Why do bicycles stay upright when in motion?

Brussels Sprout

What is the physical reason behind the fact that it is extremely difficult to stay upright on a stationary bicycle as opposed to the relative ease of staying upright when that same bicycle is moving?

thecornflake

Brussels Sprout wrote: »

What is the physical reason behind the fact that it is extremely difficult to stay upright on a stationary bicycle as opposed to the relative ease of staying upright when that same bicycle is moving?

I was thinking the same thing today (great minds think alike), I assume it is just like a spinning top or the way a bullet spins when it leaves a gun. Things are more stable that way, however i wouldn't take my word on that as i am tired and haven't really thought about any proper physics questions in a long time.

mawk

gyroscopic motion of the wheels?

blorg

The gyroscopic effect is not significant. You stay up through steering. Note incidentally that steering a bike is primarily done by moving your weight, not dramatic use of the handlebars, e.g. it is entirely possible to steer a bike 'hands free.'

http://www2.eng.cam.ac.uk/~hemh/gyrobike.htm

http://arstechnica.com/science/news/2011/04/moving-bikes-stay-uprightbut-not-for-the-reasons-we-thought.ars

Brussels Sprout

blorg wrote: »

The gyroscopic effect is not significant. You stay up through steering. Note incidentally that steering a bike is primarily done by moving your weight, not dramatic use of the handlebars, e.g. it is entirely possible to steer a bike 'hands free.'

http://www2.eng.cam.ac.uk/~hemh/gyrobike.htm

http://arstechnica.com/science/news/2011/04/moving-bikes-stay-uprightbut-not-for-the-reasons-we-thought.ars

Brilliant-that makes perfect sense. Thank you.

(I have an engineering degree and feel less embarrassed now that I didn't know this as it seems from that second article that it has confounded a lot of others as well)

Squashie

Has very little to do with steering, and a lot more to do with the gyroscope effect and the angular momentum of the spinning wheel. A cyclist can lean quite significantly to either side on a bike (thus imbalancing weight) without falling when he's moving fast, yet can't if he's moving slow.

A spinning wheel has angular momentum L = Iω (I moment of inertia, ω angular velocity) which points in a well-defined direction. Just like how forces are required to shift the direction of a body with linear momentum, forces are required to shift the direction of rotating bodies with angular momentum. That's why your bike stays up – The spinning wheels resist angular momentum direction changes. The faster you cycle, the more angular momentum you give the wheels, the more resistance to directional changes, the harder it is to lean or fall over.

Morbert

Squashie wrote: »

Has very little to do with steering, and a lot more to do with the gyroscope effect and the angular momentum of the spinning wheel. A cyclist can lean quite significantly to either side on a bike (thus imbalancing weight) without falling when he's moving fast, yet can't if he's moving slow.

A spinning wheel has angular momentum L = Iω (I moment of inertia, ω angular velocity) which points in a well-defined direction. Just like how forces are required to shift the direction of a body with linear momentum, forces are required to shift the direction of rotating bodies with angular momentum. That's why your bike stays up – The spinning wheels resist angular momentum direction changes. The faster you cycle, the more angular momentum you give the wheels, the more resistance to directional changes, the harder it is to lean or fall over.

The gyroscope effect would not be strong enough to keep balance. To test this, constrain the handlebars of your bike such that they're pointing forwards, and push your bike with no rider. It will quickly fall over.

Squashie

I understand that. Obviously a person needs to balance, but it doesn't change the fact that the resistance to the wheel's directional angular momentum changes are what provides the means for a person to balance upright. Angular momentum (and not balance) is the answer to the original posters question.

Also, your bike will only quickly fall over in your situation because friction causes it to slow and the wheel's angular momentum to decrease, thus decreasing the resistance to imbalance. The harder you shove your bike however; the more angular momentum that is given to the wheels; the greater the resistance to imbalance; the further it will travel before falling.

blorg

A bike with locked handlebars will fall over a lot sooner than one without. Friction causes both bikes to slow, but the locked one will fall first. This will hold with or without a rider.

Squashie

You aren't arguing physics, you're arguing intuition. If hypothetically you had a perfectly symmetrical bicycle, could lock your bicycle handlebars to set its wheels in a perfect forward-facing position, could launch your bicycle a number of times through a controlled atmosphere with an initial set very tiny imbalance to one side, the time taken for the bicycle to fall to the ground after each controlled launch would be a function of inital velocity.

