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Hefty parts? Full Dura-Ace? That frame/fork combo is very light in the broad scheme of things. O.K. maybe those wheels are somewhat heavy.IUbike said:Thor is a monster, so he needs the heafty parts. As for the Colnago, their lightest bike comes in at around 1200 grams for a frame... not that light when you can get a Scott, Parlee, etc under 900 grams.

K

The Colnago is claimed to be under 1100gr. Very respectable. There aren't too many frames over say 56cm that weigh under that(actual weight), let alone 900gr.

I guess that was my point-a Look 585 with Dura-Ace is "hefty"? We get our senses dulled by all the lightweight bling on this forum. I enjoy all the lightweight builds on here, including yours, but I guess I have to draw the line at calling a $6000-$8000 17lb bike "heavy" when it's basically being compared to a 13lbs Scott or a 14lbs Merlin with carbon tubulars and lightweight bars&stems&saddles that most people wouldn't ride.

Don't get me wrong, my lightweight bling for my Colnago are Hyperons and a Selle Italia C64 and Syntace F119 stem. I ain't riding that saddle on a century and if you think I would bomb down descents on Hyperons like I do with my Fulcrum's, you've got another thing coming.

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Fignon's Barber said:

It would be easier to go up the hill with the 14" lb bike, but I'd like the 18" lb bike for going downhills.

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Most pro's want a bike that is reliable to the point of being bulletproof. Plus at this point in the season many riders are riding in shitty weather and they are getting used to new kit and tweaking their set up. Later on they'll get the trick kit out.

If you look at both the CA & Milram bikes, they are fitted with clinchers not tubs. Not sure about Milram as Panaracer don't do tubs, but CA will more than likely be racing on tubs from Continental. So expect at least a couple of hundred grammes to be lost.

Then there's also sponsor limitations. PRO bars & stems aren't light by any stretch of the imagination.

as for these frames, it is possible there is extra carbon in the lay-up, esp. for the down tube.

this being said, i need to go now and swap my perfectly fine skewers with some salsa ti's, color coordinated of course...

2wheelsport said:

http://www.cyclingnews.com/tech/2006/probikes/?id=milram_colnago_becke

http://www.cyclingnews.com/tech/2006/probikes/?id=credit_agricole_look_hushovd

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Fignon's Barber said:

While I agree with the first part of your statement regarding the overstated importance of bike weight I think the theory about road contact is is dubious at best. If these bikes had nobody on them I'd say yes but we know that this isn't the case in the real world. Though I bet my bike could beat your bike in a riderless race any day.

Furthermore, the importance of frame weight and wheel weight are generally of equal importance unless we're talking about sprinting. The whole unit or system weight comprised of the rider, frame, components, wheels, tires and etc. is what's important in regards to your theory and your opinion.

...man you're triggering deja vu. Didn't you have this sort of agrument with someone else here recently?

I know I'm new here and don't want to seem like a weirdo, but wheel weight matters wayyy more than body weight or frame weight.rocco said:Furthermore, the importance of frame weight and wheel weight are generally of equal importance unless we're talking about sprinting. The whole unit or system weight comprised of the rider, frame, components, wheels, tires and etc. is what's important in regards to your theory and your opinion.

Here's the Purdue Mechanical Engineering sophomore's approach:

Angular momentum; if your body moves 10 feet forward it moved 10 feet, if your wheel moves 10 feet forward, a point that starts on the bottom first moves back, then up, then forward, then down, then sits still, then moves back again, 1.43 revolutions for 10 feet of travel of a 700c wheel with 23c tires.

Now think of each piece of rubber, the tire is 10 feet in front of where it was, so it moved 10 feet, but it also moved all of that around, which meant it changed direction, which means it accelerated, which means force was applied to it.

Calculations / long answer:

A bike wheel sliding on its side not rotating (say on ice or something) has kenetic energy, in metric, .5mv^2 joules where m is mass in kilograms and v is velocity in meters per second. So if that wheel was sliding 20mph, or 8.9 meters per second, and it weighed 1500 grams (1.5 kg) it would have .5*1.5*8.9^2 or 59.4 joules of kenetic energy.

