[Rollin' Thunda]

Originally Posted by Sutripta

^^^
Hi,
In a bike, would the fact that one can compare one wheel only against one other, that too with its independent braking circuit, cause any problems?

No, for when the wheels are rolling normally there will be no difference in the their rotational velocity, even if the brakes are being applied on one wheel and not on the other.

In a car, if on a turn one applies the handbrake (operating on the rear wheels), one has, well, the handbrake turn. Certainly doesn't straighten out the car. Why the expected differences between car and bike?

In a car, the 'hand-brake turn' occurs AFAIK because of differences in the braking force being applied to the two rear wheels, which causes a torque around the vertical axis. (This could be because of skidding of one or both wheels).

In the case of a bike, which has only one rear wheel, directly behind the center-of-mass, this does not happen to any significant extent.

I have given the answers against the questions you asked.

[Sutripta]

Originally Posted by Rollin' Thunda

I have given the answers against the questions you asked.

Hi,
In a bike, would the fact that one can compare one wheel only against one other, that too with its independent braking circuit, cause any problems?

No, for when the wheels are rolling normally there will be no difference in the their rotational velocity, even if the brakes are being applied on one wheel and not on the other.

Hi, not what I was asking. Scenario: - Rider jams on both brakes.

In a car, if on a turn one applies the handbrake (operating on the rear wheels), one has, well, the handbrake turn. Certainly doesn't straighten out the car. Why the expected differences between car and bike?

In a car, the 'hand-brake turn' occurs AFAIK because of differences in the braking force being applied to the two rear wheels, which causes a torque around the vertical axis. (This could be because of skidding of one or both wheels).

In the case of a bike, which has only one rear wheel, directly behind the center-of-mass, this does not happen to any significant extent.

Don't get it. . Could you explain it in a bit more detail please.

Regards
Sutripta

[Rollin' Thunda]

Originally Posted by Sutripta

Hi, not what I was asking. Scenario: - Rider jams on both brakes.

What I meant is that when both wheels are rolling (as normal) their angular velocity will be w=V/R, where V is the velocity of the bike (and both wheels) and R is the radius of the wheels. So in normal rolling motion both wheels will have the same angular velocity. However, when the wheel starts skidding (not rolling), then the angular velocity will be lower than what is give by that formula (and could even become zero even while the bike is in motion). So there will be a difference in the the angular velocities of the skidding wheel and the rolling wheel. ABS senses this difference and releases the brake of the slower wheel.

Originally Posted by Sutripta

Don't get it. . Could you explain it in a bit more detail please.

Regards
Sutripta

When you apply the hand-brake on a moving car, it is likely that at least one of the rear wheels will lock. The moment the wheel locks, the friction forces (exerted by the ground on the wheel) on it changes abruptly (sliding friction versus static friction). Thus the forces on the two wheels become different, which exerts a torque, which caused the car to turn.

[dhanushs]

Originally Posted by Rollin' Thunda

The reason hard braking with the front brake is not recommended during turning is because it tends to de-stabilize the bike. The braking force from the front brake on a turn does not pass through the (horizontal plane projection of the) center-of-mass of the bike and rider, and thus tends to spin the bike around its vertical axis and de-stabilizing it, as Mpower learnt on the track. Braking with the rear does not have this problem, and just starightens the bike. But the rear-brake is not so powerful, so the best thing to do on a turn is to first straighten the bike (by using the rear brake and the steering) and then applying both brakes as normal, as Mpower said.

Yes, actually, whist on a turn, the application of front brake will tend to form a couple with the center-of-mass of the bike and rider. The point of action is the contact patch of the front tyre and road and the effect is to move the front tyre away from the turn, which in effect is 'understeer'

ie, application of front brakes makes the bike to move away from the turn.

Where as, application of rear brake is slightly more preferred because :

as the point of action of the couple formed is at the contact patch formed between REAR tyre and road, the effect of this couple is to move the rear tyre away from the turn, which inturn behaves as 'oversteer'.

ie, application of rear brakes makes the bike turn more into the turn.

This is not as unfavorable as 'understeer' in a bike whist a turn. Quite manageable as the bike remains inside the turning circle.

Well, both couples also tend to overturn the bike. Hence braking while turning is not advisable. However, the frictional forces between the tyre and road, act in the opposite direction of this couple, and thereby canceling it. So as long as you keep the static friction(traction) you can achieve deceleration whist braking in a curve. but as someone mentioned, the 'bit' of brake input is very important and comes only from experience.

Originally Posted by Rollin' Thunda

When you apply the hand-brake on a moving car, it is likely that at least one of the rear wheels will lock. The moment the wheel locks, the friction forces (exerted by the ground on the wheel) on it changes abruptly (sliding friction versus static friction). Thus the forces on the two wheels become different, which exerts a torque, which caused the car to turn.

