1
$\begingroup$

I know the theory behind bending inward while taking a turn. It is done to make a component of friction act as centripetal force and prevent from us from getting thrown out due to centrifugal force.

The main question begins here: When we have entered the turn and also tilted the vehicle,if we want to change our direction a little bit,we need to apply pressure on the opposite side on the handle i.e turn handle left for going right and vice versa. And same applies to steering of a car. What is the concept behind this? Please, if possible, include the laws of physics acting here in the answer. This question is not a duplicate of this post because there everyone has discussed about centripetal and centrifugal forces and not about the position of handle and steering during this process. Whereas I have clearly mentioned that know the concept of tilting and the forces acting the vehicle.

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/20730/2451 and links therein. $\endgroup$ – Qmechanic Oct 29 '18 at 10:43
  • $\begingroup$ If you believe the question is not a duplicate, please describe your reasoning. It's not effective to simply claim that they aren't the same. $\endgroup$ – user191954 Oct 29 '18 at 12:15
  • $\begingroup$ Thanks for editing the post to reflect why it isn't a duplicate! $\endgroup$ – user191954 Oct 29 '18 at 12:34
  • $\begingroup$ Hint: It is about controlling the yaw rate of turn. $\endgroup$ – ja72 Oct 29 '18 at 13:24
  • $\begingroup$ Are you asking about motorcycles or cars? $\endgroup$ – EL_DON Oct 29 '18 at 13:58
1
$\begingroup$

bicycle and motorcycle steering behavior has been extensively studied and modeled in a subfield of mechanical engineering called tracking vehicle dynamics.

The short answer to your question is that the rider of a bicycle controls its direction and lean angle by manipulating the position of the bike's center of mass relative to the position of its tire contact points (the "center of support") on the road. For example, to enter a right turn or to sharpen it, the rider turns the handlebars to steer the front wheel to the left. this causes the bike to start tracking off to the left and places the center of gravity to the right of the tire contact points.

in this configuration, gravity then produces a rolling moment towards the right, causing the bike to begin falling to the right. to prevent the bike from going down, the rider then follows the fall and stops it by turning the bar to the right, thereby establishing a bank angle that geometrically cancels the rollover moment.

to stop turning to the right, the rider pushes the bar to rotate the front wheel towards the right, which has the effect of steering the center of support back underneath the center of mass, and the bike comes upright and tracks in a straight line again.

this "backwards" behavior is why people say that to turn a bike to the right, you first have to steer it to the left, and vice versa.

Note that the explanation given above is modified by whether you are traveling slow or fast and whether you are steering into a gradual or a sudden turn.

$\endgroup$
  • 1
    $\begingroup$ I like your answer. But I don't think the explanation is modified at slow or fast speed. It is applicable at all speeds. At low speeds it can be observed easily and at high speed its not easily observable because everything happens in very short interval of time. $\endgroup$ – baba Oct 29 '18 at 18:37
  • $\begingroup$ you will find in the tracking vehicle literature three distinctly different speed regimes in which the steering dynamics shift in well-defined ways that also depend on how suddenly the rider initiates the turn. It is complicated, but it is a solved problem inasmuch as computer programs have been written years ago that simulate it accurately. $\endgroup$ – niels nielsen Oct 29 '18 at 23:50
0
$\begingroup$

Most everyone has played this game. Sit on a motorcycle or bike at zero speed and see how long you can balance in the upright position. You and the bike will fall over unless the rider is really skilled at shifting his or her weight quickly. Add speed and the task of staying upright becomes easier. Add a lot more speed and the opposite becomes true. Now you cannot even lean the bike even if you hang off the side of the bike. The heavier the bike the more noticeable this is. An 800 pound Harley will not lean at highway speeds due to a rider leaning. To lean the bike the rider “trips” the bike by counter steering. When you’re in a turn the bike wants to stand up. Counter steering “tripping the bike” is necessary to maintain the lean. As far as the car goes the counter steering is only needed when the traction at the rear of the car is lost and it starts slide out of the turn. Turning into the slide can stop this. I don’t see the how this is related to the bike situation.

$\endgroup$
  • $\begingroup$ With your help I am very near to my answer. Can you just also name the law of physics applied in this. ( Is it angular momentum, directional intertia or some kind of right/left hand rule) $\endgroup$ – baba Oct 30 '18 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.