mikeybikes

You mentioned 4" was outside the experiment. What's the largest that was inside the experiment? If you don't know, how do you know 4" is outside the experiment?

mikeybikes

Now that's a good kind of racing!

Jaywalk3r

The largest was around 35 - 40 mm, IIRC. Testing 1 4" tire would be problematic, due to frame compatibility.

mikeybikes

So why don't road racers ride 35mm tires?

trailmix

More train please.

Jaywalk3r

For racers, the extra air resistance is a bigger factor than for typical riders. Their speeds are higher and they've made as many reductions elsewhere as they can. That being said, many of them do race on tires that some here would consider wide, or at least at the upper limits of narrow. [Source 1] [Source 2] There's no reason to believe that the trend of using wider tires won't continue if tire manufacturers start making their best racing tires in larger sizes, and frame manufacturers return to building frames with reasonable tire clearances. Currently, few, if any, racing frames will fit a 35 mm tire.

spare_wheel

are you actually trying to argue that 25 mm tires are wide??/?

lol.

and the funny thing is that the main reason pro teams started switching to 25 mm tires was because they reduce drag on wider toroidal carbon fiber wheels. in other words it's all about CFD modelling and wind tunnel testing and has absolutely nothing to do with the randonerd tire line heine pimps based on biased testing of a handful of non-randomly selected race-level clinchers.

Jaywalk3r

Not me. I think anything under 35 is skinny. Others upthread, on the other hand …

Of course, what racers use is of little importance to commuters, since we don't have support vehicles follow us around town.

Like I said. For racers, air resistance is a bigger consideration than rolling resistance. The reduced rolling resistance is an added benefit.

mikeybikes

That's actually what I mostly was curious about. Thanks for clarification.

PaulRivers

That's the reason they give, but it doesn't make much sense.

In Time Trials, aerodynamics are fairly important, and bikes specifically designed for time trials are more aerodynamic.

In races like the Tour De France, riders don't ride those more aero bikes. Because they spend a lot of their time in packs with other riders where aerodynamics don't have as big of effect because they're drafting behind other riders. Sometimes it's packs of riders competing against each other, sometimes it's in pack with their team mates where they taking turns drafting each other so that everyone benefits. But either way, the effects of aerodynamics are much smaller in racing with a team vs racing with a time trial. If a 42c tire was actually faster, Lance Armstrong would have had one on his bike and simply drafted behind his team mates for at least some of the stages of the Tour De France, you'd think.

Even the reason for going from 23c to 25c isn't clear cut either. They make tires differently than they did 20 years ago, with different manufacturing and slightly different materials. It may have been that change in how tires are made have changed, so that 20 years ago a 23c tire was faster, but in modern tires it is not.

You can read about the numerous tests racers have done to look for any way to gain an advantage - clothing material, meal timing, nutrition, etc etc. Sure, Lance Armstrong was on steroids, but he still had to beat the other top 15 contenders who were also on steroids. They're all looking for any advantage they can get. Bike manufacturers have also created entire new frames for riders to suit their preferences. If fatter tires were actually faster, they'd be riding them in serious races. I don't think the aerodynamics explanation is actually the reason they ride skinny tires. If fatter tires had better rolling resistance but were less aero, you'd still see racers riding them and drafting their team mate on a regular basis.

Bike Gremlin

Just to make one thing clear, regardless of the whic is faster:

When racing, a lot goes down to tactics. You need to be able to CHANGE your speed quickly (mostly accelerate) - so you stay where you want to be within a group. That's where lighter wheels are a benefit - rolling mass's effect is a lot higher than static when changing speed. On a TT race, fatter (hence slightly heavier) tyres wouldn't be that much of a problem - except when climbing, of course. Since most races involve some group competing and some climbing, it is clear why a fat and heavy wheel is not used.