Originally Posted by Sutripta

Don't get it. Could you explain it in a bit more detail please.

When you apply the hand brakes in a car, the effect is almost similar to a bike.

A couple is formed, between the center of mass and the resultant braking force + resultant centrifugal force, which tends to move the rear wheel's away from the turn. but as long as traction is maintained its canceled out, but the moment traction brakes, the couple comes into play and moves the rear wheels out which 'steers' the car.

Originally Posted by Rollin' Thunda

What I meant is that when both wheels are rolling (as normal) their angular velocity will be w=V/R, where V is the velocity of the bike (and both wheels) and R is the radius of the wheels. So in normal rolling motion both wheels will have the same angular velocity. However, when the wheel starts skidding (not rolling), then the angular velocity will be lower than what is give by that formula (and could even become zero even while the bike is in motion). So there will be a difference in the the angular velocities of the skidding wheel and the rolling wheel. ABS senses this difference and releases the brake of the slower wheel.

Rollin' Thunda: Imagine a situation where you are applying only the rear brakes.

From the moment the brakes are applied, there will definitely be a difference in angular velocities of both wheels(Reason: On hard braking even if the wheels are rotating, they are also skidding). So if ABS reduces the brake pressure on the wheel wont it increase the stopping distance?.

I think, unless the ABS kicks in when the wheels are 'locked' dθ/dt=0 (no steerabiltiy) stopping distance is reduced (?)

[SPIKE ARRESTOR]

Originally Posted by dhanushs

I think, unless the ABS kicks in when the wheels are 'locked' dθ/dt=0 (no steerabiltiy) stopping distance is reduced (?)

What is this dθ/dt=0 ? How is the behavior of ABS outside / inside the friction circle limits (wheels are locked and there is no steering) ?

Spike

[Rollin' Thunda]

Originally Posted by dhanushs

Rollin' Thunda: Imagine a situation where you are applying only the rear brakes.

From the moment the brakes are applied, there will definitely be a difference in angular velocities of both wheels(Reason: On hard braking even if the wheels are rotating, they are also skidding). So if ABS reduces the brake pressure on the wheel wont it increase the stopping distance?.

I think, unless the ABS kicks in when the wheels are 'locked' dθ/dt=0 (no steerabiltiy) stopping distance is reduced (?)

ABS does not just release the brake when a skid occurs, it repeatedly releases and reapplies the brake at a rate of 30-40 times a second, so even if there is skidding, it is not allowed to go out of control, as traction is regained even before the skid travels a fraction of a meter. Yes, theoritically ABS can increase the braking distance, but practically, except for international level racing champions, few riders actually do beat it on a proper road. [ABS performs poorly off-road on dirt tracks, so most off-roading motorcycles don't have ABS AFAIK].

BTW, skidding does not occur from the moment brakes are applied, it occurs only if the braking force exceeds the maximum friction force that can be developed between the road and the tire.

[vivekiny2k]

Let me explain the physics behind the "hand brake turn" or the oversteer turn. I analyzed it in high school

At any turn, the vehicle experiences strong centrifugal force (outward). Counteracting is the centripetal force provided by "sideways" friction between the ground and the wheel (inward).

when the rear wheel loses static friction with the ground (either by locked brake or by application of too much power causing "powerslide"), the centripetal force is lost, hence the centrifugal force sends it outwards. The front wheel is still stuck to the ground, hence the vehicle experiences an over-steer. The opposite happens when the front wheels lose traction (under-steer).

If the wheels where designed with lateral grooves providing minimal sideways traction, you will see this happening all the time.

To make things complicated, the front wheel of a bike is also responsible for balancing it, and so the front wheel losing traction almost always results ina crash, before one can even visualize under-steer.

Anybody hitting a wet/slippery patch while skating can experience it very well (including your truly), since the skaters always gain their force from sideways traction from rear feet.

[Sutripta]

Hi guys,
Been away for a week, and see lots of interesting discussions in the meantime.

I think we have consensus that in a car negotiating a turn, application of extra force, either braking or power, to the rear wheels will increase oversteer. Do we have any such consensus regarding bikes?

Also, in a car, abs is about maintaining control, steerability. How similar, or different, is steering a bike compared to a car. I mean an analysis of the forces involved, and their origin.

Regards
Sutripta

[dhanushs]

Originally Posted by SPIKE ARRESTOR

What is this dθ/dt=0 ? How is the behavior of ABS outside / inside the friction circle limits (wheels are locked and there is no steering) ?

Originally Posted by Rollin' Thunda

ABS does not just release the brake when a skid occurs, it repeatedly releases and reapplies the brake at a rate of 30-40 times a second, so even if there is skidding, it is not allowed to go out of control, as traction is regained even before the skid travels a fraction of a meter. Yes, theoritically ABS can increase the braking distance, but practically, except for international level racing champions, few riders actually do beat it on a proper road. [ABS performs poorly off-road on dirt tracks, so most off-roading motorcycles don't have ABS AFAIK].

