a_thomasmr23

I don't know for sure but I suspect that the deflection from pedaling by a mortal (non pro) can't even be detected by a human. I got a real nice .1 to 5000 lb transducer now I just need a strain gauge to go with it and then its bring on the science

Has anyone on here gone through the trouble to do a statics like problem on a bike? I'm engineering student and I've really started to think about some of that stuff as it relates to bikes. I think this winter I'm going to start to work one up but if someone has done it before and has what they did I would love to get a look at those numbers.

ElJamoquio

Are non-pros more ethical than pros?

jonestr

the watts per kilogram of pro cyclists is ridiculous, but there are probably some big boys on here who put out more torque and raw power than some pr tour riders

interested to see what types of experiments you are going to design though.

Flatballer

duh, everyone knows that pros have no conscience.

big john

There has been a ton of discussion here on frame flex and whether it has an effect on power. The general consensus seems to be that frame flex doesn't cause real power loss. Personally, I have had a frame flex so much that other riders thought it was broken at the bb, and I'm as moral as the next guy.
If the search was working, you could find the threads.

Flatballer

I don't think there's any question that frames flex (I know mine does, the brake rub tells me so). I think what really needs to be answered is how much hysteresis loss frames exhibit. If they're soaking up 20W each stroke, but giving 19.9W back on the other side, it's not a huge deal. But if the frame soaks up 20W, and only gives back 10W, suddenly it becomes a big deal.

Not sure how you could design an experiment just using a strain gauge that would quantify the hysteresis loss of different frames, but I'm just an EE, so there might be a way.

StephenH

I'm not sure what you're proposing to do with a strain gauge.

The general problem with bike engineering is you don't know the loads applied, so it's hard to do much meaningful in the way of analysis.

a_thomasmr23

The idea for the strain gauge is to ride the bike with them on to get real time real world data to start from there. They are pretty cheap and the circuits can be made for next to nothing. Data logging on a the bike is the pickle I'm in right now.

a_thomasmr23

Strain measurement then would be replicated by applying loads in a stand then giving some ball park loads

Racer Ex 

Before you comment further, watch this:

https://www.xtranormal.com/watch/5710775/

Your suspicion doesn't need a transducer or a strain gauge to be proved or disproved; you just need to blind test people on a variety of frames that have different measured amount of flex under specific loads.

Having been laboratory rat for testing of this nature with forks, I can tell you that my conclusions about which flexed more dovetailed very closely to the flex under load numbers they generated on a lab jig.

Or just go push on the end of a pedal with the crank at BDC and see how much you can make things move in funny directions.

If you want to test power loss, it's been done, see below.

The general consensus is wrong. I've seen real wheel dyno losses in motorcycles from changing frames using the same engines, and they are measuring in HP not watts.

One of the english based cycling mags did a PM based study on old school steel vs. new carbon and the power loss was substantial, cracking double figures in percentage for all the riders (women and climbers included) in the study.

Most of the flex is horizontal (side to side), the double triangle being a pretty rigid engineering shape in a vertical, straight line plane.

Very little to none of the return "spring" will help rotate the crank or drive the rear wheel, being on a different plane than the power producing/transferring structures.

None of this is terribly important to a recreational cyclist who prizes comfort over performance, but the math over a three week stage race is impressive.

Mark Kelly 

Yeah there is and you don't need a strain gauge.

The law of conservation of energy requires that any lost strain energy show up as heat. All you have to do is measure the temperature of the frame at the bottom bracket before and after climbing a big hill. Assuming bearing heating is minimal at these speeds, any heat gain is presumably caused by hysteresis loss in the farme.

For comparison, put a small light bulb inside the bottom bracket shell and find what power consumption gives you the same temperature difference - you'll probably neeed to supply some external air flow to equalise cooling.

I was planning to perform this experiment in the next couple of weeks, I'll keep you posted.

