What if i told ya that the stopping ability of your car never increases with wider tyres, nor with extra weight above?
IE: take this for example (If everything was all under control, the tyres don't disintegrate from heat or pressure).
There is a mini minor with bicycle wheels weighing far under a tonne (ie: < 1000kgs for sure). And we have a fully loaded semi trailer weighing 20
tonne, with about a dozen wide tyres. Both tyres are made of the same compound. If both were travelling at the same speed in the same direction, and
at a particular point, at the same time they both locked up all their wheels, the stopping distance will be EXACTLY the same (in a perfect world not
talking into account drag forces etc).
They will have the same stopping distance, despite the difference in tyre width, or pressure/weight those tyres are taking. The only similarity that
gives each vehicle the same stopping distance is the same road surface, and the same tyre compound.
This is what we learnt in first yr physics today :P all repeat of high school stuff, and heres the formulae to prove it:
Function 1: F = m x a (Force = mass x acceleration)
Function 2: f = u x n (f, being the friction force. I couldn't find the funky u symbol for the coefficient of friction, however thats what the u
stands for. & n being the net force downwards due to gravity. IE: weight. Hey!!! i mentioned weight, but u watch, the mass, which is required to
have 'weight' in a sense, will be eliminated in the functions via substitution)
Because friction = u x n, and Force = mass x a,
The value of n, (the weight force), cause its a force, it to can equal to m x a. Because the acceleration downwards of an object is only that of
gravity on a flat road, n therefore equals m x 9.8. (gravity being a value of 9.8m/s/s)
Therefore, f (friction) = u x m x 9.8. This coefficient of friction differs depending on the material, it is a constant however for a particular
material. pretty much a number that means how rough a material is. (IE: a sticky dry road could be say 1.2, a wet road could have a constant of
0.7)
Now, for the finale,
Function 1: F = m x a
Function 2: f = u x m x 9.8
Because F is force, and f is friction force, both are forces, are are therefore the same thing in concept, F = f
SO: F = f
m x a = u x m x 9.8
(Mass on both sides are able to be eliminated)
therefore a = u x 9.8
IE: Acceleration = Coefficent of friction x gravity. Acceleration in this case, being the deceleration of a vehicle, that is the stopping force. 2
objects of same deceleration with the same initial velocity are going to slow down at the same rate, and therefore stop with the same stopping
distance.
Because a = u x 9.8, this deceleration therefore does not depend on the surface area of a tyre, nor the weight above it, nor even how many tyres you
have, but simply the gravitational force and the particular material of the tyre used.
Tripped the hell outta me,
And dad calls me a liar. :P
But try telling that to the phD lectures
REMEMBER: this is tru if every factor were controlled, and there were no drag forces, and other interrupting forces didnt take place. Its just a
simplified version of a real world system, but this concept is honest and tru. IE: VERY REAL.
So for all u nuts wanting more traction, get urselves a good sticky soft tyre
So all of that physics 'Jargon' just to tell us that we need a softer compound tyre to increase traction?
Did you not play Gran Turismo 1 on drift mode?
-Staggers.
lmao :P
nah, i played GT 3 :P gotta luv the sound track too...
Just wanted to save money for all those peeps who should know that vw skinnies can do a great job rather than some fat oldskool cookie cutters or
sumfin... Aesthetics is a whole different story though, but stopping ability is no different with same tyre compound.
OK Professor Chris Sumner Miller was that to baffle us with
brilliance or bull
Its all truth!
I learnt it today, and am still trying to come to terms with it.
The equations prove it, the doctors teach it,
but in real world concept, its hard to bite.
My 4th yr uni brother had to think twice but found it refered to in his text books, and we tried explainin to dad, but he didn't believe it.
Brenton and i are having a great discussion about it. Like what he has been saying, there are so many other factors that play in the game though.
Ha, nice mention of Professor Julius Sumner Miller, Pod! 'Now why is this so??'
Yeah, I do love these discussions, Chris. I do believe you and I know what you're trying to say. I learnt about the co-efficent of friction and what
not in my 1st year of my apprenticeship.
