# Folding Bikes - Rotating vs non-rotating mass.

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View Full Version : Rotating vs non-rotating mass.

Speedo
02-09-07, 11:48 AM
The issue of rotating vs. non-rotating mass came up in the Downtube thread. I posted about it in the Swift thread once. It's an interesting topic. I haven't found a thread devoted to it; why not in Folding Bikes?

I looked at this when James_Swift quoted the old saw about a pound of rotating weight being the same as a pound on the bike. I wasn't sure if I believed it, and did the following calculation:

The energy required to raise a mass M to a speed V is the familiar:

Et=0.5*M*V^2

The energy required to rotate an object with moment of inertia I to a rotaional rate omega is:

Er=0.5*I*omega^2

If the mass M is distributed at the radius R, the the moment of inertia is:

I=M*R^2

For a wheel moving forward at speed V, the rotation rate is:

omega=V/R

so Er=0.5*M*R^2*V^2/R^2=0.5*M*V^2

So a non-rotating mass requires energy

Et=0.5*M*V^2

While a rotating mass requires both the translational energy and the rotational energy to get up to speed.

Et+Er=0.5*M*V^2+0.5*M*V^2 = M*V^2; twice the non-rotating energy.

So, there is some truth to the pound of rotating = 2 pounds on the bike. I was a little surprised at the time that the result is independent of the radius of the wheel. Prior to the calculation I would have guessed that small wheels get a bit of a free ride.

This result only applies in accelerating the bike to speed. Once at speed, on a level road, you have to apply a torque to overcome the friction in the bearings (Edit: and rolling resistance and aerodynamic drag). That torque is not a function of the rotating mass. (Edit: What ultimatly limits your speed is mostly aerodynamic drag.)

So, while I agree in general that lighter is better than heavier, unless quick acceleration is critical for you, there's no particular reason to fuss over rotating vs non-rotating mass.

I know that there are other opinions out there!

Speedo

Rincewind8
02-09-07, 12:55 PM
I was a little surprised at the time that the result is independent of the radius of the wheel. Prior to the calculation I would have guessed that small wheels get a bit of a free ride.

The difference usually is that larger diameter wheels (and tires and tubes) are heavier. But in the end they have some drawbacks that make the difference negligible.

See:
http://hea-www.harvard.edu/~fine/opinions/wheelsize.html

makeinu
02-09-07, 12:58 PM
hmmm, omega=V/(2*pi*R), no? So it is much less than double the energy: about 108% rather than 200%. Also, you have to remember that it's only 108% of the energy for the wheel, but the energy put into the wheel is only part of the total energy required to move the bike.

Furthermore, all else being equal (tire width, spoke material, etc) the smaller wheel will have a smaller mass and thus a smaller moment of inertia. Since most of the rotating mass is due to the tire, let's assume the entire rotating mass is cocentrated at the circumference. Let the mass per unit circumference be m0, then the total rotational mass is 2*pi*R*m0.

Combining all of the above refinements we obtain the following expression for the energy required to move the wheel:
Ew=(R*m0*V^2)*pi+(R*m0*V^2)/4*pi

and for the entire bike:
E=Ef+2*Ew=(M/2+R*m0*2*pi+R*m0/2*pi)*V^2, where
R is the radius of the wheels,
2*pi*R*m0 is the weight of the spokes, tires, and rims of each wheel,
and M is the weight of the rest of the components (hubs, frame, etc)

Anyone want to plug some realistic numbers into this equation?

makeinu
02-09-07, 01:06 PM
The difference usually is that larger diameter wheels (and tires and tubes) are heavier. But in the end they have some drawbacks that make the difference negligible.

See:
http://hea-www.harvard.edu/~fine/opinions/wheelsize.html (http://hea-www.harvard.edu/%7Efine/opinions/wheelsize.html)
Thanks for the link. He derived an equation similar to mine and when he plugged in some realistic numbers he calculated that less than a 1% difference in energy can be achieved due to shrinking the wheels. So I guess that settles it.

Speedo
02-09-07, 01:22 PM
hmmm, omega=V/(2*pi*R), no?

No. omega=V/R. Let T be the period of rotation, then omega = 2*pi/T. V=2*pi*R/T, divide both sides of the equation by R, and you get V/R = 2*pi/T = omega.

Also, you have to remember that it's only 108% (edit 200%)of the energy for the wheel, but the energy put into the wheel is only part of the total energy required to move the bike.

I haven't forgotten that. The analysis seeks to give you a feel for the impact of rotating vs non-rotating mass. Maybe we are agreeing that the mass of the complete package, your M, which really has to include the bike and the payload (rider and any junk they are carrying) swamps everything else.

makeinu
02-09-07, 01:30 PM
No. omega=V/R. Let T be the period of rotation, then omega = 2*pi/T. V=2*pi*R/T, divide both sides of the equation by R, and you get V/R = 2*pi/T = omega.

I haven't forgotten that. The analysis seeks to give you a feel for the impact of rotating vs non-rotating mass. Maybe we are agreeing that the mass of the complete package, your M, which really has to include the bike and the payload (rider and any junk they are carrying) swamps everything else.

Ok.

jur
02-09-07, 04:36 PM
*applauds*

I LOVE it when myth gets debunked by solid reasoning and even better, by math proof.

SO the equations show that to get up to speed, the approx. 1.5-2kg of wheel mass would be equivalent to 3-4kg of "plain" mass. That still is perhaps only 3-4% of total bike+rider mass, so unless you are racing, that won't make any difference. And even when you're racing, hardly any time is spent accelerating. When Robbie McEwen dashes for the finish line at the end, then it would come into play, but perhaps a stronger wheel would be more important to withstand those huge torques that the back wheel has to withstand. I remember from the 2005 TdF that Rasmussen (?) on the last TT day wrecked back wheel after back wheel trying to get up to speed after taking a tumble.

jur
02-09-07, 04:44 PM
Further to this, there are those who have somehow been convinced that riding uphill, rotating mass has to be "lifted twice" and therefore is doubly important to get wheel mass down. I found that little gem at the Audax Australia site.

I suppose those people don't ride their trainers on anything less than a concrete floor for fear of crashing through the floor when spinning that wheel uber-fast. :D

jur
02-09-07, 05:10 PM
Well, the spinning wheel does affect translational motion because the bicycle (by its very purpose) couples the rotational motion of the wheel to the translational motion of the bike. I know you want to consider the physics of the translational motion by itself, but the translational motion does not exist by itself with a bicycle.You are quite correct; however it is possible (and in fact simpler) to treat the 2 different kind of motions entirely separately and add the effects at the end, like in the OP. E.g., you can picture a moving bike as a bike with stationary wheels sliding along, plus a stationary bike with spinning wheels.

