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Note that my question is not why do you tilt your bike when on a curve. It's about the reduction in turning radius when one tilts the bike inwards.

bike

Short to-the-point answers are welcome.

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    $\begingroup$ physics.stackexchange.com/q/24 $\endgroup$ – Nilay Ghosh Jul 26 '18 at 4:03
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    $\begingroup$ My great-uncle, a physicist by training, had to give up cycling in his 80s and bought himself a tricycle instead. He had great fun letting his grandchildren have a go on it, because the steering was so counter-intuitive and they always went round in circles. Some of the complexities are explained here: eland.org.uk/steering.html $\endgroup$ – Michael Kay Jul 26 '18 at 8:23
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    $\begingroup$ Most of the answers below miss the critical element of angular momentum. A complete analysis would be challenging for most physicists and difficult to understand for pretty much everyone. $\endgroup$ – J... Jul 26 '18 at 14:38
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    $\begingroup$ I wonder if anyone else has noticed that the wheels of the bike in the picture are in a straight line? $\endgroup$ – jamesqf Jul 26 '18 at 16:09
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    $\begingroup$ @jamesqf the wheels aren't in a straight line. Just the angle is too small to be seen in the picture, because the curve radius is so much bigger than the wheelbase. $\endgroup$ – leftaroundabout Jul 27 '18 at 7:43
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It is actually not the tilting that causes more turning. You could in theory turn just as much while being straight up by just turning your steering wheel.

But if you do that straight up, you fall. The torque produced by the friction from the turned wheel is unbalanced and will flip you over. When you turn on a bike, your body automatically leans in and tilts slightly - not because that tilting does the turning, but because it keeps your balance.

By tilting your bike, you move your centre-of-mass sideways, so that gravity causes a larger and larger torque in you. When this gravity-torque counteracts the friction torque exactly, your turn is stable and you won't fall while turning.

So, the tilt doesn't cause sharper turning, but it allows for sharper turning without you falling over.

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  • $\begingroup$ A comment discussion has been moved to chat. Further comments which aren't specific suggestions for improving this answer will be removed without notice. $\endgroup$ – rob Jul 26 '18 at 22:25
  • $\begingroup$ The rider's body has only marginal effect on the lean angle (ignoring steering inputs from the rider's body). I know that from personal experience. You can lean in, you can stay straight up and down in the plane of the bike, or you can lean out. You will only shift the lean angle slightly. Check out MOTOBOT. His body is rigidly connected to the bike. He doesn't lean it at all, yet he is a very skillful rider. This is answer is misleading because it makes it sound like body weight is causing the leaning. $\endgroup$ – Dangph 2 days ago
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Notice that when the biker tilts more, the torque due to gravity increases(taking pivot to be any point other than point of contact or center of mass as then centrifugal forces would be needed to balance torques in an accelerated frame). Also notice that it is friction that is the counter/opposing torque to the torque by gravity. Hence as the biker tilts more, the frictional force must increase. And as frictional force is the centripetal force: $$F=\frac{mv^2}{r}$$ $$r=\frac{mv^2}{F}$$ So as $F$ increases, the radius of the turn must decrease. We can derive further with the reaction force having components $Rsin(\theta)$ equal to gravity so $$Rsin(\theta)=mg$$while the frictional component is $$Rcos(\theta)= \frac{mv^2}{r}$$. From these we can derive directly the relationship between the angle of tilt and the radius of the turn for a constant speed. It follows that $$\frac{Rsin(\theta)}{Rcos(\theta)}= tan(\theta)= \frac{gr}{v^2}$$ Thus it is again clear to see that the radius,$r$ is proportional to $\theta$ which is angle with the horizontal so the lower the angle, the more the tilt and hence lower the radius.

Note: Of-course the causality does not go on that bending more means shorter turns but rather, for stability as has been assumed in this derivation, bending more leads to shorter turns as quoted in previous answers or smaller $r$.

