Wide vs. Skinny (5 Viewers)

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spartan said:
If only they made the AT REVO in a 285 95 16 size.


37" ?? :eek:
 
expeditionswest said:
Sorry Nay, I missed your post. No Troopy dreams for me quite yet...

You bring up an interesting observation, regarding lateral traction. That might warrant some testing...

In the meantime:

1. My article was not intended for "rock crawling", but for the expedition traveler (which requires a balance). The values of a narrow tire is much more important for what my requirements are than the potential (as this has not been tested) of greater lateral traction.

2. The research I have conducted and the resources I use for data do prove the lateral traction advantages of a wide tire (deflection, roll, scrub, heat control, etc.), but these are all under high speed tests on highly tractive surfaces (CF greater than 1). It will be interesting to test these theories in a very slow, technical trail environment with a CF of much less than one.

3. Lateral traction is affected by suspension dynamics and design, gearing, lockers, etc. Just think of the design effort that goes into a race car, to allow it to turn fast... A vehicle with a front locker engaged will favor understeer.

I will say, that in my experience, lateral traction on technical terrain has never been observed as an issue for narrow tires. I have run most of the well known 4 rated trails in the west with a 10.5" wide tire.


Agreed - there is more to lateral traction than tires...it is just something that is so often left out in offroad builds as we always focus on "forward", and not as easily quantified (we don't say words like gears, lockers, and articulation very often when talking about lateral capability). In my experience, an aggressive focus on increasing forward capability must be matched by an aggressive focus on increasing lateral stability and predictability, or you end up with a "built" rig that may be less capable than a much more modest design. Tires of course have a lot to do with this.

I sometimes think the single biggest difference I ever saw in all of my mods was moving to a tire with good lateral traction (trxus MT). That was the day that almost all pucker factor went away for me in Colorado/Utah wheeling (and Colorado highway snow driving). Clearly part of that is just a change in tire type, and part of it is a change in tire size/width.

I'm all about balance, and when forward and lateral capability are both predictable and balanced then you've got a good build that a good driver can really leverage. This is where "one size fits all" on any offroad component stops making sense.

That is one of the things I love about the trxus MT - even at relatively low pressures (28-30) the tire is designed so that the contact patch is the center 10" or so of the tire, and the outer 1" or so of the outer lugs really don't make contact except when well aired down offroad, conforming to obstacles, or in snow.

So maybe I really do like the skinny tire...just with better lug spacing and more lateral offroad grab when you air it down. Best of both worlds :grinpimp:

Nay
 
expeditionswest said:
Two tires of the same carcass construction, yet different widths will have different footprints with the same pressure. The shapes will also have very close to the same length. However, the wider tire will have a greater width dimension.

two tires of the same carcass construction and different widths inflated to the same tire pressure MUST have the same total footprint AREA, regardless of shape. that only follows from basic physics. i will explain more in a bit. wider tires will have wider footprints than skinny tires, but the total AREA of tire touching the ground MUST remain the same, assuming both tires have no internal rigidity or carcass strength.


expeditionswest said:
Let me try to explain this in different terms by removing the air component for a moment. Think of a solid rubber wheel: One that is 1" wide, and one that is 20" wide. The rubber is solid, and has no measurable deflection. The 20" wide wheel will have 20 times the area in contact with the ground compared to the 1".

if you remove the air, the critical component of this argument goes away. solid rubber tires have significant internal structural rigidity which means you cant measure internal pressures on the structure of the tire. if you put a solid rubber tire under a truck, the footprints will be different from skinny to wide tires, but we wont be able to level the playing field by requiring the internal pressure on each tire be the same. we would then be arguing apples and oranges. assuming the 20" and 1" tires have the SAME internal/carcass strength across the entire dimension of the tire, the total AREA in contact with the ground MUST be the same for both tires. but since we cant fix that pressure/strength within the tire to be uniform, we cant measure and compare the two in the real world.

expeditionswest said:
Now lets reintroduce air, and go into detail on this: Air pressure inside the tire is only ONE component of what gives a tire shape and load capacity. The other is the tire itself. Have you ever tried deflating an E rated Super Swamper bias ply tire? On a light vehicle, the carcass wont even really start to deform until single digit numbers. This is the critical point. On that E rated Swamper, what is doing most of the job in supporting the Jeep? The carcass. Don't forget that the sidewalls are integral in supporting the weight of the vehicle. Where a passenger rated, single ply tire will start to deform much sooner as it relies more on air to support the vehicle.