Also, I think it is a little misleading to say that only the acceleration is affected by the rotational mass. In an ideal physical sense this is true, but in an ideal physical sense all the energy put into the pedals will result in acceleration. However, the bike will slow to a halt if the rider stops pedaling. So it is necessary to "reaccelerate" the wheels to compensate for deceleration due to other factors. I'm pretty sure the smaller wheels will increase the efficiency of this process, even though the actual speed of the bike is not increasing, but I'm not that kind of engineer, so perhaps you can clarify.Only a tiny tiny bit misleading: If a wheel rotational speed is constant, then there is no additional effect and I am 100% correct. However, as you pedal, the torque is not constant so there is in fact a small variation in speed, which is a small positive followed by a small negative amount of acceleration. This effect is larger when you ride up a very steep hill. So in that case I am indeed a tiny bit misleading. :)

I suppose this could be the reason why spinning with as constant a torque as possible is harped on a being best - you minimise those tiny accelerations.

Speedo
02-10-07, 07:15 AM
*applauds*

My blushes, thank you.

... so unless you are racing, that won't make any difference. And even when you're racing, hardly any time is spent accelerating. When Robbie McEwen dashes for the finish line at the end, then it would come into play, ...

I wouldn't go quite that far. Even people who are recreational riders and like to ride fast in pace lines may care. The ability to jump fast can make the difference between hanging on to the wheel in front of you, and dropping off the back. I have plenty of experience being dropped off the back! :D

geo8rge
02-10-07, 06:48 PM
Small Dia wheels rotate faster, larger have more mass at a further distance. So the angular momentum comes out almost the same.

noahj
02-10-07, 10:20 PM
You know, I've seen endless thrashes about this, and I really don't think it matters. You're not accelerating wheels, you're accelerating (and decelerating) the combined mass of bike and rider, of which the wheels represent a fairly small fraction of both the total mass and the rotating mass.

In terms of how fast you go for the amount of energy you put in, the only thing that really matters is the energy that's lost to friction of various kinds. So... forget masses and distances, let's see someone prove the issue in terms of thermodynamics.

ahorner1946
02-11-07, 08:32 AM
So are you guys saying I would be wasting 400 bucks on a set of Velocity Tracian wheels, as far as improving my climbing and average speed while riding my RANS Stratus, or any other bike for that matter?

Thanks guys, you saved me 400 bucks. That will buy me a DT 8H.

Al

dalmore
02-13-07, 03:59 AM
As someone who flunked more college calculus classes than I passed (4 to 3), I feel like I'm bringing a knife to a gunfight. :o

Let's try looking at this from a different perspective. Consider that all the power to propel a bike forward is applied at the hub of the wheel. This power is transfered by the spokes through the tire to the road surface. Essentially we are looking at a long lever (spokes) with the fulcrum at one end (hub) and the load at the other end (tire).

The mass on that lever and it's distribution along the length of that lever will have a direct influence on the power transferred from your feet to the road service. More mass - particularly mass nearer the load end (rim/tire) of the lever - will require more power to move the lever. The power required to move the lever is power that is not transferred to the road surface and used to propel the bike forward. So a heavier wheel and tire will be diverting power from the process of moving the bike forward.

Having documented my lack of math skills, I'll leave it to someone else to do the math. :)

Speedo
02-13-07, 07:01 AM
... So a heavier wheel and tire will be diverting power from the process of moving the bike forward.

Having documented my lack of math skills, I'll leave it to someone else to do the math. :)

Sort of, but not quite. Initially some of the engine's (that's you) power will have to do toward spinning the wheel up, but at speed you only have to add power to overcome friction, rolling resistance and aero drag.

Here's a thought experiment for you. What if, instead of really light wheels, you have really, really, really heavy wheels. Titanium spokes with a rim and rolling surface make of spent uranium. These wheels are so heavy you can't lift them, and can barely provide the energy to start them rolling. Now, the claim above is that the problem with heavy wheels is that they slow your acceleration, but don't appreciably limit your speed. What if we were able to magically skip the acceleration step. Somehow you get started at 20 MPH on a level road with these mondo heavy wheels. What will happen?

If the claim is true that the heavy wheels slow you down, then you would expect that you would slow and stop immediately. But these wheels act like flywheels, they are really energy storage devices. Even just coasting the energy stored in the rotational momentum of the wheel will easily overcome the rolling resistance, bearing friction, and aero drag. You will gradually slow down, but it would take much much longer to slow down than it would for light wheels, which would have stored less energy.

I hate to do this to you, but here's an expression to expand your science education:

Lagrangian Mechanics

In Lagrangian Mechanics you analyze the motion of a system by following the energy. You can think about what happens when you ride a bike by thinking about where the energy goes. The engine (that's still you) puts energy into the system. From a start (on level ground) that energy goes into translational kinetic energy, rotational kinetic energy, bearing friction (heat), rolling resistance (heat), aerodynamic drag (still more heat). The system will accelerate until the input (you) is balanced by the rest of the system. When you are no longer accelerating none of your input power is going into accelarating the system (duh). It is going into friction (heat), rolling resistance (heat) and aero drag (heat). Most of it will go to aero drag. The power that goes into overcoming aero drag is proportional to the cube of your speed(!!). You haven't lost the energy that went into accelerating the system. It is stored as translational kinetic energy, rotational kinetic energy (and if you climbed a hill, potential energy). If you start to coast then the stored energy, gets turned to heat, and you will slow down and eventually stop.

The exclamation points on the aero drag comment are to point out that the cube of your speed is a very very steep curve. It's aero drag that will limit your speed on level ground. You want to go faster? Before you spend \$400 on some fancy wheels, spend \$40 on aero bars.

Speedo

dalmore
02-13-07, 07:54 AM
Sort of, but not quite. Initially some of the engine's (that's you) power will have to do toward spinning the wheel up, but at speed you only have to add power to overcome friction, rolling resistance and aero drag.

...

Before you spend \$400 on some fancy wheels, spend \$30 on aero bars.

Speedo

While I don't find any fault with the logic you present, it doesn't explain the physical observations I have made. Try this physical experiment for yourself. Get up to speed on your bike using any means you desire. Without changing your front chainring, shift to your largest rear cog and gauge how much resistance you find on the pedal. Then while still at speed and again without changing your front chainring, shift to your smallest rear cog and gauge how much resistance you find on the pedal. There was more resistance when you were on the smaller cog yes? Friction, aero drag and rolling resistance are practically unchanged. I argue this supports my theory of the wheel as a lever. (If you are going to argue that it's the effects of gearing - look at how gearing works and see if that brings right back to the lever.) The larger cog is applying the force is a position that is more advantageous effectively shortening the lever. So if the lever theory holds to this degree, why does it suddenly not apply when trying to calculate the effects of varying the load at the end of the lever - ie more mass at the rim and tire?

