Originally Posted by DannoXYZ
Heh, heh... I see Rory O'Reilly out on just about every ride I do. Interesting to see that he still holds the 500m record.
He doesn't. 25''850 ARNAUD DUBLE (FRA) 10.10.2001 LA PAZ (BOL) His time is still one of the fastest ever though. The UCI need to tidy up that page up and update it. The Aussie Teams pursuit record from nearly 18 months ago isn't on there yet! |
Originally Posted by formulaben
My 2 cents: the built up energy (spring effect) is recoiled or absorbed back into your leg (backwards from the direction of rotation...meaning its bad) when you go from the high-torque position (horizontal or 3 o'clock position) down towards the vertical dead spot. The energy that is built up in the flex as your legs reaches its maximum leverage is lost as it goes towards the dead spot. I'm only talking about the energy that is involved in flexing the frame and drivetrain.
Unless you suddenly stopped pedaling or started pedaling backwards, the torque from the spring system is transmitted back to the drive train. The only way it would be "absorbed" would if the leg was not applying enough pressure on the pedal to overcome the torque of the spring system. |
Originally Posted by hiracer
Unless you suddenly stopped pedaling or started pedaling backwards, the torque from the spring system is transmitted back to the drive train. T
Originally Posted by hiracer
The only way it would be "absorbed" would if the leg was not applying enough pressure on the pedal to overcome the torque of the spring system.
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Originally Posted by formulaben
That would be true if you kept constant torque all the way around the pedal stroke, but explain what happens at the deap spot.
Although you might be placing downward pressure, the rotational torque of the recreational rider is probably zero, or very close to it at the dead spot, and this is where the built up "spring energy" is lost. At the deadspot nothing happens because the spring's energy has already been put back into the drivetrain. It doesn't just sit around waiting for the deadspot. In fact, it the lack of constant torque that allows the spring to rebound during the second half of the power stroke when the torque is decreasing. Once pressure on the pedals starts decreasing immediately after the zenith of the power stroke, the spring's energy starts feeding back. The spring has sprung long before the deadspot. Yours is a common misconception. |
Originally Posted by hiracer
The point of my first post is that the spring system is way, way faster than the cadence of pedaling.
At the deadspot nothing happens because the spring's energy has already been put back into the drivetrain. It doesn't just sit around waiting for the deadspot. In fact, it the lack of constant torque that allows the spring to rebound during the second half of the power stroke when the torque is decreasing. Once pressure on the pedals starts decreasing immediately after the zenith of the power stroke, the spring's energy starts feeding back. The spring has sprung long before the deadspot. Yours is a common misconception. |
Just like with designing a lot of things, it's about compromises. Notice that many bikes use shaped tubings for their bikes? They're shaped for a reason. Round tubes behave similarly in all directions, but shaped tubes might be extra stiff in one direction and flexy in another. For all the parts of the bike that are involved in power transmission, you want them to flex as less as possible. For example, you don't want your BB spindle to be doing a see-saw motion when you're sprinting-->you want that displacement to go toward turning the crank (that's why more and more cranks are going with outboard bearings). Similarly for the front area (if you care about the handling of the bike). You want the bike somewhat flexy when it comes to having shock/vibration transmitted from the road to the rider. It's unlikely you'll see suspensions on road bikes, so you address this issue by trying to address the shock/vibration closes to the source (the parts connecting the wheel to the frame)-->use carbon fiber fork on the front, and things like S-bends on the seat stays, carbon seat stays, etc. So, frame flex is good or bad depending on where it is on the bike.
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Originally Posted by mayukawa
They're shaped for a reason.
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Originally Posted by formulaben
So if I were to use a set of bamboo cranks that really built up some energy, it would have no net effect on the power output to the cranks? If all this energy from flexing were not wasted, then why would all the component manufacturers be so eager to tout stiffness qualities?
Moreover, any point taken to its extreme can be perverted and thereby disproven. My remarks relate to existing, typical bicycles. * * * Ironic that you reference Keith Bontrager in your posts. He was one of the first bike personalities to question the received wisdom that frame flex is bad. His article on frame flex, that flex is not necessarily bad, is still floating around the net somewhere, last I checked. |
Originally Posted by hiracer
My remarks relate to existing, typical bicycles.
http://www.calfeedesign.com/images/bamboo.jpg Anyway, I still have yet to see anything to convince me that the energy that goes into flexing the frame/drivetrain eventually goes back into it as you continue the pedal stoke. BTW, if you're going to say "yours is a common misconception", you'd better have something better than your opinion to back it up. |
Edit. Never mind.
http://66.102.7.104/search?q=cache:D...ame+flex&hl=en Cached paged. Bontrager has apparently disowned Keith, except for marketing purposes. There are four previous pages. Anybody got a link to them? BTY, where's the bamboo crank? And am I understanding you to say that is a typical bike? I was aware of Calfee's bike when I made my earlier statement, hence my use of the term "typical." -------------->"Anyway, I still have yet to see anything to convince me that the energy that goes into flexing the frame/drivetrain eventually goes back into it as you continue the pedal stoke." <------------- Well I guess that about settles it. You told me, yessiree. I don't know. I'm pretty comfortable with Keith Bontrager in my corner. |
Originally Posted by hiracer
Yes, but is the reason marketing or engineering? In the case of ti frames, that is not an idle question.
