Steel wheel "flywheel effect"
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Steel wheel "flywheel effect"
In trying to think of some slight benefit to steel wheels, I've wondered whether their added mass allows them to act a bit like flywheels, relative to lighter aluminum wheels. They're heavier, so it takes more energy to accelerate them, yes? But doesn't that also mean they also tend to stay in motion longer once they're up to speed? Wouldn't make a significant difference, I guess, but it must make some, right? Or is my ignorance of basic physics showing?
Maybe the "flywheel effect" accounts for the terrible braking, too.
Maybe the "flywheel effect" accounts for the terrible braking, too.
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I've been wondering this myself. I've heard it said that an ounce in the wheels is worth a pound in the frame but why exactly? What are the forces we're talking about?
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From Wikipedia:
"Rotating mass
Due to the fact that wheels rotate as well as translate (move in a straight line) when a bicycle moves, more force is required to accelerate a unit of mass on the wheel than on the frame. To accelerate a wheel, total wheel mass matters less than the moment of inertia, which describes the inertial effect of the mass resisting acceleration (inertia) based on its location with respect to the axis of rotation (the center of the wheel hub/axle). In wheel design, reducing the rotational inertia has the benefit of more responsive, faster-accelerating wheels. To accomplish this, wheel designs are employing lighter rim materials, moving the spoke nipples to the hub or using lighter nipples such as aluminum. Note however that rotational inertia is only a factor during acceleration (and deceleration/braking). At constant speed, aerodynamics are a significant factor. For climbing, total mass remains important. See Bicycle performance for more detail."
Saving weight in the hubs would have negligible effect...rims, nipples, tubes, tires...the farther form the center of rotating you lose the weight the better.
"Rotating mass
Due to the fact that wheels rotate as well as translate (move in a straight line) when a bicycle moves, more force is required to accelerate a unit of mass on the wheel than on the frame. To accelerate a wheel, total wheel mass matters less than the moment of inertia, which describes the inertial effect of the mass resisting acceleration (inertia) based on its location with respect to the axis of rotation (the center of the wheel hub/axle). In wheel design, reducing the rotational inertia has the benefit of more responsive, faster-accelerating wheels. To accomplish this, wheel designs are employing lighter rim materials, moving the spoke nipples to the hub or using lighter nipples such as aluminum. Note however that rotational inertia is only a factor during acceleration (and deceleration/braking). At constant speed, aerodynamics are a significant factor. For climbing, total mass remains important. See Bicycle performance for more detail."
Saving weight in the hubs would have negligible effect...rims, nipples, tubes, tires...the farther form the center of rotating you lose the weight the better.
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The Wikipedia information quoted by khatfull is quite right, qualitatively. In quantitative terms, mass concentrated at the axle centreline is identical to any other non-rotating mass on the bike. Mass at the extremity of the tire (i.e. where it contacts the pavement) is equivalent to 4 times the same amount of non-rotating mass. Mass between the axle centerline and the tire extremity falls in between these extremes, as a function of where it falls radius-wise between these two end points.
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I = m·r^2 Rotational Inertia of Hoop = mass · radius^2.
Example accelerating to 20mph from a stop in 10 seconds, bike weighs 25 lb, rider weighs 170 lb, each rim weighs 500 grams (1.10 lb) and each tire weighs 250 grams (0.55 lb). Together they have their mass concentrated at the bead diameter of 622mm or 311mm radius (12.2 inches or 1.02 ft) . Using 700c x 25mm tires, outer circumference = 2,110mm (6.92 ft or 1.10 ft radius). Yes I made the numbers up, but bear with me.
Vo = 0 ft/sec, Vfinal = 29.3 ft/sec,
Average acceleration rate = DeltaV / DeltaT = 29.3 ft/sec / 10 sec = 2.93 ft/sec^2 (0.091 G's).
Tire must accelerate at AccelerationAverage / Circumference = 2.93 ft/sec^2 / 6.92 ft/rotation = 0.423 rotations/sec^2.
Knowing that 1 rotation is 2 · Pi radians, the acceleration rate becomes 2.66 radians/sec^2 (known as alpha, in the rotational equivalent to F = m · a, which is T = I · alpha).
Rotational Inertia of wheel & tire hoop (since we're considering it to be a hoop) = m · r^2 = W / g ·r^2. (1.10 + 0.55) lb / 32.2 ft/sec^2 · (1.02 ft)^2 = 0.0533 slug · ft^2.
Torque to accelerate each wheel & tire assembly, T = I · Alpha = 0.0533 slug · ft^2 · 2.66 radians/sec^2 = 0.142 slug · ft^2 / sec^2 or 0.142 lb · ft.
