Tire width and rolling resistance.
Daniel Connelly
Hans-Joachim Zierke <ha...@quijote.in-berlin.de> wrote:
>Note: follow-up set
>
>> Jobst Brandt's experimental data, however, clearly shows that
>> narrow tires have lower rolling resistance. It therefore appears
>> that the profile of the narrower tires allow them to deform
>> more efficiently (more elastically), offsetting the greater
>> work required.
>
>No.
>
>More air volume puts more stress on the casing. Either you need stronger
>casing (-> thicker -> more loss) or you have to reduce the pressure (->
>more deformation -> more loss).
I think we are in agreement; I consider casing thickness to be one
aspect of the profile (meaning cross-section).
>for road use, 8.5 is all you may need. You won't gain much advantage
>beyond that, and for the same pressure, the 28-622 has less
>rolling resistance than the 20-622 (but maybe more air drag).
Thanks for the info.
From Jobst's data, averaging over all tires which made it to 9.0 atm:
Pressure d ln|Rolling Resistance
-------- -------------------------
atm d ln|Pressure|
=================================
3.74166 -0.672808
4.24262 -0.656592
4.74343 -0.632024
5.24403 -0.561641
5.74454 -0.572118
6.24499 -0.506997
6.74533 -0.483471
7.24564 -0.487317
7.74595 -0.444221
8.24618 -0.471219
8.74647 -0.223447
(note the pressures listed are the geometric means of the
data at half-atmosphere increments)
Increasing pressure yields diminishing returns on rolling
resistance. You may be right about the elbow at 8.5, but it is
unclear from the data. Clearly though the infinite-pressure
limit is not zero RR, so the elbow must occur somewhere.
Even if the "elbow" doesn't occur, high pressures still may not
be desirable because of the associated rough ride, though.
>I think that super-narrow road tires should be a thing of the past. 28-622
>as a standard road tire would dramatically reduce the number of riding
>partners who refuse to take a shortcut via a trail on weekend rides. But it
>would be 80 g more...
I agree narrow tires should be racing only, but they do seem to corner
better. Shortcuts aren't an issue in road races, or shouldn't be :).
Dan
--
Patrick McNally
> I agree narrow tires should be racing only, but they do seem to corner
> better. Shortcuts aren't an issue in road races, or shouldn't be :).
>
> Dan
> --
I disagree. I like the feeling of the 23mm tires over the 18s or the 20.
The 23's seem to not be subject to chattering over the surface
of the road so much.
______________________________________
‹ Patrick McNally, Campagnolo bigot
Jim Papadopoulos
>Even if the "elbow" doesn't occur, high pressures still may not
>be desirable because of the associated rough ride, though.
or measure SUSPENSION losses (whether in tires themselves, or in
the load being transported). For hard tires, these will be large.
JP
bui tho xuan
>Note: follow-up set
>
>> Jobst Brandt's experimental data, however, clearly shows that
>> narrow tires have lower rolling resistance. It therefore appears
>> that the profile of the narrower tires allow them to deform
>> more efficiently (more elastically), offsetting the greater
>> work required.
>
>No.
>
>More air volume puts more stress on the casing. Either you need stronger
>casing (-> thicker -> more loss) or you have to reduce the pressure (->
>more deformation -> more loss).
How does this work, Hajo? for the same pressure (in psi), the stress on
the casing is the same (also in psi). What creates the deviation from
this simple ideality? I can't quite see how an increase in volume
(for same pressure) would increase the stress. If anything, I can
_almost_ see the opposite effect.
tho
Jobst Brandt
> Increasing pressure yields diminishing returns on rolling
> resistance. You may be right about the elbow at 8.5, but it is
> unclear from the data. Clearly though the infinite-pressure
> limit is not zero RR, so the elbow must occur somewhere.
I don't know why you believe there should be an elbow in these curves.
This is a continuous function that asymptotically approaches zero with
increasing pressure. Because the dominating loss is from internal
friction from flexing, and this effect is continuously reduced by
increasing pressure, the curves have no elbow.
Jobst Brandt <jbr...@hplabs.hp.com>
Jobst Brandt
> How does this work, Hajo? for the same pressure (in psi), the
> stress on the casing is the same (also in psi). What creates the
> deviation from this simple ideal? I can't quite see how an increase
> in volume (for same pressure) would increase the stress. If anything,
> I can _almost_ see the opposite effect.
The stress in a tube is proportional to its cross section diameter.
You can arrive on this by separating the cylindrical cross section
into two halves at the diameter and computing the pressure at that
separating section. It is the length of the diameter times a unit
length of the cylindrical tube. The axial component is just the
cross sectional diameter of the tube that increases as the square
of the minor diameter. Equations for the longitudinal forces of
bias ply tires can be found in "the Bicycle Wheel".
Stress increases with tire cross section.
Jobst Brandt <jbr...@hplabs.hp.com>
bui tho xuan
>Bui Tho Xuan writes:
>> ....[about stress and tire cross-section]...
>> ....[a very very very dumb question (for an engineer)]...
>
>....[a clear and patient explanation]....
Urgg. For once, I wish Jobst would rake me over the coal for this
dumb question--stuff I used to know and learned as a 2nd year undergrad.
The frequency of my dumb questions are getting too high. Apologies to all.
tho
Jobst Brandt
> Urgg. For once, I wish Jobst would rake me over the coal for this
> dumb question--stuff I used to know and learned as a 2nd year undergrad.
> The frequency of my dumb questions are getting too high. Apologies to all.
Not to worry, it isn't obvious and you saved someone else the problem of
asking. I have answered this one before... a few times. It's just that
some will think you are a straight man for some of my ramblings, a foil
so to speak.
Jobst Brandt <jbr...@hplabs.hp.com>
Alexander Harth
:
: Continental now produces a tire that uses the same casing for 20mm and 28
: mm road racing tires, and is rated for the same pressure (120 psi / 8.5
: bar). Sure this means that the 20 mm can pumped up harder than 8.5, but
: for road use, 8.5 is all you may need. You won't gain much advantage
: beyond that, and for the same pressure, the 28-622 has less
: rolling resistance than the 20-622 (but maybe more air drag).
:
Could you tell us to which tires you refer, i only know 23-622 with 8.5 bar
( grand prix, ultra )
Alexander
Hans-Joachim Zierke
In reply to djco...@leland.Stanford.EDU (Daniel Connelly)
I did not know about Jobst's data, but it looks rather familiar. I think
that
> 8.74647 -0.223447
must be an error. The curve should flatten more and more.
It should also get rather steep at 2 bar, the typical "Berlin bicycle
commuters with fat tires don't care" pressure.
> Even if the "elbow" doesn't occur, high pressures still may not
> be desirable because of the associated rough ride, though.
The curve is specific to the tire. The data I thought of is rather old. For
tires like the IRC Paperlite, it should be at higher pressure. But tires
get "jumpy" at very high pressure. This is not an issue on the track, but
on normal roads, pressures above 8-9 bar might not be desirable.
> I agree narrow tires should be racing only, but they do seem to corner
> better.
This will stay true. But it also means that less pressure makes the tire
corner better.
> Shortcuts aren't an issue in road races, or shouldn't be :).
Not? That's sad. It would be my only chance to compete! :-)
hajo
--
Der Journalismus dient nur scheinbar dem Tage. In Wahrheit zerstört er die
geistige Empfänglichkeit der Nachwelt. (Karl Kraus)
Mike Fabian
> In reply to djco...@leland.Stanford.EDU (Daniel Connelly)
>
> > I agree narrow tires should be racing only, but they do seem to corner
> > better.
>
> This will stay true. But it also means that less pressure makes the tire
> corner better.
This seems indeed to be the case in my experience too.
But why do narrower tyres corner better?
And why does less pressure make the tire corner better ???
Mike
--
------------------------------------------------------------------------------
Mike Fabian fab...@apollo.ph1.uni-koeln.de
Universitaet zu Koeln
I. Physikalisches Institut
Zuelpicher Str. 77
50937 Koeln
Germany
------------------------------------------------------------------------------
Jobst Brandt
> But why do narrower tyres corner better?
I don't see any evidence to support that narrow tires corner better.
In fact I and my riding companions have found the opposite and it
seems reasonable as well.
> And why does less pressure make the tire corner better ???
If you ride a narrow tire, and have it highly inflated, road roughness
and irregularities in the surface are amplified and reduce effective
ground contact while a softer tire absorbs these features and presents
an averaged surface to the wheel. Lower pressure works down to the
point where loads off the plane of the wheel (as in deep cornering)
cause the tire to deform laterally. At this point the tire begins to
walk off the road sideways. A fatter tire at lower pressure will have
enough casing stress to retain shape AND absorb irregularities. For
this reason, I am certain fat tires corner better than thin ones.
They may not roll as well but that's not the subject here.
Jobst Brandt <jbr...@hplabs.hp.com>
Hans-Joachim Zierke
ha...@connery.rz.uni-frankfurt.de (Alexander Harth) wrote:
> Could you tell us to which tires you refer, i only know 23-622 with 8.5 bar
> ( grand prix, ultra )
I use the Conti Super Sport 100 in 28-622 for my racebike, rated for 8.5
bar.
I would prefer to use a Conti Grand Prix in 28-622, but they don't make it.
The Grand Prix gives better puncture protection, and the sidewalls are
manufactured to tighter tolerances.
And the white sidewalls of the Super Sport are ugly, at least to my taste.
hajo
Hans-Joachim Zierke
Mike Fabian wrote:
> > > I agree narrow tires should be racing only, but they do seem to
> > > corner better.
> >
> > This will stay true. But it also means that less pressure makes the
> > tire corner better.
>
> This seems indeed to be the case in my experience too.
>
> But why do narrower tyres corner better?
>
> And why does less pressure make the tire corner better ???
The contact patch is longer both for the narrower tire and for less
pressure. On a real-world road, there are small sandy spots and
imperfections, and a longer contact patch will give better cornering under
these real-world conditions.
But the difference is very small. Therefore I assume that most of the talk
about narrower tires cornering better is voodoo, and I suspect that the
talk would still be there if you added enough pressure to the narrower tire
to make the contact patch as long as the contact patch of a wide one.
Jim Papadopoulos
>The contact patch is longer both for the narrower tire and for less
So is vertical stiffness of tire on the ground.
Only wheel radius and pressure are important.
(I'm speaking of smooth treads, not knobbies.)
This is the result of a calculation, whose details I will explain
if anyone wants.
Jim
Joshua Moss
wrote:
> Mike Fabian writes:
>
ide a narrow tire, and have it highly inflated, road roughness
> and irregularities in the surface are amplified and reduce effective
> ground contact while a softer tire absorbs these features and presents
> an averaged surface to the wheel. Lower pressure works down to the
> point where loads off the plane of the wheel (as in deep cornering)
> cause the tire to deform laterally. At this point the tire begins to
> walk off the road sideways. A fatter tire at lower pressure will have
> enough casing stress to retain shape AND absorb irregularities.
> Jobst Brandt <jbr...@hplabs.hp.com>
Do you suppose that it makes any sense to overinflate the tires slightyly
then, and to let a bit of air out when you go offroad?
Hans-Joachim Zierke
Jim Papadopoulos wrote:
> Contact patch length is independent of tire narrowness.
For the same pressure?
> This is the result of a calculation, whose details I will explain
> if anyone wants.
Sure!
Jim Papadopoulos
>Jim Papadopoulos wrote:
>
>> Contact patch length is independent of tire narrowness.
>
>For the same pressure?
Bicycle tires are special among all tires --
skinny enough that the contact patch is several times as
long as the tire width; large enough that the contact
patch is somewhat shorter than the wheel radius; and pressured
enough that each tire cross section is deformed rather little.
In such a special case, it's legitimate to consider the
tire made of a sequence of non-interacting stations, each
a few mm long. (Like slices of bread.)
When a wheel is pressed to the road, each slice near the bottom
is compressed an amount depending on its position and wheel
displacement. Like a series of springs, we can add up
the forces of each slice, using calculus, to determine
the total force holding the wheel up for each amount
of vertical displacement.
The key point concernsthe 'spring stiffness' of each slice
as it is pressed towards the ground. Actually, the force arises
from contact area times pressure (I have verified this
experimentally), so the real question is how
the contact AREA of each slice depends on its proximity to
the ground..... MOre simply, how does the contact WIDTH vary
with vertical displacement??
It happens that contact width, as a function of displacement,
is initially a linear function, which eventually deviates
from a straight line.
That initial slope (of contact width versus vertical displacement)
is completely independent of the scale of the cross section,
i.e. 28 mm or 280 mm or 2800 mm.
This is rigorously true when the rim size is exactly scaled to
the tire size (I mean WIDTH in both cases); and it's approximately
true even if the rim width is fixed while the tire width varies.
The foregoing is an outline of the mathematical proof,
that the stiffness of each slice is independent of tire
fatness. But in that case, wheels with different thickness
tires all pumped to the same pressure
will all sink down towards the ground (due to a load) by the
same amount, and will all have the same length contact
patch.
I compared results from this model to CALSPAN
plots of cornering stiffness, and they agreed surprisingly well.