As far as upgrades ... I for one can not magically reach 20 mph where the rotational mass of the tire become a flywheel and aerodynamic drag is my biggest enemy... I have to reach that 20 mph point first by exerting energy. And with the rolling terrain around here, I spend more time accelerating up hills than I do gliding down them in a tuck position. So for me, \$400 on lighter wheels to get better acceleration might be a more cost effective upgrade than \$30 on aero bars.

Speedo
02-13-07, 09:09 AM
Try this physical experiment for yourself. Get up to speed on your bike using any means you desire. Without changing your front chainring, shift to your largest rear cog and gauge how much resistance you find on the pedal. Then while still at speed and again without changing your front chainring, shift to your smallest rear cog and gauge how much resistance you find on the pedal. There was more resistance when you were on the smaller cog yes?

Yes, you apply more torque, but at a lower rotation rate, so the power is the same. The energy you put into the bike is the same in either case. Your engine might run more efficiently at a particular torque/rate combination, so you will find you prefer a gear, but the energy you put in is the same in either case.

The wheel is a "lever" as you put it. It factors into the gear sense that you experience. That's why Gear Inches includes the diameter of the wheel. But don't confuse the mechanical advantage of gearing with where the energy goes when you put it into the bike.

As far as upgrades ... I for one can not magically reach 20 mph where the rotational mass of the tire become a flywheel and aerodynamic drag is my biggest enemy... I have to reach that 20 mph point first by exerting energy. And with the rolling terrain around here, I spend more time accelerating up hills than I do gliding down them in a tuck position. So for me, \$400 on lighter wheels to get better acceleration might be a more cost effective upgrade than \$30 on aero bars.

When you go up a hill you, the engine, are increasing your potential energy. Light wheels, being lighter, are easier to carry up the hill. But as you crank up the hill, even the little changes in speed from the crank pulses are a subtle dance between the energy you put in, the energy stored as translational, and rotational kinetic energy, friction and drag, and potential energy. The energy that goes into spinning up a wheel, you get back; it is conserved.

Look, I've always conceded that lighter is better. But the point is to understand what you are, and what you are not, buying. I'm someone who has bike priorities, and believes that money spent on bikes is money well spent. If you want to spend \$400 (\$500, \$1000!) on wheels and your only reason is that they look cool and make you feel fast, that's a good enough reason in my book. If you have a set of modern alloy rim wheels, with good quality hubs then, upgrading for lightness is not going to dramatically improve your overall speed.

Speedo

spambait11
02-13-07, 11:53 AM
=The energy that goes into spinning up a wheel, you get back; it is conserved.
I guess it depends how you define "conserved" because someone else has argued that this is not accurate. He says the bike is a conservative system, but muscles are not. Thus you lose energy every time you pedal; energy does not all of a sudden go back into your leg. The discussion is in terms of frame flex, but it does get to this issue in a roundabout way.

Speedo
02-13-07, 12:30 PM
I guess it depends how you define "conserved" because someone else has argued that this is not accurate. He says the bike is a conservative system, but muscles are not. Thus you lose energy every time you pedal; energy does not all of a sudden go back into your leg. The discussion is in terms of frame flex, but it does get to this issue in a roundabout way.

No, but the energy does go toward motion. See the example of the heavy flywheel wheel. The energy is stored in the wheel and goes toward overcoming friction, drag, and rolling resistance. In a normal bicycle wheel, the amount of energy stored is small, but that's the point.

If you have to hit brakes, it will go toward heating the rims.

Speedo

makeinu
02-13-07, 01:11 PM
No, but the energy does go toward motion. See the example of the heavy flywheel wheel. The energy is stored in the wheel and goes toward overcoming friction, drag, and rolling resistance. In a normal bicycle wheel, the amount of energy stored is small, but that's the point.

If you have to hit brakes, it will go toward heating the rims.

Speedo

Yeah, but the same holds true apart from the rotating wheel considerations. For example, you won't lose any energy riding a 100 pound bike instead of a 30 pound bike. However, a 100 pound bike is obviously undesirable.

The important quantity here is the power required, not the energy.

Speedo
02-13-07, 01:31 PM
Yeah, but the same holds true apart from the rotating wheel considerations. For example, you won't lose any energy riding a 100 pound bike instead of a 30 pound bike. However, a 100 pound bike is obviously undesirable.

The important quantity here is the power required, not the energy.

Looking at the energy is useful because, regardless of the absolute value, it reveals interrelationships between the mechanical components. I started down the energy path because some posters seemed to be arguing that rotating wheels were a constant sink of energy, and hence consumed an inordinate amount of the available power. Following the energy reveals that that is not the case.

I'm always happy to concede that light is better than heavy. My only claims are that:

1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

Speedo

dalmore
02-15-07, 03:50 AM
Looking at the energy is useful because, regardless of the absolute value, it reveals interrelationships between the mechanical components. I started down the energy path because some posters seemed to be arguing that rotating wheels were a constant sink of energy, and hence consumed an inordinate amount of the available power. Following the energy reveals that that is not the case.

I'm always happy to concede that light is better than heavy. My only claims are that:

1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

Speedo
Sorry to have taken so long to reply - been busy. Looking at energy can also be misleading. Consider that whether a biker rides up a hill or pushes the bike up the hill, the energy is the same but the experience is quite different.

I think the weight of wheels and tires have a constant effect on bikes that is greater than the effect that same amount of weight would have elsewhere on the bike. I tried to explain the idea with the example of a lever and failed. :o (I still maintain that's valid but I'll drop that for now.) Let me try again with a different aproach.

We all agree that rotating mass has an effect on acceleration to a greater degree than non rotating mass. We all agree that the forces of friction (from various sources) and aerodynamic drag are acting to prohibit forward movement of the bike. These forces are constant - if you stop applying power, these forces cause a bike to slow down.

Now consider that power from the biker is not being applied constantly like gravity or a rocket engine. Instead power is being applied in pulses. So the bike is in a constant cycle of acceleration and deceleration.

Perhaps you think the pulses are so short that they don't matter. Well regardless of the cadence, a bike with a standard crank and plaform pedals is only under power about half the time. 1/4 of the time from the left foot pushing down on the pedal from about 90 to 180 degrees and 1/4 of the time from the right foot pushing down on the pedal from about 90 to 180. I think that does matter.

jur
02-15-07, 04:29 AM
Perhaps you think the pulses are so short that they don't matter. Well regardless of the cadence, a bike with a standard crank and plaform pedals is only under power about half the time. 1/4 of the time from the left foot pushing down on the pedal from about 90 to 180 degrees and 1/4 of the time from the right foot pushing down on the pedal from about 90 to 180. I think that does matter.I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.

dalmore
02-15-07, 06:43 AM
I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.

