jmaccelari

OK. I assumed 36 (18 of each) spokes' date=' since this was not specified. I also assume nipples (36) on the inside of the rim. You didn't give me the tyre/tube weight, so I measured one of my own (0.280kg). I(hub) is about 4.65x10-6 k/m^2 (minimum bound is 2.73x10-6, maximum is 9.80x10-6) You were given I=m*r^2 / 2 (Note that I don't have the exact geometry of the hub, so I calculated the maximum and minimum values, which are both so small we can actually ignore the hub entirely. The value given is a ratio approximation from measurements off one of my Shimano hubs.) I(spokes) is about 5.53x10-3 kg/m^2 You were given I=m*L^2 / 3 I(nipples) is about 12.531x10-3 kg/m^2 You were given I=m*r^2 I(rim) is about 197.759x10-3 kg/m^2 You were given I=0.5*m*(r1^2 + r2^2) I(tyre) is about 114.849x10-3 kg/m^2 You were given I=0.5*m*(r1^2 + r2^2) Total is about 330.677x10-3 kg/m^2. To get this we simply "integrate". This means in plain, simple English that we add all the little bits up... Now that wasn't all that "complex" was it? You didn't need to go the numerical integration route (you would not be able to do it with a pen and paper, either. And, I must admit, neither would I). Why then the nonsense about how difficult it is? So the hub is about three orders of magnitude smaller in effect than the rest of the wheel. As GoLefty!! says we can ignore it as it is less than the rounding error. Spokes are about 1.3% of total MOI Nipples are about 3.7% of total MOI Rim is about 60% of total MOI Tyre is about 35% of total MOI The equivalent "ideal wheel" radius is about 480mm. So if we reduced your "real in the hand" wheel to an ideal wheel of the same mass, the radius we would be working with would be 480mm. If course. we could keep the radius constant and reduce the mass to 950g which would have the same effect. Or we could choose any one of the infinite combinations... So the question now is: what is your point? Or are you just playing a little game and wasting my time? As I have stated before: the actual wheel is immaterial, so this exercise was pretty pointless unless you are still trying to convince yourself of something... Surely you mean "bated"? "Baited" could be a little smelly.

Good lord! I concur!

The bearings have nothing to do with the argument since they do not affect the rotational energy requirements of the wheel (they are part of the frictional component we are ignoring for the comparison). You are quite correct that the hub has a smaller effect because of the smaller radius. This is the exact reason. The radius (the "moment arm") is proportional to the all important MOI. The smaller the radius, the smaller the MOI and the smaller the effect the component has on the system. That is why nipples on the hubs are better than ones on the rim. BUT: the effect in this case is so small you won't notice it. I would assume the aerodynamic saving is much more significant in this case... jmaccelari2010-01-09 04:49:00

I gave you the equations for the components. As I told you and you obviously have problems understanding: measure the components yourself and then add up the bits. You think you are a mathematical, brain - then stretch yourself by doing the simple sums. As I stated, we have proved that for ANY wheel that the energy saving on rotational mass on the wheel is greater than the energy saving on the frame per gram. I am not being patronising. I am just tired of you harping on about irrelevancies either because you refuse to admit you are wrong or because you really do not understand the fundamentals. My opinion is that it is the first option. If it's the second send me: The mass and outer radius of the hub. The mass and length of one spoke. The mass and inner and outer radius of the rim. The mass and inner and outer radius of the tyre (with tube). The mass and radial distance of a nipple. I will then compare the MOI to the ideal wheel we have been discussing and you'll see there is a difference in the absolute figures, but not the final conclusion. The more accurately modelled wheel will be the equivalent of an ideal wheel as we have been discussing, but of a smaller radius (see if you can figure that out!). If you do this, please remember you are going to be further embarrassed as I'll will show how easy it is. And you'll be supply the ammunition to shoot yourself down with again...

No they are very simple. And by "very" simple' date=' I mean "very, very simple". Moment of inertia for a rotating object is defined by: I = integral(m * r^2) It doesn't get much simpler than that. Cut your wheel up into bits. Measure r for each bit (the distance from the hub), weigh it and add (m(bit) * r(bit)^2) up for each bit. The smaller the pieces the better the approximation. It may be tedious, but the calculations are primary school level. "Complex" my arse! I have to laugh at that one! Don't try to intimidate people by using the word "complex"! I state you do not need to do a numerical integration. You do need to do an analytical one. Without an integration of some form you do not have a moment of inertia to start with! The numerical integration is a lot of work and will not provide much improvement over an analytical one. You do understand the difference between a numerical and an analytical integration' date=' don't you? By an analytical approximation. Duh. I'm repeating myself again...

