Tyre wear and fuel load

[CatharticF1]

Does anyone know the rate of tyre wear of an F1 car full of fuel compared to empty?

I once read that vehicles do damage to roads proportional to the cube of their weight. If that applied to an F1 car's tyre use, they'd get 17 odd laps near empty compared to 10 near full.

Any references for this?

[brafield]

--- damage to roads, or damage to tires, you mean?

[Kerry Montgomery]

"CatharticF1" <rasf1...@gmail.com> wrote in message
news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...

Cathartic F1,
Sorry, no references. Did your calculation include the effect of aero
downforce?
Thanks,
Kerry

[CatharticF1]

No - downforce doesn't work like weight in that it has no inertial effect aiui.

[Bobster]

Well F1 cars consume fuel at a rate way beyond that of most road cars,
so the percentage differential between full and empty is going to be
much higher.

Also remember that tyres are not tyres are not tyres. They're obliged
to use two different compounds with different wear rates, and they
usually end up having to use at least one set of used tyres during a
race.

[Bigbird]

I make it closer to 10/14 based on Bahrain figures. What was your calc?

The tyre wear will be resultant of the forces applied and slip. I doubt
that there is such a consistent relationship to weight.

[CatharticF1]

Actually I think I was low. I used fuel of 120kg. But let's say it's 170kg, and take 150 as the average difference between the first and last stint.

790^3 / 640^3 gives around 1.9 times the wear

But that presumes the wear is proportional to the cube.

[Michael Press]

In article
<12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,

With no aerodynamic down force a car can turn at (say) 1 g.
With aerodynamic down force a car can turn at (say) 3 g.
The turning force is provided by the tire contact patch,
so increased aerodynamic down force can make for increased
rate of tire wear. What you intend to convey by saying it
is not an inertial effect escapes me.

--
Michael Press

[News]

I think he's pointing out that "weightless" aero downforce has none of
the centripetal force component comparable to an equivalent added mass.

[Bobster]

On Apr 30, 6:48 pm, News <N...@Groups.Post> wrote:

> I think he's pointing out that "weightless" aero downforce has none of
> the centripetal force component comparable to an equivalent added mass.

OK. I'm confused! What is weightless aero downforce when it's at home?

[News]

Watching TV or ticking off the honey-do list, like everyone else?

[build]

On May 1, 2:48 am, News <N...@Groups.Post> wrote:
> On 4/30/2012 12:30 PM, Michael Press wrote:
>
>
>
> > In article
> > <12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,

> >   CatharticF1<rasf1pos...@gmail.com>  wrote:

>
> >> On Tuesday, April 24, 2012 1:43:18 PM UTC+10, Kerry Montgomery wrote:

> >>> "CatharticF1"<rasf1pos...@gmail.com>  wrote in message

> >>>news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...
> >>>> Does anyone know the rate of tyre wear of an F1 car full of fuel compared
> >>>> to empty?
>
> >>>> I once read that vehicles do damage to roads proportional to the cube of
> >>>> their weight. If that applied to an F1 car's tyre use, they'd get 17 odd
> >>>> laps near empty compared to 10 near full.
>
> >>>> Any references for this?
>
> >>> Cathartic F1,
> >>> Sorry, no references. Did your calculation include the effect of aero
> >>> downforce?
> >>> Thanks,
> >>> Kerry
>
> >> No - downforce doesn't work like weight in that it has no inertial effect aiui.
>
> > With no aerodynamic down force a car can turn at (say) 1 g.
> > With    aerodynamic down force a car can turn at (say) 3 g.
> > The turning force is provided by the tire contact patch,
> > so increased aerodynamic down force can make for increased
> > rate of tire wear. What you intend to convey by saying it
> > is not an inertial effect escapes me.
>
> I think he's pointing out that "weightless" aero downforce has none of
> the centripetal force component comparable to an equivalent added mass.

I am not often surprised in the group but you just surprised me. Yes
as far as my knowledge goes you have worded it better than I would.
Perhaps you could expand that for others. I'd say something like aero
has vertical forces but none of the overhead of lateral forces which
would motivate Bird to abuse.