It is hard to grasp the concept of it all to start off with, but you and I both know that this only applies to the mechanical link between the road
and the tire.
As for...
'Just wanted to save money for all those peeps who should know that vw skinnies can do a great job rather than some fat oldskool cookie cutters or
sumfin...'
If all of your calculations are correct, then by rights if you had 4.5 inch tires you would then be able to steer as effeciently as a 6 inch tire?
Now I'd much rather fat old school cookie cutters as opposed to skinnies when is comes to braking and steering...
Why?
Because your formulae only take into account straight line stopping, but as we all know, when stopping fast you're not always travelling in a
straight line.
-Staggers.
Ok, so as I sit here I think about when I was 4WDing a while back and had to steer and brake at the same time. I was in mud, with 295 70 R16 equiv.
A/T tires... when the wheels were locked up in the mud steering was impossible. So in this instance 295 tires did just as much as a 185 would.
My question is, would a wider tire lock up or skid sooner or later than a narrow tire under hard braking?
-Staggers.
keep an anchor & chain in the passenger seat, it'll stop the car for ya when need be
but ur right dude.
Its a good topic, we need baja wes in here as well.
Quote: |
ie: they would lock up at the same time under hard braking cause they involve the same friction.
Yes, but they do not have the same mass.
-Staggers.
I don't believe that is correct in Our world..
maybe OK in a book and perfect world??
Talk to someone running 135s or 145 front tyres on their beetle and see if they can pull up as good as with 165 tyres on the front...
especially if the road is wet...
I owned back in the 70s a Renault R10 which had 4 wheel disc brakes and 135 x `15 Michelin tyres standard.
I had the rims widened and fitted 165 x 15 tyres as the narrow ones used to scare the hell out of me...
so, I still don't believe it...
narrow tyres will lock up much easier than wide ones...
the road surface also comes into it...
smooth or rough - wet or dry??
its NOT a perfect world out there...
Lee
what day is it? A=april F = fool
luv causing a stir
WTB (Vic) Bicycle wheel rims in 5 stud pattern with 15" diamater to suit type 3.
I would also consider some 22" rims if they came with white wall tyres. Send U2U
just make sure u get good a good tread compound
I have a doctors certificate for temporary insanity. Will this suffice??
thing that is being missed is that it is generally better toi stop without skidding - much quicker then with skidding! the math also estimates the friction as a 'point load' rather then taking into account the contact area of teh tyres, which again would vary depending on tyre pressures
Nice work Chris, had a big think and came up with some questions:
Wouldn't that solution actually only be expressed with regards to the kinetic energy potential of the differently massed vehicles, and therefore only
apply if no external force other than gravity was applied: i.e if the vehicles rolled to a stop without external braking force, wind resistance
etc.?
Tyre to ground friction is a tiny component of the friction force in this 'lock up' context, most of the stopping force is applied by friction in
the highly different mechanical systems of the vehicles brakes themselves, so wouldn't that friction force be far more accurate as an equational
variable? If a braking force (delta v) is applied, then wouldn't this equation be more applicable:
If an object with mass, m, traveling at speed v, is opposed by a constant force F (delta v again so a negative number), then F = ma (Newton's law).
So the acceleration is F/m and the mass will stop when v- (F/m)t = 0 or when t= mv/F. Therefore, surely the objects' mass is directly proportional
to the time it takes to stop, even in a 'perfect system'?
Surely mass can be cancelled out in a 'perfect system' with regards only to kinetic potential and not inertial mass?
Confusing, good fun, but!
Skidding is nothing more than drifting for cars lacking in horsepower but making up for it with their drum brakes.
The mathematics is well beyond my level of comprehension. Logic is more preferable for laymen and plebs that can't do the math...
That is nearly as good as the Michael Schumacher story of him returning to Ferrari for 2 seasons from next race for $100,000,000 per year
Just one thing, if I had two sets of tyres... one had compound of 0.5 and was 10 inches wide, and another set of tyres which had a compound of 1.0 but
were 5 inches wide... Wouldn't that be an equal rating?
Yes? No?