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    $\begingroup$ About what point do the torques in your diagram act? If it is the point of contact with the road, the centripetal force (friction) has no torque about this point. If it is the centre of mass (=centre of gravity) then gravity has no torque about this point. $\endgroup$ – sammy gerbil Jul 25 '18 at 12:30
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    $\begingroup$ Notice how when you take the center of mass then the normal reaction force will provide a torque just as weight would if you take the point of contact to be the pivot $\endgroup$ – Tausif Hossain Jul 25 '18 at 13:22
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    $\begingroup$ Editing answer in-order to make it more clear. $\endgroup$ – Tausif Hossain Jul 25 '18 at 13:24
  • $\begingroup$ The torque described as "torque due to centripetal force" is actually the torque due to the reaction to centripetal acceleration, which some call centrifugal reaction . $\endgroup$ – rcgldr Jul 27 '18 at 0:24
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This is called countersteering. see an in-depth physics explanation here.

As you turn, the top part of the bike wants to go keep going straight (because of inertia), whereas the wheels are bound to the ground by friction. This is experienced by the bike as a torque trying to flip the the bike in the direction opposite to the turn (i.e. if you turn left there's a counterclockwise torque).

It's basically something very similar to the imaginary "centrifugal" force one experiences while spinning a bucket of water.

An effective way to counteract this torque is to push down the bike on the opposite side. This will put it off-balance and, if the bike was not turning, it would fall down. If you think about it, gravity acts as a torque in the opposite direction as the centrifugal torque above. Gravity is a force, but since the wheels are bound by friction it acts as a torque with respect to the center of mass.

An experienced driver will twist the bike just enough so the two torques cancel out. In this manner, all the centripetal force is focused pushing the wheels down on the asphalt -- this is great for racing because

  1. gravity makes the bike turn without any other force, and
  2. there is increased weight on the wheels making the bike stick to the ground better in the turn

Note that unlike what the top answer says, it is indeed this tilting that makes the bike turn. In fact, due to the geometry of wheels, bikers have to unintuitively turn the wheels left to turn right (countersteer) while applying this technique -- otherwise the bike does not go down as it should.

See a biker demonstrating this quite clearly on this YouTube video

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  • $\begingroup$ Yes, in effect, if the rider steers to one side it will “trip” the bike to the opposite side. The greater the speed the harder one must push to create the lean. You steer left to turn right and right to turn left. $\endgroup$ – Lambda Jul 26 '18 at 15:45
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    $\begingroup$ "While this appears to be a complex sequence of motions, it is performed by every child who rides a bicycle." Wow. I had no idea. $\endgroup$ – Infiltrator Jul 26 '18 at 23:14
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    $\begingroup$ I'm not sure I agree here. OP is asking why leaning the bike increases the turn ratio. Whether or not leaning affects turn ratio, countersteering is a technique to make the bike turn faster, it's not a reason to why leaning the bike increase the turn ration. Note that unlike what the top answer says, it is indeed this tilting that makes the bike turn. It's actually the other way around, as they show in the video you link to. $\endgroup$ – Alex Jul 27 '18 at 13:04
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    $\begingroup$ By turning the wheel, you induce a tilt. This tilt can then be used as a proper lean for an opposite turn. So it's initiated with a counter-turn, not by a tilt. And if you don't follow up with steering at the "right" direction directly after this initial counter-turn, when you have achieved desired lean, you go down. $\endgroup$ – Alex Jul 27 '18 at 13:04
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    $\begingroup$ @Alex leaning makes you turn. The OP is wrong when they say it makes you turn sharper, so there's no "why". If you look at the picture posted by the OP, the biker... is not really steering at all. $\endgroup$ – Sklivvz Jul 27 '18 at 15:12
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There is an effect is due to geometry. Turn the front wheel of a bike to the left by thirty degrees, say. Now lean the bike into the turn. For simplicity, suppose that you could lean the bike completely on its side while the tires maintain contact with the ground. Having done so you will find that the turning radius is now roughly the one that would be achieved by turning the front wheel to ninety degrees while keeping the bike upright. (If the front wheel can't be turned to ninety degrees, this turning radius is not even achievable without tipping the bike.) This shows that tipping the bike affects its turning radius.