I will try to answer any questions or objections; if someone
really wants the full analytical derivation I will try to
find it. (But I may have to ask for some help to cover
photocopying and postage!)
Jim
Jobst Brandt
> Do you suppose that it makes any sense to overinflate the tires slightyly
> then, and to let a bit of air out when you go offroad?
Only if you enjoy snakebites.
Jobst Brandt <jbr...@hplabs.hp.com>
Jobst Brandt
> The foregoing is an outline of the mathematical proof, that the
> stiffness of each slice is independent of tire fatness. But in that
> case, wheels with different thickness tires all pumped to the same
> pressure will all sink down towards the ground (due to a load) by the
> same amount, and will all have the same length contact patch.
> I compared results from this model to CALSPAN plots of cornering
> stiffness, and they agreed surprisingly well.
But did you compare the results with real tires? It seems to me that
you may have chosen an indirect and unreliable way to assess this
phenomenon. Geometrically speaking we are looking at the compression
of a toroid against a plane. It is evident that the width of this
contact is larger for a fat tire than a slender one. Hence, it seems
to me to require the length of the contact patch to be different for
the two tires at the same inflation pressure and load.
> I will try to answer any questions or objections; if someone
> really wants the full analytical derivation I will try to
> find it. (But I may have to ask for some help to cover
> photocopying and postage!)
Just plot the results and put them in a GIF is that is what it will take.
Meanwhile I think I'll go to my ink pad and paper method and make some
impressions that can be measured. For instance, I can take a 2" wide MTB
smooth tire and make a 1" wide contact patch. I cannot do the same with
a 7/8" wide tire. This seems to contradict your findings.
Jobst Brandt <jbr...@hplabs.hp.com>
Lee Green MD MPH
which I signed on this newsgroup. My wife and I ride a tandem with 26x1.5
road tires. Can we realistically expect any difference in performance
(particularly speed) by reducing width to 26x1.25 tires? How much? Thanks
for any info -Lee
--
Lee Green MD MPH Disclaimer: Information for general interest
Family Practice and discussion only. I can't examine you via
University of Michigan the Internet, so you should ALWAYS consult your
gre...@umich.edu personal physician. These posts are my personal
doings, not a service of nor the responsibility
of the University of Michigan.
Jim Papadopoulos
>phenomenon. Geometrically speaking we are looking at the compression
>of a toroid against a plane. It is evident that the width of this
>contact is larger for a fat tire than a slender one. Hence, it seems
the contact patch by imagining the UNDEFORMED tire 'penetrates'
the ground slightly, and finding the intersection between ground
plane and tire. This doesn't work very well for a slender
bicycle tire, which can compress quite a bit with only a narrow
contact patch.
To get a better idea of what's happening, think of two hoses
(2-d tires) lying on the ground -- a thin one and a fat one.
If you squash each hose between two platens, or between a
rim section and the ground, you will find that the growth
of the contact width as a function of 'squash' is linear,
with the same constant of proportionality for either hose.
(This is a simple exact result for the two-platen case; it is
an excellent approximation for the rim/ground case, where there
is an eventual deviation from linearity which is different
for slender and fat.)
For the two-platen case, contact width is equal to
(pi/2)*(distance of squash), independent of initial radius.
So I don't agree that contact width is greater for a fatter tire.
>Meanwhile I think I'll go to my ink pad and paper method and make some
>impressions that can be measured. For instance, I can take a 2" wide MTB
>smooth tire and make a 1" wide contact patch. I cannot do the same with
>a 7/8" wide tire. This seems to contradict your findings.
I'm not in a position to secure any wheels to test. But this
is my claim:
If you inflate both tires the same (e.g. 80 psi)
and if you load them the same but not excessively (e.g. 60 lbf)
then they will have exactly the same contact patch length and width,
and they will sink down the same distance.
If you got a 1" wide contact patch, you must have had a very high
value of (load/pressure).
If it will heighten you interest in checking this, I am willing
to bet money on it.....!
Jim Papadopoulos
andrew cooke
Contact patch length is independent of tire narrowness.
So is vertical stiffness of tire on the ground.
Only wheel radius and pressure are important.
(I'm speaking of smooth treads, not knobbies.)
i assume that this is true only for tyres which are
wider than the contact patch you calculate?!
or does it mean that a tyre that thin would let the rim
touch the ground?
ok, i'll bite!
andrew
QRman
wheels size carries connotations that border on the religious. Could be
fun. Anybody care to tackle the question of 700c vs 26"? (road, I'm
talking about).
QRman
Jim Papadopoulos
>fun. Anybody care to tackle the question of 700c vs 26"? (road, I'm
>talking about).
depends on how the rider's body reacts to bumps.
Rolling resistance on a perfectly smooth plane or
roller is one thing, measurable in several different ways.
But as soon as bumps are involved, you get concerns
about what is pressing the wheel to the ground -- a spring?
A spring plus shock? A mass? A flexible, viscous mass (person)?
So I don't put a lot of stock in rolling resistance measurements,
an d I cerainly wouldn't trust a calculation !
Best I think is coasting tests side-by side, switching wheels
and pressures and riders, to see if there are ANY reliable
repeatable results.
IMHO........ JP
Jobst Brandt
> Jobst Brandt writes:
>
>> phenomenon. Geometrically speaking we are looking at the compression
>> of a toroid against a plane. It is evident that the width of this
>> contact is larger for a fat tire than a slender one. Hence, it seems
>
> It is a common mistake among tire researchers to approximate
> the contact patch by imagining the UNDEFORMED tire 'penetrates'
> the ground slightly, and finding the intersection between ground
> plane and tire. This doesn't work very well for a slender
> bicycle tire, which can compress quite a bit with only a narrow
> contact patch.
I made no reference to such a geometric intersection. My contention
is that a 2" diameter cross section tire will have a larger contact
patch than a 1" tire under the same vertical displacement of the rim.
Taken to the limit with the tire completely flattened the width is
exactly 2:1, the 2" tire being pi inches wide and the other pi/2.
> I have used the ink pad method before, and trust it. Unfortunately
> I'm not in a position to secure any wheels to test. But this is my
> claim:
> If you inflate both tires the same (e.g. 80 psi) and if you load
> them the same but not excessively (e.g. 60 lbf) then they will have
> exactly the same contact patch length and width, and they will sink
> down the same distance.
Somehow this does not seem reasonable. As I said, taking things to
extremes often gives insights to such problems. If I use a large tire
and a very small one that have, for instance, a 2 inch and a 1/2 inch
cross section the deflection of the small tire is greater than the
large one because it must get enough contact to support the load at
that pressure and to get the same area it seems that it must deform
more.
> If you got a 1" wide contact patch, you must have had a very high
> value of (load/pressure).
>
> If it will heighten you interest in checking this, I am willing
> to bet money on it.....!
I'll have to get with the ink pad and investigate. I realize that the
deformations are complex and work in non obvious ways. I am only
aware that wet tire tracks on dry pavement are variably wide and that
wide tires are not usually inflated to the same pressure that thin
ones are. Because the condition of equal pressure is not the norm
for large and small tires, I have no experience with this condition.
More later.
Jobst Brandt <jbr...@hplabs.hp.com>
andrew cooke
I'm pretty sure you shouldn't be betting money on that,
but the change may be so small you don't notice...
First, I'll separate the argument into two parts:
i. The area of the contact patch is equal (with the correct
units) to the ratio of bicycle weight to tyre pressure.
ii. The geometry of the patch depends upon the geometry of the
surface of the tyre, to a first approximation only the local
geometry is important.
Now I'll consider two simple cases:
i. The long tubes in your example.
Geometry: a straight line.
Area: for an infinite tube with a finite weight the
width would be zero. For a finite tube (or one with
a specified weight per unit length) the width will be
proportional to load.
ii. A sphere.
Geometry: a circle.
Area: again, proportional to load. So the radius of
the contact patch is *independent* of the radius of the
ball.
Both these cases support the idea that the radius of the tyre
is unimportant. Unfortunately that is because they are limiting
cases of the geometry.
Now consider an `oval' shape, like a (american or rugby)
football. The contact patch will be an oval. I'm not going
to prove it - I'm not sure I can - but it seems obvious.
Note that the shape of the patch will not change if the ball
is scaled in all directions. The geometry is constant and
the are of the patch depends only on pressure and load.
Finally conside a bicycle tyre. The region of the tyre
local to the road can be described as a surface with two
different radii of curvature, in orthogonal directions.
One is the radius of the tube and one is the radius of the
wheel. r and R, for simplicity. If one takes a leap of faith
based on maths being generally well behaved it seems likely that
the oval contact patch is going to have axes in the ratio of
R to r. Is this reasonable?
Yes. For the tubes, r was about a centimetre and R was infinite
(they were flat), so the ratio is 0:1, or a straight line.
For the sphere the ratio was 1:1, a circle, as expected.
So it seems very reasonable that the contact shape has axis in the
ratio of the two radii describing the local surface. Maybe it
is not the ratio, but the ratio to some power. This is possible
and makes patches either behave more like circles or more
like lines, but does not remove the effect altogether. Moreover,
it is clear that the effect is not strong if one imagines a football.
OK, so finally, what would we expect for two tyres on a 70cm
diameter wheel?
A tyre of radius 0.7cm will have a contact patch of length:
width ratio 50:1
A tyre of radius 1.4cm will have the ratio 25:1. For the same area
as the first, write this as 35:1.4 and it is clear (I hope)
that the fatter tyre, for the same pressure and load, has a
wider contact area by a significant amount.
I hope this is clear. I even used capitals where necessary
so that people wouldn't be influenced by aesthetics ;-)
Andrew
Phillip N Price
this discussion has led.... the question is, given thin and fat
tires AT THE SAME PRESSURE, how will the contact patches vary?
We have on person (Papadopolous) who claims that the contact
patches will be identical, unless the downward forces are so
great that deformations are nonlinear.
We have another person (Brandt) who thinks the larger tire will
always have a larger contact patch.
Then, there's me. I think the contact patches will be different
in shape but will have about the same AREA. Why? Well, consider
just one tire for the moment. There's a downward force, from the
weight of the bike and rider and rim etc. This is counteracted
by an upward force from the contact patch with the ground. These
have to be exactly equal---the bike neither accelerates downward
nor upward. What is this force? Well, it's the integral of
(upward force on tire from ground) over the area of the tire.
This upward force is not equal over the entire contact patch---it's
nearly zero at the very edge of the patch, and it's equal to the
tire pressure at the middle of the patch---but over most of the
patch it's VERY CLOSE to the total tire pressure, if the pressure
is high enough (for lower pressures, when the contact patch gets
really big and the tire cross-section is a long way from circular,
the approximation is not as good). To first order, this force is just
(Area of patch) * (tire pressure). Width of tire, and indeed
shape of contact patch, don't come into it until we go to the next
order in the analysis. So my prediction is, the contact patches
will have nearly the same area, but will be different in shape:
At high pressure, the wide tire will have a wide-but-short contact
patch, while the narrow tire will have a narrow-but-longer one.
Don't know how to test this, though: All the wide tires I've seen
have substantial tread. What we really want is a comparison between
wide and narrow slicks.
--Phil Price
Jim Papadopoulos
>I made no reference to such a geometric intersection. My contention
>is that a 2" diameter cross section tire will have a larger contact
>patch than a 1" tire under the same vertical displacement of the rim.
If the wheel radii are the same, and they are displaced vertically
the same, and the treads are 'thin' (not built up like a slick
or covered with knobs), then the contact patches will be identical.
Actually, this is an approximation which is most exact for
small amounts of vertical displacement, and
shows increasing deviations as displacements increase.
However I have confidence in it for normal wheel loads
and normal road tire pressures.
>Taken to the limit with the tire completely flattened the width is
>exactly 2:1, the 2" tire being pi inches wide and the other pi/2.
totally breaks down.
>More later.
Good....
Jim Papadopoulos
Jim Papadopoulos
>We have on person (Papadopolous) who claims that the contact
>patches will be identical, unless the downward forces are so
>great that deformations are nonlinear.
a great fraction of tire width that patch width becomes nonlinear.
It's not more 'right', but removes issues of force and pressure.
>This upward force is not equal over the entire contact patch---it's
>nearly zero at the very edge of the patch, and it's equal to the
>tire pressure at the middle of the patch---but over most of the
>patch it's VERY CLOSE to the total tire pressure, if the pressure
casings a
and no built-up tread which might 'touch down' even with zero load.
>order in the analysis. So my prediction is, the contact patches
>will have nearly the same area, but will be different in shape:
>At high pressure, the wide tire will have a wide-but-short contact
>patch, while the narrow tire will have a narrow-but-longer one.
toward the ground, the contact patch lengths will be the same.
Why? Assuming successive sections of tire are relatively decoupled,
any tendency to intersect the ground results in contact.
So you can calculate the length of arc over which contact occurs.
The second part of the argument depends on the linearisation
of the vertical displacement/contact width relation for a tire
cross section. If such a linearisation exists (separate argument)
it must be scale independent.
(Rigorously, the linearisation will be different if rim and tire
do not scale up equally. But I think they do approximately,
and anyway differences in the linearisation are modest.
...over to you!
