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Q: Why can I ride my bike ? ( Answered ,   14 Comments )
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 Subject: Why can I ride my bike ? Category: Science > Physics Asked by: gruffgareth-ga List Price: \$4.00 Posted: 14 Jun 2002 11:55 PDT Expires: 21 Jun 2002 11:55 PDT Question ID: 25883
 ```The bit I'm interested in is the gyroscope effect of the wheels. If I get a puncture and take the wheel off, hold the axle in my two hands and spin the wheel, it is very hard to alter the direction of the axle if the spin is fast. Can anyone explain this (gyroscope) phenomenon in simple English with little or no maths ?```
 Subject: Re: Why can I ride my bike ? Answered By: thx1138-ga on 14 Jun 2002 12:48 PDT Rated:
 ```Hi gruffgareth and thanks for the question, When you hold your bike wheel and spin it, the resistance you can feel when you try to alter the angle of the spindles is called “gyroscopic inertia” as Sir Isaac Newton explained “ a body tends to continue in its state of rest or uniform motion unless subject to outside forces.” In other words something will stay still or continue moving unless something else affects it, in this case you, trying to alter the angle of the spinning wheel. For a simple explanation using your example of the spinning bicycle have a look at this website: http://www.exploratorium.edu/snacks/bicycle_wheel_gyro.html For an explanation of "gyroscopic inertia" have a look at this website (near the bottom) http://www.probertencyclopaedia.com/SG.HTM#GYROSCOPE "How Gyroscopes Work" is a detailed description without too much maths. http://www.howstuffworks.com/gyroscope.htm and for the full maths version have a look at this website: http://www.geocities.com/CapeCanaveral/2188/main.html#index Hope this doesn´t set your head spinning too much. THX1138``` Clarification of Answer by thx1138-ga on 14 Jun 2002 12:50 PDT ```Sorry I forgot to mention the type of searches I used to try and explain your question: Search Strategy: gyroscope science explanation ://www.google.com/search?hl=pt&as_qdr=all&q=gyroscope+science+explanation&lr=```
 gruffgareth-ga rated this answer: `nice googling !`

 Subject: Re: Why can I ride my bike ? From: sdk-ga on 14 Jun 2002 15:16 PDT
 ```Although you limited your question to gyroscopes, I'd like to add something about gyroscopes and bicycles. The way your wheel is connected to your frame is more important for balance than gyroscopic action. If you look at the frame of your bike, you'll notice that the axis of rotation for the front wheel is not vertical. This is called caster. As you start to tilt right (don't fall), the wheel will turn right. This is because the wheel is touching the ground behind the axis of rotation. When the bike turns right, your body wants to keep going straight, pulling you upright again. This web page describes this process very well. It also has a good picture showing caster and trail. It even has references. http://www.dclxvi.org/chunk/tech/trail/```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 17 Jun 2002 16:02 PDT
 ```I'd also like to add that the resistance to the change in rotational axis due to gyroscopic effects is *not* directly holding you up. The tire only weighs a couple pounds and you weigh 50-100 times more. People always cite the spinning tire in the hands physics experiment, but the fact that you *can* still move it with your hands is proof that it's not a significant direct effect. We've spun bicycle tires up to over 100mph and still were able to force them to change their rotational axis. What actually happens is that gyroscopic precession (ambly described elsewhere) causes the handlebars to turn (easier than trying to describe the right angle nature of the change in the tire's rotational access) the same direction that you lean. As the bike starts to fall to the right, the gyroscopic precision turns the front tire to the right and that drives the tire back underneath the center of gravity, keeping the bike from falling over. At all times the bicycle is attempting to balance by means of the gyroscopic precession, plus the above mentioned castor and rake, trail (and tire profile) which all cause it to try to keep the front tire underneath the Cg causing the bike to remain either upright or at least at the same lean angle it started at. What's more interesting is how you actually initiate a turn on a bicycle or any balancing vehicle. It's called countersteering and it means that you actually have to counter the natural balancing effects above to cause it to turn. Practically speaking it means that if you want to turn right, you actually apply pressure to the handlebars such that they briefly turn left. Simpler said, push on the right handlebar to initiate a turn to the right, and left bar to initiate a