I prefer 28s to 25s or 23s, because they are MUCH more comfortable on rough pavement. I didn't notice any speed loss, not significant anyway. Wouldn't call them faster... but on cobblestones they definitely are - the worse the pavement, the faster fatter tyres are.

cyccommute 

You cannot have force independent of mass. Simply can't happen. The definition of force is mass times acceleration. Without both of those items, you don't have force. The larger frontal area has a role but you still have to provide force to keep the bike moving. More importantly, you have to provide force to move the air out of the way. A larger frontal area requires more force from the rider to move the system. Since the force needed to move the air is equal to the force needed to move the bike and the bike/rider system has mass, a larger mass requires more force to move it and more force to maintain constant speed.

cyccommute 

Try again. Air resistance is related to frontal area but the bicycle is being propelled through the air and the bicycle has mass. The engine, i.e. the rider, has to work harder to maintain speed for a heavier load. Assuming a constant frontal area, a 90kg rider on an 8kg bike requires 113W of power at 30 kph to overcome drag. A 70 kg rider on the same bike requires 109W of power to over come drag at 30kph. If the bike weight is increased to 18 kg (40 lb), the 90kg rider requires 162W to maintain speed. Add another 10kg of luggage weight (but keep the frontal area the same) and the 90 kg rider needs 167W to maintain speed.

You can go play with the numbers here for your own entertainment.

I'm not talking about getting both wheels off the ground but about lofting the wheels over a pot hole.

You may want to further edit the above. If you "assum[e] the same surface area", the surface area is the same. My calculations above are assuming the same surface area which is obviously not the case but works for simplification.

But acceleration and force aren't zero. Drag is a force. You have to put power into the system to overcome it. The power required to overcome drag is "work" over time and "work" is a force applied over a distance. In the absence of drag, you wouldn't have to put any power into the system but, on the surface of the Earth, you are never in a situation where you have an absence of drag. Stop providing power to the system and what happens? The whole systems slows down and stops. If the velocity = 0, then acceleration and force equal zero. But if the system is moving in an atmosphere, there is always an acceleration which may be negative.

It's actually pretty easy (although the math is complicated). If you account for resistances and you want to maintain constant speed, you have to apply constant force to balance the negative acceleration. The air may not have much mass but you have mass and thus mass is part of the equation.

Andy_K 

This is incorrect. Mass times acceleration is the most common formula used in connection with force, and it is always relevant, but there are a good many forces which are not dependent on mass.

The formula for drag force is this:



You'll notice that none of those terms is mass, not even the mass of the air. The closest you get is the rho which is the density of the air.

As I said before, this is the force which must be counteracted to achieve constant velocity. If this force is not completely matched, you decelerate and then, as a previous poster pointed out, your "F = ma" does indeed come into play as a more massive body will experience less deceleration for the same difference in force. As long as the forces are balanced, however, mass is not a factor.

Jaywalk3r

You're forgetting that the heavier load is not slowed as quickly by the air resistance. Acceleration is lower for the heavier bike in the negative direction, also.

My mistake. I thought you were talking about being able to bunny hop.

cyccommute 

To put up a bunch of letters without defining what the letters stand for is meaningless. If you go and look at the formula, FD is the drag force, v is the speed of the object relative to the fluid, A is the cross section area, CD is the dimensionless drag coefficient, and ρ is the density of the fluid.

A density is expressed as a mass per volume. Notice "mass" is part of the equation. You, as the cyclist, are pushing the air out of the way. Air may be light but it isn't massless. And drag is expressed in newtons, N, which is defined as kg-m/s^2. Since that "kg" stands for kilograms, there is a mass element in the equation, as there is anytime a force is present.

You aren't the first to say that there are "forces" which aren't dependent on mass but I have yet to see an example. From microscopic, molecular forces to universe size forces, mass is always part of the equation. Otherwise it wouldn't be a "force".



Part of the problem is that people take the wrong view about aerodynamic "drag". Yes, it is present and, yes, it has consequences. But what is flowing? When you ride a bicycle (or drive a car or fly a plane), the air flows around the vehicle but the air isn't moving. The vehicle is moving. Some motive force has to be put into the vehicle to move it through the air. If the two forces...negative aerodynamic force and positive motive force...are balanced, the overall force of the system is zero and the speed is constant. But you aren't moving through the same packet of air all the time. The forces on the air isn't balanced. As you move, you are moving another packet of air that you are accelerating up to your speed. Each time you encounter a new air particle, you are throwing the system out of balance.

