the rider's torque can stretch the spoke by 1/16 inch. It makes the diagram easier to see because you're not trying to work with lines that are visually too close together to see. So we get this sort of effect that's going to make the wheel feel more rubbery at the pedals. At least my brain was this awake.... :D

+1
+1 again

Now that's some interesting numbers! Thanks for cranking them out.

+1 Damn right

But this is a bike, a moving object. Once you start cranking the bike begins to move, so it's not that easy to say how much torque and tension increase the wheel really will see. You might be getting "much more" in the radial wheel but it can still be within limits for the design.

You have to consider it at the stage of peak torque. Like when the big guy stands up and both pushes and pulls with both feet. The spokes will see the full measure of that moment and that "worst case" is what they need to be able to withstand. They need to be able to withstand this force even though it's only present for 4 or 5 cranks. That's still a significant time and the spokes, indeed EVERY component in the wheel has to be able to withstand these peak forces. It's not about the average.

Danno, those are some good numbers but let's back up a little and look at your final analysis.

I'm with you up to the final total pulling force of 147 kg and that it being split between 16 pulling spokes for a linear TANGENTIAL PULL AT THE SPOKE HOLE OF 9.2 Kg. But that's where it breaks down. From there you have to consider the direction the spokes are angled at to determine the tensile load increase in the spoke. For a true tangential spoke the added tension in the spoke is at 9.2 kg. But as the angle of the spoke changes towards radial you need to get the sins and cosines out and figure out the difference. And when the spoke is not directly in line with the force there is a mechanical leverage of forces that increases the tension in the spoke to a value higher than it is when the spoke is in line with the force. Your otherwise fine set of data doesn't take that final factor into account.

And since my mind is awake now I was able to do up this vector analysis. Note that the angle isn't even close to radial in the second diagram but you can see how as the spoke angle approaches the true radial how the force in the spoke will multiply rapidly to extreme values. In fact if the spokes did not have the ability to stretch at all then a truly laced radial wheel would see the increase in spoke tension hit infinity. In real life a radial laced rear wheel would defect to some angle (as in wind up) and allow the spokes to come to an angle where they would generate the required triangle of forces. But the leverage on such a spoke at this very small angle would result in HUGE increases in tension that are way beyond what even the static tension of the spoke is. I think at that point you'd see either the spokes fail or the hub flange fail.

And this is why a disc brake or rear wheel should not be done in a full radial lacing.

EDIT- for the heck of it I reduced the spoke angle to 5 degrees from radial and 2.5 degrees. The resultant spoke tension increases at that point were 105 kgs and 210 kgs. At 2.5 degrees we're probably looking at an angle that would typically occur when our strong rider wound up the wheel and stretched the spokes from the true radial angle. I can only see very bad things occuring from such a situation.

The thing is, spokes ONLY take loads in tension. It's not a solid object where lateral forces can be broken up into two independent vectors. The forces are translated into the direction of the spoke and are applied fully in tension. For example, if you hang a weight on a rope and push sideways on teh rope, does it increase tension on the rope at all. No, it just push the object sideways and it swings, but the tension on teh rope is always the object's mass*G.

So the initial wind-up of the hub actually results in mininal-increase in tension on teh spoke on a radial wheel. Lets magnify the results for comparison. Lets say a 3x wheel winds up 1-degrees for 5.58kgm torque at the rear-hub. You can do an integration from 0-1 degrees and figure out the tension build-up for each spoke. When all the spokes' tension increases enough to counteract the torque, the hub no longer winds up. A radial wheel would end up winding say... 5-degrees for that same increase in tension.

A spoke, like a rope, ONLY experiences tension IN-LINE with itself. Lateral forces do not result in tension. You can test this out yourself. Clamp a rear wheel by the rim in a vise or a door. Get a long chain-whip, wrap it around a 15t cog and hang a 15kg weight from it. Measure the spoke-tension increase due to torque and you'll find that its the same regardless to numbers of crosses or teh tangential-angle of the spoke leaving the rim.

Another way to calculate this is the actual stretch amount. Plug into Young's Modulus for amount of stretch based upon length and cross-section of spoke and you'll get the amount of lengthening and teh resultant tension. You'll find that direct tangential lacing has the least amount of wind-up angle necessary for the amount of lenghtening needed. With a radial lacing, you'll need a lot more angle of wind-up for the exact same elongation. There's a direct relationship between tension-increase and elongation and you'll find that both tangential and radial-lacing results in the same elongation amount for the same tension-increase. Just that to generate that same amount of elongation, you'll need to twist the hub more.