turn to the left. What actually happens is the tire turns very slightly to one way, the bike begins to fall the other way, and then all the above stabilization mechanisms kick into place to hold the bike at a particular lean angle which will cause the bike to go around in a circle. The harder you countersteer, the quicker or further it turns. You also countersteer to stop a turn by turning the bars back toward the direction you're leaned which causes the bike to stand back upright. Because of the greater mass of motorcycles most motorcycle riders are exposed to the concept of countersteering eventually, many right from the beginning. It's critical to making an effective turn. However most people are never taught that countersteering works on bicycles but their bodies simple learn it. The process of learning to countersteer is basically the "steep learning curve" part of learning how to ride a bicycle and most people never know they're doing it. Watch a kid who's just learning when they're presented with an obstacle to avoid. You'll often see them develop wild wobble where they manage to traverse 30-40 feet and smack right into the thing they're trying to avoid. The problem is their brain (and possibly their parent) is telling them to turn the handlebars one way, while their body, having just learned countersteering knows it need to go the other way. The conflict is displayed as wobble and in the end they average a straight line toward the obstacle. Try countersteering yourself. Ride your bike, apply a gentle pressure to the right bar (as though you want to turn it to the left), and you will end up turning right. Try twisting the bar with one hand right at the stem. You'll see the same effect. ian```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 17 Jun 2002 16:31 PDT
 ```One further thought. The original question was how to explain the gyroscopic effect in simple terms without math. I think I can do it without using the words "angular momentum", "torque" and "gyroscopic precession". :-) Imagine someone fires a bullet horizontally past you from some distance away. For all practical purposes the bullet travels a sraight line toward you (ignore gravity for now). As it passes you, you reach out and very quickly push down on the bullet for a fraction of a second. You've now deflected the bullet onto a new course angled down toward the ground. It was going horizontal and now it's angled down. The change in angle depends on how fast the bullet was travelling, how heavy it is, and how much force you applied to it as it went by. One straight line, with a bend in it followed by another straight line. Ok, so now someone ties a string to a second bullet and attaches the other end of the string to a post with a pivot on it. They stretch the string taut point the gun perpenticular to the the string and fire fire the bullet and (again discounting gravity and drag) the bullet travels horizontally. But now because it attached to the string it follows a curved path in a horizontal plane. A circular path around the central post, round and round. Again as the bullet passes by you, you reach out and push the bullet downward. Again it is deflected from its horizontal path toward the ground, but the string still has a hold of it so it simple swoops down closer to the ground reaching its lowest point one quarter turn (90 degrees) from the place where you deflected it, and then continues on around reaching its highest point 1/2 turn (180 degrees) beyond that and when it returns back to the place where you deflected it it occupies almost *exactly* the same space, just at a different angle. If you continue to delfect it downward each time it comes by the low side will get lower, the high side higher and ultimately it continues to travel around and round the post in a circle. The plane of that circle is simply tipped more and more away from horizontal and where you deflect it, is actually "noticed" 90 degrees forward in rotation. THAT is gyroscopic precession. Any tipping force you apply to a spinning object is equivalent to deflecting the path of that bullet (every atom in the spinning object is like the bullet on a string around the spinning axis), and it reacts to the force by tipping 90 degrees in rotation beyond the point you applied the force. For a simple home experiment, tie a tennis ball to a 2-3 foot long string and then spin it around vertically. Once you have it going, stick your foot out and let it deflect the ball a little bit (careful not to bean yourself). Notice which way the spinning circle tilts. If you spin the ball and string around clockwise, and push outward with your foot at the bottom of the circle, the circle will tilt out on the left, in on the right, instead of out at the bottom as you might originally expect. Try it. Gyroscropic precession need not be so mysterious. ian```