Now take the same 1500g wheel, put it on a bike rolling 20mph, it is still moving forward 20mph, so there is still 59.4 joules of "translational" kenetic energy, but now the wheel has "rotational" kenetic energy as well. Now assume a wheel has all of its weight at one radius, almost at the tires, but a little inside since there is a hub and spokes, say the average is 10" from the center, or .254 meters. Rotational kenetic energy is .5Iw^2 where I depends on the shape, and is mr^2 for a hoop, and w is the rotational velocity in radians per second where one radian would be one radius worth of tire, or .3397m. So substituting this secodn equation for I into the first equation for rotational kenetic energy we have RKE = .5(1.5*.3397^2)26.2^2 = 59.4 joules of ratational kenetic energy.

Now add the ratational kenetic energy to the translational, and your total kenetic energy in the wheel is 118.8 joules, or double that of the wheel just slipping along not spinning.

The double factor just worked out for some weird reason with my estimate of the average place of mass of the wheel, if It is further outward than 10" from the center there would actually be more than double the energy in a spinning wheel compared to a sitting wheel.

Spend twice as much for a lighter tire than you would for a lighter handlebar. This also applies to anything else that spins like cranks or shoes or pedals, but not as much since they do not spin as fast (in the equations the rotating speed gets squared, as well as the radius, so a crank that is spinning slower around a smaller circle would have less rotational kenetic energy than a wheel on the same moving bike that weighed the same as the crank)

The wheel didn't have a force applied to it's angular component if it's spinning at a constant angular velocity.mikbowyer said:Quick answer:

Angular momentum; if your body moves 10 feet forward it moved 10 feet, if your wheel moves 10 feet forward, a point that starts on the bottom first moves back, then up, then forward, then down, then sits still, then moves back again, 1.43 revolutions for 10 feet of travel of a 700c wheel with 23c tires.

Now think of each piece of rubber, the tire is 10 feet in front of where it was, so it moved 10 feet, but it also moved all of that around, which meant it changed direction, which means it accelerated, which means force was applied to it.

Remember rotational kinetic energy is KE=(1/2)*I*omega^2. Work done is d(KE)/dt=d((1/2)I*omega^2)/dt=I*omega=torque. By definition, then, if omega remains constant, then torque=0. Moreover, if d(omega)/dt is small--as it is on a bike--then the torque will be small.

Your conclusions are erroneous because all you've looked at is energy. What is important to the cyclist is dE/dt=work. Over flat ground dE is mainly determined by losses due to drag, rolling resistance, and other losses in the drivetrain. Going down hill, dE/dT can be considered the same. Going up hill dE/dT has two components: the first is d(PE)/dt which is solely dependent on the mass of the system and dH/dt, where dH is change in height from t(0) to t (For simplicity, I've not included drag, rr, or other drivetrain losses here). For an 80kg bike/rider system, moving at roughly 16 mph up a 6% grade (approx. 7m/s) that gives a dH/dt of 0.06m/s. Over ten seconds d(PE)/dt=PE(final)-PE(initial)=mg(h(f)-h(0))=80kg*(9.8m/s^2)*(10s*0.06m/s)=470.4 J of work done. Now look at torque. First, we'll say that r(I)=622mm=0.622m since this is the diameter at the bead seat. Also for both rims we'll remove the weight of the hub since this is at the center of the wheel system and it's moment of inertia is negligible. So for that wheel system of 1500kg we'll subtract 300g for the hubs (100f, 200r....we'll assume the rear hub is really light) so the wheel system now has m(I)=1200. As stated earlier, d(RKE)/dT=torque="angular work"=I*omega. For our wheel system we then have T=2*I(rim,spokes)*omega=2*m*R^2*omega=2*m*R^2*delta omega. Note here that the factor of 2 can be dropped if the rims are identical....i.e., you can just use m=1200g and the same radius, R, as well as the same omega. For the wheel system to do the same work as the rider does in changing elevation (i.e., 470.4 J), delta omega would have to be 470.4/(1.2Kg*0.622m^2)=delta omega=1013 s^(-1). That's a HUGE change in angular velocity, changing the angular velocity of your wheels by 1013 revolutions per second over the course of 10 seconds. You can safely assume that the actual change in angular velocity per second would be 0.01 of that ( a decent approximation). By that, then, the contribution to total work done by changes in angular velocity, thus torque, are very small compared to changes in energy due to changes in elevation (i.e., positional energy or potential energy.).mikbowyer said:Calculations / long answer:

A bike wheel sliding on its side not rotating (say on ice or something) has kenetic energy, in metric, .5mv^2 joules where m is mass in kilograms and v is velocity in meters per second. So if that wheel was sliding 20mph, or 8.9 meters per second, and it weighed 1500 grams (1.5 kg) it would have .5*1.5*8.9^2 or 59.4 joules of kenetic energy.