It's the emperical data that will show one way or another whether an arm-chair physicist or racer's idea is valid. How many people have actually built-up or ridden a rear radial wheel anyway? Have you guy actually ever measured the angle wind-up or spoke-tension increases? Thinking in your head without looking at the experimental data is just goofy. It's like those guys saying that generating hydrogen using electrolysis from your car's alternator and pumping it into the intake will magically increase gas-mileage. It sounds plausible, but the actual emperical data will show you something else. ;)

Another note, spokes rarely EVER fail from over-tension or overcoming its ultimate-strength. Go ahead and calculate how much tension you need on a spoke to break it, you'll be amazed. Regardless of which one of our models you choose, the increase in tension doesn't even come close to the tension needed to break a spoke.

What causes the vast majority of spoke failures is fatigue. That's due to loading and unloading cycles over time. And the major problem with this is spokes that go completely slack at the bottom of the rotation due to inadequate tension.

In reference to this picture, you need to actually use an integration from 0-5.58kgm of torque to find out the exact twist-angle.

http://www.bikeforums.net/attachment...1&d=1215283027

where actual increase in TENSION = TORQUE/LEVER * SIN(X)

The tension on a tangential 90-degree spoke ALL goes into stretching the spoke. On a 0-degree radial spoke, none of the initial tension goes into the spoke (like pushing sideways object hanging on a rope). All of the torque goes into twisting the hub. YOU Can test this by twisting a hub with very loose spokes. You'll find that ALL of your hand's torque goes into twisting the hub, and you can twist it quite a bit, but none of that force goes into stretching any of the spokes. As the angle increases more and more, more of the torque goes into stretching the spoke. At some point where the stretch balances the applied torque, the hub stops twisting.

If you add Young's Modulus and the cross-section of the spoke to the above equation and integrate from 0-5.58kgm torque, you can determine the exact amount of tension, resultant stretch and angle of wind-up.

I agree, spokes do only support a load in tension. But that's why it's valid to use vector analysis to resolve the tensile loads in the spoke. Vector diagrams are a valid way of analyzing angular forces by breaking them down to components. In particular when the resistive force (the spoke) is not lined up with the applied force due to retaining factors. In this case the rim and the hub.

In your rope example if you have a weight hanging on a rope and you push the weight sideways in fact the tensile load in the rope DOES increase. It becomes a vector addition of the downward force of the weight plus the sideways force and it adds up the same as shown in the triangle of forces for the spoke.

In fact your rope example is the whole basis of the increase of tension in the spoke due to the shift away from a true tangential angle that I show in my diagram. I suggest you find go find a mechanical engineer and ask them about your rope example and see what they say about the increase in the tension in the rope since you don't seem to like my explanation.

On this vector picture, I see where you're getting confused:

http://www.bikeforums.net/attachment...8&d=1215283027

What you didn't clarify was the source and direction of the additional-tension force. In your example, what it shows is resultant force at the hub from pulling on the spoke nipple at the rim!. That's why you would need 29.7kg pull on the nipples to generate 9.2kg tangential-force on hub. Where is this force coming from?

You need to reverse the model and start the force at the hub. Your model goes from X-angle of wind-up to 0-degrees. And at 0-degrees, you have infinite tension. Does that mean a real radial wheel would break all the spokes the instant a rider pushes on the pedal? Calculate your model for 1 degree vs. 5 degree and you'll see it's reversed. When a radial-wheel is at 0-degrees, there's no torque, it's at rest and there is no tension-increase. When torque is applied...starting from teh hub, that torque adds zero-tension to the hub , all that force goes into winding up the hub. Only as the angle starts to increase, then more and more of te torque goes into pulling on the hub. The function starts at 0-force and 0-angle and increases as forces increase.

Let's do some testing and see if spoke-tension is infinite at 0-angle and decreases as we wind up the hub with torque. The ultimate test of a mathematical model is if the empirical data matches.

Let's back up a little. Let's look at how when you use a tensile element at an angle to the applied force the manner in which the forces in the tensile element can be multiplied. Because that's the core of this misunderstanding.

THis is a long one but I hope it's worth it.

