What is the truth behind bike weight? Does it really help THAT much?

Crankyfeet

I have been mentioning this point in half a dozen or so posts since a few days ago. Thanks for eventually reassuring me that I wasn't making a huge conceptual error.

However this is not a minor little error. I don't know how you were able to determine theoretically the proportion of the input torque that went into the frame, and how much went into the drivetrain (which is the efficiency in any case). You gave us a few equations then a few assumptions, then gave us this 0.05% figure. I can't for the life of me work out how that efficiency percentage would be attainable without knowing the stiffness modulus of the complete frame in infinite axes of rotation, notwithstanding the different interactions of various truss elements with differing properties. Then you would need to know the resistance of the drivetrain, to know how much is transferred to the frame. Notwithstanding that the percentage is going to increase at higher torques as well.

Let me repeat an example I used in an earlier post: if you could hypothetically be in a 300 X 10 gear on your bike, you may not be able to accelerate the bike at all at maximum torque. All the energy would be lost before it transferred to the tire/pavement interface. As you reduce the gearing ratio, gradually there is less resistance in the drivetrain, and energy is slowly transferred more and more to the wheel. It starts at 100% loss hypothetically in the gear that is impossible to move. What is the energy loss in the frame at a 53/11 gearing? Best way to find out would be experimentally. Because each frame's behavior under load is too complex to model theoretically (and one would definitely not trust the accuracy of the manufacturers' stiffness data).

Your proposed experiment, John, would not give me any confidence whatsoever in the data. The only thing worse than not having any data is having bad data and not realising it. In the test you suggest, a 1 degree vibration in the camera on its mount (like going over a pebble) will create the illusion of nearly a 1 cm deflection at the BB assuming the camera to LED light distance is 50cm. Need I say more. Garbage in = garbage out. And "Awesome" would be the last word to describe how you would feel.

Furthermore, the effect of a quadratic relationship of torque to energy loss means that there is 12.25 times the absolute loss at 1400 watts as there is at 400 watts. 12.25 times. But when a sprinter is accelerating (say from a slow start coming out of a sharp turn in a crit) the maximum torque would probably be applied at a much lower cadence than 90rpm. If you assume it was 60rpm at the beginning of acceleration, then the peak torque quotient compared to 400W (90rpm) will be

1400/400 X 90/60 = 5.25 times.

The ratio of energy loss at 1400W (60rpm) compared to 400W(90rpm) will be

5.25^2 = about 27.5 times.

That means there would be 27.5 times the amount of energy lost. And 27.5 x 400/1400 = almost 8 times the percentage loss that occurred at 400W and 90rpm. Notwithstanding that there are only a handful of people in the world who can sustain 400W at anaerobic threshold in any case.

Now we are starting to see why sprinters put a high value on stiffness. It might be that they are correct. Conventional wisdom is that stiffness is a factor. If one wants to make extraordinary claims that stiffness is irrelevant, and that Cat 1 sprinters, for example, may as well buy a cheaper "noodly" frame, then the evidence should be extraordinary, and there should be respect for accuracy in the data and the conclusions.

If people aren't at Cat 3 or above IMHO, then stiffness is probably not worth it. If you are doing 30 miles a week on bike trails to keep fit, it is irrelevant. A smooth comfortable ride is definitely the most important consideration. However, it is POSSIBLE, and actually seemingly likely to me, that stiffness could be a significant factor if you are accelerating quickly at high torque and you are competing at a high enough level. Obviously, one would need good data to back this up. And the experiment would probably need to be expensive in its set-up to control the many complexities in the system and make it precise.

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TheDarkLord

Cranky, this information (efficiency of the drivetrain) is definitely there on the internet. I don't have the time to give a link right now, but if I find something, I can post later. You can do a search too. The bottom line is that in a derailleur system, as long as all the components are well maintained, the efficiency is very high - upper 90s. Hub gear systems have lower efficiencies.

Crankyfeet

I'm not referring to the efficiency of the drivetrain. I was referring to the RESISTANCE of the drivetrain, which is dependant on the gearing that you are in. To illustrate my point - try standing up out of the saddle and applying torque on the flat when you are in a 34/25 gear ratio, then switch to a 50/11 gear and try to apply torque, and note how much more force you can apply to the pedals in high gear (but less power probably as the cadence will be S-L-O-W). Also note how much less power may be delivered to the wheel in high gear. This lesser output of power is happening because energy is stored in the system. Some of it in the frame, but also in other areas like the increased friction and the elasticity of components etc.

This is a distraction from the point which is that the frame will absorb torque (and hence energy - which is likely lost) when it can't be absorbed or released easily in the drivetrain. These conditions occur in high gears, low cadences, when you are applying maximum thrust (and maximum torque).