Perhaps but I don't think so. Angular momentum is considered to be a small percentage of the force at work here with most of the force at work being aero drag at speed. Since aero drag continues to get greater with more speed, I think any advantage a heavier wheel gains from angular momentum during the deceleration phase would get smaller and smaller as speed increases.

Speedo
02-15-07, 06:47 AM
I wonder if this doesn't come out in the wash: Yes you are correct that there is small variations in speed due to varying pedalling torque in a full revolution (see post #9); so a wheel does indeed have to be re-accelerated. However, a heavy wheel's bigger angular momentum will reduce the slowing by a small amount compared to a lighter wheel, so has to be re-accelerated by less.

Agree with all this except "re-accelerated less". In the case of pulses of speed as you piston crank up a hill, with each up pulse you accelerate the wheel adding energy, but as it slows that energy is given back. With each piston stroke you add add kinetic energy (translational and rotational) and as you rotate the pedal (without putting a lot of power in) preparing for the next big down stroke that kinetic energy is transformed into potential energy. i.e. you went a little further up the hill.

makeinu
02-15-07, 07:12 AM
I think you guys are getting distacted. Obivously energy is always conserved. Apart from certain loss processes (drag, friction, etc) energy will always remain within the system.

However, the body is primarily restricted by the power it can deliver, not the energy. It doesn't take any more energy to move an elephant up and down a ladder than it does a paperclip (zero in both cases), but your body will waste a lot of energy in order to muster the necessary power to move the elephant.

Furthermore, it's better to retain control over acceleration/deceleration with your legs. If you invest energy in getting a large wheel up to speed then you risk losing that energy if you need to brake, but if you keep the energy in your muscles then you don't risk as much. On average the higher risk should translate to more energy lost due to braking.

Speedo
02-15-07, 07:28 AM
Looking at energy can also be misleading. Consider that whether a biker rides up a hill or pushes the bike up the hill, the energy is the same but the experience is quite different.

Even more different if (since this is in folding bikes) the biker folds the bike puts it on his back and carries it up the hill. But in any of those cases examining the energy is useful. It's just a technique, it reveals relationships in the system. How the rider experiences the system, may be misleading. It's subjective.

I think the weight of wheels and tires have a constant effect on bikes that is greater than the effect that same amount of weight would have elsewhere on the bike. I tried to explain the idea with the example of a lever and failed. :o (I still maintain that's valid but I'll drop that for now.) Let me try again with a different aproach.

We all agree that rotating mass has an effect on acceleration to a greater degree than non rotating mass. We all agree that the forces of friction (from various sources) and aerodynamic drag are acting to prohibit forward movement of the bike. These forces are constant - if you stop applying power, these forces cause a bike to slow down.

Well, no and yes. None of the friction and drag forces are "constant". Bearing friction will be proportional to rotation rate, and hence speed. Aerodynamic drag is proportional to the square of the relative air speed. I think that tire rolling resistance is proportional to speed, but don't know for sure. But, yes I agree that on level ground, in no wind, if you stop applying power they will cause the bike to slow down.

Now consider that power from the biker is not being applied constantly like gravity or a rocket engine. Instead power is being applied in pulses. So the bike is in a constant cycle of acceleration and deceleration.

Perhaps you think the pulses are so short that they don't matter. Well regardless of the cadence, a bike with a standard crank and plaform pedals is only under power about half the time. 1/4 of the time from the left foot pushing down on the pedal from about 90 to 180 degrees and 1/4 of the time from the right foot pushing down on the pedal from about 90 to 180. I think that does matter.

Okay, we are at the point where we agree that the bike is "up to speed". In getting the bike up to speed I paid a penalty for rotating vs non-rotating mass. Now I'm up to speed, and I deliver power to the system in a non-constant way. In the subjective way that I experience this, unless I am piston pedalling up a hill, I don't experience this as pulses. Why not? Well, we need to look at (sorry :rolleyes: ) energy. The inertia of the system, that is, the energy stored as kinetic energy smooths the motion over the non-constant input power. We aren't really in a steady state, but have small variations around a given speed. With each power pulse the energy delivered is distributed between heat (friction, and drag) and kinetic energy (we speed up a bit). In the between pulse period, the kinetic energy goes to heat (friction and drag) and we slow down a bit.

All you need to recognize is that kinetic energy is kinetic energy. It can be stored as forward motion, and it can be stored in a rotating object. It doesn't matter which. Consider the example of the bike with massive flywheel wheels. At speed, with each power pulse I would put most of the energy delivered into rotational kinetic energy. Very little would go into translational kinetic energy, so I would experience very little speed change, but I have stored the same amount of energy. Between pulses the very slightest of slowing the the flywheel will deliver enough energy to overcome friction and drag. So I would experience very little slowing between pulses. Note, though, that there is not more power going into the system then there is in the light wheel case.

Note to all respondees: please don't get on my case as advocating giant flywheel wheels. I'm not. They are just a specific example that is useful because they are so extreme!

Speedo
02-15-07, 07:50 AM
I think you guys are getting distacted. Obivously energy is always conserved. Apart from certain loss processes (drag, friction, etc) energy will always remain within the system.

However, the body is primarily restricted by the power it can deliver, not the energy. It doesn't take any more energy to move an elephant up and down a ladder than it does a paperclip (zero in both cases), but your body will waste a lot of energy in order to muster the necessary power to move the elephant.

Furthermore, it's better to retain control over acceleration/deceleration with your legs. If you invest energy in getting a large wheel up to speed then you risk losing that energy if you need to brake, but if you keep the energy in your muscles then you don't risk as much. On average the higher risk should translate to more energy lost due to braking.

Uh, sorry, but you don't seem to be clear on energy, power, or the realtionship between them.

To move a mass M, up a ladder, increasing it's height by h, in a gravitational field with gravitaional acceleration g, will take energy M*g*h. So, if an elephant is a mass M and a paperclip is mass m, then the difference in energy required would be (M-m)*g*h.

Power is energy per unit time. It is the rate of energy delivered, or extracted from a system. You are right in that a rider will be limited in how much energy he can deliver per unit time. His power does have some finite limit.