I'm a bit gob smacked ... or did you google that ... I'm a skeptic.

regardless, well said.

beers,

[build]

On May 1, 2:48 am, News <N...@Groups.Post> wrote:

> On 4/30/2012 12:30 PM, Michael Press wrote:
>
>
>
> > In article
> > <12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,

> >   CatharticF1<rasf1pos...@gmail.com>  wrote:

>
> >> On Tuesday, April 24, 2012 1:43:18 PM UTC+10, Kerry Montgomery wrote:

> >>> "CatharticF1"<rasf1pos...@gmail.com>  wrote in message

> >>>news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...
> >>>> Does anyone know the rate of tyre wear of an F1 car full of fuel compared
> >>>> to empty?
>
> >>>> I once read that vehicles do damage to roads proportional to the cube of
> >>>> their weight. If that applied to an F1 car's tyre use, they'd get 17 odd
> >>>> laps near empty compared to 10 near full.
>
> >>>> Any references for this?
>
> >>> Cathartic F1,
> >>> Sorry, no references. Did your calculation include the effect of aero
> >>> downforce?
> >>> Thanks,
> >>> Kerry
>
> >> No - downforce doesn't work like weight in that it has no inertial effect aiui.
>
> > With no aerodynamic down force a car can turn at (say) 1 g.
> > With    aerodynamic down force a car can turn at (say) 3 g.
> > The turning force is provided by the tire contact patch,
> > so increased aerodynamic down force can make for increased
> > rate of tire wear. What you intend to convey by saying it
> > is not an inertial effect escapes me.
>
> I think he's pointing out that "weightless" aero downforce has none of
> the centripetal force component comparable to an equivalent added mass.

OK, you did so well with that that I'm tempted to ask for a
description of the difference between airfoil or wing induced
downforce and ground effect downforce. And I stress that is a request
not a challenge as to me the reply would be a novel not a sentence or
even a paragraph.

beers,

[build]

Ummmm, Bob,
He said what it is and he said it very well. Read it again.

beers,

[Bobster]

No, just trying to understand you're point. I'm not saying you're
wrong, I'm saying that I don't understand.

[build]

On May 1, 2:30 am, Michael Press <rub...@pacbell.net> wrote:
> In article
> <12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,
>
>
>
>  CatharticF1 <rasf1pos...@gmail.com> wrote:
> > On Tuesday, April 24, 2012 1:43:18 PM UTC+10, Kerry Montgomery wrote:

> > > "CatharticF1" <rasf1pos...@gmail.com> wrote in message

> > >news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...
> > > > Does anyone know the rate of tyre wear of an F1 car full of fuel compared
> > > > to empty?
>
> > > > I once read that vehicles do damage to roads proportional to the cube of
> > > > their weight. If that applied to an F1 car's tyre use, they'd get 17 odd
> > > > laps near empty compared to 10 near full.
>
> > > > Any references for this?
>
> > > Cathartic F1,
> > > Sorry, no references. Did your calculation include the effect of aero
> > > downforce?
> > > Thanks,
> > > Kerry
>
> > No - downforce doesn't work like weight in that it has no inertial effect aiui.
>
> With no aerodynamic down force a car can turn at (say) 1 g.
> With    aerodynamic down force a car can turn at (say) 3 g.
> The turning force is provided by the tire contact patch,
> so increased aerodynamic down force can make for increased
> rate of tire wear. What you intend to convey by saying it
> is not an inertial effect escapes me.
>
> --
> Michael Press

in·er·tia: the property of matter by which it retains its state of
rest or its velocity along a straight line so long as it is not acted
upon by an external force.

regards,

[News]

Thanks. I've been doing this since the 1970s, and have had time to
practice the theory and explanations...

[build]

Hey Mate,
Please take Bob's reply literally and continue you explanation. You
have done so very well.

Can I suggest expanding on centrifugal, with an example of a lead
weight as opposed to vertical only forces of aero. BUT more eloquently
stated.

beers,

[News]

m = mass = of chassis/engine
v = vehicle speed
r = radius of curve

Fc = centripetal Force = m * centripetal acceleration = (v^2)/r =
m((v^2)/r) = required friction "grip" to retain vehicle on radius,
provided by sum of mass and downforce acting through contact
patch/steering/camber/suspension dynamics

Now compare the effects of adding 3000 lb aero downforce (via wing or
ground effects) versus adding 3000 lb mass to chassis for equivalent "grip"

3000 lb (negative) lift will entail adding significant induced drag,
requiring significantly more power at any relevant corner entry velocity
v, but no added centripetal or inertial effect (proportional to vehicle
mass = unchanged)

3000 lb mass will induce no additional drag but require more power to
accelerate to relevant corner entry velocity v, and add to centripetal
and inertia forces proportional to the quadrupling of mass (4000lb vs.
1000lb)

In recent times, DRS permits the immediate shedding of significant
downforce and drag, with no inertial effect

[News]

No difference in centripetal/lateral force contribution, assuming equal
mass of mounted wing(s) or chassis ground effects (some other
differences due to the higher center of force likely to occur with the
mounted wing(s) vs the G/E).

