TRaffic Jammer

it's the rider that basically holds you "up" while you ride. That's why you had to "learn" to ride a two wheeler, and once you got it.. the mechanics of the act of balancing was practically hard wired into your brain. My newfew spent his toddeler year son a tag-along bike... complete bike minus frnt wheel and the toptube stretched out to attach to the seat post in front. He was well versed in balance from going around with his Mom, and when it came time to get him on his own two wheels....we plopped him on and he was riding like he had been for years. SO for your paper the "holding up effect" is done by the body.

Bald Student

True, the best thing for Parker to do would be to include an explanation of both theories in his report and then give his own interpretations on whether each one is correct.

Mchaz

Agreed.

Without getting into complicated physics discussion, Try this demo:

Take a bike wheel, spin it, and hold it with both hands on either side of the hub. Now try to tilt it side to side. You will feel a resisting force. At high wheel velocities, you will feel a rather big one. Are people seriously arguing that the resisting force has no effect in keeping a bike balanced? C'mon.

The angular momentum of the wheels is certainly not the only thing keeping a bicycle balanced, but most likely it is a major contributing factor at speed. The gyroscopic theory would also explain why it's so hard to trackstand, and why inexperienced riders wobble all over the place at low speed. In the absence of a gyroscopic force the rider must utilize steering inputs, and weight shifting to keep the bike up.

In the case of fixed gear trackstanding, the rider turns his wheel sideways. The force of gravity pulling the rider to the side is counteracted by moving the bike side to side by pedaling. With the wheel turned, pedaling forwards or backwards is converted into sideways motion (well, technically a crazy helix arc motion, but sideways enough at the small angles involved).

Another bit to ponder. If you hop off a bike while it's going, and let it "ghost ride", what keeps it up? Why does it eventually fall over? Could the wacky swaying from side to side when the bike is slowing down before it falls over be caused by precession of the giant gyroscopes we call wheels?

TRaffic Jammer

How one can say there is no gyroscopic effect at all on a moving bike, simply hasn't spent much time on a bike. This is why you tell the kids to PEDAL faster when they are getting the slow down wobbles in learning their riding.

cny-bikeman

I'm sorry but the above info is incorrect, as pointed out by several previous posts. Not only does centrifugal force (or even gyroscopic force) have nothing to do with balance - there actually no such thing. Intertia tends to keep things going in a straight direction. A force that pushes one into a circle is call CENTRIPETAL FORCE. What people feel as centrifugal force is actually a result of centripetal force acting against inertia. Think of it this way. If you were tied by the waist to a rope secured by a loop to a pole and tried to run at a right angle to the rope, it would pull you into a circle instead. You would feel pressure on your waist on the opposite side to the pole. But there is nothing pulling outward. Inertia is going forward, the rope inward and the result is a circle. PLEASE use citable references for your work, not well-meaning opinions.

Bald Student

The best work on the subject was published in the American Journal of Physics in Dec. 1982 link but you need a university ip address (or be a subscriber yourself) to view it.

There's a good summary of the idea from Georgia State University here.

cny-bikeman

OK, here's the experimental proof that many of the assumptions about how a bicycle balances (including gyroscopic or "centrifugal" force) are incorrect: https://www.phys.lsu.edu/faculty/gonz...9no9p51_56.pdf

Bald Student

Firstly, gyroscopic and centrifugal forces are completely different things so it's not correct to say 'gyroscopic or "centrifugal" force'.

Also, if you don't mind me quoting from the paper you linked:
"a falling bicycle can be saved by proper steering of the front wheel. The theory explains, for example, that the ridability of the bicycle depends crucially on the freedom of the front forks to swivel (if they are locked, even dead ahead, the bicycle cannot be ridden), that the faster a bicycle moves, the easier it is to ride (because a smaller steering adjustment is needed to create the centrifugal correction)"
It entirely agrees with me and is actually cited in the paper I linked to (though, I'll accept that you might not be able to see that because it's behind a password). The significance of Dr. Jones' paper you cited is that he showed how the angle of the front fork uses a form of mechanical feedback to apply centrifugal forces to the bike and keep it stable. It is the very phenomenon I had in mind when I wrote "Let me know if you understand that and if you do I'll tell you some more" in my first post. I didn't mention it because I thought introducing it into a discussion on high school physics would be confusing and also because it first requires one to understand the role centrifugal forces play in keeping a bike from falling over.

Sir Stuey

Quote:

This test completed the ingredients for a more complete theory of the bicycle. In addition to the rider's skill and the gyroscopic forces, there are, acting on the front wheel, the center-of-gravity lowering torque and the castoring forces; the heavier the bicycle's load the more important these become.

Ok, so everyone's right? For unriden bikes, gyroscopic forces dominate. For very heavy riders, human balancing skills dominate. For all others in between, there's a combination of factors.

Well look at that. Nowhere does it say that centripetal forces are responsible.

