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Old 20th November 2010, 22:18   #271
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What about aerodynamic downforce?

Wouldn't the maximum deceleration on an F1 car at 250kmph, be more than at 25kmph?

I've heard that an F1 driver needs to reduce braking force as the car slows down, since the lack of downforce reduces the braking force needed to lock up the tyres.

Shan2nu

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Old 21st November 2010, 00:13   #272
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Originally Posted by Shan2nu View Post
What about aerodynamic downforce ? Wouldn't the maximum deceleration on an F1 car at 250kmph, be more than at 25kmph ?
Yes, what you say is true. However, the discussion so far has not factored in aerodynamic drag & downforce since the issue being debated is whether the maximum braking force (or retarding force) a tire can generate is independent of velocity or not. And even in a car, the rate of deceleration, not velocity, influences the retarding force (of a tire) only because of weight transfer (an increase or decrease in the normal force).

If you draw a simple force diagram of a car that's braking, you'll find 4 moments that are essentially trying to "tip-up" the car. This is basically where the weight transfer under braking comes from. The greater the height of the CG, the greater the moments, & hence greater the weight transfer (for example SUVs). The axis along which the car pitches is called the kinematic pitch axis (I think). During aerobraking, this weight transfer does not take place as the center of pressure is usually at or below CG height. Downforce, however, would influence braking since it essentially influences the normal force - if downforce is involved, the braking force a tire can generate would be dependent on velocity.

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Originally Posted by Shan2nu View Post
I've heard that an F1 driver needs to reduce braking force as the car slows down, since the lack of downforce reduces the braking force needed to lock up the tyres.
This is because as a F1 car scrubs off speed, the downforce reduces, which in essence can be thought of as the car losing weight (normal force) - as the normal force reduces so does the retarding force a tire can generate.

Last edited by im_srini : 21st November 2010 at 00:22.
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Old 21st November 2010, 00:18   #273
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Originally Posted by Sutripta View Post
Used the term the way it is popularly used. ...
Can't fault that logic, sir - "earth is flat" was also once held true by the same logic!

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Originally Posted by DerAlte View Post
... Though this is tangential to the rotor rotation, it can be modeled as angular force ...
b. The momentum of the vehicle (speed dependent) is being transferred as angular force to the brake rotor, ...
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Originally Posted by Sutripta View Post
Linear Momentum is M.V. (=P). Angular Momentum is M.w. F= dP/dt. Different dimensions.
Definitely, since we are talking of 2 different things.
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For this particular case, I find it easier to think in terms of F=M.Acc, rather than F= dP/dt.
Unless the thinking models are the same, it is difficult to share / understand concepts! There are no ambiguities in engineering.

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... the maximum force (result of static friction) between tyre and ground, ...
LOL Static Friction plays a role when an object that needs to be accelerated is static. Which object are you referring to here which is static?

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Originally Posted by Sutripta View Post
... First order terms/ approximations. Not a full modelling of braking behaviour in which we start factoring in higher order terms. ...
Sir, first order terms / approximations cannot be in a different direction than higher order terms, right? Peace!
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Old 21st November 2010, 01:00   #274
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b. The momentum of the vehicle (speed dependent) is being transferred as angular force to the brake rotor, +ive since it is along the direction of rotation of the rotor
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- the brake force (a above) exceeding, especially when brake is pressed real hard - the force transferred from the vehicle momentum (b above; speed dependent)
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- b suddenly becomes 0 because tyre loses adhesion so momentum is no longer transferred (depends on coeffiecient of tyre friction with road surface; skidding if incidentally there was an sandy, icy or oily patch at the time of braking)
Hi, Der Alte, the transfer of momentum you've mentioned in all the cases above is itself happening only because of the frictional force available at the contact patch, isn't it ? And, frictional force is independent of velocity...