Jim
Jim Papadopoulos
>If the wheel radii are the same, and they are displaced vertically
>the same, and the treads are 'thin' (not built up like a slick
>or covered with knobs), then the contact patches will be identical.
'built up like a slick' should have read 'built up to a non-circular
cross section'. My mind was on the old Orange Krate / Apple Peeler
'racing slick' tires which I recall as almost flat, not
modern smooth tires which I guess might also be called slicks.
By the way, I never rode, nor hankered for, a Stingray.
(Wouldn't want anyone to get the wrong impression about me!)
JP
Jim Papadopoulos
>jim wrote:
>Contact patch length is independent of tire narrowness.
>So is vertical stiffness of tire on the ground.
>Only wheel radius and pressure are important.
>(I'm speaking of smooth treads, not knobbies.)
>
> i assume that this is true only for tyres which are
> wider than the contact patch you calculate?!
which brings the rim towards the ground a distance up to
10%? 20%? 30%? of the tire width, contact patch shape
is essentially independent of tire width.
Deviation from this rule occurs at displacements which
are substantial fractions of tire cross section.....
Hope this is clearer??
JP
Jobst Brandt
> The only flaw in your method is the large weight you used. About
> half that weight would conform more to what I had in mind (and be
> closer to actual front wheel loading). I am guessing from the
> lengthXwidth numbers that you applied your weight gently, and didn't
> let the wheel tilt or roll....
It was done carefully with both brakes on and an assistant holdiing the
bicycle.
Jobst Brandt <jbr...@hplabs.hp.com>
Phillip N Price
|> In article <xGzxPx6.b...@delphi.com>,
|> Jim Papadopoulos <bicyc...@delphi.com> wrote:
|> >Here's an argument: for wheels of equal size, displaced equally
|> >toward the ground, the contact patch lengths will be the same.
|> >Why? Assuming successive sections of tire are relatively decoupled,
|> >any tendency to intersect the ground results in contact.
|> >So you can calculate the length of arc over which contact occurs.
|>
|> contact width will be the same since the area is constant for
|> constant pressure and load.
|>
|> your assumption is not valid, and without it, assuming that the
|> tyre is locally an ovoid, you get the results i posted earlier.
|>
|> andrew
We seem to have reached the point of diminishing returns in this thread.
I agree with Andrew, the assumption that "succesive sections of tire
are relatively decoupled" doesn't make any sense. Consider a tire
(not in contact with the road), inflated to some pressure. The forces
on some part of the wall are: outward force due to the pressure,
inward forces due to pull from the adjacent portions of the tire.
If the "successive sections of tire are ...decoupled", there's
nothing left to hold the tire circular! Well, I guess it depends
on what you mean by "relatively" decoupled---the forces on a spot,
due to the portion of tire that's on a given line through the spot,
are proportional to the curvature of the tire along the line (assuming
the tire material is isotropic). The curvature ACROSS the tire IS
much higher than the curvature ALONG the tire: for a 2-inch tire,
the ratio of radii is about 1/13, and for a 1-inch tire it's about
1/26. To the extent that both of these numbers are "small", you're
right that the sections are decoupled...that's equivalent to
approximating a tire as an infinitely long cylinder, like a garden
hose. But that's also one level of approximation too far, you've
thrown out the term that contains the first-order variation with
respect to radius!
The fact is, the 1-inch tire is TWICE as tightly curved across the
tire than is the 2-inch tire, but they both have the SAME curvature
ALONG the tire. [APPROXIMATION ALERT...or something: The above would
seem to predict that the contact patch would be 13 times as long as
it is wide, or something like that. I AM NOT CLAIMING THAT! That
would be true, I think, IF the tire were a big elliptical "balloon",
with no rim: just a pressurised hollow rubber thing that's much more
curved in one direction than another. I don't know what the result
is for a torus, but it's probably different, so I don't have a
quantitative prediction. But it sure as hell ain't the same as
an infinite garden-hose type cylinder! As Einstein said, "In making
our models, we must simplify nature as far as possible. But no
farther."]
Anyway, I'm sticking with my original prediction: both tires will
have (nearly) the same AREA of footprint, but the thinner tire will
have a thin, long footprint compared to the wide but short footprint
of the fat tire (assuming same tire pressures, of course).
What we need, of course, is experimental data!
--Phil Price
Phillip N Price
|>
|> In article <hCwWouH.b...@delphi.com>, Jim Papadopoulos <bicyc...@delphi.com> writes:
|> |> Jobst Brandt <jbr...@hpl.hp.com> writes:
|> |>
|> |> >clean white sheet of paper putting my whole body weight on the wheel
|> |> >to make a contact patch print. The print is 120 mm long and 19 mm
|> |> >wide. Then I found a bicycle with a 7/8" smooth front tire and
|> |> >inflated it to 60 psi, inked and repeated the process. This time the
|> |> >contact patch was 136 mm long and 16 mm wide.
|> |>
|> |> Thank you for making the effort. The results
|> |> present me with a challenge: are my calculations
|> |> somehow wrong, or were the test conditions wrong?
|> |>
|> |> [snip]
|> |> JP
|>
|> Jim I don't think your calculations are wrong, your are just not looking at
|> things in the right perspective. If I understand what you calculated, your
|> calculations are based on equal DISPLACEMENT of the wheel. Mathematicaly, if
|> any two wheels of the same radius are displaced 5mm downward at the axle then
|> yes, all contact patches have the same length. The catch is, it will take
|> different amounts of force to displace wheels of different crossection
|> because the support from the tire is proportional to the contact patch area.
|> Since the wider tire will have a wider contact patch for the same displacement
|> as the smaller tire, the patch will have a greater area and the load force for the displacement will be greater. Since Jobst is displacing the wheel with the
|> same FORCE (his body weight) this will produce a smaller displacement of the
|> wheel for the larger crossection and hence a shorter contact patch.
[snip] Rick Radko
Rick, thanks for pointing out a flaw (the flaw?) in Jim's calculations.
I believe you have, indeed, discovered a problem with his model.
The rest of us should all feel embarassed about not noticing this
point, nice job.
Jobst, thanks for doing the experiment. I'd like to point out that
both patches have almost exactly the same area, within better than
1% according to Jobst's figures (wouldn't take the exactness too
seriously, though: was the contact patch really 16mm wide, or 16.1?)
Anyway, I feel completely vindicated: different shapes, same area.
BWAH ha ha ha ha. Just kidding.
This has been a kinda fun discussion.
--Phil
Jobst Brandt
> Anyway, I'm sticking with my original prediction: both tires will
> have (nearly) the same AREA of footprint, but the thinner tire will
> have a thin, long footprint compared to the wide but short footprint
> of the fat tire (assuming same tire pressures, of course).
That's a pretty safe thing to do in light of my posting yesterday that
I'll repeat here, lest it never got to your site.
------------------------------------------------------------------------
Well... I went to the bike shop and found a bicycle with 1.75" slicks
with the front tire still smooth and round but without mold flash down
the middle. It was inflated to 60 psi, inked and pressed it on a
clean white sheet of paper putting my whole body weight on the wheel
to make a contact patch print. The print is 120 mm long and 19 mm
wide. Then I found a bicycle with a 7/8" smooth front tire and
inflated it to 60 psi, inked and repeated the process. This time the
contact patch was 136 mm long and 16 mm wide.
Fat Thin
120 x 18 136 x 16
------------------------------------------------------------------------
> What we need, of course, is experimental data!
Is that what you had in mind?
Jobst Brandt <jbr...@hplabs.hp.com>
John Frampton
: Jobst Brandt <jbr...@hpl.hp.com> writes:
:
: >clean white sheet of paper putting my whole body weight on the wheel
: >to make a contact patch print. The print is 120 mm long and 19 mm
: >wide. Then I found a bicycle with a 7/8" smooth front tire and
: >inflated it to 60 psi, inked and repeated the process. This time the
: >contact patch was 136 mm long and 16 mm wide.
: Thank you for making the effort. The results
: present me with a challenge: are my calculations
: somehow wrong, or were the test conditions wrong?
Jim - shouldn't Jobst's data be taken as confirmation of your proposal?
They seem to show that tire patch shape is only weakly dependent on tire
size at the weight he used, with the particular tires he used. Are you
claiming more than that?
JF
QRman
Let us assume the same tire width, pressure, etc., but 571 bead diameter
(road 26") vs. 700c. Do I assume that you still think the footprint would
be the same? Secondly, does this represent the sole parameter regarding
rolling resistence, or are there other factors?
One question I'd like to ask is this. Is the inside angle created between
the ground, and the tangent of that theoretical point of the circle just
in front of the contact patch, more obtuse for a 26" wheel than for a
700c? If so, would that make a difference to any of you regarding rolling
resistence?
By the way, when I say "circle" I am talking about the wheel as if it was
a circle, which it clearly isn't because of the existence of the contact
patch. I don't think that screws up my model too much.
I'd be obliged if you guys'd take a crack at this.
QRman
Jim Papadopoulos
>Jim - shouldn't Jobst's data be taken as confirmation of your proposal?
>They seem to show that tire patch shape is only weakly dependent on tire
>size at the weight he used, with the particular tires he used. Are you
>claiming more than that?
I had expected EXACT correspondence at low (but still
reasonable) loads. To the extent that there's (minor) disagreement,
I would love to know what in my approach is slightly wrong.
But yes, thanks for pointing it out, the results show that
I am roughly right at least!
JP
John Frampton
:
: >Jim - shouldn't Jobst's data be taken as confirmation of your proposal?
: >size at the weight he used, with the particular tires he used. Are you
: >claiming more than that?
:
: Well, yes, you're right. I'm just slightly troubled because
: I had expected EXACT correspondence at low (but still
: reasonable) loads. To the extent that there's (minor) disagreement,
: I would love to know what in my approach is slightly wrong.
One easy experiment would be to repeat what Jobst did, then do it over
again applying applying less downward force. If the results were in
closer agreement with your proposal, it would show that the basic idea
is correct, but that non-linear effects (whose identity is still to be
determined) played a role at smaller deformations than you expected. If
the results are not in better agreement, it would seem to indicate a
more serious problem with your proposal.
JF
andrew cooke
Fat Thin
120 x 18 136 x 16
actually, this is rather in support of jim. the geometric argument
would change the thickness by much more for a doubling in tyre
diameter. his approximation seem to be better than my simple-minded
geometry.
btw it's nice to see that the area of these patches are almost equal.
andrew
Jim Papadopoulos
>In fact it should be easy to convince yourself that the ratio of width to
>length is actually the square root of the ratio of the two radii
>(approximately).
>A few simple calculations show that for a 27"x0.5" (radii) tyre the patch
>dimensions will be 4.6"x0.6" to carry a load of about 200lb at 90psi, with a
>vertical deflection of 0.1" (I'm assuming that the patch is oval and that the
>area of an oval is pi.a.b where a, b are the two radii). For a 27"x0.25" tyre,
>the same area comes from a patch of 6.0x0.47" with a deflection of 0.16". The
>patch is a bit longer and thinner, and the deflection is much greater.
Could you please clarify -- is your calculation based on the
idea of 'intersecting' or 'slicing' the undeformed
tire (an approach I believe to be quite wrong), or
on deforming the tire (and if so by what relationship?)?
Thanks -- JP
Phillip N Price
Now, wait just a second here. Jim, you were quite clear in your
earlier stuff that you could PROVE that the contact patches would
be identical, you were even willing to put money on it (wish I'd
made that bet)...then Jobst does the experiment, uses two different
widths of tires (radii differ by a factor of two), and finds that
the contact patch is TWENTY PERCENT WIDER and TWELVE PERCENT SHORTER
for the wide tire than for the narrow tire, and you're going to
claim that you are "roughly right"??!!
Other people (like me, for instance) claimed that although the area
of the contact patch would be the same, the wider tire would have
a wider patch. You disagreed---contact patches would be identical,
you said.
Now, I am very glad you started this discussion, which has been
interesting and informative, but I am not going to let you claim
that the experiment supports your original theory!
--Phil Price
Phillip N Price
description of a bike wheel contact patch:
We are, by now, hopefully agreed that the area of the contact
patch must be such that Pressure*Area = Weight (=Mass*g).
So the question is, what shape of the patch gives us the
area that we need?
Suppose we had a bike tire in which the membrane had no
stiffness---we make it out of thin rubber like a balloon or
a condom. Also suppose that we keep the applied weight low
enough that the vertical displacement is much less than the
radius of the tire (not to be confused with the radius of the
wheel). If the tire is pressed against a flat surface, and if
the displacement is small compared to the radius of the tire,
then only the section very near the contact patch is affected.
The area (and shape) of the patch can be calculated from the
intersection of a torus (the bike tire) and a plane (the ground).
[Jim P, you seem to disagree with this, although I don't see why:
surely you must agree that it's true if the displacement is
small enough. Or are you worried about the stiffness of the
rubber?]
It's actually possible to calculate the area of such an
intersection, although the solution to the integral involves
arcsines of messy functions. Much easier is to assume that
(1) the contact patch is elliptical, or (2) the contact
patch is rectangular. Neither of these is actually correct,
so I'll (arbitrarily) choose something about midway in
between. An elliptical patch has an area of pi*a*b, where a
and b are the major and minor radii; a rectangular patch would
have 4*a*b, where a and b are the half-lengths of the sides.