Both induce less centripetal/lateral force requirement versus the same
vertical force component due to mass alone.

No difference in induced drag contribution assuming same L/D (although
it should be possible to produce more efficient bare wing(s) than G/E
planform).

[Alan LeHun]

In article <jnmfps$jbc$1...@dont-email.me>, Ne...@Groups.Post says...

> I think he's pointing out that "weightless" aero downforce
>

itym mass-less.

Extra "weight" is the raison d'etra of aero downforce.

--
Alan LeHun

[News]

No shyte, Sherlock. The full quote is:

[Alan LeHun]

In article <jnmnd8$3dl$1...@dont-email.me>, Ne...@Groups.Post says...

> > itym mass-less.
> >
> > Extra "weight" is the raison d'etra of aero downforce.
> >
>
>
> No shyte, Sherlock. The full quote is:
>
> "I think he's pointing out that "weightless" aero downforce has none of
> the centripetal force component comparable to an equivalent added mass."
>

So weightless downforce has no mass? What /does/ it have?

--
Alan LeHun

[News]

As stated plainly, aero downforce has none of the centripetal (lateral)

[Mower Man]

I've always been under the impression that a given quantity of ground
effect induced downforce produced less drag than aerofoils? After all,
via the Bernoulli effect the air aft of the "step" helps as it loses
density helps the high pressure air (incoming flow) to pass under the
car?

--
Chris

'Fashion is a form of ugliness so intolerable that we have to alter it
every six months.'

(Oscar Wilde.)

[Mower Man]

Downforce. No mass to add to the car per se - which, having the same
mass as when it was stationary, is able via the downforce to behave *as
if* it weighed more but has no more mass when cornering. Does an
aircraft weigh less when airborne?

[Bobster]

LOL.

I have lots of physics to catch up on

[Alan LeHun]

In article <jnmos8$85c$2...@speranza.aioe.org>, chris...@nospamf2s.com
says...

> > So weightless downforce has no mass? What /does/ it have?
> >
>
> Downforce. No mass to add to the car per se - which, having the same
> mass as when it was stationary, is able via the downforce to behave *as
> if* it weighed more but has no more mass when cornering. Does an
> aircraft weigh less when airborne?
>

Ok. We are all singing from the same sheet as it were but your last
example was a poor choice. Under some definitions, the weight of an
aircraft can and does change when it is airborne.

--
Alan LeHun

[News]

Only as if burns off fuel(or jetisons/drops loads).

[Timmy]

News wrote ...

What if it's a glider ?

[Mower Man]

Please elucidate?

[Mower Man]

Same deal, silly. ;-) Good try.

[Mower Man]

On 30/04/2012 9:39 PM, Mower Man wrote:
> On 30/04/2012 9:29 PM, Timmy wrote:
>> News wrote ...
>>
>>>
>>> On 4/30/2012 4:13 PM, Alan LeHun wrote:
>>>> In article<jnmos8$85c$2...@speranza.aioe.org>, chris...@nospamf2s.com
>>>> says...
>>>>>> So weightless downforce has no mass? What /does/ it have?
>>>>>>
>>>>>
>>>>> Downforce. No mass to add to the car per se - which, having the same
>>>>> mass as when it was stationary, is able via the downforce to behave
>>>>> *as
>>>>> if* it weighed more but has no more mass when cornering. Does an
>>>>> aircraft weigh less when airborne?
>>>>>
>>>>
>>>> Ok. We are all singing from the same sheet as it were but your last
>>>> example was a poor choice. Under some definitions, the weight of an
>>>> aircraft can and does change when it is airborne.
>>>>
>>>
>>>
>>> Only as if burns off fuel(or jetisons/drops loads).
>>
>>
>> What if it's a glider ?
>>
>
> Same deal, silly. ;-) Good try.
>

As in take a dump, throw out of window?

[News]

Gliders can jettison/drop loads. Pilot relieves him/herself?

