Motorcycle Racing - Why Don't The Bikes Tip Over On The Corners?
October 28, 2013 5:10 AM   Subscribe

Just curious why the motorcycles don't completely tip over when they corner in the motocycle races?

I was flipping the channels over the weekend and landed on a MotoGP race. I have always been amazed at these races and riders and the skill involved and one of the most interesting things in when they corner. They get so low, its amazing that they just don't completely roll over. A couple of other people commented who were also watching with me and it got me thinking - why don't they completely roll over??? I know it has to do with physics, but can someone provide an explanation? Is it because they are going so fast? Is it because they touch their elbows or knees? Just curious....
posted by dbirchum to Sports, Hobbies, & Recreation (32 answers total) 3 users marked this as a favorite
 
I assume you want to know why they don't completely fall over to the inside of the turn? Basically, because all of the force in the turn actually wants to make them fall over to the outside of the turn. By accelerating and leaning in, they are actually using the grip of the tires to use that centripetal force (that wants to make them tip to the outside) like a wall, or a bank.

That's why you always accelerate through a turn on a motorcycle. Decelerating can make your tires lose grip - which will make you tip over. To the outside of the turn, though.
posted by Benny Andajetz at 5:31 AM on October 28, 2013


Notice how those riders do not only not fall over, but that they have to really work to bring the motorbikes down into a turn?

This is because a rolling wheel creates an angular momentum along the axis of the axle. This makes a splnning wheel act like a gyroscope.

You may have seen the demonstration where a person holding the axle of a spinning bicycle wheel is spun in circles as the wheel is tilted. You can try it yourself, it takes some effort to tilt a spinning wheel.

If the wheels stop spinning, the angular momentum disappears and the bike falls over as you would expect.
posted by three blind mice at 5:36 AM on October 28, 2013 [1 favorite]


As the rider approaches a left hand turn, they'll push on the left side of the handlebar and pull on the right side. This is called countersteering, and it's how every motorcycle and bicycle in the world works. The wheel will turn slightly to the right, and the bike will start to lean to the left.

Now the rider will continue the left lean and will turn the bars to the left, which will decrease the radius of the turn, and if they hold everything constant, they'll make a perfect circle at that radius. At this point, you've got a force tipping them left (because they're leaning to the left) and a force tipping them right (centrifugal force from the left turn). If they lean farther left, they'll turn more to the left. If they lean less to the left, and/or give the bike more throttle, they'll lean less to the left.

Balancing all this out are the tires. The tires are resisting the forces that want to make the bike slide to the left. The tires have a finite amount of grip, and if they exceed that friction window they will slide out. They'll go as fast as they can until 100% of the available traction is being used on turning, so that if they touch the brakes at all, they will slide out (because 0% of traction is available for braking).
posted by spikeleemajortomdickandharryconnickjrmints at 5:38 AM on October 28, 2013 [2 favorites]


If you swing a bucket of water around you (with you pivoting on the floor - the bucket stays level with the ground), the water stays in the bucket right? Because of your rotation there is a force parallel to the ground and pointing away from the center of rotation (i.e. you), holding the water in the bucket. Hold that thought - that when going around in a circle, there's a force pointing away from the center of rotation.

Now take a simple box, taller than it is wide. Tip it more and more, until it tips over. It tips over when the center of gravity of the box is past the corner of the box that is still on the ground. That's the bike, leaning.

So, why does the bike keep from falling over? The center of gravity is certainly past the "tipping corner". Well, there's also that force from going 'round the bend. Think of that like a string pulling horizontally in a direction that tends to right the box.
posted by notsnot at 5:40 AM on October 28, 2013 [1 favorite]


At such speeds the horizontal forces outward, away from the center of the turn, are much larger than the forces of gravity, so the vector of the two forces acting on the bike is closer to horizontal than vertical. The center of mass of the bike and rider and the tires' contact patches must stay in line with that vector.
posted by jon1270 at 5:42 AM on October 28, 2013


Don't think of them laying on their sides, think of them completely vertical driving across as steeply sloped road.
posted by blue_beetle at 6:04 AM on October 28, 2013 [3 favorites]


This might help - it covers the physics of bicycle handling, but is applicable to motorcycles as well.