Sympathies, Ian. Hope you recover quickly!!!

**** knows - I never got that far!!! My guess is that a tubular sits in a low cavity/dip around the rim which is easy to manufacture. The dip has no structural requirement apart from providing a surface to glue the tubular on. A clincher will need a strong side wall on the rim to hold the clincher tyre bead in. This, because of the high pressures road tyres are under, would require a lot of design and manufacturing effort to make light but still strong enough to hold the tyre. My opinion. No maths involved.

Ouch! I hope he's OK...

What I mean is that at the rim we might save (for example) 16J of rotational energy - which is a small overall saving. At the hub' date=' we will save 0J due to rotational energy (there is no rotation exactly on the axle so no rotational energy is saved there). So if we compare 0J to 16J the relative saving is much more (you can't get much more than infinite!), but the overall saving is still small - around 1% for our example. In the real world we never will get exactly 16J (all exactly on the rim) or 0J (all exactly in the axle), so our real figures will fall within this range. The important statement is that the further out, the more energy we save. Are you trolling here? The rotational energy and kinetic energy savings in a bicycle bearing are MUCH smaller than the small savings for a wheel (they are lighter and smaller, so less kinetic, rotational and potential energy in the system and thus to lose). My guess is that we might be able to lose 1g on a bearing and this will save us 0.1J in the best case for our example. I'll side with JB you'll never notice that!! This is about 0.0025% of the overall energy budget. Remember the wheels best case is about 1%. Ceramic bearing have other advantages discussed on other threads that I will not go into now... jmaccelari2010-01-06 05:42:33

Correct. Again we have hills (potential energy) and so it's outside the scope of this discussion. Why has no on asked why am I so pedantic about excluding potential energy? This is because potential energy has no dependence on any rotation of the body. It only depends on the mass and height of the body (and the gravitational acceleration of the point in the gravity field it is at at that point and place in time, but let's not go there ). So when we are talking about energy saving due to wheel rotation (weight saving on the wheels) there is no difference with respect to potential energy. It does not give a rat's bum where on the bike the weight is lost! jmaccelari2010-01-06 04:50:07

No. As soon as we are talking hills, we are talking potential energy, which is another discussion. I want to try and keep this one as simple and as to the point about wheel/frame weight loss during accelerations. This discussion is about kinetic energy only. Lighter wheels will provide an advantage over a similarly lighter frame by a goodly factor. We are talking about the RELATIVE wheel/frame energy saving - not OVERALL. Losing a couple of hundred grams on a bike is a small amount, so the overall effect is small, but cumulative (it also makes climbing easier, but again that is a potential energy problem). The results of this discussion (flat road, no wind or friction) are: - a couple of hundred grams weight loss makes a small difference to energy usage; - this energy usage is cumulative over a race and may become worthwhile worrying about (or may not depending on your opinion); - losing the weight off the wheels results in greater energy saving than off the frame (this loss is still overall small compared to our entire energy usage); - losing the weight from the rim results in a relatively MUCH larger energy saving than losing the weight nearer to the hub (but it is still small overall). (thanks SwissVan for the comments) jmaccelari2010-01-06 05:47:20

I'm posting a second response for this so we keep this separate from the energy discussion to avoid confusion. The effect we notice on the flat with less deceleration into the air due to a heavier rider is due to the higher inertia of a heavier rider. He takes more energy to get to the speed and thus has more energy to free wheel. He does not get anything for free - he pays for it in the extra energy he requires to get up to speed. This assumes the drag coeffient ("size") of the cyclist is the same in both cases. A big, fat cyclist will lose proportionally more energy to drag than a smaller. more aerodynamic one. This is the same argument as GoLefty preferring his heavier wheels. They cost more in energy to get going, but once going have more inertia - resistance to change in their speed - which means they roll more nicely. It is also used as one of the claims for 29er superiority over 26 inchers. But let's not go there... jmaccelari2010-01-06 04:03:10

No. On a flat the energy we generate is converted solely into kinetic energy. You are talking here about "recovering" kinetic energy. For potential energy we have: dE(potential) = m * g * dh where dE (delta Energy) is the energy we need to put in to climb dh metres (dh = delta height). Since dh is zero (flat) dE is zero. What you see here is the energy you put in to accelerate being used to overcome friction and air resistance. The energy is not saved, but expended in overcoming these factors. We have specifically excluded wind and friction losses in our discussion to keep to the point of weight savings on the wheel. In our ideal example you would continue coasting at the speed you accelerated to until you clapped on brakes at the end of the race (sounds lovely, doesn't it?). This is Newton's First Law of Motion you would have learned from school. To regain kinetic energy, we would require a KERS system similar to those on Formula 1 cars. As cyclists in the real worls all we can do is lose this kinetic energy to heat (moving the air, overcoming friction and braking) and then keep on putting it in so we keep on going. This is why we have to keep on pedalling... jmaccelari2010-01-06 03:51:25

I reckon there's sstill some tension there. We saw the signs last week and also had an incident with a speeding motorist in a big BMW X5 who hooted at every cyclist he saw no matter how they were riding. I'm sure he'll get over it...