Whether a wheel locks or not is essentially decided by which frictional force is greater - A) frictional force generated by the brake pads against the rotors, & B) frictional force generated by the tire at the contact patch. Given a car of particular weight, a particular tire, & a particular surface, the maximum force you can generate at each contact patch is given by the equation for frictional force - the only thing that velocity (hence momentum or kinetic energy) decides is distance (braking distance).
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Old 21st November 2010, 01:01   #275
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Yes, what you say is true. However, the discussion so far has not factored in aerodynamic drag & downforce since the issue being debated is whether the maximum braking force (or retarding force) a tire can generate is independent of velocity or not.
If this is like one of those "Bugatti Veyron, in an ideal physics environment needs 800 odd bhp to push it's mass over a 1/4 mile in under 10 secs" discussions, then i guess velocity alone should not affect braking force.

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Old 21st November 2010, 03:12   #276
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Originally Posted by Shan2nu View Post
What about aerodynamic downforce?

Wouldn't the maximum deceleration on an F1 car at 250kmph, be more than at 25kmph?Shan2nu
The aerodynamic force in an F1 car is produced courtesy a spoiler or wings. The spoiler is always at an angle to the horizontal and hence the aerodynamic force also acts at an angle.
Split this force into two vector forces, perpendicular and parallel to the horizontal. The perpendicular force would be aerodynamic downforce also referred to as grip while the parallel force would be drag or decelaration force The angle at which the spoiler is set up determines the strength of the aerodynamic forces. If the spoiler is almost parallel, then hardly any grip or drag would be produced - best for tracks with more straights and less turns
Now the total decelaration force in a F1 car would consist mainly of three factors, all equally important
1. Aerodynamic drag - increases in proportion to square of velocity
2. Engine braking - I have heard commentary during an F1 race stating that the braking effect produced in an F1 car by just removing the foot from the accelerator is equal to the effect of applying full panic braking on a topline BMW/ Merc
3. Decelaration produced by the brake system.

Considering these three factors (and their different combinations due to varying spoiler setups, gear in which braking) could result in multiple scenarios, it would be difficult to answer whether decelaration is more at 250 or 25 kmph.

However, this topic is more suited for an F1 thread than ABS

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LOL Static Friction plays a role when an object that needs to be accelerated is static. Which object are you referring to here which is static?
I have come across quite few surprises on braking and ABS on this thread. But nothing comes close to finding out that that the friction between a rolling tyre of a car in motion and the ground is not kinetic friction but static friction
The contact patch of a rolling tyre is always stationary with respect to the ground. As the wheel rolls, a new contact patch comes in touch with a new piece of ground. At no point does a contact patch slide along the ground resulting in kinetic friction.
During wheel lock up, the static friction at the contact patch changes to kinetic friction as both surfaces are sliding with respect to each other.
Even though this sounds counterintuitive, it happens to be correct.


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the only thing that velocity (hence momentum or kinetic energy) decides is distance (braking distance).
Nicely summed up
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Old 21st November 2010, 21:33   #277
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Thanks to the inputs from all, I think we have reached consensus (at least majority) on the two points I had raised.

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Sir, first order terms / approximations cannot be in a different direction than higher order terms, right? Peace!
Something which cannot be answered with a simple yes or no. But if I'm forced to choose between the two, I'd go with 'No'. Should be part of a Modelling (and Simulation) thread.

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Originally Posted by Shan2nu View Post
If this is like one of those "Bugatti Veyron, in an ideal physics environment needs 800 odd bhp to push it's mass over a 1/4 mile in under 10 secs" discussions, then i guess velocity alone should not affect braking force.
Shan2nu
An F1 'car' at speed is more like an aircraft flying inverted. In normal cars speed normally brings about a problem of lift, not increasing downforce. In any case, for basic understanding, we should look at it as a mechanics rather than aerodynamics problem.

For understanding/ feel of a subject, it is always better to start from the simplest. = Strong foundation.

Peace
and
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Old 22nd November 2010, 00:09   #278
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Quote:
Originally Posted by im_srini View Post
... And, frictional force is independent of velocity...
Didn't see anyone mentioning that!

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Originally Posted by im_srini View Post
... Whether a wheel locks or not is essentially decided by which frictional force is greater - A) frictional force generated by the brake pads against the rotors, & B) frictional force generated by the tire at the contact patch.
Errr... that's what I had written, but didn't realize the focus has subtly been shifted here to discussing friction at the tyre patch area, as opposed to comparing everything at the brake rotor as one rightly should!