For my next trick I'm going to try to convince one of the local bike shops to let me measure the resonant frequency of several bike frames at the bottom bracket. Acoustic theory says that the frequncy of the primary resonant mode is proportional to the stiffness to weight ratio of the structure (in fact resonant probing is used to estimate the elastic modulus of microstructures in MEM engineering)

Racer Ex 

You're assuming that the lost energy accumulates and stores itself as heat in the bottom bracket, that there's no dissipation to other parts of the frame or drivetrain, and that there's no heat loss to atmosphere over the duration of the climbs.

Those assumptions would not withstand any peer review.

Resonant frequency...I'm not sure what you're looking for here as it relates to anything practical. There's a huge amount of variability in damping; material, layup. shaping.

Whatever. I'm just an idiot bike racer.

zzzwillzzz

i don't think that there is any loss or if there is it is to small to matter. it control that you are losing. compare 70/80's superbikes with modern superbikes, the issue wasn't with power with the old bikes it was getting the power to the ground through the spindly, flexy frame and fork and skinny tires. it's the same with bicycle frames, you don't lose power through losses in the frame you just cannot apply as much power without having difficulty controlling the bike and keeping it headed in the direction you intend

Mark Kelly 

No, you've missed the point. Of course some of the heat is transmitted through the frame and dissipated to atmosphere but this will be proportional to the amount of heat being generated. The light bulb idea is to give me an idea of this proportionality.

I don't need it to withstand peer review, I'm not an academic, just an idiot bike designer.

BTW there's an assertion in one of your previous posts that I do not understand. You say that the frame resiling cannot possibly return energy to the rider because the plane of the motion is not coincident with the plane of pedalling. Since the return motion is simply the reverse of the initial deflection (minus any (as yet unproved) hysteretic losses), it seems to me that if the force on the pedals can cause the inital deflection the frame resiling from this deflection must be able to transmit force back to the pedals.

I think the factor that you are missing is the amount of energy stored in strain in the achilles tendon. It has been successfully shown that a running track can be tuned to assist runners if the rebound time of the track is made to be roughly the same as the resilience time of the achilles tendon - obviously these times are very different for a 100m sprinter and a 10k runner. It seems to me to be logical that the same thing can happen with a bicycle frame. This is part of the purpose of the experiments with resonant frequency at the bottom bracket.

Pedaleur

Double digit total drive train loss? Or double digit loss to extra flexing? That's a huge number.

Any chance you have a handy link or reference to this? An (admittedly short) search didn't find anything scientific.

The key here is the inefficient human muscle. Very little (a scientific term I've pulled out of my ass) of the energy that goes into flexing the frame gets dissipated as heat, but the work that the frame does as it unloads goes into both the drive train (good) and the muscles (which aren't known for their perfect spring qualities). I would love to see someone's numbers.

I can't wait until I get an all-carbon bike so I can start racing in the Tour...

Pedaleur

A bicycle frame with a 2 Hz resonant frequency, which is what you'd need to get this "rebound" effect, would be best made out of linguini, not carbon fiber.

Not to poo-poo the idea of investigating it all -- it sounds like a great topic for a project -- but I think you'll find the resonant frequency of a bike frame orders of magnitude higher than what it would need to be.

Metzinger

I am also interested in seeing a source for this double digit loss claim.
40W loss on a 400W input sounds high.
But it could explain why the paint around my BB is always blistering and flaking off.

Mark Kelly 

Well there are several things going on here.

Firstly, we are not trying to match the resonant frequency of an unloaded frame to the rider's cadence.

We are looking at the resonant frequency as a shorthand method of estimating the rebound speed of the frame when loaded then unloaded by the rider.

We do not know whether the best return of energy occurs when this matches the timing of the "unload" phase of the pedal stroke, it may be that it is best for the frame to be faster. The one thing we know from the running track studies is that if it's slower that's the worst of all scenarios.

The "unload" phase of the pedal stroke is a small part at the btottom of the cycle so the effective rate is much faster than the cadence. It appears to be determined by the spring constant of your achilles and the inertia of your lower leg, fortunately it's pretty easy to estimate these last by a couple of simple body resonance experiments.

My best guess is that this will all prove to be moot and that what actually happens is that the frame rebound simply assists the achilles to initiate the upstroke of the leg at the end of the pedal stroke. That's what it feels like when I try to concentrate on what my achilles is doing while pedalling.

