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



Crankyfeet

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TheDarkLord said:
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.
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.
 

TheDarkLord

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Crankyfeet said:
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.
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

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TheDarkLord said:
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.
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.]
 

Bigbananabike

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




Do the arithmetic. Saving 2 pounds on a 200 pound package, rider and bike, is 1%. Not significant, not meaningful, not really anything of note. WAY to much emphasis on the bike and trying to lose grams, not pounds. 4 things make a SIGNIFICANT difference in cycling performance.
-Fit-does your bike fit ya
-Fitness-are you fit?
-Fat-lackthereof on you where you may be able to lose a SBW(standard bike weight-20 pounds)
-Finesse-riding and racing smart

Not the frame, not the wheels, not the machine really at all. Whether the bike is 18 pounds or 16 pounds, the bike only has a significant impact on riding performance is if something breaks, then it slows your ride down a lot.

But will somebody spec 32h wheels, even if they are .1 offa ton++? Just to save those 28 grams, yep. Will somebody buy a chain with holes in the plates and pins to save grams, yep.....so it goes, the daffy bike market.

Latest 'craze' is ceramic balls. Most engineers that have tested this say 1-2% more power..if it's worth the $400 or so for ceramics all the way thru-go ahead. I don't think so.
 

Bigbananabike

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There must be SOME difference in speed between a heavier bike and a lighter ('faster wheel') bike.
For training I ride my steel framed single speed or my alloy Felt F80 with R500 wheels. My normal hilly training rides give me an average speed(or lack of speed !:eek: ) of 25kph.
Racing - I use my race only bike (Fausto Coppi alloy with Ksyriums and 20mm tyres, not a lot lighter than the Felt). A race example(sure I spent most of the time in the middle of the bunch) of my speed is 36 kph for 48 kms. A race the other week 72kms (a lot of it by myself) at average 31 kph.
There must be a bike component in there somewhere.




sogood said:
Problem with light and stiff bikes is that the feel disappears quickly as you get used to it. Then it's that old slog up the hill. You need to have a heavier and mushier training bike to remind you the difference.
 

TheDarkLord

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Bigbananabike said:
There must be SOME difference in speed between a heavier bike and a lighter ('faster wheel') bike.
For training I ride my steel framed single speed or my alloy Felt F80 with R500 wheels. My normal hilly training rides give me an average speed(or lack of speed !:eek: ) of 25kph.
Racing - I use my race only bike (Fausto Coppi alloy with Ksyriums and 20mm tyres, not a lot lighter than the Felt). A race example(sure I spent most of the time in the middle of the bunch) of my speed is 36 kph for 48 kms. A race the other week 72kms (a lot of it by myself) at average 31 kph.
There must be a bike component in there somewhere.
Do you train alone? The big difference between riding alone vs riding in a group/race is the drafting that reduces the air drag significantly. About the second race you mentioned above - maybe the extra adrenalin in the race accounted for the difference? And yes, there will be SOME difference between a heavy and light bike. But it will be very small, and probably unnoticeable, especially if your fitness is improving since the latter will swamp the former.
 

Bigbananabike

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Hi.
Yep, you're right - I do train alone(mainly) so its just me plodding into the wind....:eek:
I knew that'd be the biggest variable + riding with others inspires one to go faster (even if they're behind).
My fitness would be the same - even in our race season my training pace doesn't get any faster but I get faster racing.
Strange huh?!:)


TheDarkLord said:
Do you train alone? The big difference between riding alone vs riding in a group/race is the drafting that reduces the air drag significantly. About the second race you mentioned above - maybe the extra adrenalin in the race accounted for the difference? And yes, there will be SOME difference between a heavy and light bike. But it will be very small, and probably unnoticeable, especially if your fitness is improving since the latter will swamp the former.
 

artemidorus

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Bigbananabike said:
There must be SOME difference in speed between a heavier bike and a lighter ('faster wheel') bike.
Of course there is - on a steep ascent, the difference in speeds is directly proportional to the weight difference of the total system. If I trimmed my 8.7kg bike to the UCI limit (and I have no intention of doing that!) I'd be about 2% faster on a steep climb. On a 20min climb, this would be 24 seconds - a pretty fair margin. Even if you just want to hand your mates a caning, this'll help.
On the flat, there is no difference in steady speed. On a descent, a light bike is a slight disadvantage.
 