Well, for modeling wheel lock, one assumes that B is much greater than A to start with. The driver exerts more and more force (psychological - impending danger; the car doesn't brake itself) at the break pad. At one point (the knee point if one draws a curve of the wheel speed), A becomes greater than B - and the wheel locks as B is now close to 0 and is no longer able to rotate the wheel. Coefficient of friction is variable here, since beyond the knee point, either
- the vapors produced by the high heat at the road contact surface further reduce friction, or
- if ice, the top surface of the ice melts, or
- if sand / gravel, the rolling of the particles make it behave like a liquid for all practical purposes

Any (or a combination) of the above suddenly reduce the co-efficient of friction of the tyre-road contact area, and the vehicle can no longer be slowed down by braking action. Steering becomes ineffective, since the steering force also relies on friction with the road surface.

The whole purpose of ABS (of any kind) is to monitor the wheel speed for the knee condition (which indicates skid onset; one monitors the rate of deceleration), at which point the brakes are modulated so that B always remains > A, i.e. since A is the only thing that one can actively control, A is not allowed to exceed B.

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Originally Posted by im_srini View Post
... the only thing that velocity (hence momentum or kinetic energy) decides is distance (braking distance).
Very correct sir. That is an inherent property of a vehicle measured and published since after Ralph Nader's time (mostly by reviewers) as "Braking distance at X speed" (assuming dry tarmac)! ABS serves to reduce this distance on any road surface.

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Originally Posted by arun4242 View Post
... The contact patch of a rolling tyre is always stationary with respect to the ground. As the wheel rolls, a new contact patch comes in touch with a new piece of ground. ...
Perfect F.E.M., sir. However, it has to be read a bit differently:
* The momentum of the vehicle at a significant speed has enough PE to exceed the static friction should the patch remain in one place for a short time (the C.o.Static Friction is significant here to prevent the 'overcoming')
* However, since the wheel is rolling, the event of 'overcoming' is obviated despite the possibility
* When the wheel just stops moving on braking (the whole objective of ABS is to prevent this), and the momentum PE is still significant, it takes only a very short time (proportional to C.o.Static Friction) for the Static Friction to be overcome. If the momentum PE is not significant, normal braking takes place
* Shortly after wheel lock (when the patch is moving linearly w.r.t. road surface and static friction has been overcome) C.o.Dynamic Friction now prevails since the patch is no longer stationary

Last edited by DerAlte : 22nd November 2010 at 00:11.
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Old 22nd November 2010, 00:13   #279
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An F1 'car' at speed is more like an aircraft flying inverted. In normal cars speed normally brings about a problem of lift, not increasing downforce. In any case, for basic understanding, we should look at it as a mechanics rather than aerodynamics problem.
I think what you're trying to say is that if you apply the same braking force on a car, at 2 different speeds (without considering any other factor), then the rate of deceleration remains the same, irrespective of velocity

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Old 22nd November 2010, 20:35   #280
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^^^^
I have a feeling that there is a 'gotcha' involved somewhere, so I think I better cover my backside.

As a first approximation I'll say yes. And the maximum deceleration possible (threshold braking), which is limited by the tyres, not brakes, will remain the same irrespective of velocity. This is based on what we know of friction. If we actually measure the performance of tyres (which are quite complex), we find that its grip in the direction of its travel, metriced by mu, actually decreases with speed. This is, explained very unscientifically, because the tyres have less time to 'grip' the road. Which means that maximum deceleration possible at a higher speed is less than at a lower speed. I don't recollect the actual figures. Maybe tyre design has changed so drastically in the last few years that it is no longer the case, but I very much doubt it.

This question of modelling. At what level of refinement should we factor in the suspension. As anyone knows, it has quite a influence on braking performance.

Or to use another example which has been brought up in this thread: static vs kinetic (using the term in its popular sense) friction. When do we get the most grip from the tyres. Once again, going by what we know of friction, choosing between locked and not locked, we go by not locked. Our first approximation. In actual fact, maximum is when the tyre has 15 - 20% slippage (IIRC). Do we straight away start our modelling based on this.