Braking is an issue. In the other discussions I was arguing that in near steady state, you get the energy out of the system, in overcoming drag and friction, that you put in. If you hit the brakes, then all that energy goes to heat. But remember, hitting the brakes you give up kinetic energy from ALL sources. That includes the translational kinetic energy. The translational kinetic energy of a 150 pound biker on a 25 pound bike dwarfs the rotational kinetic energy in the rotating wheels. If you are doing nothing but start and stop, your wheels are a small part of your worries!

Speedo

dalmore
02-15-07, 08:06 AM
Well, no and yes. None of the friction and drag forces are "constant". Bearing friction will be proportional to rotation rate, and hence speed. Aerodynamic drag is proportional to the square of the relative air speed. I think that tire rolling resistance is proportional to speed, but don't know for sure. But, yes I agree that on level ground, in no wind, if you stop applying power they will cause the bike to slow down.

They most certianly are constant in the sense that they are not pulsing.

Okay, we are at the point where we agree that the bike is "up to speed".

It's obvious that you are not getting it. The bike is never at a constant speed. Never. It's either speeding up or slowing down.

makeinu
02-15-07, 09:07 AM
Even more different if (since this is in folding bikes) the biker folds the bike puts it on his back and carries it up the hill. But in any of those cases examining the energy is useful. It's just a technique, it reveals relationships in the system.

Yes, it reveals precisely that you are leaving out the most important energies. It reveals that smaller wheels, like gearing, weight, and almost everything else about a bicycle are not meant to reduce mechanical losses, but biomechanical losses. It reveals that we are better off considering the power than the energy.

Okay, we are at the point where we agree that the bike is "up to speed". In getting the bike up to speed I paid a penalty for rotating vs non-rotating mass. Now I'm up to speed, and I deliver power to the system in a non-constant way. In the subjective way that I experience this, unless I am piston pedalling up a hill, I don't experience this as pulses. Why not? Well, we need to look at (sorry :rolleyes: ) energy. The inertia of the system, that is, the energy stored as kinetic energy smooths the motion over the non-constant input power. We aren't really in a steady state, but have small variations around a given speed. With each power pulse the energy delivered is distributed between heat (friction, and drag) and kinetic energy (we speed up a bit). In the between pulse period, the kinetic energy goes to heat (friction and drag) and we slow down a bit.

It smooths the pulse at the output. The input is still very much experiencing pulses, which affects the sources ability to generate energy with minimal losses. Matching the load seen at the input to the source is the most important factor and it is one you are completely ignoring. It's the entire reason for cycling in the first place!

All you need to recognize is that kinetic energy is kinetic energy. It can be stored as forward motion, and it can be stored in a rotating object. It doesn't matter which. Consider the example of the bike with massive flywheel wheels. At speed, with each power pulse I would put most of the energy delivered into rotational kinetic energy. Very little would go into translational kinetic energy, so I would experience very little speed change, but I have stored the same amount of energy. Between pulses the very slightest of slowing the the flywheel will deliver enough energy to overcome friction and drag. So I would experience very little slowing between pulses. Note, though, that there is not more power going into the system then there is in the light wheel case.

Yes, there is more power going into the system then in the light wheel case. Assuming constant gearing, every turn of the pedals needs to provide energy to the flywheel in addition to linear acceleration of the bike. The linear acceleration of the bike is not magically reduced to compensate. The speed remains the same. The additional energy stored in the flywheel comes from your legs. As you keep saying, you get it back later, but your legs don't care about later. You can't convince your legs to let you jump into outer space by promising that you won't walk for the next 10 years.

Speedo
02-15-07, 09:18 AM
They most certianly are constant in the sense that they are not pulsing.

If what you say below is true, that the bike is never at a constant speed then they must be pulsing. They are proportional to the speed, or the square of the speed. As the speed changes, they change.

It's obvious that you are not getting it. The bike is never at a constant speed. Never. It's either speeding up or slowing down.

No, no, yesterday was Valentines Day. I did!

I'm not claiming that the bike is at constant speed. I'm only saying that if you look at what happens as the bike speeds and slows with the pedal pulsing there is energy that goes into kinetic energy on the speed up, that comes out on the slow down. If the bike had no kinetic energy, then it would stop dead as soon as no power was put in. But the kinetic energy of the bike goes toward overcoming friction, rolling resistance and aero drag, so the bike continues to move forward (slowing as it goes as the kienetic energy is turned to heat). Some of that kinetic energy disappated is the rotational kinetic energy of the wheels.

Speedo

makeinu
02-15-07, 09:28 AM
Uh, sorry, but you don't seem to be clear on energy, power, or the realtionship between them.

What do I need to clarify? As you seem to know, power is energy per unit time. The body is unconstrained in the energy it can produce, but it is constrained in the power it can produce. Producing a large amount of power makes you tired. Producing a large amount of energy only makes you tired if it corresponds to a large amount of power.

To move a mass M, up a ladder, increasing it's height by h, in a gravitational field with gravitaional acceleration g, will take energy M*g*h. So, if an elephant is a mass M and a paperclip is mass m, then the difference in energy required would be (M-m)*g*h.

That's just to go up. As you argued in the case of the wheel, the energy is returned when you come back down. However, the energy lost by the source (your body) due to heat, etc is much greater with the elephant. The reason more energy is lost is because the maximum power is greater when lifting the elephant.

The energy required to go up the ladder is proportional to the maximum power. That is why you intuitively considered only the energy required to go up the ladder and not the total energy at the end of the day. Formally this is the maximum power.

Likewise, in the case of bicycle wheels, we need to consider the energy required to get up to speed (not the total energy at the end of the journey) because this is proportional to the peak power provided by your legs.

Power is energy per unit time. It is the rate of energy delivered, or extracted from a system. You are right in that a rider will be limited in how much energy he can deliver per unit time. His power does have some finite limit.

Not only is the power limited, but, more importantly, higher power results in greater losses. Your body is less efficient when its required to deliver too much power.

dalmore
02-15-07, 09:37 AM
If what you say below is true, that the bike is never at a constant speed then they must be pulsing. They are proportional to the speed, or the square of the speed. As the speed changes, they change.

I'm not sure where the disconnect in our terminology blies. But the air causes aerodynamic drag does not vanish for an instant while I'm not appliying force on the pedal. That air is still resisting the forward movement of the bike and causing it to slow. If that's not "constant" perhaps "consistent" is a better word.

No, no, yesterday was Valentines Day. I did!
:D

I'm not claiming that the bike is at constant speed. I'm only saying that if you look at what happens as the bike speeds and slows with the pedal pulsing there is energy that goes into kinetic energy on the speed up, that comes out on the slow down. If the bike had no kinetic energy, then it would stop dead as soon as no power was put in. But the kinetic energy of the bike goes toward overcoming friction, rolling resistance and aero drag, so the bike continues to move forward (slowing as it goes as the kienetic energy is turned to heat). Some of that kinetic energy disappated is the rotational kinetic energy of the wheels.