It turns out that the angular momentum from the wheels is not a very significant factor in keeping a bike upright (although the link does mention that the higher moments of inertia of motorcycle wheels may make those more significant). There was a study I saw recently (and unfortunately my search skills are failing me at the moment) where researchers built an angular momentum-free bicycle to show what the link says - the geometry of a bike causes it to turn into a fall and keep the bike upright.
posted by backseatpilot at 6:32 AM on October 28, 2013


Yep, a two-wheeled vehicle is essentially self-righting, to a point. If you push a riderless bicycle and then let it go, when it begins to slow the front wheel will turn in the direction of the bike's lean. But the rest of the bike wants to fall the other way. So the bike will circle (and remain upright) until there's not enough forward momentum to support it.
posted by Benny Andajetz at 6:44 AM on October 28, 2013


Cars work on the same principle and can be seen on speedways as well as highways. Did you ever notice the superelevation on every single highway interchange ramp? The faster the vehicle goes on a curve, the steeper the ramp needs to be to maintain control. The coolest thing ever is a velodrome like this.
posted by JJ86 at 6:48 AM on October 28, 2013 [1 favorite]


It turns out that the angular momentum from the wheels is not a very significant factor in keeping a bike upright

Is completely incorrect. Watch this for a more thorough demonstration of angular momentum.

"None of this is intuitive."
posted by three blind mice at 6:53 AM on October 28, 2013


Is completely incorrect.

Sorry, but I must disagree. I found the study I was looking for. A bicycle/motorcycle can be stable with zero gyroscopic effects.
posted by backseatpilot at 6:58 AM on October 28, 2013


A bicycle/motorcycle can be stable with zero gyroscopic effects.

Well, yes, but that doesn't mean it doesn't help, no?

The angular momentum keeps the bike stable at whatever angle it is at in a given time but doesn't produce a force to move it back to the upright position. For an example, see the video three blind mice links to where the weight is added - the wheel takes a set at an angle but retains that angle when the weight is removed. The reason the wheel doesn't fall over when placed on the rod in the first place is because the wheel wants to retain it's orientation due to the spinning effect. Not because it wants to stay 'upright'.

So the angular momentum stops the wheel from falling over more, but it doesn't keep it 'upright' as such so it sounds to me like two different people saying the same thing from both sides, here. The angular momentum doesn't 'hold the bike up' at all, but it makes it less likely to move from the angle you have set it at from other means. It helps it hold the lean, if you will.

Cornering force is the biggest single factor, though. Centripetal force (Centrifugal effects are not a force, but a resultant) wants to force the bike to go straight. The turning effect of the bike means you have to lean in the opposite direction (ie left for a left hand corner) to counteract this. Watch a big SUV corner. It will lean to the outside of the corner as it goes around (top of the car moves away from the inside of the turn). The bike is just pre-empting that force effect. The more the bike would lean away from the turn (like the SUV) if the rider tried to corner in a vertical position, the more he can (and must) lean into the turn to maintain his speed.
posted by Brockles at 7:19 AM on October 28, 2013


Agree. But isn't the angular momentum vertical until you're actually in the turn?
posted by Benny Andajetz at 7:30 AM on October 28, 2013


Response by poster: Ok - I think I have it now (please correct me if I am wrong), and I will try and state this in the most basic terms (for my benefit).

And yes, to clarify, I should have stated, when a rider is going around a right hand turn (for example) and they are significantly leaning into the right, why don't they tip over (to the RIGHT), i.e. fall into the turn, vs. falling on the outside of the turn (as Benny above tried to clarify for me - sorry about that Benny...my fault).

Ok, so if a Motogp rider is heading into a right hand corner and swings his body weight and leans into the right (the inside of the corner), the reason why he doesn't fall over the the right (the inside of the corner), is because the force of the turn is wanting his to go straight (i.e. to the outside of the corner), so he/she actually has to place a lot of effort to lean into the corner as to not be vertical?

In basic terms, is that what is going on? (the suv corning example really made sense to me - I do notice how they lean towards the outside of the corner).

Ok, IF I have that correct (and please let me know if I don't), is there a point (maybe what notsnot called the 'tipping corner') that regardless of the force trying to pull the rider vertical (to the outside of the corner), that they will just lean over?? I mean, yesterday, that had a "angle graphic" that showed that in some cases, the rider would lean 60 degrees INTO the corner! That is insane. What happens if they leaned 62 degrees? Or 67 degrees? Would they eventually fall into the corner? Or does this depend on speed?