[Mower Man]

As I said - but a potentially messy business, given the air pressure
outside vs. inside the cockpit.

[Timmy]

Mower Man wrote ...

> >>
> >> What if it's a glider ?
> >>
> >
> > Same deal, silly. ;-) Good try.
> >
> As in take a dump, throw out of window?

You tried taking a dump in a glider?

[Timmy]

Mower Man wrote ...

Gliders ain't pressurised, but you do notice the blast when the window
slides open.

[Alan LeHun]

In article <jnmt9i$hv2$4...@speranza.aioe.org>, chris...@nospamf2s.com
says...

> > Ok. We are all singing from the same sheet as it were but your last
> > example was a poor choice. Under some definitions, the weight of an
> > aircraft can and does change when it is airborne.
> >
>
> Please elucidate?
>
>

Well you can't weigh a flying aircraft (or a moving F1 car) by Newtonian
definitions of weight, because that definition insists that the object
be at rest. Other definitions are also problematic as best shown by the
ISS which may have weight by a non-relativistic scientific definition
(gravity is still acting upon its mass) or may be weightless by a
mechanical definition. If you try and include relativity in a definition
of weight then it can be "proven" that weight is a perceptive illusion
and does not exist.

By mechanical definitions it can be argued that an aircraft that is
climbing could be said to have a negative weight.

Essentially, it all comes down to your frame of reference which brings
relativity into again.

In some scientific disciplines, "weight" has become a bit of a taboo.

--
Alan LeHun

[News]

Relief tubes are pretty well understood.

[News]

In flight, an aircraft's mass does not change in the short-term (other
than via fuel burn and load jettison, as mentioned) but thrust, lift and
drag often do, which produce acceleration, climb, deceleration, descent
dynamics.

[Mower Man]

Nevah.

Buuurt.... there's a 1st time for everything!

[Mower Man]

Exactly. Does the poo go out or come back in, d'you suppose?

[Mower Man]

On 30/04/2012 10:06 PM, Alan LeHun wrote:
> In article<jnmt9i$hv2$4...@speranza.aioe.org>, chris...@nospamf2s.com
> says...
>>> Ok. We are all singing from the same sheet as it were but your last
>>> example was a poor choice. Under some definitions, the weight of an
>>> aircraft can and does change when it is airborne.
>>>
>>
>> Please elucidate?
>>
>>
>
> Well you can't weigh a flying aircraft (or a moving F1 car) by Newtonian
> definitions of weight, because that definition insists that the object
> be at rest.

For all practical purposes the aircraft has a given mass, which is
unaffected by altitude or rate of climb.

Other definitions are also problematic as best shown by the
> ISS which may have weight by a non-relativistic scientific definition
> (gravity is still acting upon its mass) or may be weightless by a
> mechanical definition. If you try and include relativity in a definition
> of weight then it can be "proven" that weight is a perceptive illusion
> and does not exist.

That's utterly irrelevant, specious at best, bollocks at worst.

>
> By mechanical definitions it can be argued that an aircraft that is
> climbing could be said to have a negative weight.

Oh? How? Because it's climbing? Explain?

> Essentially, it all comes down to your frame of reference which brings
> relativity into again.

Into again? WTF?

>
> In some scientific disciplines, "weight" has become a bit of a taboo.

Mass has for many, many years been the dog's.

[Timmy]

Depends on whether you've bagged it or had to use your hands...

[Alan LeHun]

In article <jnn16b$tod$1...@speranza.aioe.org>, chris...@nospamf2s.com
says...

> For all practical purposes the aircraft has a given mass, which is
> unaffected by altitude or rate of climb.
>

I'm sorry. I thought we were discussing weight.

--
Alan LeHun

[Michael Press]

In article <jnmfps$jbc$1...@dont-email.me>, News <Ne...@Groups.Post> wrote:

> On 4/30/2012 12:30 PM, Michael Press wrote:
> > In article
> > <12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,

> > CatharticF1<rasf1...@gmail.com> wrote:
> >
> >> On Tuesday, April 24, 2012 1:43:18 PM UTC+10, Kerry Montgomery wrote:

> >>> "CatharticF1"<rasf1...@gmail.com> wrote in message

> >>> news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...
> >>>> Does anyone know the rate of tyre wear of an F1 car full of fuel compared
> >>>> to empty?
> >>>>
> >>>> I once read that vehicles do damage to roads proportional to the cube of
> >>>> their weight. If that applied to an F1 car's tyre use, they'd get 17 odd
> >>>> laps near empty compared to 10 near full.
> >>>>
> >>>> Any references for this?
> >>>
> >>> Cathartic F1,
> >>> Sorry, no references. Did your calculation include the effect of aero
> >>> downforce?
> >>> Thanks,
> >>> Kerry
> >>
> >> No - downforce doesn't work like weight in that it has no inertial effect aiui.
> >
> > With no aerodynamic down force a car can turn at (say) 1 g.
> > With aerodynamic down force a car can turn at (say) 3 g.
> > The turning force is provided by the tire contact patch,
> > so increased aerodynamic down force can make for increased
> > rate of tire wear. What you intend to convey by saying it
> > is not an inertial effect escapes me.
> >
>
>

> I think he's pointing out that "weightless" aero downforce has none of
> the centripetal force component comparable to an equivalent added mass.

Okay. How does added mass come into it?
What is the purpose of added mass?
Added mass does not per se increase cornering ability.

As for aerodynamic downforce, it does increase tire wear
and road surface wear; even though it does not add mass.

--
Michael Press

[Michael Press]

In article <MPG.2a090f1da...@news-europe.giganews.com>,

The force concept is useful in many practical endeavors.
Theoretically it is elusive and ultimately a derived concept.
A force can be calculated by taking the time derivative
of a momentum, or by taking the gradient of a potential energy.

--
Michael Press

[Michael Press]

In article
<9b6d07c5-11d4-4dda...@h10g2000pbi.googlegroups.com>,
build <bui...@gmail.com> wrote:

> On May 1, 2:30 am, Michael Press <rub...@pacbell.net> wrote:
> > In article
> > <12549741.266.1335242058351.JavaMail.geo-discussion-forums@pbbox1>,
> >
> >
> >
> >  CatharticF1 <rasf1pos...@gmail.com> wrote:
> > > On Tuesday, April 24, 2012 1:43:18 PM UTC+10, Kerry Montgomery wrote:

> > > > "CatharticF1" <rasf1pos...@gmail.com> wrote in message

> > > >news:24480039.175.1335234231590.JavaMail.geo-discussion-forums@pbvd8...
> > > > > Does anyone know the rate of tyre wear of an F1 car full of fuel compared
> > > > > to empty?
> >
> > > > > I once read that vehicles do damage to roads proportional to the cube of
> > > > > their weight. If that applied to an F1 car's tyre use, they'd get 17 odd
> > > > > laps near empty compared to 10 near full.
> >
> > > > > Any references for this?
> >
> > > > Cathartic F1,
> > > > Sorry, no references. Did your calculation include the effect of aero
> > > > downforce?
> > > > Thanks,
> > > > Kerry
> >
> > > No - downforce doesn't work like weight in that it has no inertial effect aiui.
> >
> > With no aerodynamic down force a car can turn at (say) 1 g.
> > With    aerodynamic down force a car can turn at (say) 3 g.
> > The turning force is provided by the tire contact patch,
> > so increased aerodynamic down force can make for increased
> > rate of tire wear. What you intend to convey by saying it
> > is not an inertial effect escapes me.
>

> in·er·tia: the property of matter by which it retains its state of
> rest or its velocity along a straight line so long as it is not acted
> upon by an external force.

Read what I wrote. I am not asking for a definition of inertial.

--
Michael Press

[Frank Adam]

On Wed, 02 May 2012 14:58:36 -0700, Michael Press <rub...@pacbell.net> wrote:

>> > > No - downforce doesn't work like weight in that it has no inertial effect aiui.
>> >
>> > With no aerodynamic down force a car can turn at (say) 1 g.
>> > With    aerodynamic down force a car can turn at (say) 3 g.
>> > The turning force is provided by the tire contact patch,
>> > so increased aerodynamic down force can make for increased
>> > rate of tire wear. What you intend to convey by saying it
>> > is not an inertial effect escapes me.
>>
>> in·er·tia: the property of matter by which it retains its state of
>> rest or its velocity along a straight line so long as it is not acted
>> upon by an external force.
>
>Read what I wrote. I am not asking for a definition of inertial.
>

The 'l' is silent you dweeb. :)

I don't really know, but there are two things that come to my mind.
1) With inreased downforce, the rolling friction would increase, which will
indeed lead to more tyre wear, especially if slip exists.
2) However, slip, which IMO is the major part of wear, would be greatly
reduced overall and thus so would be tyre wear.