<--- Increasingly getting confused despite being well versed in introductory, advanced, and quantum mechanics.

This is getting difficult to analyze. I'm comparing the bike to a motorcycle which isn't a far fetched analogy except for the fact that a motorcycle has a fairly low center of mass and the rider does not have as great an influence on the vehicle's motion.

My opinion remains fixated that angular momentum is mostly responsible, but I have opened up to the possibility that other factors come into play.

The articles says that if steering is disabled, the bike would be impossible to ride and balance. I don't trust that. Slap on a reasonably fast electric motor, get the wheel spinning, and then let it go. It should balance by itself for quite a bit before it topples. Again, the rolling disc visualization.

Bald Student

The article goes into more detail on that. He experimented in the university car park and found that a bike set off without a rider would stay up once it was at a reasonable speed, even in the bike he especially designed to negate the effects of the centrifugal forces. He put this down to the gyroscopic effect since the bike he designed to remove the gyroscopic effect fell right over.

When he put weights on the bike to simulate a rider's weight the gyroscopic forces weren't enough and the bike fell. The regular bike performed better but the mechanical feedback in the front forks caused the centrifugal forces to push it from side to side and it eventually wobbled so much it fell. His conclusion was that both played a part but the skill of the rider was still needed to keep everything balanced.

gcl8a

Gyroscopic forces do play a part, and they are of course a good explanation for the stability of an isolated wheel (or a rolling quarter or disc), but on a moving bicycle, the effect is small, since the mass of the wheels are small compared to the overall mass of the system.

This is why there is still an argument even with the counter-rotating (zero-angular momentum) bikes. They're perfectly ridable, but, depending on the level of the cyclist, the difference may or may not be all that noticeable.

I have read this on the internet, so it must be true. Case closed.

(Actually, I'm too lazy to look it all up, but there was a thread not too long ago on this where several websites (from MIT and other schools) were linked, with papers)

LóFarkas

1. Gyroscopic effect is not the main force keeping a bike upright. Otherwise bike B with twice as heavy wheels as bike A would be twice as easy to ride, which it is not. Also, (theoretical) bikes with weightless wheels would be unridable, which is absurd. Also, people would not need to learn how to ride a bike. Getting up to speed and not interfering would be enough, and it obviously isn't.


Edit: Also, those who think gyroscopic effect is what keeps "ghost riding" bikes going, will you please
1) look at a few videos and see how bikes lean to a side, have the wheel turned because of that, corner - sometimes pretty sharply - because of that, and get pushed back upright by the centripetal force. Just like a rider would do.
2) Look up bike steering geometry: trail. Trail is what stabilizes the steering of a bike: build a bike with no trail and it will be almost impossible to ride, and impossible to "ghost ride". So where is the gyroscopic effect then?


2. Wider tyres do _not_ necessarily have better traction at all. On loose and uneven surfaces they do, but that's not strictly traction.

cny-bikeman

OK, I need to clarify some things. I said centrifugal or gyroscopic because they ARE different, not as alternative terms. I believe it is others who are confusing the concepts.

Secondly I thought someone would jump on the use of centrifugal force in the article. But tell me, please, where is the source of that force? (Apparent) centrifugal force pulls outward, but the force referred to in the article is pushing the bicycle inward to its previous path. If anything centripetal force would have been more appropriate. I believe the author is using the term incorrectly. His summary gives centrifugal force little or no significance in stabilty.

Nobody has yet shown me any mechanism by which centrifugal force (supposed energy directly outward from the center of a circle) would contribute to bicycle statbilIty. As for gyroscopic effect I will accede that it may have an effect on a riderless bike but that is more an academic excercise than a relevant discussion, as there is no function served by a riderless bike.

p.s. For those who still believe in centrifugal force, try this: Tie a string to a rock, strong enough to hold it but weak enough to break under modest pressure. Find a clear space outside and start to twirl the rock faster and faster. When the string breaks note that the rock does not fly directly away from you but rather at a right angle. This is because two forces were at work – intertia that would tend to keep the rock moving forward and the string which kept forcing it inward. Remove the string (centripetal force) and the rock will travel at a tangent (right angle) or “straight ahead.”

soloban

How about looking into the rolling weight of components and how lightening the rotating weight decreases....take a look at moments of inertia

willtsmith_nwi

Ahh, lets look at it from the standpoint of energy. All the brakes we use (except magnetic brakes on hybrid cars) work by converting kinetic energy to heat. The amount of energy converted depends on the coefficient of friction between the braking surface and the pad, the force applied to the pad and the speed at which the braking surface passes under the pad.

Disc brakes use a harder pad on a disposable rotor. This provides much more friction. Rim brakes use a rubber compound that are "grabby" but that avoid actually roughing things up as that would pre-maturely destroy the wheel rim. The rim brakes compensate by having their braking surface at the far edge of the wheel. There the linear speed of the wheel is harder. Moving the surface makes more heat. Try rubbing your hands together fast and slow. See when you get the most heat.