I've done up another couple of diagrams (OH NO! NOT ANOTHER ONE! :D) that hopefully shows where my spoke diagram thinking came from. In this a simple weight is suspended on a shop wall mounted cantilever beam for moving the weight around. In this case a free pivoting beam (C) is supported by an angled rope (A) and there's a second shorter rope (B). As you can see the (B) rope has a simple 100 lbs of tension in it. If we compare this to my spoke diagram rope (B) is the equivalent of the tangential spoke In other words the spoke is in line with the force being applied so the spoke only sees the simple load.

But when you force the rope off at an angle like with rope (A) and keep it there with a beam (C) it gets more complex. Now you end up with the load in the ropes trying to compress the beam. But we're also supporting the load with the rope (A) which is at an angle. What happens is that it's like using a lever to move a heavy object with a long arm and a pivot in close to the load. The weight is your effort on the long side of the lever and the angled rope is on the short side. The force is multiplied in the angled rope because the beam is translating the direction of the load.

Now we can certainly do this wiht trigonometry and drag out the sin's and cosines' but for this example the force of the weight is vertical, the compressive forces of the beam horizontal and that forms two sides of a nice convienient right angle triangle. And since the tensile force in rope (A) is obvioiusly in line with the rope we can use that angle for the third side. So we draw a triangle with the vertical side at a scale of 100 units and a horizontal line from one end of that line and an angled line of the same angle as rope (A) from the other end to form a triangle. That's shown above the diagram. From there we scale the lengths of the other two sides compared to the known 100 units side and we get the numbers shown. But note that both of these are significantly higher than the original 100 lbs. There's no additional forces being introduced. However note that a lot of the rope's tension is balanced by compresive forces in the beam. But the rope is still under 333 lbs of tension to support that 100 lb weight. The tension in rope (A) and compression in beam (C) are strictly due to the geometry of the suspending arrangement and the weight it is lifting. In effect I've just sketched out your hanging rope example where you push the load on the end of the line to one side.

This diagram ties in directly to the spoke loading diagram. The only difference being that there's no physical beam to hold the offset spoke at the angle shown. However that "wall and beam" are still there but it's the other spokes and the rim that hold the one shown in my diagram in place at the angle shown. In the triangle of forces I showed this support from all the other spokes and the rim would all resolve to the 28.6kg virtual line that is extended off the 9.2 kg vector at the right angle.

However in looking at the diagram I will admit that I put the arrow for that vertical "beam" compoment on the wrong end if you see it as a counter to the spoke force and being conatained by the rest of the wheel. On the other hand it's fine as it is if you look at the arrows in the triangle as broken down components of the tensile increase in the spoke.

So for the spoke angle shown can you now see how when you torque the hub with 9.2 kg per spoke and the spoke is at an angle to that force just why the mechancial "leverage" due to the angle of the spoke to the force is going to induce a tension increase in that spoke by so much?

Let's say the hook on the wall was too high up and for some reason the boss chose to make us lower it so the distance from the beam pivot to the rope eye ring is only 1/3 as much. This compares to our spoke getting closer to a true radial angle where the spoke is at or very close to 90 degrees to the tangential torque force of the hub. Note how the tensile load in the rope (A) just more than tripled to over 1000 lbs needed to support only 100 lbs of weight.

Why? Look at the arm length of the beam and the distance between the pivot and the rope(A) mount. Mechanical leverage. It's the same with our spoke that's trying to resist the forces that are not in line with the spoke. The spoke is acting just like rope (A) in the diagram. Actually as you mentioned first off we have 16 spokes all working together to resist the load but we broke it down so only one is being looked at. The "braking" spokes and the rim form the supportive wall and beam that hold the 16 pulling spokes at the angle we lace the wheel up for. In the case of my original spoke diagram the angle shown is probably not far off what a 1 cross pattern would be.

Danno,

You've said a great many sensible things in this war of egos here, but the above sentence is what stressed me out.

Tension does not balance torque. Tension is a force, and a force must be multiplied by a distance (the "moment arm") to make a torque. To revise your statement:

What happens is ALL hubs will wind-up until the increase in tension, multiplied by the moment arm (perpendicular distance from the axle), equals the applied torque. (Perpendicular distance means perpendicular to the spoke.)

There we go. Now the radial has a tiny perpendicular distance from the axle, generated solely by the wind-up, so a huge tension increase is needed to equal the applied torque. The tangential has a healthy perpendicular distance built into the structure, so only a modest tension increase is needed.

I think that's what BCRider is trying to get at, though I'm blearily having trouble deciphering those posts myself.