The initial linear velocity of the bike is coupled with associated wheel-radius-dependent angular velocity on the wheels and hence angular momentum L created. The initial tiny imbalance causes a torque dL/dt on the angular momentum. The torque would also have an associated time derivative as it would also be increasing with time. As the imbalance increases, so would the torque. How long it takes this simplified bike to fall (assuming negligible friction) depends on the initial angular momentum of the wheels and the associated torque on that angular momentum.

That's the basic functionality of the bicycle. Unlock the handlebars and sit a person on the bike, and the whole thing becomes about conserving angular momentum, and shifting centres of gravity and torques to balance and steer and make the bicycle do what you want to do.

blorg

The argument is that a bike, even without a rider, self-corrects thanks to its steering.

Push two bikes the same speed without a rider and the one with unlocked steering will stay up longer.

In the articles I linked above you can see that the theory that gyroscopic forces has been tested and found not to be significant.

Squashie

I've only had a quick glance through that article, but it seems to me that Hugh Hunt of Cambridge University needs to go back and redo his 1st year mechanics.

"and this is the forced precession rate of the front wheel acting as a gyroscope. At its peak, the couple required to achieve this precession motion, due to gyroscopic effects, is
Θ = J ω Ω = 0.1 * 20 * 1 = 2 N m
The bike and I weigh, say, 100 kg = 1000 N, so the gyroscopic effect will only help me if I don't tilt more than 2 mm from being perfectly upright (1000 N * 0.002 m = 2 N m). This doesn't give me much safety margin."

Since when does the entirety of his and the bicycles weight act along the axis causing precession? The vast majority of the weight will act vertically and follow Newton's third, while only a small component will act radially causing imbalance. His maths is crap.

rccaulfield

The physics gets much more interesting on motorbikes at higher speeds with big lean angles and no steering input.

vampire of kilmainham

great im going for a gyroscopic spin on my bike tommorrow

Morbert

Squashie wrote: »

I've only had a quick glance through that article, but it seems to me that Hugh Hunt of Cambridge University needs to go back and redo his 1st year mechanics.

"and this is the forced precession rate of the front wheel acting as a gyroscope. At its peak, the couple required to achieve this precession motion, due to gyroscopic effects, is
Θ = J ω Ω = 0.1 * 20 * 1 = 2 N m
The bike and I weigh, say, 100 kg = 1000 N, so the gyroscopic effect will only help me if I don't tilt more than 2 mm from being perfectly upright (1000 N * 0.002 m = 2 N m). This doesn't give me much safety margin."

Since when does the entirety of his and the bicycles weight act along the axis causing precession? The vast majority of the weight will act vertically and follow Newton's third, while only a small component will act radially causing imbalance. His maths is crap.

While heuristic calculations can be misleading, you are equally guilty of hand-waving. The article lists experiments where the gyroscopic effect is removed by rotating wheels. It has been shown that the gyroscopic effect makes no significant contribution to bike stability.


Some of the links are broken, but here are two:

http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/vol59no9p51_56.pdf

http://bicycle.tudelft.nl/schwab/Bicycle/index.htm

stevenmu

As an experiment you can try easily at home (and one which I tried as a kid), turn your bicycle upside-down so that it's resting on it's handlerbars and saddle (or if you can just it's saddle). Use your hands to turn the pedals and spin up the back wheel as fast as you can (you can get it pretty fast since you don't have to push the whole bicycle + rider).

Now see how easy it is to knock the bike over

As a control, also knock the bike over without the wheel spinning.

Capt'n Midnight

There is also the castor effect

if you extend the head tube to the ground it will be AHEAD of where the wheel touches the ground so it's actually stable

Tea_Bag

Didnt they just recently just disprove the gyroscopic effect theory?

i didnt look much into it when i read the article a while back, but its there.


EDIT: article



look up "two-mass-skate bicycle"

VNP

http://youtu.be/Cj6ho1-G6tw there must be some interesting equations to explain this guys cycling

clearz

Tea_Bag wrote: »

Didnt they just recently just disprove the gyroscopic effect theory?

Please don't say the physics of a bicycle in motion is not understood in this age of superstrings and supersymmetry.