agreed, and i think i understand what you mean by carcass strength now.

what is a tire? it is really just a steel-reinforced balloon. the carcass strength really governs how much air is needed to support the tire, not the truck. rigid tires dont need as much air to support the weight of a truck, right? that is why half the guys in my local club run their MT/Rs at 13 psi, while i run my BFG ATs at 50 psi on the road, even though our trucks weigh the same. stiff tires have more internal structure and can support a load without as much assistance from internal tire pressure. that makes sense.

conversely, we are debating footprint areas between tires of different widths that are inflated to equal tire pressures with the same carcass structure. that is different than what you describe above. equal internal/carcass structures and tire pressures negate the relevance of sidewall deflection.


expeditionswest said:
Area and shape are different. A wide and narrow tire of the same construction must have different areas and different shapes.

yes. area and shape are critically different. area has dimension, shape has description. a wide and narrow tire of the same construction and pressure MUST have different SHAPES, but under an equal load, MUST have the SAME area exerting force on the ground. hopefully this will become clear in a bit.

expeditionswest said:
Think again in the extreme. Do you believe that a 39x18 and 38x11 Bogger will have the same total area in contact with the ground? It is physically impossible, as they are within less than 10% of the load rating of each other, and within 5 psi of max pressure.

you just described an apples and oranges situation. you have described two tires that have radically different carcass strength, right? if the tires are not of the same construction, which is implied because you said they dont have the same load rating, the tires cant be compared equally. we are trying to eliminate the structural integrity of the tire from the equation to discuss total footprint area of wide vs. skinny tires at equal tire pressure.


expeditionswest said:
Carcass strength and air pressure is what supports the load (vehicle and payload). Without the carcass, air has nothing to contain it.

true. and, without air pressure, the tire must support the load only on merit of sidewall strength. the point of adding air pressure is so that you relieve the vertical stress the sidewalls have to support. it is much easier to support an outward pressure, perpendicular to the sidewall, than a shear load, so air pressure is needed.


expeditionswest said:
Look at the side of a 255/85 and a 285/75. The sidewall is the same height. Now think of those tires without a rim; just resting against the garage wall. The tires maintain form because of the carcass construction. The heavier the load rating, the more they maintain form under load. Carcass' do not stretch (much), so if the air pressure in both is the same, and the only difference is width, why would the narrower tire elongate significantly more than the wide one? It wouldn't.

(i hate metric sizes... i cant ever remember what the actual dimension of the tire is... :o )

the narrower tire MUST elongate. this is REQUIRED by a ground pressure force, which i will discuss below.

expeditionswest said:
Please elaborate, thanks.

ground pressure is the force exerted on the ground by the vehicle. we all know this fairly clearly. if you put your foot under a tire, you dont feel the entire weight of the vehicle on your foot. you feel a fraction of it, that part that was supported by the tire from under which your foot now would be resting.

if, for round numbers, your 40 weighs 4000 lbs, regardless of tire size and shape, must exert 4000 lbs of force on the ground. the earth exerts 4000 lbs back up. your truck does not move. if the forces were not equal, the truck (and the earth to some very small degree) would move. this is very well described by newtons third law of motion.

now, if we remove sidewall strength and examine only the dimension of the tires that support the vehicle, we can examine the area of the tire required to exert the force needed to provide that support. you stated we are discussing tires of equal sidewall strength when it comes to total tire footprint area at equal tire pressure.

lets say that our four tires in the truck above (4000 lbs) equally load all four tires. (in the real world, the front usually has a bit more of a load because the center of gravity tends to be closer to the front of the vehicle.) if this is true, each tire must exert a force of 1000 lbs on the ground. if you lift one tire off the ground, the remaining three must then exert 1333.33 lbs each on the ground to keep the vehicle from accelerating up or down.

lets now say that each of our four tires are inflated to 10 psi and that each tire on the truck are of different widths. remember, sidewall strength is negated here, we are only going to discuss total area required to support the truck.

if each tire exerts 1000 lbs on the ground, if we do the math, that means the area of tire that touches the ground and exerts that force must be 100 si. (square inches - 1000 lbs divided by 10 lbs/square inch = 100 square inches) negating carcass strength, that is the REQUIRED area needed to exert force on the ground so that the truck does not accelerate up or down.