An all-radial rear might feel a bit rubbery, and I don't doubt that Danno has ridden one. It's almost certain that the spokes can take the large tensions involved. The flanges might blow up on you, though. It's a nonstandard way to build a rear wheel and for good reason.

This is all I'm going to say. I'm not going to flash my credentials, re-explain this concept in new words, or engage in any below-the-belt sniping. After all, I do have work in the morning.

Please put the first part into numbers and equations and you'll see that you have exactly the same equation I presented earlier.

As for the 2nd part, compare the "perpendicular distance" from the axle between the radial and 3x and tell me what that distance is in numbers. You too are assuming that the force that's increasing the spoke-tension is being applied at the nipple. It's not, it starts at the hub.

The thing here is that we need emperical data to see if the model in our heads fit with reality. I'll conduct BCriders rope configuration with real physical structures and measure some real data. Then we can continue the discussion with common assumptions and workable models.

1. The mistaken assumption BCrider and you have is that there is some sort of mechanical-advantage going on with the angle. In reality, there isn't, and the empirical data on the rope-experiment will borne that out. Spokes and ropes work only in tension and do not have lateral force components. Tension-A in the upper rope-section will be exactly the same as the vertical section directly attached to the weight.

2. Vector-analysis is incorrect. What vector diagrams show is how to divide a larger diagonal force (hypotenuse), into two X-Y right-angle vectors. It can break up the diagonal into two smaller right-angle components. But it doesn't work the other way around with a smaller lesser component multiplying into a bigger diagonal. We'll deal with that in a different topic.


BTW - regardless of your credentials, does the airplane on the conveyor belt take off or not? You'd be surprized at how many people with "credentials" got that one wrong. ;)

While you're waiting for me to conduct the experiment, I suggest you read this article: Dean's World - The Brainteaser That Changed My World

Search around. Someone may have the correct size. And you don't have to have the exact size that the calculator tells you. A 1-2 mm difference isn't that big a deal. Shorter is usually better than longer.

I lost a friend (well, he wouldn't have been a very good one) over that very brainteaser when I was 17. Conditional probability is tricky, and yes, you do have to switch. I got that one wrong, but since then I went to college, I became a little quieter and more thoughtful, and I learned never to mouth off over something theoretical until I've followed the other person's logic exactly.

Well, I lost that friend both over that problem (often called the Monty Hall problem) and also a physics problem regarding the motion of binary star systems. I was right about that one. I appealed to a professor to back me up, which he did. I wonder what I meant to do by prolonging the argument: make him look like an ignorant loser? I knew, by the force of my logic and the weakness of his, that I was right. That should have been enough.

That summer was the one when I learned not to get into fights with people over physics (or mathematics, or mechanics, or religion, or anything else abstract) for that reason. The person who is wrong will never be convinced, and both people come off like jerks. Right now I live near the University of Chicago, and just hearing the undergrads on the bus makes me cringe, because they haven't learned this.

I will search around, cyccommute. None of the sizes listed are even close (I need 272-274 mm spokes) but someone might carry them. Thanks for your reply.

And sorry, Aeneas, for making a mess of your thread. It was a really good, germane question that lots of people ought to read. Well, maybe with this flame war we kept it near the top for awhile. ;)

While we're at it, let's not forget that nipples are available in 3 different lengths as well. With 14 and 16 mm to choose between apart from the most common 12 a 6 mm deviation from the target length is a manageable issue.

Be sure to get spoke-lengths that are at least long enough to come up to the bottom of the slot in the nipple-head. That's on the short end and is the minimum. You can go 1-2mm longer and have the spoke-end flush with the top of the slot. Problem with being shorter than the bottom of the slot is that the nipple-heads will be unsupported and bear all the load without a spoke inside. The brass in the nipples is a lot weaker than the steel in the spoke and it's not uncommon to have nipple-heads snap off if the spoke is too short.

Well, I'd like to think that this isn't so much about a collision of egos but more as both of us trying to learn something from the other. I have only been around BF.net for a while but from what I've seen from Danno's posts he's got pleanty of knowledge to offer. But one thing I've learned over the years is that the more I know the more I realize I don't know. I've learned a lot from others over the years and been shown wrong on many an occasion. If I do have an ego it was tipped off the shelf and smashed on the floor years back... :D

In any event I certainly hope that Danno is taking this all in a friendly manner and a spirit of learning. I know I am.