Speedo
Exactly!! And because of this the rotating mass is ALWAYS having more of an effect that non-rotating mass. After all you did a fine job of proving that back in Post number 1

While a rotating mass requires both the translational energy and the rotational energy to get up to speed.

Et+Er=0.5*M*V^2+0.5*M*V^2 = M*V^2; twice the non-rotating energy.

Speedo
02-15-07, 11:18 AM
I'm not sure where the disconnect in our terminology blies. But the air causes aerodynamic drag does not vanish for an instant while I'm not appliying force on the pedal. That air is still resisting the forward movement of the bike and causing it to slow. If that's not "constant" perhaps "consistent" is a better word.

Consistent I'll buy. It will change a bit around some mean value in the same way that the speed will change a bit around a mean value.

Exactly!! And because of this the rotating mass is ALWAYS having more of an effect that non-rotating mass. After all you did a fine job of proving that back in Post number 1

I think that you are just freaked out by the idea that energy can go into and out of a rotating mass. Let's look at somebody pushing a sled. No rotating mass. I'm trying to maintain my sled at a constant speed. I have to overcome drag, and aero drag. With each push I add a little kinetic energy to the system: we go faster. Between pushes that kinetic energy is disapated through runner drag and aero drag, and I slow down a bit. Is this shocking? Are we surprised? No! It's inertia. Momentum. Take your pick of terms. It was the fact that I had the kinetic energy that allowed me to move forward between pushes. The momentum of the package, me and the sled, carried me along.

Okay, I'm on the bike. I'm "at speed". "At speed" is near constant, but will vary with how I am pushing on the pedals. With each push on the pedals I add kinetic energy. I am adding kinetic energy not just in translational kinetic energy, but in rotational kinetic energy of the wheel. There is not only translational momentum, but angular momentum as well. Just as in the sled case, it is the momentum that carries us forward between pushes. The fact that some of your push energy went into angluar momentum means that at that it did not go into translational momentum, so your speed increase at that push was not as much at the push. BUT that angular momentum will now help carry you through the non-push interval, and your speed will not decrease as much.

When you are trying to maintain speed, what you put into angular momentum in little pulses, comes back to you, helping carry you forward, between the pulses. You don't lose because of it.

Speedo
02-15-07, 11:32 AM
Yes, it reveals precisely that you are leaving out the most important energies. It reveals that smaller wheels, like gearing, weight, and almost everything else about a bicycle are not meant to reduce mechanical losses, but biomechanical losses. It reveals that we are better off considering the power than the energy.

Energy is important as an analysis tool to understand the inter-relationships between the parts.

It smooths the pulse at the output. The input is still very much experiencing pulses, which affects the sources ability to generate energy with minimal losses. Matching the load seen at the input to the source is the most important factor and it is one you are completely ignoring. It's the entire reason for cycling in the first place!

The input is the cyclist adding energy in the most efficient way they can manage. I'm only showing that that energy, once added is not squandered by rotating masses.

The linear acceleration of the bike is not magically reduced to compensate.

No, it's not magic. It's physics.

Speedo
02-15-07, 11:44 AM
What do I need to clarify?

You know, at some point above I introduced the concept of massive flywheel wheels to help with the idea of angular momentum, and energy stored in the rotating mass. They were just an example. I'm not recommending that people go out and put massive wheels on their bikes. I'm sorry for any misunderstanding.

As I noted earlier:

I'm always happy to concede that light is better than heavy. My only claims are that:

1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

dalmore
02-15-07, 01:51 PM
ok I guess I see what you are saying on this point ... For the sake of clarity say I'm averaging a constant speed of 20 mph by bouncing between 21 and 19 mph. You are saying that the rotational energy change when dropping from 21 mph to 19mph is equal to the rotational energy change of going back from 19 to 21 so it's a net zero game. ok, I can swallow that.

BUT that is not to say that I agree that rotating mass is the same as non rotating mass except when accelerating. Only that I don't have the math skills to argue the point. :D

makeinu
02-15-07, 02:14 PM
Energy is important as an analysis tool to understand the inter-relationships between the parts.

Not when you leave out the most important energies, such as the energy lost by the body at a given power level.

The input is the cyclist adding energy in the most efficient way they can manage.

Which is more efficient with the smaller wheel.

I'm only showing that that energy, once added is not squandered by rotating masses.

This goes without saying. There is no electricity involved. So the only way to squander energy is by heat. The wheel obviously isn't going to heat up merely because it's rotating. So the energy will be squandered in the brakes via the excess momentum and in the heating of your muscles due to the increased power requirement.

No, it's not magic. It's physics.

The linear acceleration certainly isn't physically reduced to compensate either. At a given cadence, the bike will only go slower if you reduce the gear. Otherwise you will have to pedal harder, just like you have to pedal harder with a heavier bike, just like you have to pedal harder when going up hill. Pedaling harder and deviating from optimal cadence both waste more biomechanical energy. Therefore using a bigger wheel wastes more biomechanical energy.

You know, at some point above I introduced the concept of massive flywheel wheels to help with the idea of angular momentum, and energy stored in the rotating mass. They were just an example. I'm not recommending that people go out and put massive wheels on their bikes. I'm sorry for any misunderstanding.

I'm not saying that you're recommending that people go out and put massive wheels on their bikes. I'm saying that the fact that larger wheels do not themselves dissipate energy is irrelevant to the conclusion that a bike with larger wheels will waste more energy.

I'm always happy to concede that light is better than heavy.

Yeah, but you are so occupied with analyzing the energy that you're missing the reason why light is better than heavy. Your missing the most important inter-relationships which you claim your energy analysis is so helpful to understand.

1) When accelerating the bike it is, in fact, true that a pound in the wheels is like two on the bike.
2) After the bike is up to speed, there is no continuing penalty due to the fact that the mass of the wheel is rotating.

Ok, but the bike will most likely never be "up to speed". Most of the time riding is probably spent pedaling/accelerating and braking.

jur
02-15-07, 03:00 PM
I'm with you on this one Speedo.

Looks like it is going to require a detailed simulation to decide it. I have a simulator, just need the math to put into it. I'll see if I can squeeze in the time. Or maybe you can furnish a starting point?

We will need equations for:
forward motion, with drag from wind resistance;
model of torque supplied by legs on pedals, I am thinking a sine wave with variable amplitude and offset may do here;
angular momentum converted to linear momentum as a function of wheel diameter, with wheel moment of inertia variable;
maybe some drag from bearings;

That will probably do.