wide tires will support that 100 square inches by making a "wide" rectangle, the narrow tires will support that 100 square inches by making a "narrow" rectangle. but the total area of BOTH tires must be the same.

this is much like the argument of IFS vs. solid axle when a tire comes off the ground. force exerted on the ground cannot exceed, in this case, 4000 lbs. if one tire leaves the ground, the force on the ground remains only 4000 lbs, it is just distributed differently. a solid axle, with an unloaded spring, cannot provide more force to exert on the ground through the axle to the tire on the ground. (in the real world, an axle with one tire off the ground still supports the weight that was originally on that axle, just with one tire.) in fact, that one tire that is on the ground would actually support LESS than the weight originally applied to that axle because the rigidity of the frame of the vehicle will transfer some of that weight, albeit small, to the rear axle and its tires. IFS suspensions work the same because of frame strength.

here is another thought experiment. if you stand on a scale with two feet, your weight is a number. what if you stand on one foot? does your weight actually change? no, of course not. what does change is the pressure exerted by the remaining foot on the scale. your single foot must support the entire weight that, originally, both your feet supported. if the area of both of your feet are the same, then one of two things must happen to support your weight now on one foot. either the area of your foot touching the scale must double to support the weight at the same internal pressure originally applied by both your feet, or, the internal strength of your foot must be able to support a higher pressure, pounds per square inch, that is being exerted on the scale. fortunately, because we have "carcass strength" in our bodies, we dont have to worry about our feet crushing to double their area to support a higher load!

in the extreme, what if we supported that weight on the scale with the tip of a toe? your toe would hurt because the pressure required to support all that weight on such a small area is is much greater. but your weight on the scale never changes, right?

expeditionswest said:
Think of the example above. Just a naked tire carcass with no wheel and no pressure. It maintains its shape, and will even take a load on top of it without crushing. Once you combine a tire, a wheel (making a chamber) and air pressure the load capacity increases significantly. The narrower tire and wheel combo have the same internal pressure as the wide one, only the volume changes. So why would the narrower tire be subject to greater elongation than the wider one with the same load? It wouldn't.

well, here two issues are being confused. a tire with no wheel or air pressure is supporting itself with carcass strength. we cant compare area of ground contact if we include carcass strength because we cant quantify it. this also plays in when we include air pressure. carcass strength changes the whole equation, which i will discuss when i talk about the integral part...

expeditionswest said:
Sorry, I don't quite follow you here. Please provide more detail, thanks :beer:


ok, an integral basically takes into account the differential pressures under a tire when it is loaded. (integral, here really refers to calculus. an integral of a function is the sum of some force across an area, in this case.) because of sidewall/carcass strength, air pressure doesnt dictate what the actual force is across the entire footprint of the tire. if the tire is filled with air to a point where the sidewall does not support any of the weight of the vehicle with carcass strength, we could measure the footprint pressure on the ground and it would be equal to the air pressure in the tire, just as described above.

with carcass strength, the edge of the tire footprint underneath the sidewall actually supports more weight. this is partly why tires of different construction, with different sidewall/carcass strength, have different tire pressure recommendations. tires with strong sidewalls dont need as large an air pressure to help support the load and keep the tire flat on the ground. tires with weak sidewalls need more tire pressure to support the load.

(to be thorough, there is also a condition, called a boundary condition, that dictates that the force on the edge of the tire must be different than at the center of the tread because the conditions that govern what keeps the edge of the tire in are different, but we dont really need to discuss that right now. suffice it to say, the boundary condition is covered by the strength of the sidewall.)

in this case, the integral (or sum) of all the pressures multiplied by the areas beneath each pressure of all four tires MUST equal the weight of the vehicle. because of carcass strength, pressure is not equally distributed across the entire footprint of each tire, but the TOTAL pressure exerted by each of the tires must equal and support the weight of the vehicle. pressure under the edge is higher than in the center of the tread, but the average of all of them should be somewhere near the pressure of air inside the tire.

expeditionswest said:
Yes, true. Vertical load is vertical load. What changes between narrow and wide tires is how much load there is per square inch.