Spike, by dθ/dt=0, I meant angular velocity = 0.

IMO, ABS has no role inside friction circle limits. As long as traction is there, ABS has no role. ABS comes to play the moment the contact patch looses traction, by reliving brake fluid pressure to an extent where dθ/dt is not 0.

I think, an effective ABS doesn't 'just' pulse (repeatedly releasing and reappling the brake at a rate of 30-40 times a second) the brakes. It releases brake fluid pressure just to an extent where dθ/dt is greater than 0, where normal brake pressure again tends to make dθ/dt=0. If this occurs, ABS senses and releases pressure.

Well, 'this' procedure takes place 30-40 times a second, and hence we 'feel' it as a pulse.

Originally Posted by Rollin' Thunda

BTW, skidding does not occur from the moment brakes are applied, it occurs only if the braking force exceeds the maximum friction force that can be developed between the road and the tire.

Rollin' Thunda, what I meant is the moment brakes are applied, there may be a difference in angular velocities of both tires, even if they are not skidding. Depending on road/tire conditions.

like in a car, while taking a turn, the outer/inner wheels have different angular velocities. Also, another case is, while braking, one of the wheels may 'roll' at a slower speed than the others. So does ABS tend to cancel out the difference in Angular velocities?.



P.S - I have very limited technical knowledge. Please feel free to correct me, as and when applicable. I would be more than happy to learn .

[SPIKE ARRESTOR]

Originally Posted by dhanushs

Y
I think, unless the ABS kicks in when the wheels are 'locked' dθ/dt=0 (no steerabiltiy) stopping distance is reduced (?)

Originally Posted by dhanushs

Spike, by dθ/dt=0, I meant angular velocity = 0.

^^ Angular velocity / steerability, what exactly??

IMO, ABS has no role inside friction circle limits. As long as traction is there, ABS has no role. ABS comes to play the moment the contact patch looses traction, by reliving brake fluid pressure to an extent where dθ/dt is not 0.

Dhanush, are you sure on this? Does that mean ABS kicks in only when it realizes loss of traction? How is this sensed?

Spike

[Rollin' Thunda]

Originally Posted by dhanushs

Rollin' Thunda, what I meant is the moment brakes are applied, there may be a difference in angular velocities of both tires, even if they are not skidding. Depending on road/tire conditions.

No, if they are not skidding, then they will have the same angular velocity, even if the brake is applied only to one wheel. It seems counter-intiutive, but it's true. [In a car it is different, when the inner wheels in a turn will rotate slower.]

[quote=SPIKE ARRESTOR;2294958 Does that mean ABS kicks in only when it realizes loss of traction? How is this sensed?

Spike[/quote]

It is sensed by checking if the angular velocity of the two wheels are different. In which case, ABS kicks in and starts release-and-reapply pulses for the slower-rotating wheel, till the wheels are both rotating at the same angular velocity.

[vivekiny2k]

Originally Posted by Rollin' Thunda

No, if they are not skidding, then they will have the same angular velocity, even if the brake is applied only to one wheel. It seems counter-intiutive, but it's true. [In a car it is different, when the inner wheels in a turn will rotate slower.]



It is sensed by checking if the angular velocity of the two wheels are different. In which case, ABS kicks in and starts release-and-reapply pulses for the slower-rotating wheel, till the wheels are both rotating at the same angular velocity.

this is not true, turn the handle of a bike to almost 90 degrees and you will have the front wheel moving while rear wheel almost stationary. This is because front wheel steers and moves along the new direction, while rear wheel just trails along straight line.

[Rollin' Thunda]

Originally Posted by vivekiny2k

this is not true, turn the handle of a bike to almost 90 degrees and you will have the front wheel moving while rear wheel almost stationary. This is because front wheel steers and moves along the new direction, while rear wheel just trails along straight line.

Does that seem like a normal riding situation to you?

[vivekiny2k]

Originally Posted by Rollin' Thunda

Does that seem like a normal riding situation to you?

Does that sound like the only situation where the speeds will be different to you?

That was an example to help you visualize. speeds will always be different when the steering is not straight.

[Rollin' Thunda]

Originally Posted by vivekiny2k

Does that sound like the only situation where the speeds will be different to you?

That was an example to help you visualize. speeds will always be different when the steering is not straight.

In any situation when the bike is moving with speed, it will either (a) be going in a straight line or (b) going along an arc of a circle (instantaneously, when it weaves, the circle keeps changing radius and position). In either case, the wheels will be moving at the same angular velocity if they are not skidding. Even in the case you mentioned, if you think about it, the bike will be moving along the arc of a circle.