The efficiency of the drivetrain is not the object of the frame stiffness test (we are trying to isolate stiffness effects). In the experiment, you would have to measure frame deflections and know the stiffness moduli for various axes of deformation which is very complicated. If you just compared input power to output power, then you would have to take out all the other inefficiencies like drivetrain friction and crank arm, chain and wheel deformations etc.

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TheDarkLord

And what exactly is the difference between resistance and efficiency??? Efficiency measures how much of the input power is transferred to the wheels. If there is high resistance, then the efficiency has to be low, since that implies that much of the input power goes to overcoming resistance, and that is NOT transferred to the wheels. And I think you are wrong when you say that the resistance is higher for a 53 x 11 gear as opposed to 34 x 25 gear. In both cases, I think most of the input power goes to the drivetrain. However, what differs is that the cadence of the first for any given speed is much lower than that for the second gear, and this translates to greater discomfort to the body/stress to muscles and knee. It has nothing to do with resistance, and what you "feel" is just an illusion due to the strain to your body. If you look at the actual power input, I'll bet that they are the same (or similar). As I mentioned, people have actually measured efficiencies, and there are articles in the web.

Crankyfeet

You're talking about the efficiency of the drivetrain. I'm talking about the resistance felt by the geared up affects of the road transferred through the high ratioed (gearing)drivetrain to the pedal (or vis versa). Here is a simple illustration to counter your scenario of powers being the same but "appearing" psychologically to be different. Forces (torques) might be the same, but the power is dependent on the cadence that the system allows.

SCENARIO: If the power output was the same in the highest or lowest gear, then why if you stand up on the pedals from a standing start in 50/11 gear do you accelerate at a lesser speed than if you stand up on the pedals (applying the same torque) in a low gear (say 34/25). The effect of the lower pedal cadence should be cancelled by the gearing in your example. But the power delivered to the wheel (what accelerates the bike) is higher in the lower gear because you will accelerate faster in your 34/25 gear. Now once the momentum of the bike increases, then the optimum gear starts to go up.

Maybe my choice of word in using "resistance" is the problem causing confusion. Maybe I should qualify it with the help of dhk2 below to "resistance due to geared inertia effects"

I am using the case of a very high gear starting from a standing start to hypothetically illustrate a high resistance (through gearing) which will channel the input torque (by your legs to the pedals) to the other components such as the frame (and not to the tire/road in the form of forward motion).

And BTW what is your point? You are debating an illustration I was using to show where energy is mostly transferred to the frame because the resistance inertia effects allow one to apply more torque to the pedal. It doesn't change the math one bit.. and the point that energy losses are a much more significant factor at higher torques.

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dhk2

Cranky, agree your discussion about "resistance" of the drivetrain is a distraction. The resistance you feel at low cadence in a big gear is the inertia of the rider/bike mass, not anything magical in the drivetrain. The reason the smaller gear feels easier is of course because it provides for increased force to the tire/pavement for any given amount of torque that the rider generates.

Same principal applies of course to a manual shift car: low gears reduce the engine torque required to move the vehicle, or enable the same torque to apply more force to the pavement and accelerate the car faster.....what a deal

As Darklord says, efficiency of the drivetrain refers to power transmitted to the tire or pavement vs. that input to the cranks. I've seen figures quoted of 95-98% as "typical" for drivetrain efficiency. Believe this value is based solely on the frictional losses in the chain and cog drive system.

Crankyfeet

Oh and BTW: one difference between "resistance" and "efficiency" is that resistance has units (force delivered through to the chain) and efficiency doesn't have any units (it is expressed as a percentage). I can't imagine "resistance" expressed as a percentage.

And also BTW: I appreciate the questioning and debate. It helps to get to the truth and reduces the chance you are making an error in your judgment and/or calcs. Its hard to keep your ego out of it of course, but it is the basis of science, and if you make errors, you have to learn to accept it. I envy you, TheDarkLord, that you have made it your career.

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TheDarkLord

Ok, I see what you mean.

First point about why we stand up at harder gear: the reason has to do with the torque applied to the rear wheel (which is not the torque applied to the pedals). What is torque? It is the cross product of the radius vector and the force vector. Assuming that the two vectors are perpendicular, it becomes a simple multiplication. Let us assume that this is the case (it won't be the case if there is significant cross-chaining - i.e. low toothed gear at front and back). Ok. Now, when you are in a hard gear in the rear, the radius from the wheel center to the point where you are applying the torque through the chain is small. Consequently, the force on the chain required to provide the same torque to the rear wheel has to be accordingly larger. To apply this larger force (which translates to larger force on the pedal), you have to stand up. Note that this torque matters most when you are accelerating. When you are actually moving at constant speed, you have to supply a much smaller torque to counter the deceleration due to air drag, rolling friction, etc. Hence, the hard gear does not matter so much when you are moving at constant speed.