enter image description here

Figure 1. Here we assume a rake of 25 degrees. $\theta_x$ is the tilting angle and $\theta_z$ the turning angle with respect to the handlebars. $\phi = \theta'-\theta$, where $\theta$ and $\theta'$ are the angles the tire makes with respect to the forward direction before and after tilting the bike. (Note that due to the rake of the bike, $\theta_z\ne\theta$.)

enter image description here

Figure 2. One can see that for small turning angle that tilting the bike has little effect, agreeing with our intuition, but that for any turning angle the effect grows with tilting angle and can become a large effect.

enter image description here

Figure 3. For a turning angle of two degrees the effect of tilting 45 degrees is appreciable.

enter image description here

Figure 4. Here we assume a rake of 25 degrees and let the turning angle be ten degrees. The tilting angle varies from zero to 45 degrees.

enter image description here

Figure 5. A top view of the situation depicted in Figure 4.

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    $\begingroup$ This is technically speaking correct and might actually be what the OP had in mind. However, it's actually pretty theoretical because one never turns the handlebars by more than a few degrees whilst leaned into a curve. That would not actually result in a small turning radius, but instead in a crash. For a very narrow curve, you need to be slow and then there's not enough centrifugal force to allow leaning in a lot. $\endgroup$ – leftaroundabout Jul 26 '18 at 7:35
  • $\begingroup$ @leftaroundabout: See Figure 3 added above. Note also that it is not clear from the question statement that this is strictly about motorcycles. On a bicycle one can achieve large turning and tilting angles. $\endgroup$ – user26872 Jul 26 '18 at 13:54
  • $\begingroup$ This is a good answer. Intuitively, I think if you created the simulation using a torus, the effect from leaning would be increased. As you lean on a real tire, the point of contact does not remain at the center line of the tire. $\endgroup$ – JimmyJames Jul 26 '18 at 17:19
  • $\begingroup$ @JimmyJames: I agree that this effect would be increased with a tire a finite thickness. ;) $\endgroup$ – user26872 Jul 26 '18 at 20:05
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    $\begingroup$ This is an important effect that is not typically recognised as being part of achieving the turning effect. However it should be seen in the wider context of the centrifugal/petal lean and even the superbike side saddle seating posture (same is used on bicycles). The effect is a practical part of the fork angle effects. $\endgroup$ – Philip Oakley Jul 29 '18 at 9:19
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Another way of seeing this that you would fall faster when the bike is tilted more. So you are more in a hurry to move the contact point with the road below your center of gravity.

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I am sure the answer lies in the point of contact of the wheels with the ground.

I think you are right. You don't need torque or centripetal forces or gyroscopic forces to understand this. Two wheels also complicates this unnecessarily. The answer lies in geometry.

If you take a (e.g. lego) wheel and stick an axle through it and balance it, it will roll forward in a reasonably straight line on a flat, level surface. Now move the axle to one side so that it is leaning slightly. Push it forward and it will turn the in the direction of the lean.

Another way to think about it is like a cylinder and a cone. A cylinder rolls straight, a cone rolls in a circle. You might think that has something to do with the point of the cone but you can slice off the top of the cone and it still works. If you shortening the cone until it is just a sliver (balanced such that it's base is at the same angle,) it will continue to roll in the same manner. This situation is a reasonable approximation to the geometry of a tire that is leaning.

