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



ScienceIsCool

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I've only been half following this thread and notice that my half-assed, poorly described modelis being take to task. Fair enough. If I'm too lazy to explain myself properly then you should do your best to tear the model to shreds.

I'm still feeling too lazy to derive the whole she-bang. Again. And then try to transfer it here. But then again... a single measurement is worth a thousand theories or something like that.

So I'd like to propose a fairly simple experiment. It requires a bike, a light source like one of those knog led lights, and a digital camera that can go ito movie mode. You can either do it outside or indoors on a trainer.

First, attach the light source on the seat tube and down close to the bottom bracket. Arrange the light so that it points upwards towards the seat and make it secure so that it doesn't move around.

Second, strap the digital camera under the saddle (using the rails and some zip ties??) and arrange it so that you can video the bottom bracket area. Make sure the camera is nice and secure.

Turn on the light and the camera. Since both are rigidly mounted to the frame, the image of the light source should not move. Unless... the frame is being bent. Now go riding. When you get back, you'll have a fairly accurate measurement of how much your frame is flexing, how it varies with input, whether standing makes a difference, etc, etc. You should even be able to tell the difference between torsional flex and lateral flex.

From there you can make a fairly good calculation of how much energy/power was lost. If you have a power meter, you could correlate the results and calculate the efficiency of your frame vs input. Make a plot and see if it's linear or not. Awesome.

What do you guys think? If anyone wants to take this on, I'll help with setup, troubleshooting, analysis, etc. I'll also co-author a paper with you and put it up on my site.

John Swanson
www.bikephysics.com
 

thoughtforfood

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ScienceIsCool said:
I've only been half following this thread and notice that my half-assed, poorly described modelis being take to task. Fair enough. If I'm too lazy to explain myself properly then you should do your best to tear the model to shreds.

I'm still feeling too lazy to derive the whole she-bang. Again. And then try to transfer it here. But then again... a single measurement is worth a thousand theories or something like that.

So I'd like to propose a fairly simple experiment. It requires a bike, a light source like one of those knog led lights, and a digital camera that can go ito movie mode. You can either do it outside or indoors on a trainer.

First, attach the light source on the seat tube and down close to the bottom bracket. Arrange the light so that it points upwards towards the seat and make it secure so that it doesn't move around.

Second, strap the digital camera under the saddle (using the rails and some zip ties??) and arrange it so that you can video the bottom bracket area. Make sure the camera is nice and secure.

Turn on the light and the camera. Since both are rigidly mounted to the frame, the image of the light source should not move. Unless... the frame is being bent. Now go riding. When you get back, you'll have a fairly accurate measurement of how much your frame is flexing, how it varies with input, whether standing makes a difference, etc, etc. You should even be able to tell the difference between torsional flex and lateral flex.

From there you can make a fairly good calculation of how much energy/power was lost. If you have a power meter, you could correlate the results and calculate the efficiency of your frame vs input. Make a plot and see if it's linear or not. Awesome.

What do you guys think? If anyone wants to take this on, I'll help with setup, troubleshooting, analysis, etc. I'll also co-author a paper with you and put it up on my site.

John Swanson
www.bikephysics.com
Great sounding experiment, and actually realizing how to test it is impressive, so even in laziness, you have exceeded my capacity (not like that is a difficult thing to do). However, I do have one more question, and it goes back to my earlier question about the change in pitch of the wheel/tire to direction of movement.

This question on my part may have been ignored because you all said "that guy doesn't know what the hell he is talking about." But how do we test for that factor? The torque at the BB alters the pitch or turn of the wheel in a way that causes them to cause more friction and therefore a loss of power. Is your experiment just for the frame flex, or can it take into account this problem (if it is a problem, though it seems like it is to me)?

In case I am not being clear (and I think I am, but just in case), lets say your bike is leaned to the left and you are pulling up and pushing down on the right crank. Because of the torque, your front wheel will turn to the right, and your rear to the left. You can look at pictures of a spring to see this effect. How do you account for that loss of power?

Let me also say that, I do believe that in the long run, the guy with the most fast twitch will probably rule the day, but there has to be a loss of power, unless I have no idea of that of which I speak. (It would NOT be the frist time....even today that has happened.)
 

TheDarkLord

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Dec 24, 2007
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ScienceIsCool said:
First, attach the light source on the seat tube and down close to the bottom bracket. Arrange the light so that it points upwards towards the seat and make it secure so that it doesn't move around.

