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

[alienator]

Maybe it is just a "hunch" of alienator? Just like the opinions/hunches of other people regarding bike weight?

Simple: from the first and second order analysis of forces in bike/rider system. From the complete lack of data making any correlation whatsoever between bike stiffness and improved performance.

Math. Science. Look 'em up, and give 'em a try. People actually use it in the real world.

 

[TheDarkLord]

Simple: from the first and second order analysis of forces in bike/rider system. From the complete lack of data making any correlation whatsoever between bike stiffness and improved performance.

Math. Science. Look 'em up, and give 'em a try. People actually use it in the real world.

You know, your condescending attitude is really disgusting at times. What is even worse is how a scientist like you twists statistics or the lack of it to suit your own argument in totally non-scientific ways. Do you really publish papers with such twisted interpretations? Or are you a theorist?

 

[Crankyfeet]

Simple: from the first and second order analysis of forces in bike/rider system. From the complete lack of data making any correlation whatsoever between bike stiffness and improved performance.

Math. Science. Look 'em up, and give 'em a try. People actually use it in the real world.

So lack of data on the effect of stiffness = alienator having an opinion on the effect of stiffness.

Interesting logic for he who decries having an opinion when there is no data to support it...

 

[ScienceIsCool]

I'm more interested in the effects at 1200+ watts which is the often quoted power that pro sprinters such as Boonen are reputed to be applying in the final 250 yards of a race. And I wouldn't be surprised if the real effects are a non-linear relationship (exponential) to the force applied to the pedal. Call it instinct.. but that's how a lot of scientific tests begin... testing a hunch.

Also I'm quite interested by your ability to perform a controlled experiment and get a testing accuracy to 5/10,000 (0.05%). That's a very precise test for measuring changes in output without falling within the margin of error of the testing apparatus.

I didn't perform an experiment. I used the known values of stiffness and computed the theoretical losses. But yes, you made one of my points for me. The losses are generally too small to be measured...

This is how I calculated the losses:

- Assume the stiffness values (expressed as k = N/mm) represent a linear, spring relationship. And it should because there are no non-linear elements.
- The energy stored in a spring (and subsequently lost) is equal to 0.5kx^2 where x is the displacement.
- The amount of displacement can be estimated from F = kx and P = Tw, where P is power, T is torque, and w is the rotational speed of the cranks.
- For a given power output and cadence, you can figure the average torque on the crank.
- Assuming the instantaneous peak torque is ~ 4 times average (that's generous) and using a force diagram, you can calculate the force that goes into bending the frame (i.e., pushing on our model of a spring)
- This force gives you the energy stored in the spring as described above: 0.5kx^2 and all that.
- This energy loss happens twice every rotation of the cranks, which happens 2cadence/60 seconds.
- Therefore, the power lost is equal to Energyx30/cadence Watts.
- Divide that by the original power input and there's your efficiency.

I found that using this model, which should be accurate by way better than an order of magnitude, the efficiency of a bike frame is ~99.95%. That might change by ~0.025% depending on whether you have a super stiff or super flexible frame.

John Swanson
www.bikephysics.com

 

[TheDarkLord]

So lack of data on the effect of stiffness = alienator having an opinion on the effect of stiffness.

Interesting logic for he who decries having an opinion when there is no data to support it...

Maybe alienator is just confused with what inferences can or cannot be drawn from "lack of data". Case in point - there is no data showing that string theory is right. Hence, string theory should be wrong. Right, alienator?

 

[ScienceIsCool]

Oh yeah. And the efficiency is the same whether you're putting out 12 or 1200 Watts. The system is incredibly linear over a wide range of inputs.

John Swanson
www.bikephysics.com

 

[TheDarkLord]

I didn't perform an experiment. I used the known values of stiffness and computed the theoretical losses. But yes, you made one of my points for me. The losses are generally too small to be measured...

This is how I calculated the losses:

- Assume the stiffness values (expressed as k = N/mm) represent a linear, spring relationship. And it should because there are no non-linear elements.

I would include a damping term though, as there is bound to be some energy that will be absobed by the bike frame. The damping term could dominate the losses in the system; but I wouldn't be surprised if any loss is very small.

 

[Yojimbo_]

I'm not pretending to understand string theory; however, I believe there are 6 different theories, all requiring many more dimensions than the four we live with.

They can't all be right.

 

[TheDarkLord]

I'm not pretending to understand string theory; however, I believe there are 6 different theories, all requiring many more dimensions than the four we live with.

They can't all be right.