Some experimentation has convinced me that elliptical is a
better approximation for extremely small displacements, but
for larger ones (where the approximations aren't as good, by
the way) the rectangular patch does better.
With the above assumptions, we pretty easily find that:
1. the vertical displacement of the tire (the amount that the
hub goes down with weight) is given by
E = Mg/(2*Pi*P*sqrt(R*r)), where
M = mass of load
g = gravitational acceleration (about 10 m/s^2 in metric)
P = pressure in tire
R = Radius of WHEEL (from center to outer edge of tire)
r = radius of TIRE
2. The half-width of the footprint, at its widest, is
a = sqrt(Mg/(Pi*P)) * (r/R)^(0.25)
3. The half-length of the footprint, at its longest, is
b = a * sqrt(R/r)
Note that, in (2), the dependence of footprint width on tire
width (all other things constant) is a FOURTH root... that's
just the way it works out.
For larger displacements, replace the Pi everywhere with 4 (or
with some number between Pi and 4).
[DISCLAIMER: don't forget the assumptions that went into this.
The key one is that the displacement is small, so we can calculate
the contact patch by looking at the intersection of a torus with
a plane. That's only true if E << r ]
There are NO free parameters in this (unless you count the
'fudge factor' , which is constrained to be between
Pi and 4 (realistically), but could be anywhere in there (and
which varies with E).
I don't know Jobst's mass, nor how accurate the tire pressures
were; let's suppose a load of 75Kg (165 lbs). At 60 psi (about
414000 Newtons/m^2), that predicts a displacement of 1cm for
the fat tire, and about 0.5 cm (5mm) for the skinnier tire.
Neither of these meets the requirement of E << r, but I'll
proceed anyway.
With M = 75 Kg, g=10 m/s^2, P = 414000 N/m^2, we find that
sqrt(Mg/(Pi*P)) = 0.024 m.
So for the fat tire (r=2.2 cm, R=37cm) we get a half-width
a=0.012 m, or a full width of 0.024 m = 24mm. This compares
with Jobst's experimental result of 19mm.
For the thin tire (r=1.1 cm, R=36cm) we get a half-width
a=0.010 m, or a full width of 0.020 m = 20mm (compared to
Jobst's experimental result of 16mm).
It's hard to say whether these results are good or only fair.
Considering I didn't 'cheat' at all, I think they're pretty good.
There is a real discrepancy, but we're WAY outside the
low-displacement regime where I've got confidence in the
exact predictions.
HOWEVER, let me point out that this model predicts a
footprint width/tire width relationship that goes like
(r/R)^(0.25). That means that given a fat tire that's twice
as wide as a skinny one, we expect the footprint widths to
be in the ratio of (2^.25):1, or 1.189:1
Jobst's experiment found a ratio of 19:16, or 1.187:1
Basically, I think my model is a pretty good one. In fact,
I think it is CORRECT in the LIMIT of low loads and perfectly
flexible (non-stiff) tires.
Until someone convinces me otherwise, I'll believe in the
(r/R)^(.25) rule.
--Phil Price
First tire:
half-width of footprint = a = 0.023 * (2.2/37)^(.25) meters
= 0.011 m = 11 mm
width of footprint = 2a = 22 mm
length of footprint = 2b = sqrt(R/r)*(2a) = 90 mm
Second tire:
half-width of footprint = a = 0.024*(1.1/37)^(.25) meters
= 0.010 m = 10 mm
width of footprint = 2a = 20 mm
length of footprint = 2b = sqrt(R/r)*(2a) = 116 mm
James Annan
> [Jim P. argues that tyre patch width is independent of tyre size]
>
> I'm pretty sure you shouldn't be betting money on that,
> but the change may be so small you don't notice...
>
> First, I'll separate the argument into two parts:
>
> i. The area of the contact patch is equal (with the correct
> units) to the ratio of bicycle weight to tyre pressure.
>
> ii. The geometry of the patch depends upon the geometry of the
> surface of the tyre, to a first approximation only the local
> geometry is important.
>
>
>
> Finally conside a bicycle tyre. The region of the tyre
> local to the road can be described as a surface with two
> different radii of curvature, in orthogonal directions.
>
> One is the radius of the tube and one is the radius of the
> wheel. r and R, for simplicity. If one takes a leap of faith
> based on maths being generally well behaved it seems likely that
> the oval contact patch is going to have axes in the ratio of
> R to r. Is this reasonable?
>
> Yes. For the tubes, r was about a centimetre and R was infinite
> (they were flat), so the ratio is 0:1, or a straight line.
> For the sphere the ratio was 1:1, a circle, as expected.
>
> So it seems very reasonable that the contact shape has axis in the
> ratio of the two radii describing the local surface. Maybe it
> is not the ratio, but the ratio to some power. This is possible
> and makes patches either behave more like circles or more
> like lines, but does not remove the effect altogether. Moreover,
> it is clear that the effect is not strong if one imagines a football.
>
In fact it should be easy to convince yourself that the ratio of width to
length is actually the square root of the ratio of the two radii
(approximately).
>
> OK, so finally, what would we expect for two tyres on a 70cm
> diameter wheel?
>
> A tyre of radius 0.7cm will have a contact patch of length:
> width ratio 50:1
>
> A tyre of radius 1.4cm will have the ratio 25:1. For the same area
> as the first, write this as 35:1.4 and it is clear (I hope)
> that the fatter tyre, for the same pressure and load, has a
> wider contact area by a significant amount.
>
> I hope this is clear. I even used capitals where necessary
> so that people wouldn't be influenced by aesthetics ;-)
>
> Andrew
>
>
A few simple calculations show that for a 27"x0.5" (radii) tyre the patch
dimensions will be 4.6"x0.6" to carry a load of about 200lb at 90psi, with a
vertical deflection of 0.1" (I'm assuming that the patch is oval and that the
area of an oval is pi.a.b where a, b are the two radii). For a 27"x0.25" tyre,
the same area comes from a patch of 6.0x0.47" with a deflection of 0.16". The
patch is a bit longer and thinner, and the deflection is much greater.
Whether or not this means anything in the real world or not, is open to
argument.
--
James Annan.
Disclaimer: please feel free to correct the deliberate mistake(s) in the above,
I did it in a hurry and I'm supposed to be working.
John Frampton
: In article <JG9Xo2X.b...@delphi.com>, Jim Papadopoulos <bicyc...@delphi.com> writes:
: |> John Frampton <jfra...@lynx.dac.neu.edu> writes:
: |>
: |> >Jim - shouldn't Jobst's data be taken as confirmation of your proposal?
: |> >They seem to show that tire patch shape is only weakly dependent on tire
: |> >size at the weight he used, with the particular tires he used. Are you
: |> >claiming more than that?
: |>
: |> Well, yes, you're right. I'm just slightly troubled because
: |> I had expected EXACT correspondence at low (but still
: |> reasonable) loads. To the extent that there's (minor) disagreement,
: |> I would love to know what in my approach is slightly wrong.
: |>
: |> But yes, thanks for pointing it out, the results show that
: |> I am roughly right at least!
: |> JP
: Now, wait just a second here. Jim, you were quite clear in your
: earlier stuff that you could PROVE that the contact patches would
: be identical, you were even willing to put money on it (wish I'd
: made that bet)...then Jobst does the experiment, uses two different
: widths of tires (radii differ by a factor of two), and finds that
: the contact patch is TWENTY PERCENT WIDER and TWELVE PERCENT SHORTER
: for the wide tire than for the narrow tire, and you're going to
: claim that you are "roughly right"??!!
: Other people (like me, for instance) claimed that although the area
: of the contact patch would be the same, the wider tire would have
: a wider patch. You disagreed---contact patches would be identical,
: you said.
It seems to me that there was a debate about the effect of tire width on
patch shape - one side said that the patch length and width would stay the
same if the tire width was halved, the other side said that the ratio of
the patch length to patch width would double if the tire width was
halved. The experiment showed the ratio to increase only from 6.3 to 7.9,
rather than to 12.6 as the doublers would have predicted. Among the
theories proposed, JPs is by far the most successful in predicting the
actual results.
It is not accurate to characterize one side of the debate as claiming that
the patch would elongate "a little bit" if the tire width was halved.
JF
John Frampton
: description of a bike wheel contact patch:
<see original article for the model>
<see original article for Brandt's experimental results>
: It's hard to say whether these results are good or only fair.
: Considering I didn't 'cheat' at all, I think they're pretty good.
: There is a real discrepancy, but we're WAY outside the
: low-displacement regime where I've got confidence in the
: exact predictions.
: HOWEVER, let me point out that this model predicts a
: footprint width/tire width relationship that goes like
: (r/R)^(0.25). That means that given a fat tire that's twice
: as wide as a skinny one, we expect the footprint widths to
: be in the ratio of (2^.25):1, or 1.189:1
: Jobst's experiment found a ratio of 19:16, or 1.187:1
: Basically, I think my model is a pretty good one. In fact,
: I think it is CORRECT in the LIMIT of low loads and perfectly
: flexible (non-stiff) tires.
If we consider the length measurements as well, it doesn't look quite as
good. Your model predicts that the aspect ratio of the footprint (the
ratio of its length to its width) should vary inversely as the square
root of the tire width (for a fixed wheel). Halving the tire width
should result in the aspect ratio increasing by a factor of the square
root of 2 (about 1.41). Brandt's data (if I did the calculation
correctly) gives an increase in the aspect ratio of 1.26.
JF
Jim Papadopoulos
(You may want to read and respond offline.)
TIRE DEFORMATION, AND ROLLING RESISTANCE
Thanks to everyone for their pointed questions, calculations,
and experiments. I've been inspired to reconstruct my old
tire calculations (there's litle chance of FINDING them
over the next few days), and even take them further in
response to the input.
Hopefully, I haven't dropped any factors of 2.....
I'm posting a summary (for the two or three still
following this thread), but first a couple of caveats:
1. The whole calculation is for a wheel pressed against a
smooth plane. This problem is interesting in its way, BUT
I don't think it's all that relevant to shock absorption
while riding -- most bumps are not simply changes in level
(from one plane slab to a higher or lower one) but rather
ridges and pebbles. So the real property of interest is a
tire's ability to 'swallow' a small protrusion without
(additional) vertical or horizontal force.
2. Some of the results are expressed in terms of contact
patch width and length. I expect that the REAL contact patch
will always be a little shorter and less pointed than the
mathematical result, because tire squash near the end of
the patch will actually distort the tire a little way
BEYOND the patch's end. If this is indeed the case, the
"length" in the formulas should probably be the calculated
length, not the measured one (assuming I've provided the
correct formula).
FORMULAS FOR VERTICAL STIFFNESS OF A BICYCLE TIRE
PRESSED AGAINST A FLAT PLANE (provisional until math checked)
d0 is the amount the wheel is pressed toward the ground
('penetration' if undeformed)
D is wheel diameter (including inflated tire)
p is tire pressure, the minute volume changes due to
contact don't affect it
K is a factor of proportionality between contact width w
of a tire section, and the amount d it is squashed.
(w0=K*d0 for the bottom point of the wheel, but also
for any other tire section, which has its own value
of squash, d.) The value of K will be discussed below.
L is the total length of the contact patch
From d0, we can immediately calculate maximum contact width,
as just stated w0=K*d0. (This linear relation overestimates
w0 if d0 is substantial, correcting it is possible but messy.)
Also d0 lets us calculate the contact patch length,
L=2*sqrt(D*d0)
Quite accurately except at the patch ends, the 'penetration'
d all along the contact patch, and hence the width of the
contact patch, follows a parabolic law d=d0-(x^2)/D where
x is position along patch, measured from its midpoint.
The patch area then works out as A= (2/3)*L*w0, the
factor 2/3 comes from integrating area enclosed by a parabola.
Multiplying area times pressure, the support force F is then
F=p*(d0^1.5)*(D^0.5)*(4/3)*K
In many cases it is not so easy to work with d0, so we
use this relationship to find d0 in terms of F, and
substitute that back to determine w0 and L (etc.)
L=((6*F*D)/(p*K))^0.333
w0=(((3*F)/(4*p))^0.667)*((K/D)^0.333)
What is K?? It is defined as the derivative of
contact width change, with respect to rim height change,
for a cross-sectional slice of tire. It turns out
that K can be written in terms of the angle THETA
(in radians) subtended by one sidewall, which follows
a circular arc from ground to rim.
As the tire section just touches down, THETA is just the
angle for half the tire (something like 2.36 radians, or
135 degrees) As the tire is squashed more and more,
THETA can grow to 180 degrees (when contact width equals
rim width) or beyond.
K is different for each value of THETA, but not too
different if THETA doesn't vary over too wide a range.
Then it becomes appropriate to think in terms of a
constant 'average' value of K for that tire.
The formula for K (remember to use radians):
sin(THETA) - THETA*cos(THETA)
K = ------------------------------------
1 - cos(THETA) - (THETA/2)*sin(THETA)
If you want to calculate THETA for the undeformed tire
(instead of somehow measuring it), you need to solve the
equation: (tire)*sin(THETA) = (rim)*THETA
Here, (rim) is the straight line distance from rim edge
to rim edge where the tire last touches it; and (tire)
is the perimeter length of a cross section of tire,
likewise from rim edge to rim edge. (Part of the
circumference of a circle, for the inflated but
undeformed tire.)