Second, strap the digital camera under the saddle (using the rails and some zip ties??) and arrange it so that you can video the bottom bracket area. Make sure the camera is nice and secure.

Turn on the light and the camera. Since both are rigidly mounted to the frame, the image of the light source should not move. Unless... the frame is being bent. Now go riding. When you get back, you'll have a fairly accurate measurement of how much your frame is flexing, how it varies with input, whether standing makes a difference, etc, etc. You should even be able to tell the difference between torsional flex and lateral flex.
If you attach the camera under the saddle though, all you will see is a nice image of the top tube, and you will not see the light source at all, since it will be blocked by the top tube. If you attach it to under the top tube, there is the question of whether you can attach the camera securely enough to detect any flex. The camera will be subject to vibrations, and also some lateral motion against the tube. The systematics sources of error will probably swamp any signal from the actual flex, especially considering that the flex is going to be pretty small. Just my thoughts on your proposed experiment.
 

ScienceIsCool

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I'm pretty sure that if you mount the camera at the back, where a saddle bag would go, that you could get a fairly clear image of the bottom bracket area. The marker (light source) wouldn't necessarily need to be right beneath the camera, either. You could mount it opposite where the FD is, on the side. I'm sure something could be worked out...

Oh, and about tires wandering, etc. You could model the amount of tire deformation and add that to the losses. Essentially, it's a spring too and will add in a normal way. And if this sideways force causes your front tire to change its path? Well, you could compare the straight line distance to the elongated path taken by the front wheel. Get the tires wet and meaure the difference. Since P = Fxv (actually it's a path integral) the ratio of path lengths is your percentage change in power.

However, I'm just interested in the frame's contribution to efficiency. The camera thingy has a chance to work well for very little to no cash outlay. Okay, now back to work...

John Swanson
www.bikephysics.com
 

Crankyfeet

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Jun 5, 2007
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ScienceIsCool said:
I've only been half following this thread and notice that my half-assed, poorly described modelis being take to task. Fair enough. If I'm too lazy to explain myself properly then you should do your best to tear the model to shreds.

I'm still feeling too lazy to derive the whole she-bang. Again. And then try to transfer it here. But then again... a single measurement is worth a thousand theories or something like that.

So I'd like to propose a fairly simple experiment. It requires a bike, a light source like one of those knog led lights, and a digital camera that can go ito movie mode. You can either do it outside or indoors on a trainer.

First, attach the light source on the seat tube and down close to the bottom bracket. Arrange the light so that it points upwards towards the seat and make it secure so that it doesn't move around.

Second, strap the digital camera under the saddle (using the rails and some zip ties??) and arrange it so that you can video the bottom bracket area. Make sure the camera is nice and secure.

Turn on the light and the camera. Since both are rigidly mounted to the frame, the image of the light source should not move. Unless... the frame is being bent. Now go riding. When you get back, you'll have a fairly accurate measurement of how much your frame is flexing, how it varies with input, whether standing makes a difference, etc, etc. You should even be able to tell the difference between torsional flex and lateral flex.

From there you can make a fairly good calculation of how much energy/power was lost. If you have a power meter, you could correlate the results and calculate the efficiency of your frame vs input. Make a plot and see if it's linear or not. Awesome.

What do you guys think? If anyone wants to take this on, I'll help with setup, troubleshooting, analysis, etc. I'll also co-author a paper with you and put it up on my site.

John Swanson
www.bikephysics.com
Thanks for coming back to us John. I have read through a bit of your website and found your work and tests to be very helpful. There is very little independent analysis out there to refute manufacturers claims and gimmicks. You seem to be making a real effort to reveal some truth so consumers can get accurate data to base a decision on. I only wish I had read it earlier because our race team just expired on an offer of Neuvation wheels and I missed it... :(

I have some concerns about the experiment you propose. Best to list them I suppose.

1. Firstly, you seem to have a knowledge of fairly complex physics in your analyses on your website, but here you keep holding onto the notion that energy loss is a linear relationship to power. With the help of your equations for the energy absorbed in a spring (E = 0.5 kx^2) it is fairly simple to derive the relationship that energy loss is proportional to the deflection squared (x^2). The deflection is proportional to the Force ( F = kx ) . And torque is proportional to the force at the same lever arms that repeat in a pedal stroke. The deflection is a linear function of the torque, and the energy loss is therefore proportional to the square of the applied torque. If you could somehow keep a constant cadence in your test for differing torques, then you would find that energy loss is proportional to the square of the power.

