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.
On Apr 23, 7:23 pm, CatharticF1 <rasf1pos...@gmail.com> wrote:
> 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?
--- damage to roads, or damage to tires, you mean?
> 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
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.
On Apr 24, 4:23 am, CatharticF1 <rasf1pos...@gmail.com> wrote:
> 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?
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.
CatharticF1 wrote:
> 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?
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.
On Tuesday, April 24, 2012 4:20:28 PM UTC+10, Bigbird wrote:
> CatharticF1 wrote:
> > 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?
> 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.
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.
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 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.
> >> 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.
> >> 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.
On Apr 30, 7:01 pm, News <N...@Groups.Post> wrote:
> On 4/30/2012 12:50 PM, Bobster wrote:
> > 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?
> Watching TV or ticking off the honey-do list, like everyone else?
No, just trying to understand you're point. I'm not saying you're
wrong, I'm saying that I don't understand.
> 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.
> 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,
Thanks. I've been doing this since the 1970s, and have had time to practice the theory and explanations...
> On Apr 30, 7:01 pm, News<N...@Groups.Post> wrote:
>> On 4/30/2012 12:50 PM, Bobster wrote:
>>> 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?
>> Watching TV or ticking off the honey-do list, like everyone else?
> No, just trying to understand you're point. I'm not saying you're
> wrong, I'm saying that I don't understand.
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
> 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,
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).
>>>>>>> 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,
> 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).
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.'