Tapeworm

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artemidorus said:
Of course there is - on a steep ascent, the difference in speeds is directly proportional to the weight difference of the total system. If I trimmed my 8.7kg bike to the UCI limit (and I have no intention of doing that!) I'd be about 2% faster on a steep climb. On a 20min climb, this would be 24 seconds - a pretty fair margin. Even if you just want to hand your mates a caning, this'll help.
On the flat, there is no difference in steady speed. On a descent, a light bike is a slight disadvantage.


Sorry, just curious, why is the light bike a disadvantage on the decent?
 

TheDarkLord

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sogood said:
So? How does that translate to a disadvantage for light-weight bike? Considering just the weight, the accelerating force when descending = m*g*gradient where m is the mass, g is the acceleration due to gravity, and gradient is the gradient (well, actually gradient is the tangent of the slope angle, while the actual quantity that counts is the sine of the angle, but for gradients of the order of 10%, the difference is negligible). So, just by gravity, there should not be any dependence on mass of the bike.
 

artemidorus

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Tapeworm said:
Sorry, just curious, why is the light bike a disadvantage on the decent?
On a descent, the major retarding force is air drag, which is independent of weight. The major propelling force is gravity, which is proportional to the total system mass. The more the system mass, the higher the propelling force, and drag doesn't change.
 

TheDarkLord

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artemidorus said:
On a descent, the major retarding force is air drag, which is independent of weight. The major propelling force is gravity, which is proportional to the total system mass. The more the system mass, the higher the propelling force, and drag doesn't change.
Artemidorus, propelling force is larger for a heavier bike, but the acceleration is force divided by mass, and so it is independent of bike mass. If you drop two weights, one light and the other heavy, both will reach the ground at the same time if the air drag is the same on both. So, if the air drag is the same for both the light bike and heavier bike, I don't see how a light bike has a disadvantage.
 

artemidorus

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TheDarkLord said:
If you drop two weights, one light and the other heavy, both will reach the ground at the same time if the air drag is the same on both.
Wrong. They will hit the ground at the same time if there is no air drag, or if the air drag is the same proportion of their masses. If they have the same absolute drag, then the heavier mass will hit the ground first.
 

sogood

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Actually, mass independence is only true in a vacuum ie. In a system without resistance. For a bike with a cyclist on top, aero drag and system friction will all retard the rolling bike.
 

TheDarkLord

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artemidorus said:
Wrong. They will hit the ground at the same time if there is no air drag, or if the air drag is the same proportion of their masses. If they have the same absolute drag, then the heavier mass will hit the ground first.
Ok, I get it. The drag force is independent of mass, and hence, the drag per unit mass is less for the heavier bike. Is there any estimate on what the actual gain is?
 

artemidorus

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TheDarkLord said:
Ok, I get it. The drag force is independent of mass, and hence, the drag per unit mass is less for the heavier bike. Is there any estimate on what the actual gain is?
Anecdotally, at 90kg I can roll past most people if none of us are pedalling, often at 1-2km/h more than they are doing. Clearly there are some potential confounders with that example, but the differences are not completely trivial.
 

sogood

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TheDarkLord said:
Ok, I get it. The drag force is independent of mass, and hence, the drag per unit mass is less for the heavier bike. Is there any estimate on what the actual gain is?
Not quite. But the common reference of a heavier rider is that their aero drag is not proportional to the increased weight when compared with a lightweight rider. So heavy riders will progressively gain more advantage with increasing weight. But if you are talking about two bikes (same design without the rider but differing weight), then there's no resistive difference b/n the two, but there will be a difference in the propulsive force on a downhill.
 

TheDarkLord

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sogood said:
But if you are talking about two bikes (same design without the rider but differing weight), then there's no resistive difference b/n the two, but there will be a difference in the propulsive force on a downhill.
But what counts is force per unit mass, right? So, don't see how that alone will make a difference. Since drag is proportional to the square of velocity, if you write out the equation of motion, it will be

mg sin(theta) - kv^2 = ma,

where theta is the slope angle, and k is a coefficient related to drag, v is the velocity. So, while drag per unit mass is less, the propulsive force per unit mass is the same. So, don't see how increased propulsive force will lead to the difference here. :confused:
 

sogood

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I am no physicist and there are lots of far better qualified people here to answer this.

My understanding is that given the same bike apart from weight (same resistive forces), the propulsive force on the heavier bike would be greater due to its mass (g is a constant and same for both bikes). So as a result, the net force (propulsive) on the heavier bike is greater than the lighter bike).

Am I wrong to explain it this way? :eek:
 

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