The above results offer a host of targets for those
who want to test their accuracy. I will be glad to learn
of any errors in or confirmation of the algebra or the
model... and also naturally of any experiments which
precisely agree, roughly agree, or totally disagree
with any of the formulas. I'm confident that these
formulas, or some quite like them, will accurately
describe smooth-tread tire compression!
Two closing points:
One is a clarification of something I posted initially,
which I now understand better:
K is unchanged if you switch to a twice as fat TIRE on a
twice as wide RIM (because THETA is unchanged). BUT if
you simply switch to a fatter tire with the SAME rim,
you get a bigger THETA. Likewise, if you switch to a
thinner rim, with the SAME tire. It looks like K
decreases about one percent for each degree increase
in THETA (in the region of 135 degrees that is) ...
This kind of tire geometry change (increasing THETA,
hence decreasing K) apparently makes for a softer tire,
with a skinnier, longer contact patch.
For example, approximately speaking, let's say that K
decreased 21%. Then contact length would increase about 7%,
width would decrease about 7%,and wheel deflection under
the same load would increase about 14%.
These implications caught me by surprise, and I hope
there's nothing wrong with them!
Second point (on ROLLING RESISTANCE):
***WARNING, LOTS OF SPECULATION AHEAD***
A potential benefit of the ability
to calculate changes in tire shape in the deformed
region, is the possibility of calculating energy losses
and hence rolling resistance. For FLAT surfaces only, this
MIGHT work. (Resistance to forwards motion, due to
bumpy surfaces, is CALLED rolling resistance, but really
is a kind of *suspension loss*, likely to depend more on
the rider than on the tire). But it's by no means
certain.
Smooth-surface rolling resistance clearly has the
potential to be large (it's the main resistance in roller
riding, and probably accounts for the sluggish feel of
nearly deflated tires). But what causes it? If it's
just a matter of losses from bending the sidewall,
tread, and inner tube back and forth, then the model I
describe above could produce decent predictions of
rolling resistance. But what if the energy loss stems
more from stretching or shearing the casing? Or from the
scrub or slip of tread elements on the road? (I know these
can be huge in cars, at least during cornering when
the tires suffer large slip angles.) If so,
I expect my model will give terrible predictions, since
it ignores these motions. Maybe they can be estimated
as 'patches' to the model, but I wouldn't bet on it.
As I see it, to MODEL rolling resistance effectively
we need experiments which separate various effects,
so we can attribute losses to specific mechanisms.
For example, get rid of suspension effects with
(1) a perfectly smooth surface, and/or
(2) a wheel load that is a PURE FORCE, {not involving
inertia (mass) or viscosity (damper)}.
Once looking at true rolling resistance, how to separate
the various possible causes? (I don't know of any
data which do this, though they are
probably available *somewhere*....) One option is to
supply a lubricant to the contact patch, if it decreases
rolling resistance then scrubbing friction was significant.
Another test: if wheel load is doubled, and tire pressure
is ALSO doubled, the tire deformations are unchanged, so
their contribution to rolling resistance is FIXED.
But pressures in the contact patch are doubled, so
friction losses (if any) are doubled....
(A kind of a sanity check: if pressure is
increased tremendously for a fixed load, all the above
mentioned loss mechanisms are stilled, and the rolling
resistance should tend to zero. If not, some other factor
is at work.)
Deformation RATE may be another means of separating friction
losses (which should be fairly unaffected by slip velocities)
from deformation losses (which for polymers, would typically
be expected to show a maximum rolling resistance at one
speed, with less resistance both faster and slower).
The separation into bending and 'membrane' losses in
the tire may be possible by glueing it to a STRAIGHT
section of rim, and pressing this uniformly against a
flat surface, so deformation is due to bending effects
only. To get the rates high enough, a bouncing or
oscillating test may be necessary.
Another worthwhile test, though less definitive, is to
use similar tires with thicker sidewalls, tube, and tread
-- bending losses should be dramatically increased,
while membrane (stretching and shearing) losses should
grow only modestly.
Whoever's still reading this -- you've got real stamina!
I look forward to your comments.
JP
----------------------- WORDS TO LIVE BY -------------------------
>
>Well, what are you waiting for? Don't waste time arguing about it,
>make it happen. -- Jobst Brandt
andrew cooke
>
>Jim I don't think your calculations are wrong, your are just not looking at
>things in the right perspective. If I understand what you calculated, your
>calculations are based on equal DISPLACEMENT of the wheel. Mathematicaly, if
>any two wheels of the same radius are displaced 5mm downward at the axle then
>yes, all contact patches have the same length. The catch is, it will take
>different amounts of force to displace wheels of different crossection
>because the support from the tire is proportional to the contact patch area.
>Since the wider tire will have a wider contact patch for the same displacement
i'm suffering from slow feed on this, so if someone else has
replied to this, bear with me.
jim's model, which seems pretty sound now i understand it,
shows that a wider wheel does *not* have a wider contact patch
for the same dimension. what's more, jobst's measurements
showed that while the theory is not perfect - a tyre of double
the radius had a patch 2mm wider - it's a lot closer than
expected from thinking about pressing oval shaped balloons
against the floor.
while the result seems counter-intuitive, it is in fact the same
argument (more-or-less) as a (moderately) well-known `surprising
fact' from introductory maths courses: if you wrap a piece of
string round a golfball, then increase the length by about 6 in
it will go round a ball with a radius 1 inch larger. if you extend
a piece of string going all the way around the earth by just 6 inches
it too will be 1 inch above the ground *everywhere*.
the maths of jims model has the contact patch instead of the
extra length, and the depression instead of increasing the radius.
so whether the tyre is the size of the earth or of a golfball, the
answer is the same.
actually, all this is for a linear approximation. i did the
full case last week and confirmed what jim had calculated some
years ago.
that it doesn't fit the data perfectly indicates that the
decoupling is not correct, but the discrepancy is surprisingly
small.
andrew
(a converted doubter ;-)
andrew cooke
by a factor of sqrt(2) - from 16 to 23, not 18 mm.
i'm not going to respond to a yes-it-is, no-it-isn't
series of posts, but while you may feel personally vindicated,
the theory he suggested is better than the one people (well, me
and someone in oxford) counter-claimed. (and presumably what
everyone else was thinking of, if they had a theory at all -
the idea of an oval balloon).
so there! ;-)
andrew
Jobst Brandt
deflection from load in contrast to the original subject heading under
which all this is appearing. All this computation is moot because
slender tires are not run at the same pressure that fatter ones are
and the comparisons are merely academic besides apparently being
wrong.
I did not instrument my stamp pad experiment with which I measured the
contact patch with a height gauge, but I suspect the vertical
deflection also is sufficiently different for the two tires that the
claim that they are the same is essentailly untrue.
I don't think that any of this will convince dirt riders to switch to
18mm cross section tires or criterium riders to choose 2" slick MTB
tires, or for that matter many sizes in between. These choices will
be made on the basis of durability and individual ride experience as
has been the case before anyone proposed the difference in contact
patch size.
And you wonder why they are called ivry towers.
Jobst Brandt <jbr...@hplabs.hp.com>
Phillip N Price
|> Phillip N Price (pr...@carla.lbl.gov) wrote:
|>
|> : Other people (like me, for instance) claimed that although the area
|> : of the contact patch would be the same, the wider tire would have
|> : a wider patch. You disagreed---contact patches would be identical,
|> : you said.
|>
|> It seems to me that there was a debate about the effect of tire width on
|> patch shape - one side said that the patch length and width would stay the
|> same if the tire width was halved, the other side said that the ratio of
|> the patch length to patch width would double if the tire width was
|> halved. The experiment showed the ratio to increase only from 6.3 to 7.9,
|> rather than to 12.6 as the doublers would have predicted. Among the
|> theories proposed, JPs is by far the most successful in predicting the
|> actual results.
|>
|> It is not accurate to characterize one side of the debate as claiming that
|> the patch would elongate "a little bit" if the tire width was halved.
|>
|> JF
Hmm, well, I missed the earliest posts on this thread. I haven't seen
the "proportional" claim. One person (name escapes me) claims a
square-root relationship. I claim a fourth-root relationship (of
footprint width with tire width). Jobst's data support a fourth-root
relationship, although I agree (with everyone else) that they are
outside the low-deformations regime we'd like to stay in. They do
not support independence of patch with tire size, although I agree
that we don't have data that let us say for sure.
Until someone points out where my model is wrong, though, I'm
sticking with fourth root.
--Phil Price
Phillip N Price
|> [Phil Price's observation that Jobst's experiment found a width
change of about 20% in the contact patch, deleted]
|>
|> My, aren't we scientific. Of course we are absolutely certain that
|> Jobst performed each test multiple times and averaged out the measurements.
|> We are absolutely sure that Jobst performed each experiment with all
|> of the delicacy of hydralic presses. Yes indeed, that 20% difference
|> sure is important in an experiment like this.
|>
|> May I compliment your clever observations.
|> >
Tom,
I'll try to use short words so you can understand this.
There were several theories suggested. At the time of my post, these
were: (1) There would be NO change in patch width, (2) there would be
a 40% increase in patch width, (3) there would be some small but
unspecified increase in patch width [a suggested physical model later
predicted a fourth-root relationship of patch width with tire radius,
thus predicting a 20% increase in patch width].
There was an observed 20% increase in patch width.
Now, it is true that the experimental observations could be (if we
include experimental errors, and the fact that the tests were
outside the small-deformation regime) consistent with theory (1).
But they could also be consistent with theory (2) or theory (3)!
The fact that the observations do not conclusively disprove a
theory is NOT THE SAME as saying that those observations support
the theory!
I see this b.s. all the time in statistical applications: someone
notes "We cannot disprove the hypothesis with 95% confidence",
and then follows it up with "therefore the data support the
hypothesis."
I agree that Jobst's observations don't disprove JP's hypothesis.
BUT THEY DO NOT SUPPORT IT EITHER!
In short, Tom,
**
* *
* *
* *
** ** * * ** **
* * * * * * *
* ** *
* *
* *
* *
**************
Oh, by the way, JP has pretty much convinced me that his model is
a good one, and that at high pressures/low displacements the contact
patches are indeed independent of tire radius. But this is, as
yet, unsupported by experimental evidence.
--Phil Price
andrew cooke
old engineer...
andrew
(feeling like he's just been told off for playing football
in the car-park again :-)
Rich Feldman
width-vs.-squashing exponent by measuring contact length vs. load.
Data posted so far don't contradict Jim's analysis, where +-15%
accuracy would be pretty good for a simple model over large ratios
of tire width and load. And analytically I like the model that to
first order, patch width W is approx. proportional to squashing S
and is almost independent of little radius r.
So in an empirical formula fitted to real W measurements, I bet the
exponent on S is much closer to 1 than to 0.5, and the exponent on r
is much closer to 0 than to 0.5 or 1.
In the simple hose-between-plates case I agree with JP, W = (pi/2)*S.
------------------------
An easy way to measure S is indirectly, from the patch length.
I think we all agree that Smax is proportional to the square of the
contact length Lmax. The issue at hand is whether the simple inter-
section model, good in that plane, is reasonable in the transverse plane.
How does S vary with longitudinal position X?
S = Smax - X^2 / 2R, where R is the big radius of the wheel.
Smax = Xmax^2 / 2R = Lmax^2 / 8R. Length Lmax = 2 Xmax = 2 sqrt( 2R Smax ).
Now if the contact width W were proportional to sqrt(S), then the patch
would be an ellipse with area (pi/4)*Wmax*Lmax. Length and width would
both vary with sqrt(load), and L/W would be independent of load.
But if W were directly proportional to S then the patch would be
pointier toward the ends than an ellipse, and its area would be only
(2/3)*Wmax*Lmax. The patch would get relatively fatter as the load
increased: length goes as load to the 1/3 power, but width and vertical
deflection go as load to the 2/3 power.
Here's some spreadsheet sanity checking, using the W = (pi/2)*S formula
in lieu of more realistic handling of the rim boundary condition.
(Anybody got closed expression for W incl. rim and unloaded tire widths?)
In U.S.A. units:
R 14.00
Lmax 2.00 3.00 4.00 5.00 6.00
Smax 0.04 0.08 0.14 0.22 0.32
Wmax 0.06 0.13 0.22 0.35 0.50
Patch area 0.07 0.25 0.60 1.17 2.02
Lmax/Wmax 35.67 23.78 17.83 14.27 11.89
Pressure 90.00
Load 6.73 22.71 53.83 105.13 181.67
Don't shoot this down for having the wrong constant ratio of W/S!
Evidence that Lmax goes as 1/3 instead of 1/2 power of load
would tend to indicate that W is proportional to S instead of sqrt(S).
-rich
--
Rich Feldman Megatest Corp., San Jose, CA 95131
ri...@megatest.com {decwrl,sun}!megatest!rich
home (415) 967-6966 work (408) 441-3027
James Annan
> but the competing theory was that the diameter would increase
> by a factor of sqrt(2) - from 16 to 23, not 18 mm.