Dick R

I'm a ME student, and have been thinking about this assumption for a while. While the frame does spring back at the end of a pedal stroke, I can't see how that force can be transferred through the chain, in fact I'm thinking it may work against the rider. I'd be interested if any other engineering types agree with my logic.



The image on the left shows the forces during the cyclist's downstroke (on the leg to our left, not his left). The rounded arrow shows that the downstroke pushes the BB to the right due to the force being off-axis. The straight arrow pointing to the left is the frame's reaction force, pushing against the leg that is pedaling downward at that moment. That reaction force allows the pedaling force to be directed in the Y axis (turning the crank) versus transferring any energy into the x axis to deflect the frame. We all seem to agree on this part.

The image on the right shows the forces at the moment the cyclist has completed his pedal stroke and begins to pedal down with the right leg (again, our right, not his). The curved arrow represents the force by the leg, the red line represents the frame's deflection at that instant, and the straight arrow represents the force created by the frame wanting to spring back to it's original state.

For a moment, the frame's "spring force" and the leg will be working together to help the frame recoil back to it's original unflexed state. Only after the frame has returned to its resting position, it will then start to work against the pedal stroke so your work is directed into the crank. Therefore, in addition to losing energy by deflecting the frame during your pedal stroke, logic suggests that we also lose energy when the frame springs back. Think of it like a spring that is connected to a box. You pull on it to extend the spring, then push on the spring as hard as you can. At first, you'll be slammed forward by the spring correcting itself before your pushing force is applied to moving the box.

This seems like the logical conclusion, but then again I would have also guessed that the business end of a nozzle would be higher in pressure, when in reality it's higher velocity and lower pressure. Dynamics can be funny that way.

Pedaleur

OK, but my point was that if the resonant frequency of the system is way higher than the pedaling frequency (which I assume it to be), then the frame is in quasi-equilibrium with the system, and there's no "rebound" that you're hoping for, even if you're only looking at a small part of the pedal stroke.

But just to be clear: even though my response was cheeky, I'm not saying, "Don't bother." I will gladly admit that the above is merely what I fully expect you'll find -- I have nothing but my intuition and a couple of thought experiments to back me up. I could be wrong, but even if I'm right, it still sounds like a fun project.

Pedaleur

But the problem is that when the right leg (our right, not his ) is about to push down, the frame is _not_ deflected as you show. The dynamics of the frame are much faster than the dynamics of the legs (this is what I'm getting at with the resonant frequency discussion in my other post). Thus the frame is unloading even before the downstroke ends, and is fully unloaded by the time the leg reaches the bottom (in an idealized sense).

The idea that the frame doesn't unwind in a useful way -- generating propulsion -- is a little misleading. If you look at it from an energy standpoint, so that you don't have to worry about force vectors and moments, then it's clearer (in my mind). Some of the energy -- probably a small amount -- is dissipated in the frame. Fine. The rest has to come out as work. But even work done in propulsion has to "push" against something, and this is where the inefficiency of human muscle comes into play; your body just can't reabsorb any energy from the frame.

Here's a thought experiment: Assume a perfectly rigid bicycle -- no losses. Put a pedal at 3 o'clock (mid way through the downstroke) and lock the brakes. Put a spring on top of the pedal and compress the spring with your foot. You have just loaded the spring with a certain amount of energy. Now hold your foot perfectly still and release the brake. _All_ of the energy in the spring goes into propulsion. However, in holding your foot still, you had to resist the upward force of the spring, and even though the spring did zero work on your foot (it didn't move, remember), your body still had to expend (chemical) energy to hold it still.

You can generalize this to moving parts, and throw in all the vectors you want, but this is fundamentally (in my mind) the issue with frame flex. How important it is in the real world (both recreationally and profesionally) I don't know, which is why I'd be curious to read the study RacerEx referenced.