OK. Now let us consider the frame. Depending on the frame properties, a certain fraction of the force applied to the pedal goes into flexing the frame. At very high cadence, again one can imagine that there is power lost to the frame at the bottom of the pedal stroke. So, yes, there will be greater amount of power lost to the frame at higher power. But what about the fraction of power transferred to the frame? Is there a linear relation between the input power and the power lost to frame? I don't know the answer to that.

TheDarkLord

Yeah, the quantities are different physically. I was just saying that one is related to the other. High resistance ==> low efficiency. Of course, with respect to our discussion here, I misunderstood what you were trying to convey. Efficiency in drive train is just the fraction of the torque applied at the chain-rings that is transferred to the rear cogs, and I agree that it is somewhat irrelevant to our discussion here. While you were referring to something different; I wouldn't call it resistance.

Crankyfeet

Maybe "resisting force"? Would that cause less confusion? I'm not aware of some of the standard terminology so I might be confusiing those who are.

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Crankyfeet

The energy lost is proportional to the torque squared (T^2). Power is a function of the torque and the cadence, and is not directly proportional to the energy loss. If you have the same torque in two examples but in the second example the cadence is doubled, then the power in the second case will be doubled. The energy loss in both cases should be the same however due to the torques being the same.

It should be noted that the torque applied to the BB that propels the bike forward has a moment arm which is the crank arm length. The torque that causes lateral torsion however has a moment arm which is the lateral distance of the centre of the pedal to the center of the bike. In reality though, the torsion axis changes as the position of the pedal changes relative to the BB. The actual axis of torsion at peak torque is perpendicular to the line from the pedal at 3 o'clock on the stroke and the centre of the BB. Which just makes the calculations harder because you need to find the spring constant/stiffness modulus of the frame in this axis.

Best way to empirically find the torsional stiffness constant/modulus would probably be to apply incremental forces to the pedals in a locked state at 3 o'clock and measure the angular deflection. From there you can determine the spring constant (k). Then you can unlock the pedals and apply forces to them, this time in normal bicycle motion with the drivetrain and wheels on normal or simulated pavement. The deflections in the frame will allow you to calculate the power absorption/loss.

All these explanations are hard without the use of diagrams.

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TheDarkLord

Not resisting force either. There is no resistance. I explained the physics in my post. Here is an analogy. Consider opening a door in two ways. In the first way, you push the door near the knob, which is located near the door's edge away from the hinges. In the second method, you push the door near the location of the hinges. You will need a much larger force in the second method, because the torque is much smaller for the same amount of force applied. So, would you say that the door has more "resisting force" when pushed near the hinges?

So, frame flex is proportional to the torque applied to the frame, which is proportional to the force on the pedal. The force on the pedal relates to the torque on the rear wheel through the gear ratio, and the radius of the cogs. If you have a hard gear, you need a larger force to apply the same torque to the rear wheel, and this has nothing to do with any resistance. The side effect is that this causes a larger torque on the frame itself.

Crankyfeet

The torque to move the door is no different but the FORCE needed by your hand does change. The RESISTING FORCE I was defining as the force needed in the chain (ie applied by the pedal/crank torque) to propel the wheel. If you change the moment arm (decrease it by using a smaller geared ring on the rear cassette), then you have to increase the force in the chain. This force in the chain is the resisting force I am describing that your pedal force has to counteract.

This debate over the the definition of my use of the word "resistance" is banal and pedantic and tangential to the main point. Perhaps if you are using an electric circuit analogy, this is where the confusion is coming from on your definition of "resistance". I have explained the definition of what I meant by "resistance" or "resisting force". If you want to infer that I meant "torque applied to the wheel", then I am sorry to have mislead you. It really doesn't matter how you want to define the term "resistance"... does it? It is not a standard term in mechanical engineering AFAIK. I have defined what I meant by the term. I was just using it instead of the phrase "the counteractive force in the chain related to the moment arm applied to the wheel to overcome the inertia of the bike and cause acceleration". It is not a constant property of the bike in my definition. It is related to the gear you select. It is not a torque or moment or an inertia. In my usage of the term it was meant to describe a force.

I apologize to you and anyone else who was confused by my choice of the word "resistance" as a term to describe a force. If you're still in a quandry... read dhk2's post above. He gets it.

[Post edit - And I noticed I have talked about "torque" a lot in my previous posts before this one. In nearly all cases I was referring to the torque generated by the force applied to the pedal multiplied by the crank arm length. I am not referring to the torque applied to the wheel by the chain and the rear wheel gear selected.]

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Bigbananabike

...and Rassmussen (too many Ss?) wanted to take the stickers off his bike to save weight