I don't know anything much about it either, and I have heard some talks where some string theories could be united in a more general framework (like M-theory), but it was just an analogy (maybe a poor one) to show that the lack of data by itself should not be the basis for some conclusions.

 

[ScienceIsCool]

I would include a damping term though, as there is bound to be some energy that will be absobed by the bike frame. The damping term could dominate the losses in the system; but I wouldn't be surprised if any loss is very small.

Nah. You don't really need a damping term because you don't care *where* the losses are. It's enough to know that any energy you put into this spring system is lost. Besides, the only effect a damping mechanism would have is to improve the overall efficiency...

John Swanson
www.bikephysics.com

 

[Crankyfeet]

I didn't perform an experiment. I used the known values of stiffness and computed the theoretical losses. But yes, you made one of my points for me. The losses are generally too small to be measured...

This is how I calculated the losses:

- Assume the stiffness values (expressed as k = N/mm) represent a linear, spring relationship. And it should because there are no non-linear elements.
- The energy stored in a spring (and subsequently lost) is equal to 0.5kx^2 where x is the displacement.
- The amount of displacement can be estimated from F = kx and P = Tw, where P is power, T is torque, and w is the rotational speed of the cranks.
- For a given power output and cadence, you can figure the average torque on the crank.
- Assuming the instantaneous peak torque is ~ 4 times average (that's generous) and using a force diagram, you can calculate the force that goes into bending the frame (i.e., pushing on our model of a spring)
- This force gives you the energy stored in the spring as described above: 0.5kx^2 and all that.
- This energy loss happens twice every rotation of the cranks, which happens 2cadence/60 seconds.
- Therefore, the power lost is equal to Energyx30/cadence Watts.
- Divide that by the original power input and there's your efficiency.

I found that using this model, which should be accurate by way better than an order of magnitude, the efficiency of a bike frame is ~99.95%. That might change by ~0.025% depending on whether you have a super stiff or super flexible frame.

John Swanson
www.bikephysics.com

Sorry I haven't had time to really go through your model. But how do you determine the separate values for "k" and "x"? F= kx is different from the energy equation which needs to square the "x" value?

And how is your model able to determine the difference between a "noodly" frame and a very stiff frame. I can't see where the frame properties are accounted for in your model?

I need more time to go through and understand your model though. Thanks in any case, John, for your explanation.

 

[TheDarkLord]

Nah. You don't really need a damping term because you don't care *where* the losses are. It's enough to know that any energy you put into this spring system is lost. Besides, the only effect a damping mechanism would have is to improve the overall efficiency...

John Swanson
www.bikephysics.com

Fair point. Although I admit I don't visualise the system well enough to really follow your calculations.

 

[ScienceIsCool]

Sorry I haven't had time to really go through your model. But how do you determine the separate values for "k" and "x"? F= kx is different from the energy equation which needs to square the "x" value?

And how is your model able to determine the difference between a "noodly" frame and a very stiff frame. I can't see where the frame properties are accounted for in your model?

I need more time to go through and understand your model though. Thanks in any case, John, for your explanation.

I know that this was a quick summary and perhaps not the easiest to "digest". But hopefully I can answer your questions. The value of k comes directly from published data on frame stiffness. To be honest, I can't remember where I got my numbers, but I think it was from the famous "Tour" testing.

I then use k with the estimated F (from the Torque, power, cadence, and force diagrams) to calculate x. x = F/k

From there, I can calculate the energy loss: E = 0.5kx^2

And with a few more sums, I get the overall efficiency. I'll try to find my old, longwinded posts on the subject. If I can't find them, maybe I'll write them all out again. But don't hold your breath, I've been busy lately and I'm feeling lazy.

John Swanson
www.bikephysics.com

 

[TheDarkLord]

I know that this was a quick summary and perhaps not the easiest to "digest". But hopefully I can answer your questions. The value of k comes directly from published data on frame stiffness. To be honest, I can't remember where I got my numbers, but I think it was from the famous "Tour" testing.

I then use k with the estimated F (from the Torque, power, cadence, and force diagrams) to calculate x. x = F/k

From there, I can calculate the energy loss: E = 0.5kx^2

And with a few more sums, I get the overall efficiency. I'll try to find my old, longwinded posts on the subject. If I can't find them, maybe I'll write them all out again. But don't hold your breath, I've been busy lately and I'm feeling lazy.

John Swanson
www.bikephysics.com

Maybe you can make a document (with a scan/image of the force diagram) and put it on your website?

 

[Crankyfeet]

Nah. You don't really need a damping term because you don't care *where* the losses are. It's enough to know that any energy you put into this spring system is lost. Besides, the only effect a damping mechanism would have is to improve the overall efficiency...

