Wheel upgrade

[Cyclist33]

Please tell us what speed increase you experienced, how much nippier the ride was (please use SI units only) and how much more responsive the wheels were. Also, whilst you are at it, why not define responsive and tell us how to identify responsiveness.

Surely everyone knows that responsiveness is measured in the number and frothiness of those cartoon smoke donuts you get in Roadrunner episodes.

 

[Philh]

Well I had the first ride on the new wheels. What a difference! Smoother, no crashing on bumps, felt a lot more agile although a little more nervous at the same time. My bike feels faster
I know its all subjective but they do feel so different to the stock wheels. Maybe some of it is due to the narrower tyres, different tyre compounds etc. Or maybe it is because i want them to feel different, and special.
I like them and, you know what guys, that to me is the most important. Anything to make my cycling better for me is what really counts............

.........because..........

............in 3 weeks time i will hit the big 60. I have never been fitter, healthier and happier in my entire life. Me and my wife taking up cycling together again and running together for the first time in 40 years is wonderful. My new wheels have made it even more wonderfuller.

 

[annirak]

• Lighter wheels make the bike lighter - which is a good thing. For all kinds of reasons.
  
• Aside from a small initial benefit in acceleration, there is no significant benefit is losing weight from the wheels, comapred to any other part of the bike.
• Any disadvantages from accelerating heavier wheels will be returned in the form of slower decelleration

I think that's pretty much it - in a nutshell.

Disadvantages from accelerating heavier wheels are NEVER recovered--unless you have an E-bike with regenerative braking, but that's a whole other class of problem. In fact, you also get poorer braking performance. Whether the performance degradation is significant or not is another question.

The summary is spinning mass is worse than non-spinning mass. Not only that, but spinning mass far from the axis is worse than spinning mass close to the axis.

I don't disbelieve the 0.4% figure, since the majority of the mass and, therefore, the majority of the kinetic energy, is in the rider, but I've noticed that the feel of a bike doesn't seem to be dependent on kinetic energy. For instance, riding with paniers should have an aerodynamic penalty over using a rucksack, but I wouldn't expect it to be a big aerodynamic penalty. That being said, riding with a rucksack feels much faster than riding with paniers.

 

[Citius]

Disadvantages from accelerating heavier wheels are NEVER recovered--unless you have an E-bike with regenerative braking, but that's a whole other class of problem.

The energy expended in accelerating a heavier wheel is returned in the form of slower decelleration. That's an actual law of physics.

 

[annirak]

The energy expended in accelerating a heavier wheel is returned in the form of slower decelleration. That's an actual law of physics.

Perhaps "returned" is the wrong word to convey what you mean and I misunderstood you. Are you saying that you experience lower deceleration when coasting? Otherwise, lower deceleration is not a good thing--that's equivalent to brakes that have less stopping power.

You don't get energy back on a bicycle, except in the case of hills. You get some of what you spent to go up a hill when you go down the other side. Any other time, the energy you put into increasing your kinetic energy "comes back" as thermal energy via friction, drag, and tyre deformation.​

 

[Citius]

I'm not saying I experience anything. I'm just relaying the physics. A lighter rim may accelerate quicker - but it will also decellerate quicker. The opposite is true of a heavier rim. And in cases where everything else is equal, the two will largely cancel each other out.

What really matters in 99% of all real-world situations (especially when the road goes up) is combined weight of bike and rider as a single mass.

 

[annirak]

And in cases where everything else is equal, the two will largely cancel each other out.

Only if you neglect thermodynamics and have regenerative brakes. You're ignoring the energy flow in your system. Neglecting hills, it looks like this:
PE->KE->TE

Where:
PE = Potential Energy (chemical energy stored in cyclist)
KE = Kinetic Energy (both rotational (wheels) and linear (wheels, bike, and rider))
TE = Thermal Energy (heating from aerodynamic drag, rolling resistance--tyre deformation/bearing friction--and, most significantly, the heat generated in rims and brake pads during braking)

When you say that the slower acceleration and slower deceleration cancel each other out, you imply that the energy flow in the system is:
PE<->KE
This is just not true. When I brake, energy does not flow from my wheels into my muscles. I wish it did! Once kinetic energy is in a bike, you never get it back in any form other than heat.

No, the effects do not cancel each other out.

 

[Citius]

Nobody is talking about braking (apart from you). If you accelerate a heavier rim, with 'x' amount of force, it will roll for longer than a lighter rim, if accelerated with the same force. The lighter rim will accelerate quicker with the same input, but will also lose its speed quicker. Not really sure what your argument is.