Mark Kelly 

I'm not claiming to have the answers here (that's why I want to do some experiments) but I can help with unravelling the dynamic system. The easiest way to do it is to translate the mechanical components into their electrical equivalents and then run a simulation is SPICE (I use LTSPICE, freeware from Linear Technology). What you will find is that the system is a fourth order LCLC resonant tank, which makes the solution via ordinary methods (Laplace transform etc) very difficult.

The input waveform for the tank is muscle force vs time, the output is crankset torque vs time. The LCLC tank will determine the phase relationship between these two and also will affect the shape of the output waveform - departures from sinusoidal input are effectively higher harmonics and so could be subjected to different levels of attenuation depending on the time constant of the LC elements (the LCLC tanks acts a a fourth order low pass filter).

One intriguing possibility here is that the LP function could allow the output to be smoother than the input. I haven't crunched the numbers yet but it seems to me that smoothing would imcrease the average velocity of the output.

Racer Ex 

It was either Procycling or Cyclesport America, within the last few issues. They used a fairly broad swath of riders and I believe the median power loss was around 11%, albeit they were also using old style wheels (though not tires) and drive train.

They calc'd out that the difference in equipment performance would account for a fairly large percentage of the higher speeds seen between the era of the old bikes they were using for comparison and today.

Mark, you are correct:

As far as the plane of deflection goes, you hit maximum flex (horizontally, bottom bracket going towards the side) at BDC, the well known "dead spot". Any power that would be returned would be on the upward swinging BDC pedal side. The side of the pedal coming down from TDC is actually deflecting with the pedal stroke, not against it, another bit of power loss as you chase the frame flex.

There are plenty of pedaling efficiency studies that show little power generated by the upward pedal. And what power that is generated happens predominately in the middle of the rotation as your foot loads the cleat and shoe, at which point the frame is likely done flexing in any kind of beneficial direction.

While flex in a vertical plane will feed back into the drivetrain, the side to side can only "return" the amount of power through the rider, and you can see by the above why it doesn't to any significant degree. If you want a clear test throw on a powertap, lock the rear wheel, then load the a crank arm in each of the planes and see where you get a load reading and which one reflects the actual weight of the load.

Pedaleur

This makes little sense to me. An efficient pedal stroke should have no downward force on the pedal at BDC (and the corresponding opposite TDC for the other foot). I would find it remarkable if that were the point of maximum flex, or even significant flex.

But then again, that seems to be one of the main points of Mark's study: finding the phase differences in the system. Wouldn't be the first time I was wro -- er, mistaken.

Thanks for the references.

Mark Kelly 

Thinking out loud here, I'm not at all sure that the frame has to return the flex energy directly into the drive train. Returning it to the leg which is about to lift in such a way that it helps it lift would still contribute to locomotion.

Your leg is quite heavy and you either need to pull it upwards using hip flexors (which are in a position which makes this very inefficient) or push it upwards with the opposite leg (which robs power from that leg).

The proposed route is that the frame recoil loads the achilles tendon and the release of the achilles tendon lifts the leg.

Two problems here: inevitable energy loss in the load / unload of the achilles tendon, inevitable loss if the spring constant of the frame is greater than the spring constant of the tendon.

BTW there's another influence not measured yet - as the frame flexes it changes the apparent "Q factor" and this will have an impact on the efficiency of your pedalling mechanics.

Addendum: I've just calculated the spring constant in my achilles tendon by back calculation from my natural bounce frequency and it comes out around 16kN/m. Checking the static deflection results from Damon Rinard's frame deflection tests suggests that this is in the ball park of his figures - the frames measured cluster around 0.4 inches deflection under a 211N load which is 21 kN/m. Is this a coincidence, or is there some benefit in having the two closely aligned.

On the other hand there is no reason to suppose that my achilles tendons are representative of the world at large. Anyone who is intersted, try this test - in bare feet on a hard floor, bounce up and down as fast as you can without bending zee knees. Count the bounces in 10 seconds, multiply by 2 Pi / 10, square the result and multiply by half your weight in kilograms. Post the result. Example 30 bounces in 10 seconds x 2 Pi/ 10 = 19 s^-1, squared is 361 s^-2, times 45 kg is 16.2 kN.m^-1.