yes and no. load per square inch, or pressure, is dicated by carcass strength and air pressure, just as you have stated earlier in the thread. if you have a wide tire with very weak carcass strength, it will require MORE air pressure to support the weight of the vehicle. narrow tires, with greater sidewall strength, wont need as high an air pressure to support a load. if both tires are of equal sidewall strength and equal internal air pressure, the load per square inch MUST remain the same from one tire to the next, just as i stated earlier. but again, that would confuse to issues i am trying to detangle.

expeditionswest said:
The footprint will change, which will change the ground pressure for each square inch.

ground pressure for each square inch doesnt make sense. ground pressure is a per square inch measurement. a force per square inch. i think what you meant was pounds supported by each square inch must change, which is different. that means ground pressure, or lbs per square inch, would change, which is correctly stated.

all in all, i think your post did clarify to me what the issue was. there are posts in the thread that confuse the issues that are being discussed. sidewall/carcass strength is very different for tires of the same brand/make but different sizes, and it cant directly be compared here. generally speaking, tall 'n' skinny must have a stronger sidewall, or be filled with a higher air pressure than tall 'n' wide to support the same load. for that reason, comparing skinny vs. wide is much more complicated a venture than just discussing ground pressure. once you change the internal strength of the tire, it all goes out the window because the behavior of the tire, with or without air pressure, will change and a tire comparison from size to size is plainly a waste of time.

all in all, i have been trying to convince people that tall and skinny is the way to go anyway, so :cheers: !!
 
There is a lot to discuss from the above post, so let’s start small, and go through the basics: This discussion will likely resolve the other points.

stumpy said:
two tires of the same carcass construction and different widths inflated to the same tire pressure MUST have the same total footprint AREA, regardless of shape. that only follows from basic physics. i will explain more in a bit. wider tires will have wider footprints than skinny tires, but the total AREA of tire touching the ground MUST remain the same, assuming both tires have no internal rigidity or carcass strength.

You mention that your statement follows basic physics, so please provide the variable between the wide and narrow tire that contributes to the narrow tire being forced to elongate its contact area. No functional reasoning; please only provide the specific variable or formula.

Let’s break down the known variables.

Force:
Fv= Vertical load on the inflated tire
P = Internal Pressure, expressed in PSI (lb/in^2)

Dimensions:

A tire/wheel is expressed as a Torodial Membrane on a cylinder with the following dimension variables.

TW= Tread width
SH= Section height
RD= Rim Diameter
OD= Total Outside Diameter
V= Volume of air in (Cu Ft)
A= Area (in^2)

Other variables, value not known:
Viscoelasticity of the carcass, which is assumed to be the same for both the narrow and wide tire.

Required assumptions:

That the carcass material, construction (modulus, tensile strength, elasticity) are all the same between the narrow and wide tire with the exception of tread width.

That the two tires have the same load rating (C, D, E, etc.)

That the air pressure is the same: P

That the load is the same: Fv

The only difference is the tread width

Other important known definitions:

Structural Distortion Due to Vertical Load- The weight of the vehicle bears on the wheel and travels through the tire sidewall to the contact patch (Haney)

Internal volume of the tire/wheel NEVER changes (unless you add/remove air pressure), only the shape changes (Haney). As the load changes, the sidewalls not in the contact patch segment round out to maintain the same internal volume. A narrow and wide tire has two different internal volumes. *This point is not up for discussion. It is a well documented (and sited) fact.

Example: 33x9.5 against a 33x12.5

Fv= 5,000 lbs.
P= 40 psi

Let’s determine volume: Volume = pr2 x height

33x12.5x15: 6.41 Cu Ft, minus the rim volume of .81= V of 5.6 Cu Ft
33x9.5x15: 4.6 Cu Ft, minus the rim volume of .81= V of 3.79 Cu Ft

So, if a narrow and wide tire has different (V) values, yet the same (P), what would force a smaller (V) tire to displace its contact area to the same dimensions as a tire with greater (V)?

I will assume that you understand the compressibility of gases (in this case air/atmosphere), and that the narrower tire is a smaller container (V).

Before you answer, think about what force has changed between the two tires. Our known forces are defined and do not change: The (P) is the same in both tires and the (Fv) is the same.
 
stumpy said:
Wider tires will have wider footprints than skinny tires, but the total AREA of tire touching the ground MUST remain the same, assuming both tires have no internal rigidity or carcass strength.

I just noticed a critical point in your statement. It looks like we may agree after all.

Tires DO have internal rigidity and carcass strength, so if we assume that they don't, or have almost none like a rubber balloon than we both agree.