NO!!
This is not what I said. I said the ratio of the length to width of the patch
should be proportional to the square root of the radii. This is easily proven
based on the simple model I had assumed. The width of the patch would only
increase by the factor sqrt(2) if the displacement was such that the length
stayed the same, which it obviously won't because then the patch size would be
bigger which requires a larger downwards force.
The calculated increase would be from 16mm to 19mm which is pretty close to
that experienced, but the patches are both a bit too long and thin from my
calculations.
I suspect the reason for the inaccuracy of my simple model may be that tyres
don't have a circular x-section but in fact have a thicker layer of rubber
near the middle of the tread thus giving the (imo) suprisingly thin contact
patch.
>
> i'm not going to respond to a yes-it-is, no-it-isn't
> series of posts, but while you may feel personally vindicated,
> the theory he suggested is better than the one people (well, me
> and someone in oxford) counter-claimed. (and presumably what
> everyone else was thinking of, if they had a theory at all -
> the idea of an oval balloon).
Presumably I'm the person in Oxford you refer to - not that I am any more, but
that's what the address says.....
What is this oval balloon theory? Sounds pretty odd to me.
My idea was to calculate the shape of the intersection of a torus (the tyre) and
a plane (the ground). Seems pretty simple and obvious to me.
>
> so there! ;-)
>
> andrew
>
>
--
James Annan.
Thomas H. Kunich
> Fat Thin
> 120 x 18 136 x 16
Well, it is my observation that those two shapes are remarkably similar
and would lead me to believe that they would be even more similar if
you had place _half_ of your weight on the tire as specified originally.
In any case I would give this one to Jim.
Thomas H. Kunich
>
>Now, wait just a second here. Jim, you were quite clear in your
>earlier stuff that you could PROVE that the contact patches would
>be identical, you were even willing to put money on it (wish I'd
>made that bet)...then Jobst does the experiment, uses two different
>widths of tires (radii differ by a factor of two), and finds that
>the contact patch is TWENTY PERCENT WIDER and TWELVE PERCENT SHORTER
>for the wide tire than for the narrow tire, and you're going to
>claim that you are "roughly right"??!!
My, aren't we scientific. Of course we are absolutely certain that
James Annan
Actually, I think that Jobst's results should be taken as confirmation of MY
results that said that the ration of the width to the length of the patch
should be (approximately) equal to the square root of the ratio of the two
radii of the tyre.
But don't worry, I won't be posting in this thread (or any other) for much
longer, as this account is dying.
(Anyone who wants to know how my model is derived and who doesn't know
Pythagorus' Theorem, email me at james...@afrc.ac.uk)
--
James Annan.
Hans-Joachim Zierke
Jim Papadopoulos wrote:
> Ultimately there will be a discussion about whether
> 5% or 10% deviation from my calculation is important or not.
The differences in rolling resistance are usually 3 - 4 %. I expect
differences in tire "sink" to be in the same range. I agree that it is not
important, tolerances in the manufacturing process of the tube might be
more important than "20 mm versus 28 mm tire width". I posted the data to
get rid of the "narrow tires roll faster / tubulars roll faster" myth.
> >BTW: The load is not distributed evenly to the contact patch. According to
> >high-speed shutter photography made by tire manufacturers, the tire bulges
> >inward where it hits the road.
>
> I'm curious about what you're saying here. Do you mean the contact
> patch is concave (waisted)?
At the front of the contact patch. The magazine "tour" published a
graph showing this. Since they made their tire test at the Continental
manufacturing plant, it is rather easy to guess who was the guy telling
them, and this man is very knowledgeable.
I can't tell you more at this moment. Maybe I have an opportunity to visit
the plant in some months. Short of talking to R&D engineers, there is no
way to get knowledge.
At the Conti booth of the IFMA show, I expect knowledge to be replaced
by smiling. Very few companies send their techies to the trade shows.
> Are you suggesting that I scan my data and plots, and post them?
> I'm afraid that is far beyond my equipment capabilities at this point.
I don't have a scanner here, too. The scans that I would need for my work
are done with 50.000 $ equipment, and I can't afford that.
If I need a simple scan, I visit my neighbours, put the page on their fax
machine with resolution set to "fine", and send it to my computer fax
program. As soon as there is a bitmap, even an old PC-XT can do it.
On my system, I extract the bitmap as TIFF, import the TIFF into my image
program, convert to black and white, and save as GIF. Then the GIF is
uuencoded (because the Internet currently cannot transport binaries without
encoding to ASCII).
A simple fax program might have come with your modem. Programs for
uuencoding are usually freeware and available for all operating systems. An
image viewer might be available for free or as cheap shareware depending on
OS. Simply ask a local computer freak.
hajo
Hans-Joachim Zierke
andrew cooke wrote:
> two sets of (half-)sections are shown, for tyres with radii of 0.6 and
> 0.8 units. the rim width is 1 unit and sections are shown for no
> compression and compression by 0.1, 0.2 and 0.3 units.
Sorry, I do not understand the graph at all. I couldn't figure out what is
measured on the X-, and what is measured on the Y-axis.
hajo
andrew cooke
by oval-balloon i simply meant a surface with two different
radii of curvature, intersecting with the ground (which i
think is the same as you calculated?). i posted a description
for this giving the ratio of the length to width, to which your
article replied. when i read your post i skipped through,
presuming you'd keep the contact area the same. apologies...
andrew
(if philip price is reading this - i guess that means it's
consistent with what you posted/emailed too?)
Jim Papadopoulos
>Thanks to everyone for their pointed questions, calculations,
>and experiments. I've been inspired to reconstruct my old
>tire calculations (there's litle chance of FINDING them
>over the next few days), and even take them further in
>response to the input.
measurements and kindly forwarded them to me for comment.
Both my theory (section squashing) and the intersection theory
show so-so agreement with different features of the data.
(And I am still wondering if the data suit the purpose,
one tire has something of a tread and thicker sidewalls,
and the loads were fairly high/pressures low.)
One thing that was clear right away from this person's
use of the section squashing calculations is that
my linearisation (presented in the post referred to here)
is not too good at realistic loads, the effective "K"
factor relating contact width to squash was a good 15%
in error at a real-world load and pressure. So except for
those 60 lb riders, I need to use a quadratic approximation
to contact width (I hope that's sufficient.) Then
when I make every possible comparison to his data the
discrepancies and agreements might tell us something.
(I've urged him to post the data without my results as
yet, for those of you with other theories to test.)
Another person has proposed to supply some data too,
but I don't know how seriously.
Finally, I made an archaeological excavation of some old boxes,
and came up with scores of my own ink patches, plus some
force-deflection data (it's not linear, as the 'intersection'
modelers would have it!!!), but lacking details of rim and tire
dimension......
I'm hereby announcing my 'retirement' from posting
until I can do a deceent job presenting
various sets of data, my updated model, and a range of
tests of both section-squashing and ground-intersecting
predictions.
I've been very pleased by everyone's interest and input, and
will still be smiling if the data reveal a fatal flaw in
my approach.
Later, y'all
Jim
Rich Feldman
>The whole discussion is now about contact patch and vertical
>deflection from load in contrast to the original subject heading under
>which all this is appearing. All this computation is moot ...and
>the comparisons are merely academic besides apparently being wrong.
...
>And you wonder why they are called ivry towers.
But this is real science, almost as much fun as riding bikes, and safer
when you are sleep-deprived. So:
---***************************************---
You want experimental data, we got experimental data!
I took a stamp pad and made, measured, and processed
an array of over 60 bicycle-tire contact patch impressions.
The r,R,L,W,p,f relations empirically are well fitted to power
laws, with exponents intermediate between the competing theories
but closer to JP's.
Measurements:---------------------------
I stamped all combinations of
. Tire widths 0.95 and 1.41 inches (r ratio 1.48, r/R ratio 1.51)
. Air pressure 35, 50, and 70 psig, and 100 psig in the skinny tire
. Load 12, 17, 25, 35, 50, 70, 100, 140 lbf
with a couple points missing and a few duplicated.
I am sure the experiment is easily repeatable and extendable.
The patches ranged from about 2.0 x 0.2 up to 4.86 x 0.7 and 5.04 x 0.6.
Length measurements were very objective and easy to make.
Width measurements were fuzzy above 0.4 (skinny tire) due to
a fine, shallow herringbone pattern (.04 pitch) meeting the paper.
The wider tire has a pattern of longitudinal ridges and grooves (0.15
pitch) that touch down in discrete steps, so I only measured a
couple of the wider cases. (A friend might be getting a fat
slick tire soon.) At any rate, it is hard to be objective and
accurate about W when ink transfer across a theoretical gap of
0.0016 inches can widen even the skinny-tire mark by .08 in.,
which (for instance) is 1/4 of the 70 lbf / 100 psi patch mark.
I plotted the measurements on log-log paper and/or cruched 'em
in a spreadsheet.
Early results:-----------------------
1.
Contact length vs tire fatness, load, and pressure.
1.0
The measurement array is amazingly straight and regular in log-log plot,
so each variable closely fits a power function of other variables.
The exceptions were the wider tire at low loads (25 lbf and below),
whose patches were up to 10% longer than straight power-law extrapolation.
1.1
For identical pressure "p" = 50 to 70 psi, and load "f" = 35 to 140 lbf,
L goes as r (or r/R) to the -0.11 power
(my 1.5x wider tire patches were consistently 4% to 5% shorter).
The exponent is intermediate between 0.0 (JP model for small W)
and -0.25 power ( W/L = K sqrt(r/R) models ).
1.2
For either tire, from 50 to 70 or 100 psi and from 35 to 100 or 140 lbf,
L goes as pressure to the -0.33 power.
L goes as load to the 0.43 power
The exponent on pressure from 35 to 50 psi was about -0.37, and the
exponent on load at 35 psi was about 0.47.
All observed exponents are intermediate between +-0.33 (JP model
in transverse plane, intersection model in longitudinal plane) and
+-0.50 (W/L independent of squashing), with both models taking
L * W proportional to f / p.
1.2A
The obvious problem here is a significant failure of reciprocal relation
between load & pressure, e.g. if we double both, L increases by 7% but
was expected to be unchanged. Could this be due to a non-airpressure
component of tire resistance? I reluctantly suspect nonlinearity (!) of
cheap instruments: either actual load ratios were higher than indicated
by the bathroom scale, or actual pressure ratios were lower than
indicated by the pump gauge, or both. For example, most of the 0.10
discrepancy between f and p exponents could be explained by a -5 psi
offset in the pressure gauge, masked by the stop pin at zero. I'll try
to use better instruments next time.
2.
L/W ratio, area, and normalized shape vs load
L/W ratio changes significantly with load.
For a representative case (50 psi, 35 to 140 lbf) the ratio
goes from 10.5 to 8.5, a 21% decrease for 4x load increase.
The exponents for this (representative) pair of contact patches are:
Measured Theory A B
Length: +0.44 power of load +0.5 +0.33
Width: +0.60 +0.5 +0.67
L/W ratio: -0.16 0.0 -0.33
L*W product: 1.04 1.0 1.0
The patch area is expected to be proportional to load, so the
last term probably indicates either a) my scale DOES
underestimate the load ratio, or b) the patch shape is changing
from round-ish (theory A) toward pointy-endish (theory B),
giving 4x area increase for 4.23x increase in L*W.
The "shape" of a typical patch (about 4.0 x 0.4 at 50 lbf, 70 psi)
was measured, and is clearly and significantly non-elliptical.
Normalized to W=L=1, it is about halfway between circular
(y^2 = 1 - x^2) and the pointy-ended figure if W were proportional
to squashing (y = 1 - x^2). In any case the very ends are blunted,
explained by local failure of the intersection model (otherwise
presumed excellent IN THE PLANE OF THE WHEEL) at the non-tangent
junctions of circular arc to straight chord.
3. Other
I have not crunched the tire contact Widths, L*W, or area
vs. p and f, except for the one pair given above.
The log-log graph of W vs f for 50 and 70 psi is pretty
nonlinear, esp. at very low load. I suspect width measurement
errors, tread profile and rubber stiffness effects, and ink spot
"growth" are all significant.
4. Next
This work is preliminary. Next round should have nice smooth
tires and better force and pressure measurement, and maybe
vertical displacement measurement. Thanks for the discourse!
It has been suggested that ink mark growth can be measured by
inking a rigid pipe of same diameter as tire, and pressing
against the paper; theoretical contact geometry
is a zero-width line so whatever we get should be subtracted
from the bike-tire prints.
--
---
Rich Feldman Megatest Corp., San Jose, CA 95131
Jobst Brandt
> You want experimental data, we got experimental data! I took a
> stamp pad and made, measured, and processed an array of over 60
> bicycle-tire contact patch impressions. The r,R,L,W,p,f relations
> empirically are well fitted to power laws, with exponents intermediate
> between the competing theories but closer to JP's.
Could you put these in a table so the dimensions can be visually
compared and easily loaded into a data display program? I also found
the witdth difficult to measure but I used slick tires and there the
edge isn't all that clear either. I think we are in agreement that
the contact area is reasonably the same for the same pressure so, in
fact the length of the contact is the best dimension to report.