Demonstrate it for youself using paper. Fold them into different shapes. Trying compression, tension, torsion, and bending on the differently shaped paper-folded-tubes. Different shapes will experience the different forces differently-->that's why there will always be compromises in design in terms of stiffness and compliance. |
Originally Posted by mayukawa
Demonstrate it for youself using paper. Fold them into different shapes. Trying compression, tension, torsion, and bending on the differently shaped paper-folded-tubes. Different shapes will experience the different forces differently-->that's why there will always be compromises in design in terms of stiffness and compliance.
Why are there different shaped tubes for Ti bikes but not for steel--generally speaking? (I know some steel bikes have funny shaped tubes, but mostly they don't). There is a very specific reason why steel is round and Ti many times ain't. You may not want to know the answer if you ride Ti. (Big hint: it has nothing to do with the material's characteristics.) The engineers here should get this one, easy. And builders are disqualified from participating because it's a no brainer for them. |
Originally Posted by mayukawa
Demonstrate it for youself using paper. Fold them into different shapes.
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Originally Posted by hiracer
This is a test, repeat a test. It's pass/fail.
Why are there different shaped tubes for Ti bikes but not for steel--generally speaking? (I know some steel bikes have funny shaped tubes, but mostly they don't). There is a very specific reason why steel is round and Ti many times ain't. You may not want to know the answer if you ride Ti. (Big hint: it has nothing to do with the material's characteristics.) The engineers here should get this one, easy. And builders are disqualified from participating because it's a no brainer for them. I've an engineering background...but I'm probably stupid...so please enlighten me. :D Hmmm...wonder why they make steel I-beams and not round ones? |
Originally Posted by hiracer
I don't know. I'm pretty comfortable with Keith Bontrager in my corner.
Originally Posted by Keith Bontrager?
The information that would resolve this requires measurements. The losses due to the misalignment between the rider's pedaling force and the optimal pedal trajectory can be calculated directly from the angle between them. These are likely to be on the order of a percent or two. If the short term power at the rear wheel was plotted as a function of stiffness, the optimum may not be found to be at infinite frame stiffness. If the material damping losses are small (any bets?), then a certain amount of flexibility may actually contribute to efficiency in this case.
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Originally Posted by slvoid
Kind of.. I've done hundreds of stress and fatigue tests on instron/mts electric and hydraulic testers with some real detailed inputs coming off some really expensive extensometers and the charts always show the same pattern. The elastic graph usually looks like the one I drew on a napkin. If I were to pull on a sample 3 times using load control, and the first 2 returned the sample back to 0 load in the elastic region, notice how they return at different rates. Yes it's elastic, yes it returns back to the initial condition, but the difference in path (the area is marked by x's) is where energy is absorbed from the crosshead and expelled as heat by the sample.
Roadgator pretty much nailed it. If you've ever felt the heat coming off the center of a sample under an elastic fatigue test off a 20000rpm rotating beam fatigue tester, that difference in the chart is where the heat is coming from. Not that I recommend touching a sample at 20000rpm, especially a shot peaned sample... There is friction in things that flex! Here you go again with all that "reality" stuff ! :) |
Originally Posted by slvoid
Whether energy loss is acceptable for a flexy frame depends on who you're talking to. Lance lost a 60km (or something like that) TT by a second last year, which they said aerodynamically was the equivalent of a pencil eraser sticking out of his frame. You can bet if he loses 1/100th of a watt per pedal stroke due to flex that it'll cost him more than a second over 60km. Try and tell someone in Lance's camp that the small energy loss from flex does not matter. Then run! I like my flexible Ti frame for softening the bumps to my arthritic joints on all my century rides. Different pedal strokes for different folks. |
Originally Posted by 2manybikes
:beer:
Try and tell someone in Lance's camp that the small energy loss from flex does not matter. Then run! I like my flexible Ti frame for softening the bumps to my arthritic joints on all my century rides. Different pedal strokes for different folks. |
Originally Posted by garyhm
. This is because the power required is proportional to the square of the speed. ...
I still think most of you are missing the point. A droopy frame transmits vibrations to the rider and energy is dissipated in the body of the rider. |
Originally Posted by AlanS
Er..no.. the power is proportional to the *cube* of the speed.
I still think most of you are missing the point. A droopy frame transmits vibrations to the rider and energy is dissipated in the body of the rider. |
Originally Posted by garyhm
The histeretic loss factor for steel is in the range of 0.1% to 0.8%. For a traditional steel frame, approximately 1% of total energy per pedal stroke is stored as strain energy in the frame. So at the upper end of histeretic loss for steel 0.8% of 1%, or 0.008%. In a long climb situation this would be a direct 0.008% cost in time, or about 0.3 seconds per hour. In a flat time trial where wind drag is the primary enemy the effect is not as large. We all know that the difference in effort between 19mph and 20mph is much less than 24mph and 25mph. This is because the power required is proportional to the square of the speed. Lance uses about 500W to maintain 50kph. In that case the histeretic cost is about 0.15 seconds per hour. But Lance's bike is carbon fiber which has a higher histeretic loss of around 2%. So the carbon fiber bike has to be 3 or 4 times stiffer just to have the same histeretic loss as a traditional steel bike. If Lance's bike were 10 times as stiff as a traditional steel frame then he would only have an advantage of around 0.07 seconds per hour. But of course his OCLV is not that much stiffer.
Thanks for explaining the details, it's very interesting. But I don't think I'm going to get a new frame right now so I can pick up that precious .04 seconds on my all day rides! Maybe I'll just take some lint out of my jersey pockets before I leave the house. That might be more of a difference. :) |
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