 

[annirak]

My point is that the effects do not cancel eachother out.

 

[Citius]

My point is that the effects do not cancel eachother out.

Obviously they don't literally 'cancel each other out', but in terms of actual comparative riding feedback, the net effect is very, very, very similar.

 

[annirak]

Sorry, but since you want to talk about physics, I can say with absolute certainty that the energy requirement over an acceleration/deceleration cycle is categorically higher if the rims are heavier. I can give you the maths if you want.

 

[Citius]

In think you might be misunderstanding the context of what I mean by 'accelerating'. In 'pure' acceleration terms, lighter rims win. But, every pedal stroke effectively accelerates the wheel (and the bike) and in every dead spot (ie top or bottom of stroke) the rim (and wheel, bike, rider) effectively decelerates. The flywheel effects of wheel rims are pretty negligible anyway, as has been pointed out on more than one occasion.

However, the idea that heaver rims might require more power to keep spinning is misguided, as any energy put in will always come out again at some point - ie during the dead spots, in the form of slower deceleration.

 

[Yellow Saddle]

Sorry, but since you want to talk about physics, I can say with absolute certainty that the energy requirement over an acceleration/deceleration cycle is categorically higher if the rims are heavier. I can give you the maths if you want.

OK, you are obviously very keen to do this.

Here's the given.

Two bicycles accelerate from zero to 30 kph in three minutes. Both weigh the same with their riders - 90 kg.
However, bicycle A has wheels that are lighter than bicycle B, by 400 grams for the pair. Obviously bicycle B has the 400 grams placed elsewhere.
That weight saving is shaven of all areas of the wheel, ranging from tyres (radius 368mm) to hub (radius 36mm).
Bicycle B's wheels have a mass of 2500 grams and the mass is evenly distributed from radius 368mm through 36mm.

Calculate the energy each bike requires to accelerate up to the given velocity and also express it as a percentage difference.

Thank you.

 

[Yellow Saddle]

Obviously all else remains equal, including the aerodynamics of the two sets of wheels.

 

[annirak]

In think you might be misunderstanding the context of what I mean by 'accelerating'. In 'pure' acceleration terms, lighter rims win. But, every pedal stroke effectively accelerates the wheel (and the bike) and in every dead spot (ie top or bottom of stroke) the rim (and wheel, bike, rider) effectively decelerates. The flywheel effects of wheel rims are pretty negligible anyway, as has been pointed out on more than one occasion.

You seem to be trying to tell me that heavy rims even out the speed of the bike. This is true, though the effect is negligible. You're talking about the problem in very fine-grained terms. I'm talking about the macroscopic effects.

Can we take a step back for a moment? What I care about, as a rider, is how much energy it takes me to get from point A to point B, at speed S. There are four factors that play into this: 1) the energy it takes me to get the bike (and me!) from a stop to speed S. 2) the power it takes me to maintain speed S. 3) the distance between A and B. 4) the number of times I stop (traffic lights, etc.).

The energy it takes me to get from a stop to speed S is composed of two terms: the linear kinetic energy at speed S and the angular kinetic energy of any rotating parts at speed S.
The mass of the bike, for argument's sake is:
M = Mrider + Mbike + Mwheels
Note that Mbike is the mass of the bike minus the rims and tyres. We'll ignore the hubs and spokes since they have little effect.

Then the total kinetic energy of the bike at speed S is:
KElinear = 1/2 (Mrider + Mbike) * S^2
KEangular = 1/2 (Mwheels) R^2 * v^2
where R is the radius of the rims and v is the angular velocity at speed S, which turns out to be S/R, so
KEangular = 1/2 Mwheels * S^2
So, the total energy in the bike at speed S is:
KElinear + KEangular = (1/2 Mrider + 1/2 Mbike + Mwheels) * S^2

However, the idea that heaver rims might require more power to keep spinning is misguided, as any energy put in will always come out again at some point - ie during the dead spots, in the form of slower deceleration.

Like anything in physics, start by listing what you care about.
P(v), the power required to mantain a given velocity.
Then, list the things you know:
R(v), the rolling resistance of the wheels/tyres (in Newtons), for the avoidance of doubt, this
Dl(v), the linear drag of the bike + rider
Da(v), the angular drag of the wheels
Conservation of energy suggests that the power going into the system must be the same as the power lost by the system, or it will get very hot!
P(v) = R(v) + Dl(v) + Da(v)

I don't see a mass term. Heavy rims take exactly as much power as light ones to keep moving.