100 series performance in soft sand

divemedic

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I’m not going to get into an argument about this. I will engage in healthy debate. Saying wide tires are better in sand is categorically false as it relates to this thread.

Realistically, tire pressure effects tire performance more than tire profile (we’re not comparing arctic truck tires to tractor tires).

That said:
A 33x12-in wide tire has a wide/short contact patch.

A 33x10-inch wide tire has a narrow/long contact patch.

Both have a similar/the same contact patch surface-area. The potential for grip is equal. However the narrow/long contact patch of skinny tires has less rolling resistance due to the narrow “leading edge” of the tires. None of this is circumstantial. If you want the math I can provide it.

Some random dude’s advise based on a butt-dyno and the recommendations of Chaz at Jeep-parts-r-us is circumstantial. So if you feel wide tires are better, cool. What is the base of your assertion?

Regardless, for the OP and anyone else reading the thread, do your own research and
form your own conclusions.
 

nukegoat

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If you want the math I can provide it.
I do, please elaborate.

I mean, at x psi, sure, a skinny tire elongates the contact patch in a way that's preferable, but for a y% wider tire, one can run y% lower tire pressure. In other words, it seems you could get an equal length contact patch out of a wide tire simply by lowering tire pressure further, since the overall contact patch is wider to start. And at that point, the whole system is more bouyant etc.

But please present the math
 
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Why are narrower tires better in sand? Dune buggy tires are very wide (granted they usually have paddles). Also, arctic trucks run wider tires and driving in snow is pretty similar to driving in sand. Why would I want to cut deeper down into the sand with a narrow tire when I can float on top of it with a wider one?
 

divemedic

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I do, please elaborate.

I mean, at x psi, sure, a skinny tire elongates the contact patch in a way that's preferable, but for a y% wider tire, one can run y% lower tire pressure. In other words, it seems you could get an equal length contact patch out of a wide tire simply by lowering tire pressure further, since the overall contact patch is wider to start. And at that point, the whole system is more bouyant etc.

But please present the math
The unit for pressure (pascal) is force per unit area. One pascal is one newton per square meter. A kilopascal (kPa) is 1000 newtons per square meter. Pressure can be calculated according to the formula:

Pressure = Force / Area

Manipulating this equation, we can get a formula for area, which we will use to calculate contact patch:

Area = Force / Pressure

So contact patch is dependent on force and pressure and has no regard for dimensions. What this means is at the same pressure, a narrower tire will deform more than a wider tire to achieve the same contact patch. Any sized tire will deform just the right amount to achieve the contact patch described by the formula above.

Assume a tire is inflated to 40 psi. This translates to 275 kPa. Lets assume a weight on that tyre of 400 kg. To calculate force from weight, multiply by acceleration due to gravity (9.8m/s/s):

Force = 9.8 x 400 = 3920 N

Now we can calculate area:

Area = 3920 / 27500 = 0.0142 square meters or 142 square centimeters

So the contact patch of a tire under these conditions is about 142 square centimeters. This is independent of tire dimensions. If the tire is 20 cm wide then the length of the contact patch is about 7.1 cm (7.1 x 20 = 142). If the tire is 30 cm wide then the length of the contact patch is 4.73 cm.

What if we halve the pressure?

Area = 3920 / 1375000= 285 square centimeters

Halving the pressure doubles the contact patch. This occurs for a tire of any dimensions.

Tires aren’t perfectly flexible. These formulas assume a perfectly flexible tire where the rigidity of the tire does not contribute to holding the weight of the vehicle (why people are posting their load range). In reality the tire would contribute. For example the sidewalls of a tire can support some weight and so they would contribute to supporting the weight of a vehicle. This means the calculations are not very accurate. However the sidewall of any similarly rated and constructed tire will contribute in a similar fashion. This means in a comparison between tires of the same load range (fat vs thin), the effect of sidewall mostly cancels out and the fact that contact patch is governed by pressure essentially holds true (but the relationship is no longer linear). This leads the following conclusion:

Two tires of similar construction, of different dimensions, will have about the same contact patch when at equal pressure. But the shape of that contact patch will differ based on the shape of the tire.
 

divemedic

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Why are narrower tires better in sand? Dune buggy tires are very wide (granted they usually have paddles). Also, arctic trucks run wider tires and driving in snow is pretty similar to driving in sand. Why would I want to cut deeper down into the sand with a narrow tire when I can float on top of it with a wider one?
arctic trucks have tires specifically constructed so they have equal parts sidewall to width. They’re essentially cylindrical spheres if that makes sense? Their size is selected specifically to displace the weight of the vehicle. @cruiseroutfit (Kurt) is the only person I know on this site who has functional experience with this (excluding myself). Arctic truck tires are equally impractical as tractor tires. More importantly you’re not going to find an off-shelf tire which duplicates the floatation effects.