If I do seem "preachy" in all this it's only because I enjoy some good natured discussions and flexing my grey cell trying to show why something is the way it is in a way that others can better understand. Trust me, I had to test my own thinking before posting my earlier stuff. But it's all based on my old high school and first year university physics and my intrest in other mechanical hobbies that I've persued over the years.

Now getting back to the spokes thing. The "forces" doesn't originate at the spoke nipples. As you've said the spokes are under tension and actually the spokes are just a conductor of the forces. When the hub applies torque the spokes politely conduct the forces to the rim where the resistance to our motion is located. Being a simple conductor of the force this means that any increase in tension in the spoke is conducted equally and instantly to the spoke nipple and whatever happens at the head of the spoke shows up at the nipple and vice versa. Getting back to the examples the nipples are the equivalent to the screw eye in the wall above the arm pivot.

The arrows on the vector diagrams are just to indicate the direction of the forces resiting the original force. If there's no accelleration then "proper ettiquete" for vector diagrams says that all the arrows are supposed to fight each other such that the center point of all the vectors ends up with a net 0 force. Sort of like if you had three ropes tied together at the center and spaced out with three kids pulling on them from random angles. If the knot in the middle isn't moving and you do a vector diagram it would be three lines radiating from a common point with the arrows at the ends pointed out. All the components of these vectors would resolve to zero net accellerative force on the knot as far as motion goes for this example. The neat thing about vector analysis using diagrams drawn to scale is that it provides a very graphic view of all the forces involved. It's a lot closer to "real life" than using trigonometry calculations to figure out the alterations in loading for stuff like this. But both methods are equally as valid.

I may have a simple way for you to set up this experiment. If you use some thread a stick and a bucket you can do the whole thing and see the results. If you tie the thread to the handle of the bucket and fill it with enough weight so that the thread breaks when you try to lift it. Then remove enough weight that the thread can lift the bucket without breaking consistently. Then set up the vertical pivoting arm. You'll want to use a single pivot or guy threads to avoid sideways movement or make it a T shape with the top of the T against the fixed wall. Now pass the thread from the bucket handle a couple of turns around the end of the arm and lift the bucket up with your upper thread vertical so the arm is parallel to the floor. Start moving the upper part of the thread over to simulate the angle shown in my wall crane diagram. I think you'll find that the thread breaks at around 30 degrees from vertical. Take more weight out so the bucket is about 1/2 the weight of the breaking strength and do the test again with fresh thread. If you're holding the thread in your hand for this second one you should easily be able to feel the buildup in tension as it'll go from the light bucket only weight when the upper thread is vertical to double or more as the angle lays over closer to horizontal. When you do this you should find that if you're pretty close to the 1/2 breaking strength load for weight that the thread will now break when you lay the upper thread over to 60 degrees from vertical. The sin (or cos) of 60 being .5 so the loading increase from this specific angle is double the vertical load.

If you happen to have a digital diet scale handy place it vertically under the wall end of the beam and watch how the compressive forces in the beam build up when you tip the thread from vertical. A padding of something like two or three layers of duct tape stuck to the face of the scale pan should produce a non skid surface for the T to sit against for this.

Howzzat?

Hey, I think this is a great discussion. I'm sure a lot of people are learning from looking at the number of views. One thing I'd like to finally do is get some empirical real-world data to corroborate or dispel the common misconceptions out there. :)

Yeah, I was thinking of that scale idea. Was going to suspend the horizontal beam with some ropes so it can have a free swing in towards the scale. I'm wondering whether I should just round off the outer end of the beam or if I should install a pulley. I think a pulley would eliminate any friction at that point that may throw off the readings. And using two in-line tension-scales on the two sections of rope will allow us to take readings at two spots simultaneously.

Nix on the pulley. It doesn't replicate the issue at all. Pulleys are used to change the direction of the pull and it won't lock the beam and upper rope into a locked system. For example a rope lifting the weight and passing around a pulley produces the same tension in the rope on both sides of the pulley.

And before you shout "EUREKA! ! " :D it's not applicable to our spokes. Last I saw our wheels don't have pulleys on the spokes.... :D All joking aside our spokes are fixed in length so to replicate this in the model the upper string needs to be secured with a non slipping fit to the end of the beam. I was thinking that just a couple of wraps around the end of the beam would do it. The friction of the thread on the wood is all you need.

Besides if you did put a pulley on the end of the beam it would just let it fall down to the bucket until it jams on handle and THEN it would be back to a locked system and you can lift it and see the buildup as the angle lays over. But the bucket will tend to swivel from vertical if you do that.