Now take the same 1500g wheel, put it on a bike rolling 20mph, it is still moving forward 20mph, so there is still 59.4 joules of "translational" kenetic energy, but now the wheel has "rotational" kenetic energy as well. Now assume a wheel has all of its weight at one radius, almost at the tires, but a little inside since there is a hub and spokes, say the average is 10" from the center, or .254 meters. Rotational kenetic energy is .5Iw^2 where I depends on the shape, and is mr^2 for a hoop, and w is the rotational velocity in radians per second where one radian would be one radius worth of tire, or .3397m. So substituting this secodn equation for I into the first equation for rotational kenetic energy we have RKE = .5(1.5*.3397^2)26.2^2 = 59.4 joules of ratational kenetic energy.

Now add the ratational kenetic energy to the translational, and your total kenetic energy in the wheel is 118.8 joules, or double that of the wheel just slipping along not spinning.

The double factor just worked out for some weird reason with my estimate of the average place of mass of the wheel, if It is further outward than 10" from the center there would actually be more than double the energy in a spinning wheel compared to a sitting wheel.

Conclusions:

Spend twice as much for a lighter tire than you would for a lighter handlebar. This also applies to anything else that spins like cranks or shoes or pedals, but not as much since they do not spin as fast (in the equations the rotating speed gets squared, as well as the radius, so a crank that is spinning slower around a smaller circle would have less rotational kenetic energy than a wheel on the same moving bike that weighed the same as the crank)

btw if anyone knows of an internship at a bike company this summer, e-mail me.

You can't look at energy changes in isolation and get something meaningful. You have to look at energy, and the change in energy, i.e. the Hamiltonian for the system. Only placing one constraining equation on your system, in your case, energy is an error because the change in energy is the other important consitutive equation.

I did my analysis in a way that an avg reader can understand, I've sat through thousands of hours of statics, dynamics, physics, and thermodynamics lectures and I barely comprehend what you are saying, not necessarily because it is over my head, but more because you are trying to throw it over my head.alienator said:You've done your analysis wrong.

You can't look at energy changes in isolation and get something meaningful. You have to look at energy, and the change in energy, i.e. the Hamiltonian for the system. Only placing one constraining equation on your system, in your case, energy is an error because the change in energy is the other important consitutive equation.

What a bike rider would care about would be force*distance=work. I'll give you that, but I was pointing out to someone that a pound heavier wheelset would matter more than a pound heavier saddle. This is inarguable, and the orig. poster I replied to didn't seem to understand the concept.

Where weight matters has absolutely nothing to do with air friction, tire drag, or many other things you posted. Those may be very important factors in cycling dynamics calculations, but that was not what I was doing.

So the energy in a spinning bike wheel is still a very significant amount larger than the energy in something on the bike / rider system not spinning. And the only way to put energy into a bike is to crank on the pedals (unless you like jump out of a plane or go down a hill, but that is not what we are talking about, so we're not talking about it).

Reread the original post I replied to and quoted, possibly the length of my writeup, and the length of your reply got you off on a tangent that, though true, had nothing to do with what I was talking about.

I'm not throwing anything over your head. I was just giving you simple physics. Energy in and of itself in this model, means very little. What matter to a cyclist is the change in energy, at any given time, because that represents work that a cyclist has done or has to do. Significant contribution to roational energy by a cyclist only comes when accelerating hard from a stop and sprinting. In every other situation, changes to rotational energy are relatively small. Go here, and see for yourself. However if you've spent "thousands of hours" in lectures, then you should easily be able to quantitatively look at the appropritate differentials to see how work varies in the system. Better yet, take the problem to a physics or dynamics prof and let them tell you how little the contribution of changes in rotational energy are in the total work done by a cyclist. Moreover, as is well accepted as a result of scientific analysis that rotational weight isn't the silver bullet in a cyclist's life, the change in weight of a wheelset from say your 1500g wheelset to a 1100g wheelset is only a change of 27% so the already small contribution of work from changing rotational energy is going to be reduced by 27%. So if before the work done by accelerating the wheels was 0.01 of the work done changing the positional energy of the bike, the change in wheel weight will only save you 0.0073 times the work due to changes in positional energy. That's right: a whopping 0.27% reduction in work by going to a wheel 400g lighter.mikbowyer said:I did my analysis in a way that an avg reader can understand, I've sat through thousands of hours of statics, dynamics, physics, and thermodynamics lectures and I barely comprehend what you are saying, not necessarily because it is over my head, but more because you are trying to throw it over my head.