Then the sim should show if the terminal velocity for a steady state torque model for cases of different wheels is different.

 An approach from conservation of energy is going to be most reliable and easy.

Speedo
02-15-07, 03:21 PM
Not when you leave out the most important energies, such as the energy lost by the body at a given power level.

Nowhere have I said that the rider would have to provide more power then they otherwise would. I have no idea where you've gotten that idea from.

The rider outputs his normal power level. A bike with relatively heavier wheels acclerates more slowly, so takes longer getting to speed. Once at speed the rider is again providing the same amount of power, and riding the same speed. You can say that the overall consumed slightly more energy because the rider out put the same amount of power and thus lost time in the accelerations, or preserved the acceleration time by outputing more power. In either case it is not a dramatic difference.

Which is more efficient with the smaller wheel.

Which smaller wheel? The only thing I've said about a smaller wheel is that before I ever looked at this I would have guessed that a smaller wheel got something of a free ride. The analysis shows, and various other posters have confirmed, that the wheel diameter alone doesn't make a difference. But a smaller wheel may naturally be a lighter wheel, and I'm always happy to concede that lighter is better.

So the energy will be squandered in the brakes via the excess momentum and in the heating of your muscles due to the increased power requirement.

Which increased power requirement?

The linear acceleration certainly isn't physically reduced to compensate either. At a given cadence, the bike will only go slower if you reduce the gear. Otherwise you will have to pedal harder, just like you have to pedal harder with a heavier bike, just like you have to pedal harder when going up hill. Pedaling harder and deviating from optimal cadence both waste more biomechanical energy. Therefore using a bigger wheel wastes more biomechanical energy.

Aside from my slight aside above who said anything about a bigger wheel size?

I'm not saying that you're recommending that people go out and put massive wheels on their bikes. I'm saying that the fact that larger wheels do not themselves dissipate energy is irrelevant to the conclusion that a bike with larger wheels will waste more energy.[/QUOTETE]

Do you think that we are having a discussion about wheel diameter? I don't really have anything to say about wheel diameter.

[QUOTE=makeinu]Yeah, but you are so occupied with analyzing the energy that you're missing the reason why light is better than heavy. Your missing the most important inter-relationships which you claim your energy analysis is so helpful to understand.

Please go back and re-read the thread. I really don't have any particular argument about wheel diameter. I would guess that the specific issue of large vs small diameter is going to be lost in the specifics of the particular wheels. i.e. the rim mass, the tire mass, the tire characteristics, number and type of spokes ...

Ok, but the bike will most likely never be "up to speed". Most of the time riding is probably spent pedaling/accelerating and braking.

If you are spending all of your time accelerating and then braking then you have my sympathy. But if that's the case the energy in and out of the wheel angular momentum will be overwhelmed by the linear momentum swings.

Speedo

makeinu
02-15-07, 04:22 PM
Nowhere have I said that the rider would have to provide more power then they otherwise would. I have no idea where you've gotten that idea from.

The rider outputs his normal power level. A bike with relatively heavier wheels acclerates more slowly, so takes longer getting to speed. Once at speed the rider is again providing the same amount of power, and riding the same speed. You can say that the overall consumed slightly more energy because the rider out put the same amount of power and thus lost time in the accelerations, or preserved the acceleration time by outputing more power. In either case it is not a dramatic difference.

I got the idea that the rider would have to provide more power from the physics of the situation. If you fix the power provided by the rider then the rider's candence must be reduced to compensate for the reduced acceleration. Alternatively you can fix the cadence and the power provided by the rider must be increased to compensate for the heavier wheels. Either way the biomechanical efficiency is decreased.

Do you think that we are having a discussion about wheel diameter? I don't really have anything to say about wheel diameter.

As we've already established. The relevant difference between larger and smaller wheels is the difference in rotating weight, which is what this whole thread is about. All else being equal larger wheels are heavier wheels and smaller wheels are lighter wheels.

I'm always happy to concede that lighter is better.

Then why do you keep going on about the mechanical energy being independent of the rotating mass? Either the total mechanical energy is important and, since rotating weight does not affect the total mechanical energy, rotating weight is unimportant or the total mechanical energy is irrelevant and lighter is better. Lighter is better because the conservation of energy you keep talking about is irrelevant.

If you are spending all of your time accelerating and then braking then you have my sympathy. But if that's the case the energy in and out of the wheel angular momentum will be overwhelmed by the linear momentum swings.

Obviously small changes in weight will give small changes in performance regardless of whether the weight is linear or rotating, but braking and accelerating must be awfully popular among most cyclists. Why else would light weight and low momentum be considered so important?

Speedo
02-15-07, 05:37 PM
I got the idea that the rider would have to provide more power from the physics of the situation. If you fix the power provided by the rider then the rider's candence must be reduced to compensate for the reduced acceleration.

Why? A slowly accelerating rider is going through the same range of speeds as the fast accelerating rider just a lower rate. The range of strategies for matching available power to load (standing, shifting gears, ...) are available to be used by any rider.

As we've already established. The relevant difference between larger and smaller wheels is the difference in rotating weight, which is what this whole thread is about. All else being equal larger wheels are heavier wheels and smaller wheels are lighter wheels.

The thread is about rotating vs non-rotating mass. Some posters, yourself, included have made that claim. I would agree that if I were out to make the lightest wheels possible, I'd go for smaller rims, but that doesn't mean that small wheels must be lighter. On my small wheels I run wider tires and the rim itself is wider. I may have to try measuring, but I would not immediately conclude that my 406 wheels (and tires) are lighter than my 700Cs.

Also, there must be some other things coming into play, otherwise we's all be riding those bikes that use skateboard wheels! Time trial bikes use smaller wheels than 700C. Maybe that's the optimum. I don't really have a strong opinion one way or the other in the big vs small wheels.

Then why do you keep going on about the mechanical energy being independent of the rotating mass?

Please point me to where I said that.

Either the total mechanical energy is important and, since rotating weight does not affect the total mechanical energy, rotating weight is unimportant or the total mechanical energy is irrelevant and lighter is better. Lighter is better because the conservation of energy you keep talking about is irrelevant.

Total mechanical energy is important. The thing I've been blathering on about is that riding at steady speed the mechanical energy that goes into rotating the mass doesn't hurt you. This is even true when steady speed is (as pointed out by Dalmore) not truly steady.

Obviously small changes in weight will give small changes in performance regardless of whether the weight is linear or rotating, but braking and accelerating must be awfully popular among most cyclists. Why else would light weight and low momentum be considered so important?