This is using high school physics equations that you provided for us. It doesn't require an experiment to work out. If you get a linear relationship between torque (or power at constant cadence) and energy loss, then your experiment is invalid and the results are worthless are they not?

I am concerned that you can't see this. It is high school physics... or am I making myself look stupid by making a big error???

2. I detect a little bit of sarcasm in your one word "Awesome" sentence. I might be misinterpreting this as misinterpretation happens a lot on the internet. However if it is sarcasm, I would be wary of getting involved in doing an experiment that the experimenter has a vested interest in validating prior results. And someone who perhaps has a condescending attitude towards those who question the theory. I apologize if no sarcasm was intended, though I notice on your website that you use it a lot in your tone.

3. If your original calculations were correct (despite the claim of linear relationship to power), then 0.05% losses equates to 1/2000 of the input torque going into the frame, and 1999/2000 of the input torque being applied to the drivetrain. 1/2000 of the torque going into lossed energy in the BB is instinctively not going to produce a readable deflection, within the limits of accuracy in your proposed experiment.

4. Your experiment doesn't take into account vertical deflection of the frame. Whilst any bike needs a certain amount of vertical elasticity so that the bike absorbs bumps better (too stiff and the wheels lose contact with the pavement over even minor undulations) there has to be an optimum amount whereby energy loss is minimized while giving a smooth enough ride to keep the tires in as much contact with the pavement as possible. This vertical deflection is still a component of loss in energy in the frame and not measurable in the axis you are measuring.

5. Similar to TheDarkLord's concerns in his post above, I am not convinced that we can get the camera to mount without movement or vibration, sufficient to measure deflections in millimeters (or perhaps we even need to measure more precisely than this).

6. There are other potential elasticities in the frame as well. The torsional deflection can be absorbed in the chain and seat stays, which will tilt the wheel, but not necessarily deflect the main triangle of the frame. These deflections though are losses in the frame that need to be taken into account. In fact, given their structural size, they are probably the main area of torsional deflection.

And John... I'm in Seattle. Which is not far away. I am interested by your work. Please don't take personal offense at my concerns. I am just as interested as anyone in getting to the truth, and peer review with due respect is part of that (though I am not implying I am your peer).
 

ScienceIsCool

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Crankyfeet, sorry to make this short. I keep forgetting to address one of your issues. If Torque = P/w, and E = 0.5kx^2, then yes you are absolutely right that the losses will scale quadratically with input power. Not linear at all. I'll chalk that one up to rushing throuh my analysis and/or thinking clearly. Also another good indicator why people should question any and all claims.

That's it for now. Mea culpa and good night.

John Swanson
www.bikephysics.com
 

ScienceIsCool

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...and to answer the next obvious question. I originally assumed a power input of 400 watts and a cadence of 90 rpm.

Oh yeah. I guess my "awesome" comment sounded sarcastic, but it wasn't meant that way. I just meant that it's always nice when you can plot some data to make everything nice and clear.

John "likes data" Swanson
www.bikephysics.com
 

tallrider7

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Ah the internet... A bastion for civil, intelligent and productive discourse. There is a LOT of namecalling and character assasination that has populated this thread that undermines the potential benifits of the discussion. It is disappointing, quite honestly. I have a few questions and observations rattling around in my head when I read some of these posts:
  • Obviously these are well educated (seemingly) individuals who are probably professionals in their chosen fields. Do they treat people like this in person if they encounter a differing opinion?
  • Once a person takes a position on a subject they tend to defend it to the exclusion of all reason. Look at religion, politics, sports fans (soccer hooligans come to mind).
  • Data is vital to understanding most problems, but in my job data sources vary in their veracity. In short, “garbage in, garbage out”.
  • It is so very easy to type a message full of venom and what appears to be real hatred and then hit the enter key to post when you are not looking another person in the eye.
  • This is a forum which discusses cycling. I just wonder about what kind of riders those contributors that have been the most demeaning towards others actually are on the road. Perhaps you are all Cat 1’s and are moving to the pro ranks because you got fired from your day job for not being able to play nice with others.
  • I spent some time with one of the national team coaches many years ago and I remember something they said: “It’s not the chariot, it’s the horses”. Obviously it helps to have good equipment and some can afford better than others, particularly in the amateur ranks. Right now I’m concentrating most on my training, despite the fact that I am shopping for a new bike. I laugh, ONLY to myself, when I beat people riding $8000 bikes when I am riding a 14 year old steel bike.
  • There are some very bright people on this site with a lot to offer in the way of information on bicycle technology. However entertaining it may be to witness this intellectual **** fight, it is ugly and highly distracting from the goal which is to increase one another’s knowledge and share ideas.
I suspect that I will be the next one to be attacked for making these observations. I found this site the other day (about the time that this thread got thrown into havoc mode) and thought that I had discovered a cycling community worth joining.