So, i am also assuming that the riders have to brake to enter INTO the corner, and then accelerate through the turn?

AND, what, if any, does their know and or elbow touching the ground have an impact?

It feels like i am in a physics class! Who would of thought that I would get into this on a Monday morning!!
posted by dbirchum at 7:36 AM on October 28, 2013


Gyroscopic effects, while being a stabilizing force, don't seem to be a reason why don't they completely roll over.

I think the question is why the bikes don't continue on their way to the ground when the rider lays them to one side into a turn. To which the answer would be that the riders are controlling the turn continuously. It's the same reason a bicycle rider doesn't fall. Geometric configuration will make a bike or motorcycle somewhat self righting. But without an actual rider or means to steer, it will fall over eventually.

Ok, IF I have that correct (and please let me know if I don't), is there a point (maybe what notsnot called the 'tipping corner') that regardless of the force trying to pull the rider vertical (to the outside of the corner), that they will just lean over?? I mean, yesterday, that had a "angle graphic" that showed that in some cases, the rider would lean 60 degrees INTO the corner! That is insane. What happens if they leaned 62 degrees? Or 67 degrees? Would they eventually fall into the corner? Or does this depend on speed?

Rider skill has a great deal of effect here. Too much lean, and the bike won't recover and fall. Not enough, and the turn will be too wide, possibly causing a crash. Rider has to use judgment to call for proper speed and angle.

So, i am also assuming that the riders have to brake to enter INTO the corner, and then accelerate through the turn?

Maybe, maybe not. Rider has to make that judgment moment to moment.

AND, what, if any, does their know and or elbow touching the ground have an impact?


Probably depends how hard that knee or elbow touches the ground. Certainly it could provide some tactile feedback for the rider. Of course, if the knee or elbow provides too much friction/impact it can cause a problem.
posted by 2N2222 at 7:47 AM on October 28, 2013


In basic terms, is that what is going on?

Yes.

What happens if they leaned 62 degrees? Or 67 degrees? Would they eventually fall into the corner? Or does this depend on speed?

Yes, it depends on the speed. They are balancing the force trying to get them to go straight versus the amount they are leaning over plus the tyres grip. If they lean too much, or don't have enough speed or don't have enough grip, they will either fall over to the inside of the corner (and then slide in a relatively straight line) or the bike itself will slide out from under the rider (depending on whether not tyre grip was the limiting factor. Vertical weight on the tyre is part of that equation. If they lean too much, not enough force on the tyre means tyre grip is lot. It's a seriously compilcated balancing act.

So, i am also assuming that the riders have to brake to enter INTO the corner, and then accelerate through the turn?

In a bike, yes. Not in a car.

AND, what, if any, does their know and or elbow touching the ground have an impact?

I think they use the elbow to get an idea of how close they are to the ground (for repeatability of lean, perhaps). Also, I have seen some riders use an elbow to counteract a slide - when the bike slips it will make them lean over more (and hence fall) so I've seen them kick their elbow into the floor to knock themselves upright to retain grip (more vertical force on the tyre) and stop the slide. Which is JUST INSANE.
posted by Brockles at 7:51 AM on October 28, 2013 [1 favorite]


Response by poster: I have even more respect for this sport now. I am going to start to watch it on a more continuous basis (I know there is only one race left in the season).

It seems that motoGP is more popular in Europe and other areas vs. North America, correct?
posted by dbirchum at 7:55 AM on October 28, 2013


Response by poster: And maybe if my physics teacher in high school used MotoGP examples (or SUV cornering examples), it would have made more sense (or at least more interesting!)
posted by dbirchum at 7:56 AM on October 28, 2013


I think Superbike is the more popular racing series in the US.
posted by 256 at 8:12 AM on October 28, 2013


Glad you're going to follow MotoGP more closely. Like most sports, the more you know about it, the more enjoyable it is to watch. If you're able, I'd strongly recommend learning to ride a motorcycle. Even if you never use the skills involved for anything else, it'll increase your appreciation of 2-wheeled sports immensely.
posted by Nick Jordan at 8:38 AM on October 28, 2013


Not to threadjack, but besides MotoGP and Superbike (are they both streaming, or what channel do they air on?), are there other motorcycle events that I can watch from the U.S.? I have a heck of a time finding any motorcycle races on TV (Verizon FiOS)...I'm willing to watch internet streams.
posted by xiaolongbao at 9:18 AM on October 28, 2013


Response by poster: I haven't been on a dirtbike (have never been on a motorcycle) since I was about 8 or 9 years old. I used to have a dirtbike but it was ages ago and I never cornered like that! I spent most of the time on dirt jumps behind my house! I would love to ride a motorcycle and hopefully my budget will permit it soon.