What a bike rider would care about would be force*distance=work. I'll give you that, but I was pointing out to someone that a pound heavier wheelset would matter more than a pound heavier saddle. This is inarguable, and the orig. poster I replied to didn't seem to understand the concept.

Where weight matters has absolutely nothing to do with air friction, tire drag, or many other things you posted. Those may be very important factors in cycling dynamics calculations, but that was not what I was doing.

So the energy in a spinning bike wheel is still a very significant amount larger than the energy in something on the bike / rider system not spinning. And the only way to put energy into a bike is to crank on the pedals (unless you like jump out of a plane or go down a hill, but that is not what we are talking about, so we're not talking about it).

Reread the original post I replied to and quoted, possibly the length of my writeup, and the length of your reply got you off on a tangent that, though true, had nothing to do with what I was talking about.

There's a reason they don't give sophmore's a degree: they need to have more classes to solidify and complete their knowledge.

Actually after this semester I am done with physics and mechanics, so i'm about 8 weeks out. The second two years are more business and stuff.alienator said:I'm not throwing anything over your head. I was just giving you simple physics.

.........

That's right: a whopping 0.27% reduction in work by going to a wheel 400g lighter.

There's a reason they don't give sophmore's a degree: they need to have more classes to solidify and complete their knowledge.

And you are STILL saying things that are true, but have nothing to do with what I originally posted, and it is very obvious that you have NO interest in reading what I was replying to. The post I quoted said "Furthermore, the importance of frame weight and wheel weight are generally of equal importance unless we're talking about sprinting. The whole unit or system weight comprised of the rider, frame, components, wheels, tires and etc. is what's important in regards to your theory and your opinion."

and i said "well actually taking 10 grams out of your wheels is like taking 20 grams out of your frame"

and you keep establishing the points that

1: I have less credibility because of my engineering knowledge

-note how you have said that i have less, and not that you have any, which is a total e-lowballer way of arguing.

2: That 10 or 20 grams out of a bike don't mean crap anyway.

Congratulations you have just stated something completely new and not contradictory to what I said. When I was 9 I drank chocolate milk, but I am not posting about it since it doesn't have anything to do with the person I was quoting. What you are saying is as true as me drinking milk when I was 9, but it belongs as a reply about as much.

I am in no way trying to say anything you have said was wrong, but please do me a favor and just read what I was replying to, it has nothing to do with the end magnitude of these changes, only them related to eachother.

Only them related to eachother.

A change in a wheel weight is more important than a change in handlebar weight. Until you can say that they are of the same importance, or that a change in handlebar weight matters more, please quit trying to discredit me and e-overspeak what you are trying to say by adding credibility to your argument with overcomplicating the subject.

The ref. to a degree and more years was done with the implication that you're not done yet. I don't know exactly what kind of program your in. I assumed the typical 4-5 year engineering track. Also, while a bike is simple, the analysis of the system and how it behaves is not quite as simple, and I thought it was possible that you hadn't covered the analysis of such systems yet. But now that you've pointed out the thousands of hours you've spent in the classroom, it's evident that you're completely adept at all types of analysis. FWIW, from the earlier post the Hamiltonian is just the sum of the kinetic and the potential energy in a system. The Hamiltonian is a powerful espression that is of great importance in Lagrangian mechanics, where instead of looking at forces as in Newtonian mechanics, systems are evaluated by how the Hamiltonian changes, i.e, by looking at the Hamiltonian and its derivatives, in terms of the system's degrees of freedom. Such an analysis is mathematically identical to Newtonian analysis. However, as I said before, you cannot just understand a system by looking at energies. As in Langrangian mechanics, you have to also look at how the energies change.mikbowyer said:Actually after this semester I am done with physics and mechanics, so i'm about 8 weeks out. The second two years are more business and stuff.