Sorry I missed that qualifier...:beer:



However, in reality, the tires carcass does have internal rigidity and carcass strength (which helps support the load and nearly eliminates stretching under typical pressures). It is the carcass itself (and the pressure inside it) that prevents the narrower tire from being forced into a larger overall contact area as a percentage of its volume to match the wider tire.

Ok... I think I am done with this topic.

I have to write a spec. for an unnamed client who is building an all new, fully self contained expedition vehicle to be sold in the US (lockers, tall tires, low gears, galley, pop top, winches, etc.). That is way more exciting than "Phun with Physics", wouldn't you say :beer:
 
expeditionswest said:
I have to write a spec. for an unnamed client who is building an all new, fully self contained expedition vehicle to be sold in the US (lockers, tall tires, low gears, galley, pop top, winches, etc.). That is way more exciting than "Phun with Physics", wouldn't you say :beer:

Clearly you and they will need beta testers. I'm sure there are lots of people here that will leap at the chance to break stuff for you.
 
expeditionswest said:
You mention that your statement follows basic physics, so please provide the variable between the wide and narrow tire that contributes to the narrow tire being forced to elongate its contact area. No functional reasoning; please only provide the specific variable or formula.


F=ma

where F = force, m = mass, a = acceleration

there is no reason to get into the specifics of the torroidal forces and weight distribution to get from F=ma into something that is specifically correct. the point remains that the force remains the same on all four tires (neglecting the mass of the tire), regardless of footprint area, because the mass of the truck does not change, and the acceleration of gravity does not change.

the ultimate equation is exactly as i stated in my previous post. ground pressure = force / area

to be more precise in how to get to that, i would have to define a function that relates tire circumference and footprint area to internal tire air pressure and then relate that to volume which is well beyond the scope of this forum, and well beyond my interest in number crunching!

expeditionswest said:
Structural Distortion Due to Vertical Load- The weight of the vehicle bears on the wheel and travels through the tire sidewall to the contact patch (Haney)

ah, but we are ignoring this because we must assume that internal/carcass strength is equal between wide and narrow tires.

expeditionswest said:
Internal volume of the tire/wheel NEVER changes (unless you add/remove air pressure), only the shape changes (Haney). As the load changes, the sidewalls not in the contact patch segment round out to maintain the same internal volume. A narrow and wide tire has two different internal volumes. *This point is not up for discussion. It is a well documented (and sited) fact.

and you have just verified what i was trying to point out by stating this. SHAPE must change. the internal volume of the tire may never change, and it doesnt have to. the volume of AIR you put into the tire does change though, which is why pressure goes up when you fill the tire.



expeditionswest said:
33x12.5x15: 6.41 Cu Ft, minus the rim volume of .81= V of 5.6 Cu Ft
33x9.5x15: 4.6 Cu Ft, minus the rim volume of .81= V of 3.79 Cu Ft

So, if a narrow and wide tire has different (V) values, yet the same (P), what would force a smaller (V) tire to displace its contact area to the same dimensions as a tire with greater (V)?

well, you havent accounted for the fact that the volume of air inside the tire is NOT the same as the volume of the space contained by the tire and the wheel. if we have the same pressure, the volume of air inside both tires to get to that pressure is different. this is why, as you most certainly know, it takes a lot longer to fill a 35" tire to 30 psi than a 31" tire to the same 30 psi.

the situation described above oversimplifies the relationship between internal volume of the tire, pressure and what it actually is, and the relation of the two to footprint area. you cannot equate the volume of the tire to the footprint area without accounting for the volume of the air required to fill the tire and/or the pressure at which the tire is filled.

the fundamental point remains - ground pressure must remain the same, and excluding carcass strength, if the internal air pressure of both a skinny and wide tire are the same, the area of tire required to apply the appropriate force on the ground must remain the same for both tires.



expeditionswest said:
Before you answer, think about what force has changed between the two tires. Our known forces are defined and do not change: The (P) is the same in both tires and the (Fv) is the same.

nothing has changed between either tire, including the fact that we have assumed no internal strength. which is why the footprint area MUST be the same.


expeditionswest said:
I just noticed a critical point in your statement. It looks like we may agree after all.

i am still not sure we do. you stated the qualifier a number of posts back, which is why i was compelled to post up when you mentioned wide and skinny tires cannot have the same footprint. that is what confused me!