As to your second, well, my party line has been lighter is better. ;) How light? Damned if I know. I have a friend who insists on spending thousands of dollars every year to have the latest lightest widget. I suppose he's got the dosh, and likes to bike, so what else is he going to spend it on? I have other friends who will ride nothing but 25 year old bikes that are as much as :eek: 6 or 7 pounds heavier than a middle of the line new bike. I must admit that I learned much about the mechanics of bikes to puncture Mr. Lightbike's claims about his nanogram wheels! :D

Speedo
02-15-07, 05:50 PM
I'm with you on this one Speedo.

Looks like it is going to require a detailed simulation to decide it. I have a simulator, just need the math to put into it. I'll see if I can squeeze in the time. Or maybe you can furnish a starting point?

I don't think it will convince anyone who isn't ready to be convinced. The model is going to be a set of diferential equations. People will have to buy the model as well as the idea of numerically integrating the equations to get a result.

On the other hand, I've never been one to let practical considerations stand in the way of writing equations! I take the challenge!

I am going away this weekend. Maybe next week I can put something together. Do you want equations? Are you going to put this into Matlab or something similar? I have Matlab and can write code that can be integrated by their differential equation solver.

If Dalmore and makineu (I think they are the last who haven't been driven away by the pedantry) are interested we can post the pieces here.

Speedo

makeinu
02-15-07, 06:40 PM
Why? A slowly accelerating rider is going through the same range of speeds as the fast accelerating rider just a lower rate. The range of strategies for matching available power to load (standing, shifting gears, ...) are available to be used by any rider.

True, but when you isolate each parameter you get the same answer: Heavier bike makes a tougher ride for the cyclist and every gram on the wheel counts double.

The thread is about rotating vs non-rotating mass. Some posters, yourself, included have made that claim. I would agree that if I were out to make the lightest wheels possible, I'd go for smaller rims, but that doesn't mean that small wheels must be lighter. On my small wheels I run wider tires and the rim itself is wider. I may have to try measuring, but I would not immediately conclude that my 406 wheels (and tires) are lighter than my 700Cs.

Also, there must be some other things coming into play, otherwise we's all be riding those bikes that use skateboard wheels! Time trial bikes use smaller wheels than 700C. Maybe that's the optimum. I don't really have a strong opinion one way or the other in the big vs small wheels.

Good point, but if your 406s aren't lighter than your 700Cs then I wouldn't say it means that smaller wheels aren't necessarily lighter. It means that wider wheels are heavier or aluminum is lighter than steel, etc. Every physical statement must be qualified with "all else being equal". If all else isn't equal then almost anything could happen.

I don't think it will convince anyone who isn't ready to be convinced. The model is going to be a set of diferential equations. People will have to buy the model as well as the idea of numerically integrating the equations to get a result.

On the other hand, I've never been one to let practical considerations stand in the way of writing equations! I take the challenge!

I am going away this weekend. Maybe next week I can put something together. Do you want equations? Are you going to put this into Matlab or something similar? I have Matlab and can write code that can be integrated by their differential equation solver.

If Dalmore and makineu (I think they are the last who haven't been driven away by the pedantry) are interested we can post the pieces here.

Speedo

What's there to simulate? Everything is analytically solvable. It's just boils down to how large the various effects are when subject to real world nonidealities.

Speedo
02-15-07, 07:30 PM
What's there to simulate? Everything is analytically solvable. It's just boils down to how large the various effects are when subject to real world nonidealities.

I wondered if that were true. It might be doable if one assumed constant input power, but one of the questions to be addresssed was what happend in the case of non-constant input. The pulsing pedal stroke case. It is a periodic input, maybe Fourier analysis. Not being very smart, I often resort to numerical integration.

If you have a solution feel free to contribute it!

Speedo

jur
02-15-07, 08:58 PM
I don't think it will convince anyone who isn't ready to be convinced. The model is going to be a set of diferential equations. People will have to buy the model as well as the idea of numerically integrating the equations to get a result.

On the other hand, I've never been one to let practical considerations stand in the way of writing equations! I take the challenge!

I am going away this weekend. Maybe next week I can put something together. Do you want equations? Are you going to put this into Matlab or something similar? I have Matlab and can write code that can be integrated by their differential equation solver.

If Dalmore and makineu (I think they are the last who haven't been driven away by the pedantry) are interested we can post the pieces here.

SpeedoI have PSpice, an electronic circuit simulator. But it can take stuff like laplace transforms equations directly and produce graphical output from those. I use it extensively.

The diff. equations can be easily converted to Laplace (s-domain) format. But perhaps it is easiest, instead of constructing a diff eq model and putting that in, to construct a model with all the various pieces connected with a block diagram format.

makeinu
02-15-07, 09:42 PM
I wondered if that were true. It might be doable if one assumed constant input power, but one of the questions to be addresssed was what happend in the case of non-constant input. The pulsing pedal stroke case. It is a periodic input, maybe Fourier analysis. Not being very smart, I often resort to numerical integration.

If you have a solution feel free to contribute it!
I'm an electrical engineer not mechanical. I don't have time to struggle with figuring out what the right mechanical equations should be (if you remember my first post in this thread I didn't even know whether the equation you gave for rotational inertia should be in radians or not), but if you write down the differential equations then I will solve them.

However, I still think the most important factor is how efficiently the body can deliver power for different loads. Biomechanical losses are very tricky to analyze.

Fear&Trembling
02-16-07, 01:50 AM
As things currently stand 700c wheels are more often than not lighter than 20"/16" wheels - this is because of the r&d that goes into racing. Look at some of the weights on weight weenies (if you are really that sad!).

If you want to see heavy rotating masses check out this guy and his machine:
http://www.wolfgang-menn.de/sosenka.htm

Speedo
03-05-07, 11:00 AM
Well, I have a write up of the equations. The basic differential equation is based on solving an equation of the power sources and the power sinks for the bike acceleration. The bike acceleration can then be integrated in whatever tool you have available to get speed vs time.

The model includes a pedal power model, rotating and non-rotating inertia, a hill (if desired!), and aerodynamic drag. The model doesn't include rolling resistance and bearing friction. I found references for those terms, but thought that going without them would be better for a start. I'm not sure there is a closed form solution to the differential equation. It's complicated enought that I'm just going to go after it numerically.

What I have is a Word document that has been converted to pdf. I used Microsoft equations; this won't look very good in plain text. I'm not sure if I can paste either of those here. I think I can covert the individual pages to an image. If I can't turn it into jpegs, I'll just try sending it to interested parties...

Speedo

jur
03-05-07, 02:23 PM
I am off on my 3 week cycle tour of Tassie tomorrow, so won't be able to look at this until end March. I also came up with a block diagram (scribbled on paper), so we can compare notes when I get back.