Drag force on circumference of each tire tread to spin wheel & tire = torque on axle / lever arm (tire tread radius) = 0.142 lb · ft / 1.10 ft = 0.129 lb. For both wheel & tire assemblies X 2 = 0.258 lb
Force to linearly move bike & rider at given acceleration rate, F = m · a = (170 lb + 25 lb) / 32.2 ft/sec^2 · 2.93 ft/sec^2 = 17.7 lb [excludes aerodynamic drag, tire rolling resistance drag, and hub bearings drag].
Total % of force on bike to accelerate wheel & tire assemblies = 0.258 / (0.258 + 17.7) · 100% = 1.44%.
Make wheels and tires twice as massive, so total bike weight is increased to 28.3 lb.
Total % of force on bike to accelerate wheel & tire assemblies = 0.515 lb vs 18.04 lb or 2.78%.
My conclusion is the mass of the wheel and tire assemblies quantatively don't make that much difference, especially after you factor in the drag forces which I am not considering in my calculation.
Example accelerating to 20mph from a stop in 10 seconds, bike weighs 25 lb, rider weighs 170 lb, each rim weighs 500 grams (1.10 lb) and each tire weighs 250 grams (0.55 lb). Together they have their mass concentrated at the bead diameter of 622mm or 311mm radius (12.2 inches or 1.02 ft) . Using 700c x 25mm tires, outer circumference = 2,110mm (6.92 ft or 1.10 ft radius). Yes I made the numbers up, but bear with me.
Vo = 0 ft/sec, Vfinal = 29.3 ft/sec,
Average acceleration rate = DeltaV / DeltaT = 29.3 ft/sec / 10 sec = 2.93 ft/sec^2 (0.091 G's).
Tire must accelerate at AccelerationAverage / Circumference = 2.93 ft/sec^2 / 6.92 ft/rotation = 0.423 rotations/sec^2.
Knowing that 1 rotation is 2 · Pi radians, the acceleration rate becomes 2.66 radians/sec^2 (known as alpha, in the rotational equivalent to F = m · a, which is T = I · alpha).
Rotational Inertia of wheel & tire hoop (since we're considering it to be a hoop) = m · r^2 = W / g ·r^2. (1.10 + 0.55) lb / 32.2 ft/sec^2 · (1.02 ft)^2 = 0.0533 slug · ft^2.
Torque to accelerate each wheel & tire assembly, T = I · Alpha = 0.0533 slug · ft^2 · 2.66 radians/sec^2 = 0.142 slug · ft^2 / sec^2 or 0.142 lb · ft.
Drag force on circumference of each tire tread to spin wheel & tire = torque on axle / lever arm (tire tread radius) = 0.142 lb · ft / 1.10 ft = 0.129 lb. For both wheel & tire assemblies X 2 = 0.258 lb
Force to linearly move bike & rider at given acceleration rate, F = m · a = (170 lb + 25 lb) / 32.2 ft/sec^2 · 2.93 ft/sec^2 = 17.7 lb [excludes aerodynamic drag, tire rolling resistance drag, and hub bearings drag].
Total % of force on bike to accelerate wheel & tire assemblies = 0.258 / (0.258 + 17.7) · 100% = 1.44%.
Make wheels and tires twice as massive, so total bike weight is increased to 28.3 lb.
Total % of force on bike to accelerate wheel & tire assemblies = 0.515 lb vs 18.04 lb or 2.78%.
My conclusion is the mass of the wheel and tire assemblies quantatively don't make that much difference, especially after you factor in the drag forces which I am not considering in my calculation.
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^ But only in terms of accelerating the wheel (as in a criterium where you brake hard into a lot of corners per km and accelerate hard out of them, trying to gain an inch on your opponent.) In a hill climb or on a tour, doesn't matter where the mass is located; you just want less of it.
Edit: sorry, this post was intended in response to old's'cool, two above.
Edit: sorry, this post was intended in response to old's'cool, two above.
Last edited by conspiratemus1; 01-26-11 at 10:17 PM.
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For this hour record, he used a 3.2kg rear wheel.
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In the folding bikes forum you can find many renditions of the opposite argument. That is, in discussions of the advantages of small wheels, people like to rave about how fast you can accelerate. And it's certainly true, but I don't think the effect is very great. As for additional weight being an advantage, I'm sure it is, at certain times. Being a relatively light rider, on group rides I often find myself at the head of the pack at the top of the hills; and then while I'm pedaling down the other side all the heavier guys come coasting right by me. Any extra weight will have that effect.
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I like this. Now I can tell people my 4 pound steel wheels are a "feature". Yes.