So i dunno, someone should add up 1 and 2, and tell us the result. ;)

--

Regards, Frank

[build]

Well I tell you if you can tell me how long a piece of string is?

or I could just refer you to Redbulls tyre wear in 2011.

beers,

[Bigbird]

As we keep being told by the teams/drivers the result is less wear.

[News]

There is far less scrub because 3000 lbs of aero downforce does not come
with the need to counter the centripetal force that would go with 3000
lbs of mass.

[News]

On 5/2/2012 5:43 PM, Michael Press wrote:

[Frank Adam]

Too subtle, was i ? :)

--

Regards, Frank

[Michael Press]

What scrub comes with mass that is not equalled
with aerodynamic down force?

--
Michael Press

[Michael Press]

In article <fjm3q75gcibdtr9b4...@4ax.com>,

Frank Adam <fa...@notthis.optushome.com.au> wrote:

> On Wed, 02 May 2012 14:58:36 -0700, Michael Press <rub...@pacbell.net> wrote:
>
> >> > > No - downforce doesn't work like weight in that it has no inertial effect aiui.
> >> >
> >> > With no aerodynamic down force a car can turn at (say) 1 g.
> >> > With    aerodynamic down force a car can turn at (say) 3 g.
> >> > The turning force is provided by the tire contact patch,
> >> > so increased aerodynamic down force can make for increased
> >> > rate of tire wear. What you intend to convey by saying it
> >> > is not an inertial effect escapes me.
> >>
> >> in·er·tia: the property of matter by which it retains its state of
> >> rest or its velocity along a straight line so long as it is not acted
> >> upon by an external force.
> >
> >Read what I wrote. I am not asking for a definition of inertial.
> >
> The 'l' is silent you dweeb. :)

Mixed message.

> I don't really know, but there are two things that come to my mind.
> 1) With inreased downforce, the rolling friction would increase, which will
> indeed lead to more tyre wear, especially if slip exists.

I said that well back in this thread.

> 2) However, slip, which IMO is the major part of wear, would be greatly
> reduced overall and thus so would be tyre wear.

Reduced contrasted with what? What is going to reduce slip?
Not the drivers. They are always on the edge of their tires.
More aerodynamic down force --> more tire wear.

> So i dunno, someone should add up 1 and 2, and tell us the result. ;)

--

Michael Press

[News]

Sideload/scrub/countering centripetal force that would go with
equivalent added mass.

[Ian Dalziel]

Providing centripetal force, not countering it.

--

Ian D

[News]

Correct(ed).

[Frank Adam]

On Thu, 03 May 2012 12:34:26 -0700, Michael Press <rub...@pacbell.net> wrote:


>> I don't really know, but there are two things that come to my mind.
>> 1) With inreased downforce, the rolling friction would increase, which will
>> indeed lead to more tyre wear, especially if slip exists.
>
>I said that well back in this thread.
>

It's a miniscule amount in both scenarios.

>> 2) However, slip, which IMO is the major part of wear, would be greatly
>> reduced overall and thus so would be tyre wear.
>
>Reduced contrasted with what? What is going to reduce slip?
>

Contrasted with distance travelled.
You seem to be trying to compare tyre wear ignoring that one car is doing 1:30
laps with the other doing 1:50 or worse. Tyre wear is wear over distance.
If you are trying to tell me that a car with a better than 1:1 power to weight
ratio will *race* on its own edge of adhesion for over 25% longer time on the
same tyre compounds than one that has full downforce, then you have to be
sipping copius amounts of the good stuff.

>Not the drivers. They are always on the edge of their tires.
>More aerodynamic down force --> more tire wear.
>

Geez man, think "overall wear". Think in detail about what work a tyre does
over a lap with or without downforce.

The idea in downforce is to aid the tyres in retaining full patch contact with
the track at *most* times. The amount of time the tyre remains in full contact
greatly influences slip and the associated tyre temperature changes and as
such, downforce simply has to reduce "overall wear".
The reverse of that is no downforce, where the tyres will slip and squirm
virtually all the time. That will increase tyre temperature and wear at *most*
points of the track.