Forget about the width because that can be mathematically determined
for the curious.
If you have this data, I think it would best be displayed in rows and
columns of tire size and patch length. The major OD of the tires
should be roughly the same. This means that you should compare a 18mm
x700 tire with a 2" x 26" tire since their major diameters are closely
the same.
Jobst Brandt <jbr...@hplabs.hp.com>
Jobst Brandt
> James Annan writes:
> > Jobst Brandt writes:
> > [...]
> >> And you wonder why they are called ivory towers.
> >>
> >> Jobst Brandt <jbr...@hplabs.hp.com>
> The above are not my words but another's response to what I wrote. As
> usual the writer included my sig at the bottom of his response even
> though we are reminded weekly by Mike Eglesias to not do so. This
> kind of citation error is encouraged by the practice.
Ooops! I missed the target completely. This was my statement that
came out awhile back but I already had my finger on the trigger and
Blam, foot in mouth. Anyway, next time I'll check my target a bit
more carefully.
Jobst Brandt <jbr...@hplabs.hp.com>
James Annan
[...]
Actually, yes, I do wonder why they are called ivory towers. I never saw an
ivory tower while I was at Oxford. If it's because they are supposed to look
like ivory, then I think ebony would be a more appropriate term.
As far as mathematical modelling goes, it's far from an "ivory tower" activity.
Obviously old-fashioned experimentation has its place, but there are a lot of
experiments that you can't do (global climate change is perhaps the most
obvious) or can't afford to do (anything that used to be Govt. funded is the
most obvious example here). And even once you have done the experiments, how
are you supposed to _understand_ what's going on and make predictions without
some sort of model?
The difficulty, of course, is in constructing a model that has the necessary
simplification to enable calculations to be performed, without simplifying out
the essence of the situation.
--
James Annan
james...@afrc.ac.uk
Operational Research, n.: modelling without data.
Jobst Brandt
> Jobst Brandt writes:
> [...]
>> And you wonder why they are called ivory towers.
>>
>> Jobst Brandt <jbr...@hplabs.hp.com>
The above are not my words but another's response to what I wrote. As
usual the writer included my sig at the bottom of his response even
though we are reminded weekly by Mike Eglesias to not do so. This
kind of citation error is encouraged by the practice.
> Actually, yes, I do wonder why they are called ivory towers. I never saw an
> ivory tower while I was at Oxford. If it's because they are supposed to look
> like ivory, then I think ebony would be a more appropriate term.
Def:
1. an impractical often escapist attitude marked by aloof lack of
concern with or interest in practical matters or urgent problems.
I think that sums it up neatly. It doesn't take much metaphorical
imagination to see the ivory, intricately carved, supporting the
researcher isolated from reality.
> As far as mathematical modeling goes, it's far from an "ivory
> tower" activity. Obviously old-fashioned experimentation has its
> place, but there are a lot of experiments that you can't do (global
> climate change is perhaps the most obvious) or can't afford to do
> (anything that used to be Govt. funded is the most obvious example
> here). And even once you have done the experiments, how are you
> supposed to _understand_ what's going on and make predictions without
> some sort of model?
"old-fashioned experimentation" Oh! I wonder what all these technical
papers are about that originate in laboratories full of hardware and
samples of materials from industry if it isn't for scientific research
of the most advanced kind. Somehow putting the experimental process in
the "old-fashioned" box rings suspiciously like Def. 1. above.
> The difficulty, of course, is in constructing a model that has the
> necessary simplification to enable calculations to be performed,
> without simplifying out the essence of the situation.
It's good to see that the pitfalls of modeling are not lost here.
However there are many phenomena that either don't lend themselves
to modeling or are trivial to analyze by experimental means.
There is plenty of metaphorical ivory to go around, just the same.
Jobst Brandt <jbr...@hplabs.hp.com>
andrew cooke
duty. the curves are tyre cross-sections. if there was no
rim the centre of teh circular, unloaded, tyre would be at 0,0.
the scale is such that the rim is of unit width.
only half of each tyre is shown, with 3 loadings other than
none (sorry, clumsy english).
so, for example, take the set of curves starting at 0.5,0.3.
they are half-sections of a tyre with initial radius 0.6.
sections for increased loading are drawn with the rim at the same
position and the `floor' moving upwards. the flat contact
region is visible and the wall bulges as the load is increased.
the larger tyre overlaps with the smaller (my apologies - this is
very messy), but shows the same thing.
finally, two shorter curves, from 0.0, -0.6 and 0.0, -0.8, trace
the locus of the edge of the contact area. if you peer very closely
at the resulting mess you can see that area for the fatter tyre is
actually slightly less, for the same load, than that for the
smaller tyre. sorry, by `load' i mean `movement of floor toward
wheel'.
if that doesn't make sense + you're still curious, please email
me - i think we are a long way apart (time-wise) on the net.
finally, sorry it wasn't clearer. i now have the software to
present something a lot prettier, but work has suddenly dropped
on top of me + i have very little free time.
andrew
p.s. i've ttried using the CC bit of my newsreader, so you may get
a copy of this by email.
James Annan
> description of a bike wheel contact patch:
>
> patch must be such that Pressure*Area = Weight (=Mass*g).
> So the question is, what shape of the patch gives us the
> area that we need?
>
> Suppose we had a bike tire in which the membrane had no
> stiffness---we make it out of thin rubber like a balloon or
> a condom. Also suppose that we keep the applied weight low
> enough that the vertical displacement is much less than the
> radius of the tire (not to be confused with the radius of the
> wheel). If the tire is pressed against a flat surface, and if
> the displacement is small compared to the radius of the tire,
> then only the section very near the contact patch is affected.
> The area (and shape) of the patch can be calculated from the
> intersection of a torus (the bike tire) and a plane (the ground).
Hmmm. I didn't intend to post any more on this, so I'll just remark that it's
nice to have someone agree with me for a change.
>
> 3. The half-length of the footprint, at its longest, is
>
> b = a * sqrt(R/r)
>
That looks familiar.
[... experimental results are a good bit longer and thinner than should be
predicted]
>
> HOWEVER, let me point out that this model predicts a
> footprint width/tire width relationship that goes like
> (r/R)^(0.25). That means that given a fat tire that's twice
> as wide as a skinny one, we expect the footprint widths to
> be in the ratio of (2^.25):1, or 1.189:1
>
> Jobst's experiment found a ratio of 19:16, or 1.187:1
>
> Basically, I think my model is a pretty good one. In fact,
> I think it is CORRECT in the LIMIT of low loads and perfectly
> flexible (non-stiff) tires.
>
> Until someone convinces me otherwise, I'll believe in the
> (r/R)^(.25) rule.
>
> --Phil Price
two points:
1. Jobst's results were for fairly low pressure tyres and large displacements,
the model is based on high pressure/small displacements.
2. Even so, the patches are a lot longer and thinner than predicted by the
model.
Any offers as to why ths occurs?
James Annan
Phillip N Price
|> In article <41...@dog.ee.lbl.gov>, pr...@carla.lbl.gov (Phillip N Price) writes:
If the tire is pressed against a flat surface, and if
|> > the displacement is small compared to the radius of the tire,
|> > then only the section very near the contact patch is affected.
|> > The area (and shape) of the patch can be calculated from the
|> > intersection of a torus (the bike tire) and a plane (the ground).
|>
|> predicted]
|> >
|> > HOWEVER, let me point out that this model predicts a
|> > footprint width/tire width relationship that goes like
|> > (r/R)^(0.25). That means that given a fat tire that's twice
|> > as wide as a skinny one, we expect the footprint widths to
|> > be in the ratio of (2^.25):1, or 1.189:1
|> >
|> > Jobst's experiment found a ratio of 19:16, or 1.187:1
|> >
|> > Basically, I think my model is a pretty good one. In fact,
|> > I think it is CORRECT in the LIMIT of low loads and perfectly
|> > flexible (non-stiff) tires.
|> >
|> > Until someone convinces me otherwise, I'll believe in the
|> > (r/R)^(.25) rule.
|> >
|> > --Phil Price
|>
|> two points:
|>
|> 1. Jobst's results were for fairly low pressure tyres and large displacements,
|> the model is based on high pressure/small displacements.
|>
|> 2. Even so, the patches are a lot longer and thinner than predicted by the
|> model.
|> <snip>
|> James Annan
Ah, yes, back to this stuff.
James Parinella and others have convinced me (or perhaps I should say,
I thought about it and agreed with JP) that the "torus/plane"
model is not correct, so all of the above is junk....except that
it seems to work better than the other proposed model. Basically,
the model discussed above gives about the right behavior, but for
the wrong reasons. No one has yet posted a good explanation.
So, basically, no comment.
--Phil Price
Jim Papadopoulos
Jim Papadopoulos, Oct. 1994
The following is a long, generally boring posting about
contact patch modelling, and comparisons to data.
So far, I have no definitive answer to the question of
which model is 'right'. Both my 'squash' model and the
'intersection' model are in the ballpark on some
of the data. (I have a hypothesis which could explain
some 'squash-vs-data' disparities in terms of an
experimental error, but this is pure speculation.)
The main thing I've concluded is that real tire
contacts deviate substantially from some 'ideal'
patch, due to slight variations in tread thickness
(either intentional from manufacturing, or a simple
consequence of wear).
Additional, precise data might resolve some of the
questions. (I have not yet examined Rich Feldman's
numbers to see what they imply.) But for now, the
main thing is to take a break from all this!
PART I -- Old Jim Papadopoulos data
PART II -- Recent Bob Engel data
PART III -- Discussion of implications
********************************************************************
PART I -- Old JP data
I made these measurements about 16 years ago, while I was
an undergraduate at MIT. I had only one wheel to test,
and I can find no indication of what it was (except the
words "90 psi skin wall"). Probably it was 27 X 1.125
or 27 X 1.25.
I did some load-deflection plots on an Instron, at various
pressures. It would be difficult now to reconstruct how
to compensate for axle and rim deformations (both small).
I also made a series of 90 psi ink patches, at ten loads
under 100 lbf, and eight more up to 450 lbf. I carefully
cut out those patches in 1978 and weighed them to milligram
precision, as a way to find patch area. (Coincidentally,
I did the weighing during a summer visit to Menlo Park,
where I also met Jobst.) I plotted area versus load,
arriving at a straight line with some minor blips.
Just this last week I finally unearthed my graphs and ink
patches, plotted patch lengths and widths, and did some
curve fitting. The results follow, with some discussion.
The tire had a thin, somewhat worn tread. But even so, it's
now clear that it was not really circular in cross section.
The pattern was a smooth center band 10 mm wide, next to
some 2mm-diamond tread. At large loads, the 'elliptical'
patch shape very clearly has notches or shoulders due to
the transition from center band to pattern. And, various
plots (especially width of contact) show odd behaviour
at a contact width of 10mm.
Furthermore, even at a very slight load, the contact
patch had about 5mm width -- my interpretation is that
it had 'worn flat'.
It would be really nice, therefore, to see results
for a tire with no tread (or with a smoothly varying
tread whose thickness profile was accurately measured).
AREA VERSUS LOAD: The plot is remarkably linear, except for
some clear deviation in the range from 9.5 mm to 11 mm
contact width, and possibly also at some higher loads.
Area = (1.01)*(force/pressure) + 0.2 sq.in. , to a load
of 450 lbf (+- 0.1 sq. in.)
LOAD VERSUS DEFLECTION (del, inches): using data from the
Instron chart, force = press. * (del*sqr(50.3 in.))^1.333
This is a very accurate (about 1%) fit up to 300 lbf, after
which the deviations quickly grow to about 4%.
(Note, using psi, del should be in inches. If using pounds
per square mm, then del should be in mm, and 50.3 inches
should be converted to mm.)
That the exponent 1.3333 is clearly better than 1.5
(from my theory) can be shown by plotting del against
various powers of F, and looking for a straight line.
(I took this tack because I could be certain when F
was zero, but not when del was.) Plotting log F
against del should also work.
WIDTH VERSUS DEFLECTION (mm):
There are too few points to be really sure how to fit
these data. To me they seem best represented by a linear
relation: width (mm) = 2.1 * del(mm) + 4.6, with a gentle
falloff at higher values of del (above 7mm say).
This is qualitatively consistent with my current model,
i.e. linear with quadratic dropoff, but of course might
disagree with predictions if actual rim and tire widths
were used.
One thing that is extremely clear is the sharp, premature
climb to w=10 mm, then levelling off (and even dropping
slightly) before resuming linear behaviour.
LENGTH VERSUS DEFLECTION (mm):
Lengths fit beautifully to a square root relation, if we
except the 5-lbf load. Up to 100 lbf the error is
rarely as much as 1%; above that it reaches 2% or 3%, and
finally 4% at 450 lbf.
This is encouraging because I favor a square root model
on theoretical grounds (really, an intersection model,
for the length dimension only.)
Unfortunately, the best fit is L = 2 * sqr (del * 492 mm),
whereas the model calls for wheel diameter (700 mm or so)
rather than 492 mm. I had observed during the experiments
that the tire was still slightly flattened as it curved
away from the ground -- this necessarily leads to a
shorter contact length than expected.