So if someone had a 4.5 inch wide tyre with a compound rating of 0.7 and they wanted a 6 inch wide tyre with the same compound rating... wouldn't
increase the friction caused by the skidding tyre?
The formulae you show does not take into account surface area of the tyre so how can you prove anything??
Another thing... if the surface area didn't matter, only the friction compound... then why are brake pads and rotors becoming larger and larger?
There's no difference between the tyre and brake formulae.
-Staggers.
Although a larger area of contact between two surfaces would create a larger source of frictional forces, it also reduces the pressure between the two
surfaces for a given force holding them together. Since pressure equals force divided by the area of contact, it works out that the increase in
friction generating area is exactly offset by the reduction in pressure; the resulting frictional forces, then, are dependent only on the frictional
coefficient of the materials and the FORCE holding them together.
If you were to increase the force as you increased the area to keep PRESSURE the same, then increasing the area WOULD increase the frictional force
between the two surfaces.
It is true that wider tires commonly have better traction. The main reason why this is so does not relate to contact patch, however, but to
composition. Soft compound tires are required to be wider in order for the side-wall to support the weight of the car. softer tires have a larger
coefficient of friction, therefore better traction. A narrow, soft tire would not be strong enough, nor would it last very long. Wear in a tire is
related to contact patch. Harder compound tires wear much longer, and can be narrower. They do, however have a lower coefficient of friction,
therefore less traction. Among tires of the same type and composition, here is no appreciable difference in 'traction' with different widths. Wider
tires, assuming all other factors are equal, commonly have stiffer side-walls and experience less roll. This gives better cornering performance.
So there we have it... in laymans terms Chris is correct in saying that tire width has nothing to do with the traction meaning you'll still have the
same 'lock up' braking distance.
Get a skinny tire guy to buy a wider yet identical in compound tire and he'll have no improvement in braking, only steering.
In a perfect world Chris is right...
The really scary part is even in the real world he is actually 100 percent correct.
There you have it...
-Staggers.
Here's the formula mentioned above... which Chris should really have put in his first post...
Pressure equals force divided by the area of contact;
P=f / a
So lets say we have a tire with a contact area of 50" square, and a force, in other words wieght, is about 200 pounds...
P=200/50
P=4 pound/square inch (PSI)
Double the area...
P=f / a
P=200/100
P=2 pound/square inch (PSI)
-Staggers.
hey Brenton,
i told ya u should be a mechanical engineer with me dude
we can both end up looking like this:
very hot ...
The concept of it all still trips me out big time...
Well it was either own a business or further my education, I chose the business...
No doubt I'll learn these things along the way.
-Staggers.
AH HUH! Nice thoughts Michael. YOU ARE RIGHT!
However!
That formula works! But that force that you speak off that slows down the vehicle, is the frictional force.
F (Frictional force) = u x m x 9.8
u being the coefficient of friction
m being the mass
9.8m/s/s being gravity
Therefore when you substitute the frictional force equation in the denominator of your equation set, the mass therefore cancels out with the
numerator, and hence mass is eliminated from your set of equations. You were almost there man!
As for tyre to road friction, despite whether braking mechanisms are working or not, they all only retard the spinning of the wheels, not the car's
velocity. The only thing that slows down the car is the friction between the bitumen and the tyre, otherwise the wheels would consistantly lock up due
to the braking mechanisms and hence the car would keep sliding along the road (if friction was very small)
If the car was rolling to a stop, or braking suddenly, it will always be using the road as its only base for stopping.
Well, thats my opinion anyways man.
Your right though, confusing, but fun!
Ok just trying to remember some physics here but when talking about friction should you not put your accelleration in relation to angles of incidence,
otherwise you will be using 9.81 as a coefficent of drag not gravitational force.
sorry dont have any of my old uni books to refer to as i had to trade them otherwise i would have some fancy equasions to add.
They say that people who "Intelchooilize" things are making up for inadequies in the "Bedroom flute" department. Any truth?
If only you guys put as much effort into your cars as you have "theorising" this idea then they'd be the best cars going around!!!