No. Only in terms of total energy is what you claim true. However total energy is not the only constraint on the system. What the rider feels is the effect of a CHANGE in energy or how his POWER dissipates. And to see how those things are effected total energy is not the sole consideration: you have to define how the energy changes, so you can get work and power. Work and power are what the rider understands. An energy magnitude, say your 59.4 J of rotational kinetic energy, mean nothing to a rider. And in terms of the way the system functions, it's completely meaningless because what is more important is how you get to 59.4J or how it changes with changing dynamics in the system.mikbowyer said:And you are STILL saying things that are true, but have nothing to do with what I originally posted, and it is very obvious that you have NO interest in reading what I was replying to. The post I quoted said "Furthermore, the importance of frame weight and wheel weight are generally of equal importance unless we're talking about sprinting. The whole unit or system weight comprised of the rider, frame, components, wheels, tires and etc. is what's important in regards to your theory and your opinion."

and i said "well actually taking 10 grams out of your wheels is like taking 20 grams out of your frame"

See the above comment before you get all huffy. And the fact is that 10 or 20 or 200g of weight change in a bike don't mean diddly in terms of performance.mikbowyer said:and you keep establishing the points that

1: I have less credibility because of my engineering knowledge

-note how you have said that i have less, and not that you have any, which is a total e-lowballer way of arguing.

2: That 10 or 20 grams out of a bike don't mean crap anyway.

Nope. You didn't frame your solution correctly. In fact, you said zero re: changes in energy. You just gave energy magnitudes.mikbowyer said:Congratulations you have just stated something completely new and not contradictory to what I said. When I was 9 I drank chocolate milk, but I am not posting about it since it doesn't have anything to do with the person I was quoting. What you are saying is as true as me drinking milk when I was 9, but it belongs as a reply about as much.

I read what you said. In no way did you compare the work (dE/dt) or power (dW/dt).mikbowyer said:I am in no way trying to say anything you have said was wrong, but please do me a favor and just read what I was replying to, it has nothing to do with the end magnitude of these changes, only them related to eachother.

What you pointed out, that a pound heavier wheel would be more detrimental than a pound heavier seat, is an oversimplification. The fact is that even the pound heavier wheel won't make a great difference. Granted, again, you have to initially put energy into the wheel, but, hey, you've got to put energy into the system anyway. Other inputs to the system are very small to those large inputs over very short intervals (sprints). Even in those cases, the effect of a heavier wheel isn't that great. Energy, both translational and rotational are linear with respect to mass. So any differences in energy are going to be less than an order of magnitude. If you do a detailed analysis you'll find that over a long climb, any time saved via lighter wheels is minimal and would probably only be beneficial to cyclists at the top of the sport. In fact, I'd say wheel weight even does didly for the pros as they get a much greater effect from riding along w/ the competition.

I am off on no tangent. I am saying that any benefit from lowering rotational mass is very minimal, minimal to the point of prolly not being noticeable. My guess is that someone would never notice a drop in rotational energy of 0.27%. For your rotational energy of 59.4J (and we'll assume, for now, that the relative change is still 0.27%. It's a close enough first order approximation), that equals a change of 0.16J. With that loss of rotational energy...if you could put it into a change in potential energy that you would allow you to elevate the system a few more millimeters.

Now you can definitely make the argument that over a long day of climbing that a change in wheel weight will let you have more power in reserve at the end of the day. But guess what? You'll use less power over the course of that day with a lighter seat and lighter handlebars.

The ol' adage that a "pound off the wheels is like 2 off the bike" is pure bupkus. And the reason it is bupkus is because the EFFECT of the loss of weight at the wheels is virtually negligible compared to a loss of overall system weight or loss of weight in non-rotating parts. It's mathematically and physically true that having twice the energy as another component does not mean that the energy changes are twice as large. It is mathematically evident, in the bicycle system, that the effect of changing rotational mass is very small.

For people not interested in the science, math, or whatever, I suggest they go to AnalyticCycling.com and plug some numbers in just to see how small changes in power and work are as a result of rotational mass changes.

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