 Subject: Re: Why can I ride my bike ? From: gruffgareth-ga on 18 Jun 2002 02:39 PDT
 ```Wow folks, an impressive set of responses ! There was a Professor Laithwaite (at Manchester Uni ?) who was intrigued by spinning objects. Despite being a prof., I don't think he understood the gyroscope effect, as he was convinced that spinning an object could cause anti-gravity (like the hanging bicycle wheel). I had wondered about the gyroscope effect many years ago (when fixing a puncture) - we never did this in school physics (seems like a major omission). Since there were no teachers to ask, I imagined my own explanation, but never followed it up till I came across this fascinating Google answers site. I tried the set of web sites from thx1138-ga. Some of the sites explained it by saying it was because of "angular momentum" or "gyroscopic inertia" - which isn't any kind of explanation ! But the site at http://www.howstuffworks.com/gyroscope2.htm broke the wheel down into moving segments and was just the kind of explanation I was looking for. daemon-ga gave a similar explanation to that site using a very nice mind-illustration with bullets on pieces of string. Mention bullets and you'll have the attention of schoolboys, so this could be very good for the classroom ! Though flexible string does lose the forces that operate among the opposite sides of a rim and the axle in a rigid wheel. I had been just thinking about the gyroscope effect, but then sdk-ga and daemon-ga gave some fascinating (and new to me) insights into the other effects that keep a bicycle upright. I had wondered how those micro-scooters could possibly work with such small wheels ! I am really amazed. I already had a lot of respect for the bicycle, but now I'm thinking it may be one of the simplest, yet most sophisticated machines ever invented. Excellent for individual and planetary health into the bargain ! If it's OK, rather than select a "winner" I'd like to donate \$15 to the coming famine in southern Africa (probably thru Oxfam). Thanks again for the great responses, Gareth```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 18 Jun 2002 10:38 PDT
 ```It's not a contest for your money. thx1138 answered the question so he deserves it, as that's how this site works. I am not a Google researcher. I just provided my comments because I gain some enjoyment out of explaining everyday physics in everyday terms, particularly the study of balancing vehicles and dispelling the myth of "gyroscopic action holds you up directly" is a perpetual goal of mine. Spelling mistakes are left in as an exercise for the astute reader. :) ian```
 Subject: Re: Why can I ride my bike ? From: buddytoliver-ga on 18 Jun 2002 13:00 PDT
 ```Can anyone explain why I am NOT able to ride a bike on a tredmill? I tried to do so and it is extremely wobbly.```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 18 Jun 2002 14:09 PDT
 ```Centripedal acceleration. Or rather lack thereof. There's a simple equation which describes the relationship between your forward velocity, lean angle and the radius of a turn. It is. lean angle = 90 - atan(v^2/r/g) where v = velocity r = radius g = gravitational constant (angles in degrees not radians). When a bike is going around a turn at a specific speed it leans at a particular angle, but from the perspective of the bike it is basically still balancing upright. The forces you feel as you turn are down straight through the bike rather than sideways like in a car. This relationship is all part of the balancing act and is an important component of overall stability. The faster you go the more stable you become. At a high speed, it takes a much greater lean angle to induce a sharp turn than at a lower speed, and all the balancing effects described resist large changes in lean, thus it helps overall stability. Ok, all that said.. That equation only applies when you're actually moving and have forward momentum. When you're riding on a treadmill (and they do make special bicycle treadmills with two rollers in back and one in front), things like gyroscopic precession, rake, trail etc, all still work, but there's no forward momentum and thus no stability to be had from the relationship described by the equation above. 100% of your stability has to come from the front tire driving under the CG, and there's no centripedal acceleration to "lean against" as your speed increases. This is also why riding a unicycle is so tricky. At low speeds it's a constant wobbly balancing act, but if you actually get up some speed on a unicycle it'll become more stable, and it will lean into turns just like a bicycle (and numerous other craft, including snowboards, mono-skis (snow and water), jet-skis etc, skateboards) ian```
 Subject: Re: Why can I ride my bike ? From: buddytoliver-ga on 20 Jun 2002 13:29 PDT