Sadly, I ride mountain bikes (that's not the sad part!), so maybe I should know more about this stuff that I should??

I notice that the MotoGP is on the Speed channel every-time there is a race. I live in Canada and we get the Speed channel which I think in the US has recently been re-branded Fox Sports 1 and 2 (or something like that), but its still Speed in Canada.
posted by dbirchum at 9:46 AM on October 28, 2013


The official MotoGP Youtube channel has a series called MotoGP Classics, in which they upload complete race videos of some of the best and most notable races from MotoGP history, like this incredible race that heralded Valentino Rossi's rise to uncontested king of motorcycle racing.

If you don't mind knowing who wins (half the time it's spoiled right in the title of the video), this is the perfect way to whet your appetite during the off season.
posted by 256 at 10:00 AM on October 28, 2013


Best answer: So there are a couple of interesting snippets of physics going on in a motorcycle's turn. The gyroscopic action is largely a lean-inducing effect when the handlebars are manipulated (and this one is totally counter intuitive). On a motorcycle, if I am going straight and I push the right handlebar away from me (in such a way as to make the front wheel point to the left) it will actually initiate a lean to the right. That lean to the right will make the bike veer right. Oddly, the whole while, the handlebar wheel assembly will barely move during this series of events. An interesting Wikipedia article explaining that situation.

I think the question the OP is asking is about the effect between the wheels and the ground during a turn which has very little to do with the gyroscopic effect. Inertial forces tend to keep you going the direction you are going. In a turn, your vehicle must fight that force. You can look up free-body diagrams talking through centripetal and centrifugal forces to see how speed and radius affect the revolving body. On a motorcycle in a turn, you have the gravitational force that is pulling down toward the ground (normal to it). You also have the force keeping the rider from shooting off the side of the road (away from the center of the turn). This is a pretty solid graphic for reference. The faster a bike is going OR the tighter the turn (basically the effect of the expression above the blue vector in the link), the harder the lean must be to keep the bike from wrecking away from the center of the turn.

There are two wrecks that happen here. Either the coefficient of friction of the interface between the wheels and the road is insufficient and they slide out from under the bike. Or the center of gravity of the rider-bike system gets too high and the system pivots outward from the center of the turn about the wheel-pavement axis.
posted by milqman at 12:18 PM on October 28, 2013 [2 favorites]


Best answer: The lean angle for a motorcycle is the same bank angle of an airplane by the equation for circular motion:

arctan(angle) = V^2/(g * r) where g is the acceleration due to gravity and r is the radius of the turn.

This is modified slightly by the steering angle but it turns out that the steering angle for a motorcycle in a turn is very tiny.

Let's plug in some numbers. 50 MPH is 73 feet per second and g is 32.2 feet per second and assume a curve with a 100 foot radius.

You get an angle of 59 degrees for the lean. This assumes that the curve is unbanked.

The angle is measured from the point of contact of the tire through the center of mass of the motorcycle and rider. Looking at the formula above you can see that increasing the lean angle means you can maintain a higher speed in the turn. Notice that going into a turn the rider drops the inside leg and also hops sideways on the seat toward the inside of the turn to lower their body closer to the ground. This effectively lowers the center of gravity of the combined motorcycle and rider, which means the effective lean angle is greater, to allow a faster speed through the turn.

Why doesn't the rider fall over? It is because the forces related to the equation above balance exactly. The lean is precisely the amount so that the vertical force to match gravity and the horizontal force to produce the circular motion dictated by the velocity and radius sum up (as a vector) to the lean angle.

How does the rider maintain this precise balance? All he has to do is maintain the desired path on the curve. If he leans farther, he tightens the path. If he leans less, he lessens the curved path. So by leaning to follow the desired path, his lean precisely produces the amount of vertical force required to match gravitational force.