--

Regards, Frank

[Bigbird]

Simply not true. Listen more carefully when the drivers and
commentaters are talking about oversteer and understeer. Watch a few
laps and see how some cars move around under the driver more than
others. Notice how some driver are able to get on the gas sooner than
others because they have better grip and don't have to wait longer to
straighten up the car. That is, some drivers are having to cope with
far more slip throughout a lap than others. Cars are not driven at the
edge of adhesion throughout a lap any more than they are driven on the
red line through a lap.

Listen really carefully and you'll even hear them talk about the
drivers looking after their tyres. You are missing a lot.

[build]

On May 4, 3:03 pm, "Bigbird" <Bigbird.usenetNOS...@Gmail.com> wrote:
> Michael Press wrote:

> > In article <fjm3q75gcibdtr9b42cqp1hpv38v20s...@4ax.com>,

I absolutely agree. To say more downforce means more tyre wear is
simply ignoring the obvious. Why is it that the most aerodynamically
efficient cars win races? Is that because they wear out tyres more
quickly? ... No. More grip equals less wear, more downforce equals
more grip.

beers,

[News]

Less scrub = less energy loss = less tire wear = less time in box = faster

[Michael Press]

In article <rbb6q7hhofks0nf36...@4ax.com>,

Frank Adam <fa...@notthis.optushome.com.au> wrote:

> On Thu, 03 May 2012 12:34:26 -0700, Michael Press <rub...@pacbell.net> wrote:
>
>
> >> I don't really know, but there are two things that come to my mind.
> >> 1) With inreased downforce, the rolling friction would increase, which will
> >> indeed lead to more tyre wear, especially if slip exists.
> >
> >I said that well back in this thread.
> >
> It's a miniscule amount in both scenarios.

`Minuscule.' Contrasted with what?

> >> 2) However, slip, which IMO is the major part of wear, would be greatly
> >> reduced overall and thus so would be tyre wear.
> >
> >Reduced contrasted with what? What is going to reduce slip?
> >
> Contrasted with distance travelled.
> You seem to be trying to compare tyre wear ignoring that one car is doing 1:30
> laps with the other doing 1:50 or worse.

I'm not. That is all in your own head.

> Tyre wear is wear over distance.

I always thought tire wear is degradation of the tire.
You are defining tire wear as _rate_ of tire wear.

> If you are trying to tell me that a car with a better than 1:1 power to weight
> ratio will *race* on its own edge of adhesion for over 25% longer time on the
> same tyre compounds than one that has full downforce,

No idea why you consider this hypothesis.

> then you have to be
> sipping copius amounts of the good stuff.

Unfounded conclusion.

>
> >Not the drivers. They are always on the edge of their tires.
> >More aerodynamic down force --> more tire wear.
> >
> Geez man, think "overall wear". Think in detail about what work a tyre does
> over a lap with or without downforce.
>
> The idea in downforce is to aid the tyres in retaining full patch contact with
> the track at *most* times. The amount of time the tyre remains in full contact
> greatly influences slip and the associated tyre temperature changes and as
> such, downforce simply has to reduce "overall wear".
> The reverse of that is no downforce, where the tyres will slip and squirm
> virtually all the time. That will increase tyre temperature and wear at *most*
> points of the track.

The more downforce, the more friction at the contact patch,
the more wear on the tire. _This_ is what I am saying.

--
Michael Press

[Michael Press]

In article <a0h697...@mid.individual.net>,

They drive at the limit. They drive so that their
tires are shredded when they cross the finish line.
No point in saving the tires for the victory lap.

--
Michael Press

[News]

Mostly in scrub angle, with downforce being the predominant load instead
of centripetal side load.

[build]

On May 5, 2:14 am, Michael Press <rub...@pacbell.net> wrote:
> The more downforce, the more friction at the contact patch,
> the more wear on the tire. _This_ is what I am saying.
>
> --
> Michael Press

Michael,
I suggest you may like to read some of Carroll Smiths books. He
explains why more friction equals less wear in a plain English that
even I can understand. I doubt you will regret the small cost of the
book as you will get more enjoyment from watching the sport. You may
have to buy them second hand but I'm sure they will be out there
somewhere.

regards,

[Bigbird]

No they don't.

Regardless it is a straw man that has no relevance to your chosen
argument.

If what you argue was true a RBR would use it's tyres more quickly than
a Caterham while circulating at the same speed. They in fact circulate
much faster without using up their tyres any faster.