 ```Thanks Ian, I think you may have uncovered the answer, but I disagree with your particular explanation. When riding on the treadmill, I do have forward momentum relative to the treadmill. Thus my lean angle = 90 - atan(v^2/r/g) should be correct relative to the surface of the treadmill. After all, suppose instead of a treadmill it was a flatbed train moving in the opposite direction, etc. Eventually it is easier to visualize that the ground moving back one way is equal to the bike moving forward the other way. The key I think from looking at the equation and lies in the fact that I have to start out slow on the treadmill because of the difficulty of getting my bike on the treadmill and pushing the start button. Because I have to start out so slow, it is more difficult to balance (lower v) just as it is on the road. If I am brave enough to try, I would imagine if I turn the speed up, I could ride relatively stable. The only problem is that on the road you have a few feet to the left and right to play with where on a treadmill in my bedroom, there is considerably less space. Thanks for the equation and the time.```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 20 Jun 2002 14:50 PDT
 ```Nope, you're in the wrong frame of reference. Let's take your flatbed train, pretend it has a continuous line of cars to ride over down its entire length. If you were standing still on one of the cars while it went around a sharp corner, you would have to lean into the turn to remain balanced. Same if you were sitting still on your bicycle on top of the train car attempting to balance. (growing up in a rural western state, I spent quite a bit of time riding in the back of pickup trucks and flatbed trucks so it's something I can relate to quite easily). Now let's say the train is going 20mph through a sharp turn, and you are on the bike riding down the cars the opposite direction also at 20mph. You would remain stationary relative to the ground and not have to lean the bike at all, as the train passed under you. If you tried to lean into the turn you would simply ride off the edge of the car onto the ground. If you turned around and rode the same direction as the train you'd have to lean even further as you and the train both rounded the corner at the same time. You'd have to plug 40mph into the eq above instead of 20 or 0. The only momentum that matters for the lean angle is that relative to a fixed frame of reference (or as fixed as it can be) which in this case is the earth. On the treadmill, the only thing that has any momentum is the treadmill itself and the tires of the bike. Yes you'd have to lean a little to travel back and forth on the treadmill but that's just a simple balancing act. There's no resistance to falling over due to your foward momentum because you have none. Too bad Merry-go-rounds have been banned from all playgrounds. They were a fantastic tool for learning physics, including this little particular problem. (if you run in the opposite direction on a merry-go-ground you don't lean in at all). ian```
 Subject: Re: Why can I ride my bike ? From: buddytoliver-ga on 24 Jun 2002 14:52 PDT
 ```Thanks for hanging in there with me Ian. But I am still confused. What if my treadmill were a big concrete slab with trees and buildings on it. I cannot distinguish being on this treadmill (which also moves all the air in the room with it) from riding on the street itself. So how does my momentum all of a sudden change? Does the momentum of the treadmill (which is now considerably greater) come into play? and if so how? Also, suppose I am in a train moving at a constant velocity (Constant direction and magnitude). If I am riding my bike in the opposite direction at the same speed so that relative to the ground I am not moving, it seems my turn angle would be the same as if I was riding on the street at this same velocity. How does this differ from the situation on the tread mill? If I could use the turn angle of my bike to calculate the velcoity of the train relative to an external frame of reference, then doesn't this violate what I learned as the first premise of relativity, namely that the laws of physics are the same as observed in any non-accelerating frame, i.e. you cannot conduct an experiment inside a closed box that will tell you whether that box is moving at a constant velocity relative to an outside frame of reference. Please let me know where I have all of this mixed up. And again, thanks for your patience```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 25 Jun 2002 00:54 PDT