John Swanson
www.bikephysics.com

How do you know what proportion is put into this "spring system" and what proportion of the force is transferred to the wheel. This is the essence of efficiency.

You have a force on the crank. The only way to know how much of that force goes into deflecting the frame is to know the properties of the frame. The energy will be stored in the path of least resistence first. Its not until the force of the spring pushes back with the same force from the crank, that the drive train will start to be engaged. You need to know this F= kx of the frame, which is not the same as the F (force) of the pedal force. Correct me where I am wrong please. The "x" value is important IMO because if you have a high "x" value of the frame (high deflection ie. a "noodly" frame) then more time may be required to deflect the frame before it resists the pedal force. This will be a more significant part of the pedal rotation arc than in the case of a stiffer frame, which can push back against the pedal force quicker.

I need to think more about your model though. At first glance, I am struggling with understanding how you accounted for the stiffness modulus/properties of the frame, to get your 0.05% difference betweeen the two.

 

[thoughtforfood]

I didn't perform an experiment. I used the known values of stiffness and computed the theoretical losses. But yes, you made one of my points for me. The losses are generally too small to be measured...

This is how I calculated the losses:

- Assume the stiffness values (expressed as k = N/mm) represent a linear, spring relationship. And it should because there are no non-linear elements.
- The energy stored in a spring (and subsequently lost) is equal to 0.5kx^2 where x is the displacement.
- The amount of displacement can be estimated from F = kx and P = Tw, where P is power, T is torque, and w is the rotational speed of the cranks.
- For a given power output and cadence, you can figure the average torque on the crank.
- Assuming the instantaneous peak torque is ~ 4 times average (that's generous) and using a force diagram, you can calculate the force that goes into bending the frame (i.e., pushing on our model of a spring)
- This force gives you the energy stored in the spring as described above: 0.5kx^2 and all that.
- This energy loss happens twice every rotation of the cranks, which happens 2cadence/60 seconds.
- Therefore, the power lost is equal to Energyx30/cadence Watts.
- Divide that by the original power input and there's your efficiency.

I found that using this model, which should be accurate by way better than an order of magnitude, the efficiency of a bike frame is ~99.95%. That might change by ~0.025% depending on whether you have a super stiff or super flexible frame.

John Swanson
www.bikephysics.com

Ok, so I don't have any idea what you are talking about.


I do know this, if the flex of the frame caused by pressure at the bottom bracket causes the track of the front and rear tires to change (and it does), then the contact patch for the tires increases because of their turn, however slight, from the direction of your travel. This increases resistance and therefore causes a loss of energy. It might be slight, but in just that, it is present, and therefore a stiffer frame is less efficient in producing forward momentum.


That is totally unscientific however, though you could replicate that very easily. You don't need to test whether a larger contact patch causes loss of energy, that has been done. All you should have to show is that the flex of the frame does alter the track of the wheel/tire, and that has been done also. Therefore a more flexible frame causes a loss of energy.


Am I missing something?

 

[alienator]

You know, your condescending attitude is really disgusting at times. What is even worse is how a scientist like you twists statistics or the lack of it to suit your own argument in totally non-scientific ways. Do you really publish papers with such twisted interpretations? Or are you a theorist?

I just return condescension when condescension is distributed re: bike properties and how what properties affect what. So? Is there something intrinsically wrong with questioning because of a lack of data? Hey, I done did the analysis, and the dark, evil losses due to frame noodlism just ain't there. Sorry.

I think condescension is much more appetizing than blindly swallowing marketing spiel. Besides, condescension serves well the crowd around here.

Sorry, Cupcake.

 

[thoughtforfood]

I just return condescension when condescension is distributed re: bike properties and how what properties affect what. So? Is there something intrinsically wrong with questioning because of a lack of data? Hey, I done did the analysis, and the dark, evil losses due to frame noodlism just ain't there. Sorry.

I think condescension is much more appetizing than blindly swallowing marketing spiel. Besides, condescension serves well the crowd around here.

Sorry, Cupcake.

Ok sweetie, show your data. I mean, such an obvious mental giant like yourself should understand that when you at one point ask for data showing a finding contrary to your own, revealing your data is a logical thing to do. Put up or shut up I always say. I mean since you "done did the analysis," show your data there ****wit.

 

[thoughtforfood]

Actually, even without a larger contact patch, you still lose energy because of the manner in which a round object reacts to altering its pitch in relation to its line of travel. You are gonna lose energy, no way around it.

 

[Crankyfeet]

I mean since you "done did the analysis," show your data there ****wit.

And don't use a random insertion of Bayes Theorem again...