expeditionswest said:
Tires DO have internal rigidity and carcass strength, so if we assume that they don't, or have almost none like a rubber balloon than we both agree.

true, tires have internal strength. they must, as you have stated. which is why real world performance is very difficult to quantify from size to size and tire to tire.

if you read through the posts again, it is clear we are of a different opinion in the hypothetical model. in the real world, it doesnt matter. but for the scope of your discussion on the exp.west site, it can be misleading.

expeditionswest said:
Sorry I missed that qualifier...


no worries. sounds like we both have better things to do! :bounce2:
 
I have more glaze on my eyeballs than Krispy Kreme has on it's donuts... it is a good edjumacation but will take me a couple days to really read the posts... you are both saying you like narrow tires?

So you both like narrow donuts but disagree on the glaze?

The only easy choices are 285's, 305's & 315's. In most cases I am happy w/ my 285's but I have to admit, if I hit a rocky trail, my mind immediately goes wishing for 315's.
underaxle clearance, puffy sidewalls for traction/protect rim, better grip (I know, this point is debatable in certain situations)
 
FirstToy said:
you are both saying you like narrow tires?


i think so. but i am wierd, i still roll on 31x10.50s on a 1.5" shackle lift on my 40. and i still use armstrong steering to drive the rubicon. someday, someday, i will find a set of 33x950s that i like. i would love to get back a mud terrain, but i only own a 40, so i need good streetability too.

truxus?
 
stumpy said:
...someday, someday, i will find a set of 33x950s that i like. i would love to get back a mud terrain, but i only own a 40, so i need good streetability too.

truxus?

Cooper Discoverer STTs? Factory siped, not very loud, 3 ply sidewall...

Oops, getting OT. ;)
tire.webp
 
This needs to be the last detailed reply from me (due to time), but I felt it was important as Stumpy and I were pretty close to a conclusion.

If anyone wants to take this offline or by phone, shoot me an email: guide@expeditionswest.com

stumpy said:
F=ma

where F = force, m = mass, a = acceleration

there is no reason to get into the specifics of the torroidal forces and weight distribution to get from F=ma into something that is specifically correct. the point remains that the force remains the same on all four tires (neglecting the mass of the tire), regardless of footprint area, because the mass of the truck does not change, and the acceleration of gravity does not change.

Thanks for providing the formula, which is Newton's second law. Unfortunately, it does nothing to resolve the AREA problem in the post above. We already know what the Fv is.

stumpy said:
the ultimate equation is exactly as i stated in my previous post. ground pressure = force / area.

Yep, but the AREA is a variable. Bigger AREA means less ground pressure per square inch. Less AREA means more ground pressure per square inch. That has been my point all along.

stumpy said:
to be more precise in how to get to that, i would have to define a function that relates tire circumference and footprint area to internal tire air pressure and then relate that to volume which is well beyond the scope of this forum, and well beyond my interest in number crunching!.

That amigo is the key to the puzzle. Volume changes between the two widths. The tire carcass structure and pressure is the same. That is why a wider tire will have more AREA VOLUME on the ground and a narrower tire will have smaller AREA VOLUME on the ground.

stumpy said:
ah, but we are ignoring this because we must assume that internal/carcass strength is equal between wide and narrow tires..

It was mentioned to help maintain continuity of the post. It cannot be ignored as it is integral to how all of this works. Tires have structure.

stumpy said:
and you have just verified what i was trying to point out by stating this. SHAPE must change. the internal volume of the tire may never change, and it doesn't have to. the volume of AIR you put into the tire does change though, which is why pressure goes up when you fill the tire..

Now we are getting to the heart of this... We DO NOT disagree on shape changing, because the shapes are different (one is wide and one is narrow). What is different is the total area of the shapes.

This has been my point all along. A narrower tire has less volume for the same pressure. If you lower the pressure to increase the AREA, then you are changing a required constant variable for the comparison. A narrow tire is a smaller cylinder with less volume, which is why at the same pressure it has a smaller contact area. This is the most basic of physical laws.

stumpy said:
the internal volume of the tire may never change, and it doesnt have to.

We are getting even closer.... I agree. And I hope you agree that a narrow tire and wide tire have different volumes.

stumpy said:
the volume of AIR you put into the tire does change though, which is why pressure goes up when you fill the tire..