From the bike's standpoint, you may seem to be moving at a constant velocity but from the air's standpoint, it is accelerating. Since the whole system is dynamic, there is a constant acceleration factor in moving the air out of the way which impacts the motive force provided by the bike. Since the bike is providing force to move the air out of the way and force is related to the mass of the object moving, mass becomes a factor at all points. This is borne out by the way that we describe riding a bicycle. You do work to move the bike down the road. You encounter a drag pushing back at you. You have to use a certain amount of power to maintain your speed. Work, power and drag all involve forces in their description.

cyccommute 

A larger mass requires more force to get it up to the same speed as a smaller mass. The large mass with retain its momentum but any loss in momentum will require more force to get it back to speed. Since aerodynamic drag is always causing a loss in momentum, the amount of force needed to move a large object is always greater. That's why everyone works to reduce the mass of vehicles. The less mass that has to be pushed into the air, the less power it takes to push the air out of the way. An 18 wheeler doesn't have a frontal area that much higher than a car but it requires massive amounts of power to move it and even more power to accelerate it if it slows down. A car requires substantially more power to move the car down the road than a bicycle because of it's larger mass.

Jaywalk3r

You completely missed the point. The increased mass is balanced by a decreased reduction in velocity from the air resistance.

Andy_K 

That's an impressive bit of hand-waving, but the most you've done is elaborate on the presence of the air's mass in the density term, which I mentioned in my previous message. You're either being intentionally obstinate or you're confused.

Nothing you've said here brings the weight of the bicycle or its rider into the drag force calculation.

acidfast7

interesting discussion.

23-622s

4 LIFE

cyccommute 

Try Newton's Third Law. If the air is acting on the bicycle and rider through aerodynamic force (and it is) , the bicycle and rider are acting on the wind..

Another way to approach the drag equation is to do a "dimensional" analysis. Put in the dimensions for the variables and eliminate them algebraically. What drops out is the mass (from the density), length and time squared. If mass isn't part of the equation, where does it come from? Density definitely contains a mass element.

You said that there are "there are a good many forces which are not dependent on mass." Show some.


You, too, should look at Newton's Third Law. It takes more power to maintain the velocity for a larger mass. The calculator I linked to lets you play around with the values. Assuming the same frontal area (not a valid assumption but...) and the same bicycle weight (8kg), a 90 kg rider has to put in 234W of power while the 70 kg rider uses 224W. If you increase the bike weight to 18 kg, the needed power increases by 5W but the differential is the same...about 10W.

A heavy rider with a heavy load isn't going to be faster than a lightweight rider with a light load. Try keeping up on your 50 lb bike with someone on a 15 lb race bike. Even on the flats, you don't have an advantage.

Andy_K 

Electromagnetic forces come to mind. I've already put aerodynamic drag among forces that are not dependent on mass. Friction. The force exerted by a spring. ....

You seem to be conflating force and the effect of the force. The density of air changes with temperature, but the mass of the air does not. Thinking about this should make clear to you the difference between the drag force depending on air density and the drag force depending on the mass of the air. But you'll probably try to dispute that, so feel free to show me the force diagram you are using to understand aerodynamic drag at a molecular level.

Jaywalk3r

Right. Equally, even for bicycle/riders of different weights, if they have the same frontal area and are maintaining the same speed.

Consider two steel poles with identical external dimensions, standing side by side. One is hollow and the other is solid. Suppose the wind is blowing orthogonally to the line segment created by the two bases of the poles, such that both poles are fully exposed to the wind.

What you are claiming is that the wind senses that the solid pole is heavier, and thus blows harder against it. That's not how it works. The wind doesn't know the weight of the bicycle/rider system. It applies exactly the same force against both, and exactly the same force is required to overcome that wind to maintain speed.

Andy_K 

Maybe you should take a closer look at the calculator you linked to.

I just went there and left the default values -- notably 0% grade, 75 kg rider, 8 kg bike. Then I went to the graph it produces and saw the at 20.02 km/h the drag force is 6.08 Newtons. Next, I changed the weight of the bike to 800 kg (that's one beefy bike!) while leaving everything else the same. The graph got re-scaled a bit so I couldn't get exactly 20.02 km/h, but at 20.05 km/h it tells me the drag force is 6.10 Newtons.

So I'm pretty sure that the calculator you linked to is telling me that aerodynamic drag force is independent of the weight of the bike.

mikeybikes

I'm enjoying this discussion on physics.

It makes my brain all tingly