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^ But only in terms of accelerating the wheel (as in a criterium where you brake hard into a lot of corners per km and accelerate hard out of them, trying to gain an inch on your opponent.) In a hill climb or on a tour, doesn't matter where the mass is located; you just want less of it.
Edit: sorry, this post was intended in response to old's'cool, two above.
Edit: sorry, this post was intended in response to old's'cool, two above.
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Try this. With your bike in a stand grab a pedal and spin it up to speed. You can feel how long and how much force it takes. Now take your buddy's bike, the one with the heavier wheels, and spin that one up to speed, in the same gear of course. Feel the difference.
#16
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I think what the OP was thinking was that since steel wheels are heavier, once already spinning, they would stay spinning longer, sort of a gyro effect. I think that may well be true but only if all other aspects were equal, such as wind and rolling resistance. For instance, if I were to spin up a 10 lb iron flywheel on a spindle to say 500 rpm, and an equally sized but much lighter flywheel made say of carbon fiber weighing say 2 lbs to that same RPM on an equal spindle, the heavier iron flywheel would no doubt continue to rotate longer, long after the lighter flywheel came to a stop.
I think the only advantage to this effect would be over long, flat runs, otherwise, the only effect I could foresee is the inability to stop as well.
I think the only advantage to this effect would be over long, flat runs, otherwise, the only effect I could foresee is the inability to stop as well.
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I do way better in Crit sprints on my steel bike than I did on my carbon. Im small and light and its quite a disadvantage on all out sprints and going downhills. It just feels like when I get going on a sprint on my steel, bike keeps going. If I ease up a little on a light carbon, I slow down. Not very scientific but hey, it works for me.
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Heavy Wheel = Mass in motion... Thats why when I got back into riding at 284# riders would move away form me when going down hill fast... They were trying to avoid my gravitational field...
Now at 234# its more diminished...
Oddly the faster you go - The heaver you are...
And lets not even ask what time it is...
Now at 234# its more diminished...
Oddly the faster you go - The heaver you are...
And lets not even ask what time it is...
#20
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Is there going to be a quiz on this??
andy
andy
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There might be some confusion here about what a flywheel actually does. They were important in one-cylinder stationary steam engines to smooth out the motion of the reciprocating piston. The "spinergy" imparted to the rotating flywheel during the piston's stroke helped to carry the piston through top- and bottom- dead centre, the points where it had to stop momentarily and reverse direction while there was no steam pressure being applied to it. (In locomotive steam engines, the driving wheels themselves acted as the flywheels -- the forward movement of the train pushed the pistons through dead centres.)
On the bicycles that most of us ride, there can be no "flywheel effect" because the ratcheting freewheel prevents the rotating rear wheel from applying any "push" to the pedals. (Obviously fixies are different.) And no front wheel will behave like a flywheel. Whether flywheels, as on a fixie, do anything to "smooth" human muscular performance is an interesting question, since the action of the pedaling foot is reminiscent of a piston. Bicycles intended to be ridden on indoor tracks had the very lightest wheels possible, though, so it would seem that track riders sought to minimize the flywheel effect, not exploit it.
The effect of adding mass to the circumference of a bicycle wheel that cannot "drive" the pedals is simply to make the wheel harder to accelerate and harder to stop. In this respect it is identical to placing mass anywhere else on the bicycle, except that the effect or a given mass is somewhat more pronounced when the mass is rotating as well as being moved linearly, as Suburban Grind rigorously demonstrates.
On the bicycles that most of us ride, there can be no "flywheel effect" because the ratcheting freewheel prevents the rotating rear wheel from applying any "push" to the pedals. (Obviously fixies are different.) And no front wheel will behave like a flywheel. Whether flywheels, as on a fixie, do anything to "smooth" human muscular performance is an interesting question, since the action of the pedaling foot is reminiscent of a piston. Bicycles intended to be ridden on indoor tracks had the very lightest wheels possible, though, so it would seem that track riders sought to minimize the flywheel effect, not exploit it.
The effect of adding mass to the circumference of a bicycle wheel that cannot "drive" the pedals is simply to make the wheel harder to accelerate and harder to stop. In this respect it is identical to placing mass anywhere else on the bicycle, except that the effect or a given mass is somewhat more pronounced when the mass is rotating as well as being moved linearly, as Suburban Grind rigorously demonstrates.
#22
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I think some are confusing 'flywheel' effect with gyroscopic effect?
Heavier objects take more energy to accelerate, no one is arguing that point.
Heavier objects take more energy to stop, we all know this.
With this in mind, take a balloon and a basketball and through them down the steps, or down the hallway for that matter, if both have the same forces applied to them, the basket ball will go further and get there first.