Yep, been saying that for a few posts now.

stumpy said:
well, you havent accounted for the fact that the volume of air inside the tire is NOT the same as the volume of the space contained by the tire and the wheel.

How did I not account for this?:confused: Did you read the dimensions and volume sections of my previous post? :D I even did the math for us, which was no fun at all ;p

Again. VOLUME is different between a wide and narrow tire of the same diameter, and with the same wheel dimension.

stumpy said:
if we have the same pressure, the volume of air inside both tires to get to that pressure is different. this is why, as you most certainly know, it takes a lot longer to fill a 35" tire to 30 psi than a 31" tire to the same 30 psi.

I said this too (several times).

stumpy said:
the situation described above oversimplifies the relationship between internal volume of the tire, pressure and what it actually is, and the relation of the two to footprint area. you cannot equate the volume of the tire to the footprint area without accounting for the volume of the air required to fill the tire and/or the pressure at which the tire is filled.

You are there compadre. :idea:

I did not oversimplify. This is what makes it all work (and make sense). Internal volumes are different yet the pressure is the same. The pressure and carcass is the same, so the load support variable requirements are the same. That equates to a different AREA because the narrower tire is a smaller cylinder.

How did I not account for the volume of air required to fill the tire? I even calculated the volume in Cu Ft for us. :cool:


stumpy said:
the fundamental point remains - ground pressure must remain the same, and excluding carcass strength, if the internal air pressure of both a skinny and wide tire are the same, the area of tire required to apply the appropriate force on the ground must remain the same for both tires.

You cannot exclude carcass strength. Why would you exclude the carcass in a discussion related to deformation and contact area of wide and narrow tires? I am sorry if I am missing something.

Total ground pressure does remain the same for both tires. IT MUST (as you say), as Fv is the same. However, ground pressure per square inch is what changes because the area (in inches) changes.

Question: This might resolve the debate...

IF you included carcass strength into the variables of narrow and wide tires, do you believe that AREA would vary between the two?

stumpy said:
nothing has changed between either tire, including the fact that we have assumed no internal strength. which is why the footprint area MUST be the same.

Sorry, but I cannot see the value of debating narrow and wide tire contact areas without factoring for the container (carcass).

stumpy said:
i am still not sure we do. you stated the qualifier a number of posts back, which is why i was compelled to post up when you mentioned wide and skinny tires cannot have the same footprint. that is what confused me!

It is a fact. I have measured it myself, on the ground and with dozens of sizes.

You measure it by fitting tires to wheels and wheels to trucks. That is what I have done. Take a set of 33x9.5's and test them, measure them, take contact area imprints, etc. Then do the same with a 33x12.5"

Real world results is what we all want. When I compete in off road events, I test everything, check everything and measure everything. The Devil is in the details...

stumpy said:
true, tires have internal strength. they must, as you have stated. which is why real world performance is very difficult to quantify from size to size and tire to tire.

This statement does make me see the light at the end of this tunnel (hopefully not a freight train coming my way
guitar.gif
) for this discussion. I believe we are closer to agreement now than ever.





Stumpy,

I think we are getting much closer to agreement here. The problem is trying to maintain a dynamic discussion in a forum, and then try to weed through all of the quoting, etc. while trying to understand each others (too brief) descriptions.

Maybe we can talk about this someday around a campfire with a :beer:
 
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FirstToy said:
you are both saying you like narrow tires?

Yeah, we both like narrow tires, Toyota's and (obviously) a good, healthy debate.

I bet he also likes a stout beer :beer:

Any man who drives the Rubicon with manual steering has my respect for life:cool:
 
As for out of US goes here in Canada during the winter, the less wide they are the better traction you get on snow. But in deep snow, you'd want wider deflated tires.. Depends on what kind of surface you travel the most.

Matt
 
expeditionswest said:
This needs to be the last detailed reply from me ...


well, i wont waste time with a detailed response as well then. it is apparent from your post that you are still debating the real-world conditions which is not at all what i was trying to contend. you had made a statement about a hypothetical model that i was trying to point out didnt work in physics, and now i cant find the original statement to point it out. :doh:

in the end, it doesnt matter anyway. i believe skinny is better, so does scott. who cares.


campfire someday.... :beer:

off to OURAY!!!
 
Anyone with "skinny" tires please post pics. Looking for pics of 35x10.50, Q78, etc.
 
Man, that is one scary looking "bridge" :eek:
 

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