You are quite correct. I misinterpreted what you said. You might find there's some aerodynamic advantage (again very small), but that's beyond the scope of this argument. So we now agree on all points except whether it's worthwhile. And that's a matter of opinion!

So you don't understand analytical integration. The formulae are simple. You don't need "thousands of data points" (only if you perform a numerical integration and the wheel can be reduced to a simple approximational model that can be analytically solved) ! You don't need it.... This is taken from the Johan Bornman book. You love to wax lyrical andf then when someone comes with facts and references you accuse them of muddying the issue. Science is science and if you can handle the figures stay out of the lab! I have acknowledged that your approach works! Read my posts . What I disagree with is your blanket statement that losing weight on the wheel is more energy effective than on the frame is an old wive's tale. You and I have both shown (independently) that it is true - even if it is a relatively small differernce, it is still true. You have avoided answering me on this issue. The argument is not about whether it saves energy (it does), but whether it is worthwhile. You say it is not. I say it adds up slightly after a while. I do state, however, that other effects may be more important than weight loss on the wheel. You are black and white, I am open to opinion. You have also avoided answering my concern that you state that it does not matter where on the wheel the weight it lost. If you understood integration, you would see this immediately and not make such statements. Weight loss towards the rim has more of an effect. (Redacted sine JB answered this in a later post) Please do not cloud the issues by rehashing arguments we have discussed and that I publically state we agree on. jmaccelari2010-01-05 21:42:37

What will happen is that with a lot more accelerations the 16 or so Joules saved will add up. The significance is how much you feel. What can be said is that if you want to save weight' date=' the best place is on your wheels (the commonly accepted old wives tale that JB rubbished then proved correct ).

jmaccelari2010-01-05 08:06:22

Just a little nit picking... How the hell did a cyclist accelerate @ 200 m.s^2 when gravitational acceleration is but 9.8 m.s^2 ???? Your calculations are based on a load of bull Mr Wise Guy! Wow! No need to get this personal! If you understand the table, it's obviously a typo! "20 0.277778" I disagree with Mr Bornman's physics, but as he's not trained for it, it's a little harsh to start slagging him off for it! His gut feel is on the right track, if not his methods and logic.

Jules has made the following comments: C: Christie pointed out in that thread that an acceleration from 30kph to 60kph is extreme and is unlikely to occur more than once or twice each race. A: This is true, but as I show, a mass change makes no difference to energy change if you are accelerating from 0 to 30 or 30 to 60. They have the same energy requirement (if we ignore air and other resistances). Air resistance does, but that is outside the scope of the question. However, if we have lots of little accelerations (10kph or so) during the race, they do start adding up to a lot of extra energy required to spin the wheels up. This energy is not recoverable as it it lost to heat (breaking) or wind resistance. I do not know of a racing bike with a KERS system! C: He is of the opinion that the rotational mass story is over-hyped. A: Agreed. It's less than a 1% difference if you lose 400g on the RIM (the best scenario). However, if you're at the level where a 1% difference makes the difference between winning or losing... C: A saving of 400g is much easier to achieve on the frame than at the perimeter of the wheel. Yep - remember the "frame" includes the cyclist! C: It's also probably worth pointing out that one doesn't accelerate much on steep hills, where weight saving is going to be most important. A: Agreed. Here we are looking at potential energy and it doesn't matter where the mass is lost from (frame, wheels, cyclist, ...). Potential energy is recoverable - as we find out on free wheeling on the downhills... C: In the absence of acceleration, the weight saving is the same, no matter where it is on the bike. A: Not well put. With no acceleration, mass has no effect on energy usage (under ideal circumstances) on a flat surface. C: I reckon aerodynamics is a far more important concern than mass at the perimeter of a wheel. A: I dare not go into aerodynamical calculations (they are horrific and usually non-linear and only solvable by numerical means), but I would tend to agree with you here. jmaccelari2010-01-04 09:19:56