 ```The equation above to compute lean angle, has as one of its terms g, the acceleration due to gravity on earth. In all of your examples, big concrete slab, moving train, and treadmill, they are all operating within a much larger overriding context, which is the earth and its gravity. In addition the velocity term is along a vector always perpendicular to that gravitational acceleration. They're tied together. On the treadmill there is no velocity vector perpendicular to the force of gravity acting on the combined mass of the bike and yourself. Only the tire rims are moving and that all motion cancels out relative to the earth's frame of reference because it's rotating. If the slab/train/treadmill were all so massive that they generated their own gravity while floating freely in deep space then yes you could ride across them and compute a lean angle independent of what is observed riding over the ground on earth no matter how fast they were moving. But then the constant 'g' in the EQ would also have some other value, and the lean angle would be correspondingly different. You're right that in your black box you cannot detect that the box is moving at a constant velocity relative to an outside frame of reference from inside the box, but that's only if the box is free to move along a "true" constant velocity vector in deep space.. Velocity is a vector not just a number and it's only straight if it's free of external forces. As soon as it is acted upon by an external force (in this case gravity) you'll notice something in the box although it may take a while. If the box is travelling over the ground then you feel the *external* force of gravity inside the box, while the ground is supporting it. Inside your box you'd know immediately that the box was resting on a larger object with a gravitational pull thus you already know something about the box's frame of reference, which you didn't think you'd know. You say.. "fine, so perhaps it's orbiting around the earth in freefall, then you couldn't tell." But if it's a box with rectangular dimensions, tidal effects will eventually orient it a particular way, and you will gradually tend to float to one end of the box or the other over time, because in an orbital context all the particles of the box and you inside *want* to orbit at different velocities. (lower orbits are faster, higher orbits slower). They can't though because they're all connected to each other so stuff between the CG of the box and the gravitational source "falls" inward, and stuff outboard of the CG is "flung" outward. Some great Sci Fi based on this effect. Check out Niven's short stories. Using relativity to describe macroscopic effects can be dangerous for the above reasons. "constant velocity" simply cannot apply to an object composed of more than one particle in a big gravitational well unless it's falling directly toward the gravitational source, and that's not the case here.. Resting on the surface of that object pushes this all to the point of being a worthless comparison. But forget all that mumbo jumbo. You *know* intuitively how this works. What happens if you set a cup of coffee on the dash of your car and turn the car sharply? What if you hang a set of fuzzy dice from the mirror by a string and go around a corner? The angle that the dice hangs is the very same computed lean angle as in the EQ above. A bicycle riding in front of the car would lean *exactly* the same amount as the fuzzy dice. (note mass is irrelevant here). Just to bring this briefly back to why this is important to bicycles, all the *other* stabilizing effects of a bicycle (gyroscopic precession, rake, trail, tire profile) all are more pronounced with increased or rapid changes in lean angle. So if going faster around the same turn produces a larger lean angle due to the EQ above, it means all the stabilization effects act with all the more authority. The bike is more stable when going fast, and less so when going slow. This isn't rocket science.. Ok, well it is rocket science, but it's not hard. ;) ian```
 Subject: Re: Why can I ride my bike ? From: daemon-ga on 25 Jun 2002 00:55 PDT
 ```Oh crud. That formatting is evil. It didn't look like that until after it was posted. Sorry. What an annoying glitch. Wish I could clean it up.```
 Subject: Re: Why can I ride my bike ? From: buddytoliver-ga on 04 Sep 2002 11:19 PDT
 ```You say that there is no velocity vecotr perpendicular to the gravitational field on a treadmill. If the veolcity vector of the bike relative to the earth is the significant vector, then what happens when I ride my bike (or walk for that matter) from my seat on an airplane to the bathroom. According to the equation and this account of the appropriate velocity vector My lean angel would be enormous. Again, please help. Thanks, Buddy```
 Subject: Re: Why can I ride my bike ? From: atr-ga on 20 Sep 2003 09:00 PDT
 ```When you walk from the front to the back of the airplane, you're not even making a turn, at all. It's a straight line overlapping the plane's own straight path of travel.```