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

Crankyfeet

It's about science. Some of us are trying to use it. One of us is just spouting it.

But apologies to you Travis44 if it has gone off your original question. If you want more detail as to the effects of weight on speeds/times up hills of varying slopes, and also the often neglected effect on acceleration in races (like coming out of a corner in a crit), I can perhaps help, but it will take some time to do the calculations.

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artemidorus

Power is not irrelevant to the calculation because it is used firstly to calculate average and peak torque (and hence peak deflection of the frame and energy diverted to bending the frame) and secondly as the denominator of the efficiency fraction.

Cadence is critical to the model - lower the cadence for a given power and you increase torque and hence potential energy wastage into the frame. I'm not sure what cadence he chose for his calculation.

As I've posted earlier, I think it is quite possible that altered tracking of the rear wheel might contribute more to inefficiency than frame "spring" loading, at least in a hypothetical super-noodly frame.

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"All that we see and seem is but a dream, within a dream..."

Crankyfeet

There is no need to work backwards from power. At any given power, there are infinite torques, depending on the cadence you choose. It is the torque that deflects the frame. The energy loss is a function of varying torque (actually the torque squared since the energy loss is proportional to the deflection squared and the deflection (x) is proportional to the torque). Power introduces the irrelevant variable of cadence, which doesn't have anything to do with frame deflection IMHO.

The resulting efficiencies would need to be tabulated either as a function of peak torque, or if you want to use power, you are going to need lots of columns for different cadences.

You have just explained my point that energy loss can vary for the same power, depending on the cadence (which affects the torque). How can you talk about percentage power loss being constant with respect to power, if the energy loss can vary at the same level of power with different cadences?
It could well be very important... but it is not included in the model. As are a myriad of other factors not included, like the effect of tilting the frame from side to side on the respective moment arms and the tilting effect on lateral slippage of the tires, and the effect of other areas of elasticity in the system being loaded first before the frame starts to deflect.

It is complicated IHMO. An empirical test would be perhaps the best, if we could just design a repeating robot to replicate a human cyclist mashing a sprint.

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Jono L

Crankyfeet 15, Alienator 0

EDIT-On in depth reading it's more like more like two sets to nil

jhuskey

Great taste vs less filling.

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Sobriety is over rated!

ScienceIsCool

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

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.)

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TheDarkLord

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

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

Camilo

"Work??!?!??"

- Maynard G. Krebs

Crankyfeet

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).

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mikesbytes

OK, your convinced me, I'll buy the folding tyres for the commuter.

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ScienceIsCool

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

...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