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    $\begingroup$ No, this effect is not important, at least not for a racing motorcycle where even when fully leaned into a curve the turning curvature is neglectable compared to the “cone bevel” (i.e., the turning radius is vastly more than the wheel radius). Tire geometry does play a role, the crucial quantity is called camber, but for different reasons (statics of the rubber itself). $\endgroup$ – leftaroundabout Jul 25 '18 at 21:33
  • $\begingroup$ @leftaroundabout Look at the pictures and see if the handlebars are turned enough to make that radius. Explain that. It's not enough to just say no. What's the alternative solution? $\endgroup$ – JimmyJames Jul 26 '18 at 13:58
  • $\begingroup$ @leftaroundabout I fail to see what "(i.e., the turning radius is vastly more than the wheel radius)" has to do with what I wrote or anything at all, really. $\endgroup$ – JimmyJames Jul 26 '18 at 16:43
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    $\begingroup$ By your “cone” analogy, leaning 45° should correspond to a corner radius similar to the wheel radius. That's what a 45° cone would do. Actually though, a racing driver might well lean in 45° whilst driving a curve of radius 100 m. Conversely a trial cyclist might drive corners with radius 2m but lean in only 5°. In summary, the actual cornering forces are almost completely unrelated to any camber-thrust effects. $\endgroup$ – leftaroundabout Jul 26 '18 at 16:59
  • $\begingroup$ @leftaroundabout You are reading into things. I never mentioned 'camber thrust'. Analogies fall apart at some level. That's what makes them analogies and not models. The argument I am making is along the lines of that of user26872 which you say is "technically speaking correct". $\endgroup$ – JimmyJames Jul 26 '18 at 17:06
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Leaning and turning are independent in the case of an uncoordinated turn, but this only occurs during transitions in lean angle or perhaps a pothole dodge maneuver (described below). Uncoordinated turns are only temporary, eventually a bike will fall inwards or outwards in an uncoordinated turn.

The key point I'm making in my answer is that the linear component of this is the same regardless if the bike is falling or not. Centripetal acceleration is equal to the centripetal force exerted by the pavement on to the contact patches of the tires divided by the mass of bike and rider (force = mass x acceleration, so acceleration = force / mass), regardless if the bike is leaned properly or not (falling inwards or outwards).

The purpose of leaning properly is to keep the bike from falling inwards or outwards during a turn.

In a coordinated turn, the bike needs to lean inwards enough so that the outwards torque related to inwards centripetal force at the tire contact patches and the outwards reaction force at the center of mass is exactly countered by the inwards torque related to gravity pulling down at the center of mass and the pavement pushing up at the contact patches.

In an uncoordinated turn, such as dodging around a pot hole, a bike can be quickly turned to get the tires out from under the bike to go around the pothole, but the bike ends up leaning the "wrong" way, and this lean has to be corrected for after passing by the pothole.

More common cases of uncoordinated turns occur during transitions in lean angle during corner entry or corner exit.


In case the OP is asking how lean angle effects steering inputs, I'm adding this geometrical explanation in this part of my answer.

In a steady turn, the path of a bike is a circle, and the radius is from the path of bike to the center of the circle. Ignoring factors such as tire deformation or slippage, the radius is a function of steering angle and lean angle.

In the imaginary case where a bike was vertical and not leaned, the center of the circle would be where the imaginary extension of the front and rear axles intercept. In the case where a bike is leaned over, the intercept point is below the pavement, and the center of the circle would be a point on the pavement directly above the intercept point below the pavement, and the circle and radius would be smaller. I'm thinking that the effect of lean angle probably multiplies what would be the "vertical" radius by the cosine of the lean angle. So if leaned at 45 degrees, the radius is cos(45) ~= 70.7% of the "vertical" lean angle radius.

Note that this doesn't have anything to do with balancing of the bike, it's just the geometrical combined effect of steer and lean angles.

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With the bike upright and either no turn or a very slight one the most outer circumference is in contact with the roadway.

When you lean in a sharp turn the inward torque will balance the outward torque created by the centripetal force (preventing tipping over) and the bike moves off the outer circumference and contacts higher up with a leading vector advancing towards the direction of the turn. The portion of the front wheel's rubber after the former lowest point is no longer in contact.

With the rear wheel, which is much fatter and doesn't turn, it's simply the side that pushes and it plays no role in the turning.

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protected by Qmechanic Jul 25 '18 at 11:12

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