dune buggy’s are generally rear wheel drive and have paddles accordingly. The fronts are usually very narrow. Again this application is not practical for many more reasons.

go look at videos on YouTube of Saudi’s and Qataris slaying dunes in Land Cruisers. How many have tall/wide tires? How many have tall/skinny tires?

look at the dune sections of Dakar. Same question.
 

e9999

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one can debate endlessly about the tires...
Another big issue, though, is that there is sand and then there is sand!
It can make a night and day difference.
I have a bit of experience with some local beach sand dunes. The 100 does very well there, even at asphalt pressures. Not as capable as the 80 around steep dunes but pretty good. With minimal attention paid to dune profiles and the like, I'm a good shape and can go most places. On the flat, no problem at all. The sand is sort of coarse, more like salt than flour. I suspect most beach sand is like that.
Now, from my limited experience, fine powder sand inland can be a different beast altogether. Like very fine flour or powdered sugar, and could easily be a nightmare and take down many trucks. There you'd want to go real low pressure for example, and driving skills would be more important.
As to Polihale specifically, if it's the one at the very Northwestern end of the southern road it is such a wonderful and beautiful beach that I would rather not take my truck on the beach itself. But that is a subjective matter best not addressed here I suppose. Having said that, I remember going up to the last dunes row oceanside in sedan 2 wheel drive cars in fairly deep sand and did just fine, so I suspect the beach itself should be relatively easy too.
 

nukegoat

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The unit for pressure (pascal) is force per unit area. One pascal is one newton per square meter. A kilopascal (kPa) is 1000 newtons per square meter. Pressure can be calculated according to the formula:

Pressure = Force / Area

Manipulating this equation, we can get a formula for area, which we will use to calculate contact patch:

Area = Force / Pressure

So contact patch is dependent on force and pressure and has no regard for dimensions. What this means is at the same pressure, a narrower tire will deform more than a wider tire to achieve the same contact patch. Any sized tire will deform just the right amount to achieve the contact patch described by the formula above.

Assume a tire is inflated to 40 psi. This translates to 275 kPa. Lets assume a weight on that tyre of 400 kg. To calculate force from weight, multiply by acceleration due to gravity (9.8m/s/s):

Force = 9.8 x 400 = 3920 N

Now we can calculate area:

Area = 3920 / 27500 = 0.0142 square meters or 142 square centimeters

So the contact patch of a tire under these conditions is about 142 square centimeters. This is independent of tire dimensions. If the tire is 20 cm wide then the length of the contact patch is about 7.1 cm (7.1 x 20 = 142). If the tire is 30 cm wide then the length of the contact patch is 4.73 cm.

What if we halve the pressure?

Area = 3920 / 1375000= 285 square centimeters

Halving the pressure doubles the contact patch. This occurs for a tire of any dimensions.

Tires aren’t perfectly flexible. These formulas assume a perfectly flexible tire where the rigidity of the tire does not contribute to holding the weight of the vehicle (why people are posting their load range). In reality the tire would contribute. For example the sidewalls of a tire can support some weight and so they would contribute to supporting the weight of a vehicle. This means the calculations are not very accurate. However the sidewall of any similarly rated and constructed tire will contribute in a similar fashion. This means in a comparison between tires of the same load range (fat vs thin), the effect of sidewall mostly cancels out and the fact that contact patch is governed by pressure essentially holds true (but the relationship is no longer linear). This leads the following conclusion:

Two tires of similar construction, of different dimensions, will have about the same contact patch when at equal pressure. But the shape of that contact patch will differ based on the shape of the tire.
Yeah but who would run the exact same pressure after upgrading to a wider tire?
 

nukegoat

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You might as well add random caps To your response.
Why? I'm not following your insult, or your logic. I've been pretty diplomatic (for me, anyway) and you're ignoring my counterarguments. I thought you said you wanted to engage in a healthy debate? I thought I was going to get math? I didn't see any sort of provisions for the effect of the mating surface (i.e. the sand) w/ respect to the contact patch. You assume the sand is infinitely rigid, which is a pretty heroic assumption.

Hey, but you know what, I'll give you another chance. Please express your arguments using SAE standards:


Otherwise, for everyone else, don't run bicycle tires in sand.
 

divemedic

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Why? I'm not following your insult, or your logic. I've been pretty diplomatic (for me, anyway) and you're ignoring my counterarguments. I thought you said you wanted to engage in a healthy debate? I thought I was going to get math? I didn't see any sort of provisions for the effect of the mating surface (i.e. the sand) w/ respect to the contact patch. You assume the sand is infinitely rigid, which is a pretty heroic assumption.

Hey, but you know what, I'll give you another chance. Please express your arguments using SAE standards:


Otherwise, for everyone else, don't run bicycle tires in sand.
You’re not following my insult because there isn’t one. You are looking for one. This is reason 1 this isn’t a healthy debate.

how can I counter you’re areguments? You really haven’t made any; reason 2 this isn’t a healthy debate.