A combination empirical-theoretical approach is to use
wheel diameter in the formula, then reduce the computed
length by 16% (for this tire at least).
RELATION OF AREA TO L AND W:
For both the intersection model and the linearised squash
model, patch area works out to L * W * (numerical factor).
So I calculated W*L/A, where A came from my curve fit.
The results were not enlightening, generally from 1.16
to 1.30, with a couple higher or lower.
ASPECT RATIO (length/width):
Quite scattered, in range from 6.0 to 7.9, most of
the lower values occurring above 150 lbf.
PRESSURE EFFECT:
It's normal to expect that a tire, which is deflected
a fixed amount by a testing machine, would keep its
shape fixed as inflation pressure is varied (by pumping)
-- thus the force should be perfectly proportional to
pressure (if casing has no stiffness), or proportional to
pressure plus a constant.
I never performed that exact experiment, instead I ran
compression tests at several pressures, then looked at
force for equal deflection values. The result did NOT
look like either of the two relations just suggested --
deviations were plus or minus 5%, and there were too
few points to see if there was a smooth systematic trend.
(I'm wondering if perhaps cheap pressure gauges mightn't
be nonlinear, thus accounting for these deviations?)
*********************************************************
PART II -- Recent Bob Engel data
While all the posting and occasional calculating
was going on, Bob Engel was setting up a tire test
in his basement, to supplement the numbers provided
by Jobst. Here are his results:
"I redid the tire patch test, this time
with a more stable rig and I used ink."
Bob's rig is a pair of horizontal boards, pivoted
at one end to his workbench, holding various weights
at the other end, and pressing down on a wheel axle
in the middle. For each selected load and tire
pressure, the tire was inked and lowered to a sheet
of paper. Then it was repeatedly lifted and set down
after moving the paper, to get several impressions.
The ink patch dimensions are hopefully 'typical' or
average. (Bob faxed me a couple of the sheets, on which
most of the impressions did indeed agree.)
"I did 3 different weights and two tire pressures.
Two wheels were used:"
"Barum G10 on Fiamme rim
tire diameter = 24 mm
rim width = 21 mm
Wheel diameter = 670 mm."
This is a narrow sewup tire, finely ribbed tread, flexible
casing
"Hutchinson on Module 4 rim
tire diameter = 37 mm
rim width = 27 mm
Wheel diameter = 700 mm."
Fatter clincher (wired on) tire, stiffer sidewalls, and
a tread pattern -- 6mmX8mm diamonds. *External* rim width
is given (may not be as relevant as internal width.) I
have looked at one set of prints for this wheel, and it
was a little disconcerting to see all the space between
the tread 'blocks'. Plus the positioning of the blocks
at the narrow ends of the patch made length a little
less certain. When I measured the patches, three or even
four agreed with Bob's numbers of 108mm X 17.5, plus or
minus 0.5 mm or better. But one was closer to 113 X 19.5
and one couldn't be measured.
The following chart contains Bob's measurements, plus my
calculations from two quantitative models: SS (section
squashing, the model I proposed), and GI (ground
intersection), summarised at the end of Part II.
Wexp = measured patch width, mm
Lexp = measured patch length, mm
Wss = ss calculated width, mm
Lss = ss calculated length, mm
Wgi = gi calculated width, mm
Lgi = gi calculated length, mm
* I hope I didn't drop any factors in the calculations --
it wouldn't be the first time! I would be grateful if
someone would check a couple, and post their results. *
-------------------------------------------------------
50 psi = 0.0775 pound per square mm
pounds
Test Tire weight Wexp Lexp Wss Lss Wgi Lgi
1 Barum 65 12 85 11.6 104 14.2 75
2 Barum 108 16 113 15.7 125 18.3 97
3 Barum 156 18 131 19.3 142 22.0 116
4 Hutch 65 13 84 11.2 110 15.7 68
5 Hutch 108 18 108 15.4 132 20.2 88
6 Hutch 156 21 123 19.2 150 24.3 106
--------------------------------------------------------------
75 psi = 0.1163 pounds per square mm
pounds
Test Tire weight Wexp Lexp Wss Lss Wgi Lgi
7 Barum 108 14 95 12.4 108 15.0 79
8 Barum 156 16 106 15.4 123 18.0 95
9 Hutch 108 16 96 11.9 114 16.5 72
10 Hutch 156 18 110 15.0 130 19.8 86
----------------------------------------------------------------
I'm kind of glum about these data -- they fall roughly
half way between the predictions of the two models! I'd
LIKE to downplay them -- the loads were pretty high
compared to the pressures, and at least one tire had
thick sidewalls and coarsely patterned tread. But that's
not justified. I CAN say that it would be nice to have
more data points and, if not the actual sinkage, then
a measurement of sidewall bulging (from which sinkage
can be derived).
For any analysis of the data, (Part III) I was always
on the lookout for trends which would reveal a
systematic difference from the model, or non-trends
which would suggest some kind of error. I noticed that
in several ways the 65-lbf results deviate from smooth
trends, leading me to wonder if there's a weight error.
------------------------------------------------------
SECTION SQUASHING MODEL
The Section Squashing model assumes decoupled cross
sections, and algebraically approximates the growth
of contact width due to 'squash' of each cross section,
treated as deforming independently.
[Note: in my original posting I used a linear approximation
for width as a function of sinkage (del). But Bob's more
exact calculations showed me that this overestimates the
width on smaller tires. For his slender tire, the error
in W exceeds 10% when sinkage is greater than 3.5 mm --
theoretically, a 78 lbf load at 75 psi pressure. For the
fat tire, this error is reached only after 5.9 mm sinkage
-- 152 lbf at 75 psi. To provide increased large-deflection
accuracy with minimum effort, the following equations employ
a quadratic approximation.]
IN THIS CALCULATION, A WIDER TIRE IS NOT ASSUMED TO
CHANGE WHEEL DIAMETER (IN EFFECT, IT'S AS IF RIM
HAD SHRUNK SLIGHTLY).
Formulas:
dist= sqrt(wtire^2 - wrim^2)
THETA= arctan(wrim/dist), sidewall arc for undeformed tire
sin(THETA) - THETA*cos(THETA)
K= ----------------------------------
1 - cos(THETA) - (THETA/2)*sin(THETA)
(-0.5)*[THETA - sin(THETA)]^4
J= ----------------------------------------------------------
(THETA*wtire-wrim)*[1 - cos(THETA) - (THETA/2)*sin(THETA)]^3
K and J are linear and quadratic terms for approximating
contact width as a function of sinkage.
ap_sink is approximate sinkage:
ap_sink = (f/(p*K*1.3333*sqrt(dwheel)))^0.6666
(NOTE: pressure must be measured in the force and length
units used for other quantities.)
final calculated sinkage (again an approximation):
sinkage = ap_sink - (4/15)*(J/K)*(ap_sink)^2
Wss = K*sinkage + 0.5*J*(sinkage)^2
Lss = 2*sqrt(dwheel*sinkage)
For this model, contact area A = W*L*(0.6666), approximately.
(This is because the patch should have a nearly quadratic
variation of width along its length..... results in sharp
ends.)
------------------------------------------------------
GROUND INTERSECTION MODEL
Ground Intersection model assumes contact patch is found
by the intersection of the ground plane with the
undeformed tire, which is imagined to 'pass through' that
plane instead of being deformed.
Formulas:
F
sinkage = -------------------
p*pi*sqrt(wtire*dwheel)
Wgi = 2*sqrt(wtire*sinkage)
Lgi = 2*sqrt(dwheel*sinkage)
Note the theoretically fixed aspect ratio.
Area = Wgi*Lgi*(pi/4)
(this is because small contact patches should be ellipses)
*********************************************************
PART III -- Discussion
LOAD-DEFLECTION RELATION
The gi model predicts a linear force-deflection relation
for a tire. The ss model predicts a 1.5 power. My Instron
results were definitely nonlinear, but nearer 1.3333 power.
CONSISTENCY CHECK
What happens if both pressure and load are increased
proportionately? We can compare 108 lbf/50 psi (nominally
2.16 sq. in.) and 156 lbf/75 psi (2.08 sq. in.) for both
tires. The differences are not great. It would be nice to
have a bigger series of such comparisons. (But only
if forces and pressures are measured accurately.)
EFFECT OF INCREASING TIRE WIDTH
In my model, increasing tire width (at a fixed load,
pressure, rim width, wheel diameter) leads to a
narrower contact patch. Narrowing the rim has a similar
effect, to an even greater degree. If the tire-to-rim
ratio is maintained, then at low loads the patch
width is unaffected, while at high loads, it gets
narrower. This is exactly opposite to Bob's measurements.
(And, if we imagine a clincher tire 'acts' like its rim
should be measured internally not externally, the disparity
is even greater.)
Data for many more loads would reveal if anything funny
is happening to patch width, although that should
occur during only part of the load range.
AREA
The quantity Wexp*Lexp*p/F might be expected to stay
(relatively) constant for all loads and pressures acting
on a given tire (F/p is a theoretical contact area; and
actual patch area should be proportional to W*L if the
patches maintain shape similarity.)
W*L*p/F (p = lbf/sq.mm.)
Number Exp SS GI
1 1.22 1.44 1.27
2 1.30 1.41 1.27
3 1.17 1.36 1.27
4 1.30 1.47 1.27
5 1.40 1.45 1.27
6 1.28 1.43 1.27
7 1.43 1.44 1.27
8 1.26 1.41 1.27
9 1.65 1.47 1.27
10 1.48 1.45 1.27
In this calculation, it's odd that the experimental results
for cases 1,2,3 (equally 4,5,6) first increase then decrease.
Why should this same pattern appear for DIFFERENT tires?
The results would be much smoother if the weight DENOTED
108 lbf were actually 99 lbf. (The sequences 7,8 and 9,10
would also fall in line.)
Alternately, the weight DENOTED 156 lbf could actually be
184 lbf.
Or, that denoted 65 lbf could actually be 55 lbf.
If this last were the case, experimental lengths would show
a trend more like the ss model.
[Look for example at the ratio of Lss to Lexp:
For Barum it's 1.22,1.10,1.08,1.13,1.16
(cases 1,2,3,7,8)
For Hutchinson, it's 1.31,1.22,1.22,1.19,1.18
(cases 4,5,6,9,10)
If cases 1 and 4 referred to 55 lbf, then
case Wss Lss
1 10.5 98 ratio changes from 1.22 to 1.15
4 10.0 104 1.31 to 1.24
.. both of which are more consistent with the other numbers.]
The sequence of widths would also look better.
However, this is all a digression which may amount to nothing.
ASPECT RATIO
According to the gi theory, aspect ratio (length/width)
should be fixed for a given tire. The ss theory says it
should vary. Fixed aspect ratio is clearly a better
approximation here. (Recall 1,2,3 and 7,8 belong together.)
No. Lexp/Wexp Lss/Wss Lgi/Wgi
1 7.1 9.0 5.3
2 7.1 8.0 5.3
3 7.3 7.4 5.3
4 6.5 9.8 4.4
5 6.0 8.6 4.4
6 5.9 7.8 4.4
7 6.8 8.7 5.3
8 6.6 8.0 5.3
9 6.0 9.6 4.4
10 6.1 8.7 4.4
So..... which theory is 'right'? I still expect that
my theory will be extremely accurate for SOME
circumstances, the job is to find them and then
account for deviations.
Supporters of the intersection theory can be satisfied
that numerically they are not far off. For getting a number
relating to a standard tire, their approach is evidently
about as good as mine. (But for calculating something
like sidewall bending, it doesn't provide any answers.)
For reference purposes, I'll requote Jobst's
posted data: 60 psi, fixed load (est. 175 lbf)
Fixed wheel dia. (est. 700 mm)
tire width 0.875 136 mm X 16 mm
tire width 1.75 120 mm X 19 mm
It's possible to compare these numbers to gi model, but
for ss model need rim width(s). Agreement would be good
only if the wide tire were on a really wide rim.
Charles R. Sullivan
If it hasn't been discussed, please post your opinions. If it has, if someone
could mail me the consensus that emerged (if there was one), I'd
appreciate it.
Question:
--------
Given whichever model of how contact patch depends on tire width,
how does that affect rolling resistance?
Thanks
Charlie Sullivan char...@eecs.berkeley.edu
Jim Papadopoulos
>Question:
>--------
>Given whichever model of how contact patch depends on tire width,
>how does that affect rolling resistance?
The main issue is deciding where the main rolling resistance
arises. (Where does energy get turned into heat?)
Conceivably it could be in some or all of the followign:
rubbing/scrubbbing between tread and road
viscoelastic properties of tread and sidewalls
(and tube)
frictional rubbing of the sidewall and bead on the rim.
Vibration absorbed in the rider's body.
This is not a complete list, and maybe some items
are silly. (I think Jobst feels that sidewall and tread bending
are absorbing energy from material visco-elasticity, but
don't know his reasoning.)
Once there is a model for energy loss, any reasonably accurate
prediction of tire deformation (especially if supplemented
to include the slippages or other motions of interest)
should be able to provide an estimate of its magnitude.
So I'm saying, if someone can answer what phenomenon
causes rolling resistance, maybe we can make a good stab
at calculating it!
JP