There's no doubt that in most cases the lighter wheelset is the greater advantage but for coasting purposes, especially downhill, I could easily see where the steel wheel or heavier wheel would continue further. Acceleration has nothing to do with it at that point, it's a matter of which will continue further without any further push or energy.
Heavier objects take more energy to accelerate, no one is arguing that point.
Heavier objects take more energy to stop, we all know this.
With this in mind, take a balloon and a basketball and through them down the steps, or down the hallway for that matter, if both have the same forces applied to them, the basket ball will go further and get there first.
There's no doubt that in most cases the lighter wheelset is the greater advantage but for coasting purposes, especially downhill, I could easily see where the steel wheel or heavier wheel would continue further. Acceleration has nothing to do with it at that point, it's a matter of which will continue further without any further push or energy.
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Push on the axle or push from the axle. The rotational inertia (flywheel effect) acts to make the wheel and tire easier or harder to accerate to a higher spin speed, or easier or harder to decrease in spin speed (negative acceleration).
The gyroscopic effect is what you feel when you hold a spinning wheel assembly by the axles or QR's and rotate the axle perpendicular to the wheels' spin, such as steering with forks would do. Did you expect to have to react to the torque which tried to rotate the axle about the third axis? The appearance of twisting force about the third axis is the gyroscopic effect.
If I was an aeronautical engineer, I would describe the wheel as rotating about the pitch axis - and the steering action of the forks would rotate the axle about the yaw axis (mostly) which would cause an imparted moment into the fork which causes the wheel to try to tilt on the roll axis. Meaning the rider would have to respond by leaning or unleaning more. The front and rear wheels can also produce a gyroscopic moment even in steady turns with no steering change, it becomes more of a factor in decreased radius corners.
All of these principles are covered in dynamics engineering courses.
And someone should have admonished me for neglecting to include a weight for an inner tube in my previous calculation, and maybe some estimation of spokes/nipples. Should I guess no other C&Vers are engineers/ physicists/ or scientists?
The gyroscopic effect is what you feel when you hold a spinning wheel assembly by the axles or QR's and rotate the axle perpendicular to the wheels' spin, such as steering with forks would do. Did you expect to have to react to the torque which tried to rotate the axle about the third axis? The appearance of twisting force about the third axis is the gyroscopic effect.
If I was an aeronautical engineer, I would describe the wheel as rotating about the pitch axis - and the steering action of the forks would rotate the axle about the yaw axis (mostly) which would cause an imparted moment into the fork which causes the wheel to try to tilt on the roll axis. Meaning the rider would have to respond by leaning or unleaning more. The front and rear wheels can also produce a gyroscopic moment even in steady turns with no steering change, it becomes more of a factor in decreased radius corners.
All of these principles are covered in dynamics engineering courses.
And someone should have admonished me for neglecting to include a weight for an inner tube in my previous calculation, and maybe some estimation of spokes/nipples. Should I guess no other C&Vers are engineers/ physicists/ or scientists?
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I think there are several engineers and physicists here. You covered those omissions when you said the wheel would be considered as a hoop. At that point I had expected a short discussion of effect of how mass is distributed in the wheel, but it was mentioned in previous posts, and the omission reallyh didn't affect your conclusion.
A more detailed mass model would not affect your conclusions. No problem.
Unless you'd like to get into quantifying the inertial effects of brass v. aluminum nipples, butted spokes, Michelin tubes v. Lunar Light, large flange/low flange hubs, et cetera? Not me! Yes, there are a lot more details, but so what?
One of your best points is that all of this is routinely taught to students in college classes - it's some of the best-understood physics, and there is no magic in it. So many people act as if this is magical, esoteric stuff.
BTW, vehicle engineers use the same terms to discuss vehicle directional changes: pitch, yaw, and roll, especially when designing braking and stability systems. There we're at times flying a car along the ground. But I'm not a vehicle dynamicist, though I work with them and have stayed at a Holiday Inn.
A more detailed mass model would not affect your conclusions. No problem.
Unless you'd like to get into quantifying the inertial effects of brass v. aluminum nipples, butted spokes, Michelin tubes v. Lunar Light, large flange/low flange hubs, et cetera? Not me! Yes, there are a lot more details, but so what?
One of your best points is that all of this is routinely taught to students in college classes - it's some of the best-understood physics, and there is no magic in it. So many people act as if this is magical, esoteric stuff.
BTW, vehicle engineers use the same terms to discuss vehicle directional changes: pitch, yaw, and roll, especially when designing braking and stability systems. There we're at times flying a car along the ground. But I'm not a vehicle dynamicist, though I work with them and have stayed at a Holiday Inn.
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The biggest benefit would be that more mass in the wheel equals more mass in the legs. That is, over a period of time