If you're doing this on a small bucket and string or thread scale for ease of lifting then with the bucket on the table or floor just angle the arm down maybe 10 degrees and slip a few turns of thread around the end to lock it in place. Maybe a little notch in the end of the stick will let the final turn pass around the end and up without slipping down the length of the beam. Then when you lift just bring it up that little bit needed to reach level or near as the eye can see. That way you won't need any ropes to hold it up.

Ah.. but the hole in the hub acts as a cam...

Actually, I'm looking at the rope-and-beam picture again and I don't see it modeling spokes on a wheel. There's three anchor points: rope-to-wall, beam-to-wall and rope-to-beam. I don't see the analogous three points on a spoke. It only has two ends.

I think it would more closely resemble a wheel if we hang the beam vertically, then hang the rope and bucket from the end of the beam. Then push on joint where the rope attaches to the beam sideways to simulate the rotation of the hub.

OK. So lets rotate my spoke diagram and mirror the wall beam..... That makes it more clear?

Remember that with the spokes the "beam" is the rest of the spokes and the rim that works to hold the spoke at the angle to the tangential torque.

In fact here's why the wheel spoke has a virtual "beam". It's the other spokes that hold it at the angle to the force. In this quick sketch I've only shown the other 7 pulling spokes on the one side for clarity but you can fill in the rest. This is what I mean that the "beam" is the rest of the wheel.

OK, one last sketch for you. I don't know why I didn't think of this before. This one is PURE bicycle and it's going to be an easy one to test for yourself to prove the point and to see how the force builds up in the spoke.

Got an old hub? Mount it into a frame held somehow or make up some sort of brackt to hold it and do the testing shown below. String is fine for this. The wire is just a visual aid to help you hold the hub so the hole being tested stays level with the axle so the bucket of weights is pulling in very close to a tangential manner. Start with the upper supportive string near vertical and then shift your pull to the side to simulate moving from a 3 cross near tangential angle of spoke towards a true radial spoke. If you have a fish scale that measures with a pull you can use it to actually measure the force needed. But trust me. If you've got a couple of pounds hanging from the weight line by the time you get down to 60 degrees from vertical you'll KNOW you're pulling a lot harder.

Actually, I came up with the correct model. It's two swinging arcs that meet in the middle at 0-degrees. I'm gonna just skip the whole buckt and rope experiment because it doesn't really represent the forces on a spoke; there's no fulcrum on a wheel to leverage the spoke. I'll come up with a diagram that shows why the forces on a spoke are always in tension and then I'll build a test-jig to measure spoke-tension increase and hub-windup as a function of torque on the hub. Then test two identical wheels with different lacings.

Actually your last diagram will work for testing. We can keep the bucket and torque tangential by wrapping the rope around the hub-flange. Actually I'll use a chain-whip and hang a weight off it. The hub cannot be attached to supportive bracket, but rather should be built up into a wheel. Then clamp the bottom of the rim into a vice. This gives the strongest-torque scenario, such as in the case where our Olympic champion is sprinting with all his might while his bike is held stationary aimed into a wall.

I'll have two gauges, a tensiometer on the spoke and a dial-indicator on the chain-whip to measure hub-windup. :)

It'll be interesting to see the tensiometer results for sure. And you'll likely be the very first person to ever lace up an all radial rear because of this test.

As a short term proof of concept I really suggest the hub and strings deal of the post just before the last one. A big part of this discussion is the belief or lack of about how the angle of the spoke to the load leads to changes in the tension. You mention it's not a fair test without a rim but for this " single spoke" proof of concept your hand IS the rim. But a rim that's easily moved to alter the angle of the spoke to the hub Setting up this test with a bare hub, a bicycle frame to hold it in place, some string and a handy 2 or 3 lb weight won't take long and it'll quickly show what this whole last dozen or so posts is all about one way or the other. From there it's time to lace and get some real world numbers.

Speaking of tension from your commment about a model showing why a spoke is always in tension it suggests that I may have written something to suggest such a thing. I can't see anything I wrote that could be interpreted that way and apoligise if my wording may have suggested such a thing. Of course spokes are always in tension and they are always supporting their loads directly along the line of the spoke like any good piece of flexible wire will do. All I did with the vectors at angles to the spokes is try to show how representations of the forces can be brokend down into components and directions for study purposes only to show how the angles affect the final and only truly existing forces in the spoke.