This is very, very disappointing. It is such a cliché, but the Rodney King quote, “Can’t we all just get along” seems quite apropos.

Bottom line: You guys have more in common than most subgroups on the planet. Treat each other accordingly and EVERYONE HERE will benefit.
Right now I’m kind of disgusted, although I’m sure that is evident. Just consider the concept of community. Let’s not be like the rest of the world right now and jump for each others throats at the first chance.


By the way, I’m no stud racer, nor am I the be all end all purveyor of cycling wisdom. I ride 2-3 times a week, am a Cat 4, was a
shop mechanic for 10 years, am a highly respected bike technician, still, in the circles I travel, BUT I DON’T CLAIM TO KNOW IT ALL. I LOVE LEARNING FROM OTHERS AND AM NOT TOO PROUD TO ADMIT I’M EITHER WRONG OR THAT I DON’T KNOW SOMETHING. THAT’S WHY I’M HERE. TO LEARN FROM THE EXPERIENCE AND KNOWLEDGE OF OTHERS AND PERHAPS TO SHARE SOME OF MINE.

If anyone is interested (and at this point that would indicate stamina on your part) I will actually weigh in on the original question which relates to whether bike weight matters.


Yes, it does, but not as much as fitness. Once you have top level competition then equipment begins to play a part in determining the outcome, but ONLY when the competitors are exceptionally close to the same level of fitness. Lemond vs. Fignon in 1989 comes to mind (although that was an aero issue, but equipment related). Armstrong’s collaboration with various manufactures to come up with the best overall package for the Tour is another example. But it was secondary to his commitment to training and physical preparation. Merckx was also fanatical about his equipment (drilled everything on his hour record bike), but he was the strongest rider with a will of iron. In fact I would say his mentality had more to do with his total domination of the sport than his physical talent level, and certainly more than his equipment.

Me? I’ve lost 18 lbs in the last 6 months and plan to lose another 12-15 before June. The bike I am about to purchase will likely weigh about 5 lbs less than the one I am currently riding on. I was out of the sport for a long time. But I’ve been involved in bike racing since the mid 80’s.

I’m just so happy to be riding and competing again that it’s all gravy at this point. I did over 300 miles in the rain in January and although I did start bitching about it after 8 consecutive rain rides, I wouldn’t trade it for not riding at all.

No way.

Thanks for your time and have a nice day. Really.

Happy Trails,
Mark
 

Crankyfeet

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ScienceIsCool said:
Crankyfeet, sorry to make this short. I keep forgetting to address one of your issues. If Torque = P/w, and E = 0.5kx^2, then yes you are absolutely right that the losses will scale quadratically with input power. Not linear at all. I'll chalk that one up to rushing throuh my analysis and/or thinking clearly. Also another good indicator why people should question any and all claims.

That's it for now. Mea culpa and good night.

John Swanson
www.bikephysics.com
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.
 

TheDarkLord

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Crankyfeet said:
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).
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

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TheDarkLord said:
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.
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.
 

TheDarkLord

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Crankyfeet said:
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 - but how much less power may be delivered to the wheel due to the increased resistance. This lesser output of power is happening because energy is stored in the system. Much of it probably in the frame, but also in other areas like the friction and 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. 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.
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

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TheDarkLord said:
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.
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.
 

dhk2

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Aug 8, 2006
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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

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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.
 

TheDarkLord

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Dec 24, 2007
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Crankyfeet said:
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 53/11 gear do you accelerate at a lesser speed than if you stand up on the pedals (applying the same torque) in a low gear. 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. 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.
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

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Crankyfeet said:
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.
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

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TheDarkLord said:
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.
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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