So why use larger discs? The braking surface moves faster over the rotor and generates more heat. More heat in the brakes means more braking.

Sir Stuey

The moment of inertia of the wheels is independent of the weight of the bike components (excluding the wheels of course). But yea, that's why they tout "lite" tubes and such. Haha, the OP didn't know the mess he was getting himself into when he posed his question, did he!

Nope. Two nearly identical bikes are travelling at the same speed. The only difference in components is the rotor size. When coming to a halt, the change in velocities of the two bikes will be identical. Some of the energy "lost" will be converted to sound, but most of it will be converted into thermal energy due to the friction of the braking surfaces. The amount of thermal energy should be very similar.

Someone posted that larger rotors dissipate heat quicker than smaller rotors. I'm not sure why a cooler rotor allows for better braking, but the argument seemed valid enough.

LóFarkas

Well, bigger brake rotors give better braking simply because they have more leverage. I thought that was pretty obvious. All other things equal, it's easier to stop a wheel by grabbing it further from the centre. (Thought experiment: spin your wheel by hand, stop it by the rim. Then spin it again and stop it by grabbing a spoke, say, at the spoke crossing. The emphasis is on thought experiment... please don't break your fingers actually doing this.)

Better heat dissipation is important as brakes can overheat and fade on long downhills.

willtsmith_nwi

When coming to a full stop between two identical bikes yes, the change in velocity will be identical. And hence, the heat dissipated off the pads will also be largely identical. What you failed to take into account is the amount of distance it takes to stop.

The primary purpose of having "more powerful" brakes is so that we can stop the bicycle without squeezing the lever with all our might. Given identical pressure applied to the same caliper over two different rotor sizes. The larger rotor will generate more heat per each revolution as there is a greater rotor surface. It's a rate, not an independent quantity.

For lightweight XC folks this is not an issue. For downhill riders and heavier riders in an "all-mountain" the brakes get pretty hot. A larger rotor allows for somewhat improved heat dissipation. This is important as very high temperatures on the brakes can potentially, boil hydraulic fluid (rendering your brakes useless), melt pads and warp rotors (actually it's the rapid cooling that warps the rotors).

The primary purpose of larger rotors remains being able to slow the bike without your hand cramping up.

Those of you who ride 29ers with disc brakes, pay attention . Because the larger 29er wheel turns slower for the same linear speed, you have less braking capability then the same setup on a 26er. Depending on your size, a larger rotor could be warranted.

willtsmith_nwi

When coming to a full stop between two identical bikes yes, the change in velocity will be identical. And hence, the heat dissipated off the pads will also be largely identical. What you failed to take into account is the amount of distance it takes to stop.

The primary purpose of having "more powerful" brakes is so that we can stop the bicycle without squeezing the lever with all our might. Given identical pressure applied to the same caliper over two different rotor sizes. The larger rotor will generate more heat per each revolution as there is a greater rotor surface. It's a rate, not an independent quantity.

For lightweight XC folks this is not an issue. For downhill riders and heavier riders in an "all-mountain" the brakes get pretty hot. A larger rotor allows for somewhat improved heat dissipation. This is important as very high temperatures on the brakes can potentially, boil hydraulic fluid (rendering your brakes useless), melt pads and warp rotors (actually it's the rapid cooling that warps the rotors).

The primary purpose of larger rotors remains being able to slow the bike without your hand cramping up.

Those of you who ride 29ers with disc brakes, pay attention . Because the larger 29er wheel turns slower for the same linear speed, you have less braking capability then the same setup on a 26er. Depending on your size, a larger rotor could be warranted.

The phenomenon can be analyzed using both mechanics and thermodynamics. Both are equally valid. One is easier to compute. Remember energy doesn't just "disappear" from a system. It's necessary to find out where all of it goes to understand how the system works.

parkeredwards

Hey thanks for all the help, i could believe all the posts that this topic got, thanks! as for the information i have picked up "Bicycling Science" by David Gordon Wilson, i only have read a little bit but so far its sort of hard for me to understand the concepts, since in only in grade 10. but also the book will help me get the graphs that i need for the presentation. But i have another question... Why when you spin a wheel and then place one side of the axle on your finger it stays balanced...why does it do this and is it basically the same reason why a bike can stay up right???

Thanks,
Parker Edwards

Sir Stuey

Well, yes and no. The consensus is that angular momentum is not a major component in helping a riden bike stay balanced. BUT for an unloaded wheel - angular momentum IS the major component which is keeping the wheel balanced about your finger.

poopncow

Some other terms to think about:

brakes - Swept area
countersteering - push left go left, push right go right

parkeredwards

Ok I'm not totally sure about how the bike actually stay upright, the information you guys have posted is awesome but can someone put how this works into a diagram or something so it is easier for me to understand it would be greatly appreciated.