Car Physics Question from the Pub??
28 August 2003, 05:18 PM
#32
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Steve, not sure where the universe ends... but apparently it's getting closer 
(they think the universe is shrinking due to not enough new stars being born to keep it stable / growing !
Andy
(they think the universe is shrinking due to not enough new stars being born to keep it stable / growing !
Andy
28 August 2003, 06:07 PM
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There's a rather long thread about just this very topic here :
http://www.gtr.co.uk/forum/upload/sh...5&pagenumber=1
Including webbie and the aforementioned Mycroft, a special appearance of a Prodrive engineer and lots of confusion
Still worth a read though if you can sift through the more "personal" quibbles...
Takes a while to read and you won't remember if you were in group 1 or group 2 after reading it 
http://www.gtr.co.uk/forum/upload/sh...5&pagenumber=1
Including webbie and the aforementioned Mycroft, a special appearance of a Prodrive engineer and lots of confusion
Still worth a read though if you can sift through the more "personal" quibbles...Takes a while to read
and you won't remember if you were in group 1 or group 2 after reading it
28 August 2003, 06:19 PM
#36
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Trouble here is that school maths lessons make the approximation that dynamic friction = mu * R, where R is the reaction between road and tyres, and mu is a constant which depends on the materials used.
For an object in circular motion, such as a car going round a corner, the force required to accelerate the car towards the centre of the bend is m(v^2)r, where m is the mass of the car, v is its velocity and r is the radius of the bend.
This force is the friction exerted on the car by the road, so:
mu*R = m(v^2)r
R is the weight of the car = m * g, where g is the acceleration due to gravity,
so, mu * m * g = m(v^2)r
...and mass cancels out. It also leads to the curious result that although a car would have more weight on a larger planet, it would actually be able to corner faster!
Thing is, friction at the tyres doesn't really just depend on the rubber compound and road surface - in practise the mass of the car does matter. So, group 1 probably has the better understanding of the problem, but group 2 is actually right.
For an object in circular motion, such as a car going round a corner, the force required to accelerate the car towards the centre of the bend is m(v^2)r, where m is the mass of the car, v is its velocity and r is the radius of the bend.
This force is the friction exerted on the car by the road, so:
mu*R = m(v^2)r
R is the weight of the car = m * g, where g is the acceleration due to gravity,
so, mu * m * g = m(v^2)r
...and mass cancels out. It also leads to the curious result that although a car would have more weight on a larger planet, it would actually be able to corner faster!
Thing is, friction at the tyres doesn't really just depend on the rubber compound and road surface - in practise the mass of the car does matter. So, group 1 probably has the better understanding of the problem, but group 2 is actually right.
28 August 2003, 06:57 PM
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Weird thread, but I learned from it, believe it or not.
Although that pesky "it's way over your head" feeling never went away
Edit: didn't teach me to spell mind
[Edited by EvilBevel - 8/28/2003 6:58:45 PM]
28 August 2003, 09:35 PM
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I'm going to disagree with everyone
F = mu x R is a model designed for rigid objects and smooth surfaces, and is totally inappropriate for tyre against road. So I'll start by ignoring that.
A better model to use for tyres, which are capable of deforming to the shape of the road, is to imagine the tyre as a load of little tyrette blocks, all working together.
As the tyre deforms over the road, each little block has an amount of pressure applied to it, based on the mass of the vehicle divided by the contact patch area.
The heavier car can achieve the same level of pressure on each bit of the tyre by using a tyre with a larger contact patch, either by having a larger radius tyre or a wider tyre.
In this way each bit of the tyre is working equally on both vehicles, and generating the same level of lateral force. The total lateral force is equal to the sum of the bits.
So the answer is group 1 is correct, in steady state cornering the two cars will corner independently of mass, as long as they are both wearing the right sized rubber. Best not say that in America though, they get terribly confused.
The fatter tyre also helps to dissipate the greater heat generated by shifting a fatter car around the corner.
This isn't the end of the story, it is waaaaay more complicated, but I've already exceeded the limit of my knowledge Where's that beer I was promised?
Members of AMOC will bear witness to one-and-a-half tons of British muscle cornering every bit as quick as stripped out civic type R's
F = mu x R is a model designed for rigid objects and smooth surfaces, and is totally inappropriate for tyre against road. So I'll start by ignoring that.
A better model to use for tyres, which are capable of deforming to the shape of the road, is to imagine the tyre as a load of little tyrette blocks, all working together.
As the tyre deforms over the road, each little block has an amount of pressure applied to it, based on the mass of the vehicle divided by the contact patch area.
The heavier car can achieve the same level of pressure on each bit of the tyre by using a tyre with a larger contact patch, either by having a larger radius tyre or a wider tyre.
In this way each bit of the tyre is working equally on both vehicles, and generating the same level of lateral force. The total lateral force is equal to the sum of the bits.
So the answer is group 1 is correct, in steady state cornering the two cars will corner independently of mass, as long as they are both wearing the right sized rubber.
Best not say that in America though, they get terribly confused.The fatter tyre also helps to dissipate the greater heat generated by shifting a fatter car around the corner.
This isn't the end of the story, it is waaaaay more complicated, but I've already exceeded the limit of my knowledge
Where's that beer I was promised?Members of AMOC will bear witness to one-and-a-half tons of British muscle cornering every bit as quick as stripped out civic type R's
29 August 2003, 10:05 AM
#44
Hi All
What a great thread
As I've done about 30 pages of this on GTR, I'll try to narrow this down to bullet points.
- Mass DOES come in to it
- A lighter car can be accelerated in any direction easier than a heavier one
- Tyres are not just solid friction objects, they bend, skew, distort, twist, mould, etc so their "limit lateral acceleration" (what most of us would term as "maximum cornering grip") is not a function of it's frictional properties alone.
- Doubling the load on a tyre less than doubles the grip it can produce (for the pedants among us () - "assuming a realistic operational load for a working tyre"). This means that a heavier car has less grip available as a proportion of it's mass.
Hope that helps
All the best
Simon
What a great thread
As I've done about 30 pages of this on GTR, I'll try to narrow this down to bullet points.
- Mass DOES come in to it
- A lighter car can be accelerated in any direction easier than a heavier one
- Tyres are not just solid friction objects, they bend, skew, distort, twist, mould, etc so their "limit lateral acceleration" (what most of us would term as "maximum cornering grip") is not a function of it's frictional properties alone.
- Doubling the load on a tyre less than doubles the grip it can produce (for the pedants among us (
) - "assuming a realistic operational load for a working tyre"). This means that a heavier car has less grip available as a proportion of it's mass.Hope that helps
All the best
Simon
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