Notice that the equation above says nothing about mass. This means that the angle of lean has nothing to do with weight of the bike or rider. However mass does affect the amount of traction required by the tires to ride the curve without skidding out. The horizontal frictional force required by the tire is directly proportional to weight (mass). A lighter rider and bike are less likely to to skid, all other things equal, because there is a limit to the static friction a tire can produce.
posted by JackFlash at 12:19 PM on October 28, 2013 [2 favorites]


Oops. The formula above should have tan on the left or arctan on the right, but I guess most could figure that out.
posted by JackFlash at 3:00 PM on October 28, 2013


Response by poster: Sweet gentle....this just got complex...

So the tires also play a huge factor with the lean (the contact patch/strip), correct?
posted by dbirchum at 4:39 PM on October 28, 2013


So the tires also play a huge factor with the lean (the contact patch/strip), correct?

Depends on what you mean by huge. The formula above is good for a first order approximation for typical racing tires. On the other hand, if you instead have very fat tires, then when you lean over, the center of the contact patch is no longer in the plane of the wheel rim. The contact patch may now be several inches toward the inside of the turn from plane of the rim as you roll onto the side of the tire.

Recall above that the critical angle is measured from the contact patch through the center of mass of rider and bike. So if you keep the center of mass the same, but move the contact patch inward to the center of the turn, then you have effectively decreased the angle of lean. For a fat tire, you must lean the center of mass toward the inside of the turn the same number of inches you moved the contact patch toward the inside of the turn to maintain the same effective angle as for a skinny tire. This means the angle of the frame of the bike must be steeper for fatter tires.
posted by JackFlash at 5:33 PM on October 28, 2013


Oh, and along the same lines, you might have noticed that racers sit up straight at they go into a turn. You might think that is just to slow down using wind braking. But the more important reason is that they are raising the combined center of mass of the bike and rider in the plane of the frame. By extending the center of mass in the plane of the frame as they lean over, they have effectively increased the distance that the center of mass moves to the center of the turn. This somewhat negates the effect of the tire patch also moving to the center of the turn, reducing the necessary lean.

So they shift sideways in the seat, sit up straight and extend their knee, all of which move the center of mass towards the inside of the turn, farther out over the tire contact patch. All of these reduce the amount of lean required.
posted by JackFlash at 5:48 PM on October 28, 2013


The physics of motorcycles are insanely complex and (surprisingly) not well-studied. The most authoritative work I've found on the subject is How and Why: Motorcycle Design and Technology, written by Gaetano Cocco and published by Aprilia. The translation from Italian is a little dodgy and there's quite a bit of math, but it discusses at length the gyroscopic effects of rotating wheels and engine internals (spinning cranks, cams and and driveshafts have appreciable stabilizing forces), as well as chassis geometry.

NB: Ignore the Amazon book summary. This book is in no sense a how-to-ride manual.
posted by workerant at 7:17 PM on October 28, 2013


The physics of motorcycles are insanely complex and (surprisingly) not well-studied.


Well. Studied but not published, I'd suspect...

all of which move the center of mass towards the inside of the turn, farther out over the tire contact patch

This is contradictory. If they are moving their mass towards the centre of the turn while leaning they are moving it away from the contact patch, not to being more over it.
posted by Brockles at 6:18 AM on October 29, 2013


This is contradictory. If they are moving their mass towards the centre of the turn while leaning they are moving it away from the contact patch, not to being more over it.

Perhaps not well stated. I did not mean to imply that the center of mass is vertically over the contact patch. That would be no lean at all and the bike would go straight.

The critical factor is to maintain the proper lean angle required for speed and turn radius according to the equation for circular motion. The lean angle is measured using the line between the contact patch and the center of mass. So the center of mass in a turn must be closer to the center of the turn than the contact patch because that is what defines the lean angle. To a first order approximation the contact patch never moves. The lean defines the angle.

But if you consider tire width, when you lean, the contact patch moves to the inside of the of the turn onto the side of the tire, say two inches. This means that the center of mass must also move the same amount, say two inches, to maintain the same angle (parallel lines). You do this by either leaning the frame farther, or else by moving your body toward the center of the turn. A cyclist does this naturally by extending their knee, sliding their butt a little, and sitting up straight. By sitting up straight when the bike is leaning, your head and upper body mass move toward the center of the turn.
posted by JackFlash at 7:47 AM on October 29, 2013


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