I don’t assume anything about surface friction. They would be applied mathematically equal to both wide and narrow tires. You can add what ever variables you want you your rebuttal. We’re all still waiting for something with substance.

Thank you for “allowing” me to post on this forum by “giving me another chance”. Maybe your use to bring the smartest guy in the room, which it fine.

So why don’t you explain the diagram for me and for everyone else? It looks like a diagram pulled from the alignment to principles section of a ASE training manual. I’m not a mechanic so that’s my best guess. I honestly have no idea how it correlates to this discussion, so break it down for me if you don’t mind? it would be best if you could use it to reinforce your opinion that wide tires are narrow?
 

nukegoat

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You’re not following my insult because there isn’t one. You are looking for one. This is reason 1 this isn’t a healthy debate.

how can I counter you’re areguments? You really haven’t made any; reason 2 this isn’t a healthy debate.

I don’t assume anything about surface friction. They would be applied mathematically equal to both wide and narrow tires. You can add what ever variables you want you your rebuttal. We’re all still waiting for something with substance.

Thank you for “allowing” me to post on this forum by “giving me another chance”. Maybe your use to bring the smartest guy in the room, which it fine.

So why don’t you explain the diagram for me and for everyone else? It looks like a diagram pulled from the alignment to principles section of a ASE training manual. I’m not a mechanic so that’s my best guess. I honestly have no idea how it correlates to this discussion, so break it down for me if you don’t mind? it would be best if you could use it to reinforce your opinion that wide tires are narrow?
It's not an ASE training manual, it's from the Society of Automotive Engineers (SAE) - Fundamentals of Vehicle Dynamics

Regardless, my basic arguments are these. Let me enumerate each point:

1: Your sole argument is saying that for a given set of tires, the contact patch width is a function of the tire width, while the contact patch length is dependent on tire diameter (I assume we're holding this constant), tire pressure, and tire width. I agree with that. But why couldn't you create a longer contact patch simply by reducing tire pressure in a wider tire?

2: I believe you're ignoring that the mating surface (the sand, in this case) is a function of (essentially) bouyancy. A wider and fatter tire will "sink" less since it would require displacing more sand.

3: Taking each example to the extreme sometimes helps provide a "sniff test" to an argument. So let's try it: On the sand, assuming similar diameters, are you really saying that a bicycle tire would be preferred to the rubber equivalent of a barrel?
 

nukegoat

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Oh, and for anyone who thinks @divemedic is providing his own mathematical arguments, he literally is directly copying and pasting from "Outback Joe's" website while criticizing "Chaz at jeep parts r us."

Ridiculous.

 
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Oh, and for anyone who thinks @divemedic is providing his own mathematical arguments, he literally is directly copying and pasting from "Outback Joe's" website while criticizing "Chaz at jeep parts r us."

Ridiculous.

Not entirely true, tyres was changed to tires :)
 
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Ive always heard a tall skinny tire does better in sand. So then why does a mountain bike tire perform worse in sand vs a fat tire bike? Just curious...
 

nukegoat

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Ive always heard a tall skinny tire does better in sand. So then why does a mountain bike tire perform worse in sand vs a fat tire bike? Just curious...
Just go to outback Joe's website for all yer learnings!
 

divemedic

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It's not an ASE training manual, it's from the Society of Automotive Engineers (SAE) - Fundamentals of Vehicle Dynamics

- Good, now how does the diagram apply to your argument?

Regardless, my basic arguments are these. Let me enumerate each point:

1: Your sole argument is saying that for a given set of tires, the contact patch width is a function of the tire width, YES.

while the contact patch length is dependent on tire diameter (NO. Contact patch length is relative to the width compared to height of the tire. 33x12 vs 33x10 -or- 37x15 vs 37x12

(I assume we're holding this constant), tire pressure, and tire width. I agree with that. But why couldn't you create a longer contact patch simply by reducing tire pressure in a wider tire?

- Yes a wide tire will get longer and wider. A narrow tire will get longer and wider. The shape of their respective contact patch will be the same as their design.

2: I believe you're ignoring that the mating surface (the sand, in this case) is a function of (essentially) bouyancy. A wider and fatter tire will "sink" less since it would require displacing more sand.

They aren’t displacing different amounts of sand. They’re displacing a different shape.

3: Taking each example to the extreme sometimes helps provide a "sniff test" to an argument. So let's try it: On the sand, assuming similar diameters, are you really saying that a bicycle tire would be preferred to the rubber equivalent of a barrel?

- This is an argumentative fallacy. You’re arguing extremes. As I’ve stated before comparing an arctic truck tire to a tractor tire isn’t apples to apples, nor is it practical for our trucks.
 
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