My workflow would be Simplify, Understand, then Refine.

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Old 30th November 2010, 21:13   #281
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Pray, what is slippage? I mean does 15-20% slippage mean that the tyre is rotating 80-85% of the time and locked 15-20% of the time? Or does it mean something else? I would also love to know the experimental setup and data gathering method used to uncover this "fact".

Amazingly, this thread has gone from ABS discussion to concentrate on friction. Somebody here is bound to get a PhD in physics with specialization in friction
I wouldn't be surprised if somebody came along and invoked string theory to explain tyre dynamics. This thread is rocking!
Cheers.
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Old 1st December 2010, 12:26   #282
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Pray, what is slippage? I mean does 15-20% slippage mean that the tyre is rotating 80-85% of the time and locked 15-20% of the time? Or does it mean something else? I would also love to know the experimental setup and data gathering method used to uncover this "fact".
seems you have got the concept of wheel slip a bit wrong.
Wheel Slipage in simple terms refers to the difference in speed of a braking wheel with repect to that of free rolling wheel or in other words it the speed at the point of contact with respect to the speed at the wheel centre.
Quite simply, wheel slip occurs when the force applied to a tire exceeds the traction available to that tire. i.e. situation where the traction is too low eg. wet road, ice etc or situation when the torque is too high eg. drag racing

For this thread we need to focus mainly on longitudinal wheel slip and thus am not bringing lateral slip in the picture.

The SAE formula for wheel slip is
S= - (V - Re*w)/V where V=longitudinal speed of the wheel centre, Re=effective tire radius and w=angular speed of the tire
Slip is expressed as a % and a -ve component meaning 10% slip refers to a situation where the wheel speed is 10% less than that of wheel centre/vehicle

as for an experimental setup, I can tell you about the setup that we used for ABS testing.
  1. 4 wheel speed sensors (1 for each wheel)
  2. 1 vehicle speed sensor (can be GPS based or 5th wheel or optical)
  3. Compatible data logger
the above list is the minimum that one would needed for ABS testing though other sensors are usually added for measuring other vehicle parameters during the test.
The wheel speed sensors are fitted on each wheel and provide each wheel(s) speed independently. These measure the wheel(s) angular velocity and calculate wheel speed based on the wheel dia and its angular veolcity.
The vehicle speed sensor gives us the vehicle speed and the data logger is used to record/analyse all this data.
With the data logger getting inputs of each wheel speed and vehicle speed we can easily monitor slip on each wheel in real time.

Hope this clarifies your confusion. Do let me know if you need more info on this.

peace,
~kg
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Old 1st December 2010, 23:12   #283
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Thanks kapil, for the clarification on slippage. What I understood in simple terms is that this slippage is opposite to the 'skid' that happens when wheel is locked. Am I following it right? We would often use this opposite sense of rowing a boat to get the boat stop quicker. Is this slippage the same concept?
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Old 2nd December 2010, 00:27   #284
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Originally Posted by gostel View Post
Thanks kapil, for the clarification on slippage. What I understood in simple terms is that this slippage is opposite to the 'skid' that happens when wheel is locked. Am I following it right? We would often use this opposite sense of rowing a boat to get the boat stop quicker. Is this slippage the same concept?
Not sure if 'skid' is a technical term but I understand 'skidding' to be the phenomenon where the wheel is not rolling but simply moving in the direction of the vehicle. If I am getting this right then skid and slippage are not opposite but more like cause and effect. High slippage while braking will cause the wheel to lock and thus skidding will occur.

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Old 2nd December 2010, 00:41   #285
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Apart from the valuable discussing happening is there anyone who has experienced ABS work ?
I did. You see couple of months back while driving I looked behind before changing lanes and there was a screech in front of me. I too braked hard and I could feel the tyres rollling instead of dragging with a distinct time stamp on rotating tyres. Almost like the tyres were made to rotate after a certain interval.It was the first time I expereinced ABS at work.
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