## Please stay on the trail.

FAQ and threads for those just starting to hike the Colorado 14ers.
jdorje
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### Re: Please stay on the trail.

Sorry to rehash topics that so many have covered already, but I couldn't resist.

As far as PSI goes, certainly an inflated tire acts to distribute pressure over area. If you inflate your tire to a reasonable level while weighted, the tire deforms and the weight is distributed close-to-evenly over the area. This lowers the maximum PSI inflicted on the ground and reduces road damage. Of course if the weight* rises (most likely due to dropping down a foot, if you're riding down a mountain) the forces all rise temporarily and the PSI will increase inside the tire just as the PSI from tire to ground increases, but the distribution effect still acts (imperfectly no doubt) to lower the maximum forces. As you raise the PSI within your tire, it deforms less and there is less weight redistribution. As you lower the PSI, the weight will be more and more distributed, but your tire will also wear faster and eventually your wheel rims will just be resting on the ground. No math should be needed to understand this process; think of it like deforming a balloon in your hands.

But what I'm not sure on is why this matters all that much. Most of the time the force of a bike is straight down (no skidding) - and this surely does quite low damage to the ground. What tears up ground is friction, from a bike skidding or from a hiker's foot slipping slightly. This (along with water flow) is why the steeper trails erode so much faster, and why there's so much faster erosion in wet weather. If a bike does skid, the distributed force from the tires is going to be a lot less distributed; likewise if a hiker slips it's often with all the weight on the toes.

* By weight, I mean force, specifically the downward force of the bike on the ground. I realize this is technically inaccurate since weight/force is the force of gravity on the object which would indeed be constant. But the force between ground and bike is going to vary if any bounce is present.
-Jason Dorje Short
ajkagy
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### Re: Please stay on the trail.

jdorje wrote:This (along with water flow) is why the steeper trails erode so much faster, and why there's so much faster erosion in wet weather.

weather and water effect poorly designed and built trails more than any hiker/biker could ever effect it. All it takes is 1 big deluge to do some pretty serious damage.
tlongpine
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### Re: Please stay on the trail.

HikerGuy wrote:
Dex wrote:You are equating internal psi with external psi on the ground.

Where is Bean2 when you need him.

Internal PSI = External PSI. Simple physics, opposite and equal forces. Bean is correct in the example he uses. The pressure in a tire does not change due to load.

Following this line of reasoning I can conclude that the best way to reduce trail damage as possible is to let all the air out of my tires. Afterall, if "Internal PSI = External PSI", then 0 PSI in the tires means 0 PSI exerted by the bike upon the surface of the trail.

Right?
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It can wait forever. I cannot.
Bean
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### Re: Please stay on the trail.

tlongpine wrote:Afterall, if "Internal PSI = External PSI", then 0 PSI in the tires means 0 PSI exerted by the bike upon the surface of the trail.

Right?

Sort of. You just need a big enough tire and you can drop your pressure to almost nothing.

Otherwise you'll be cutting through with your rims.
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wildlobo71
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### Re: Please stay on the trail.

I'm reading, I'm comprehending - but I see tlongpine's statement about the 0 PSI and I am perplexed.

The required PSI of a tire is to support the weight of the bike and rider off of the tire's pressure patch area that is in contact to the ground, it has nothing to do with the force of the whole thing applied to the ground - therefore reducing the PSI in the tire to 0 is akin to... hovering? A 200 pound bike/rider assembly still has 200# of dead weight (force) applied at all times to the ground, and relatively the same force applied once it gets moving, regardless of the tire pressure. The force of the assembly is still cutting a path in the softer soil.

Please stay on the trails people (just to keep from derailing.)
You've never really stopped even if you feel like you have.
ajkagy
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### Re: Please stay on the trail.

P = F / A
Pressure = Force / Area

This should clear some things up
http://en.wikipedia.org/wiki/Ground_pressure
jdorje
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### Re: Please stay on the trail.

wildlobo71 wrote:The required PSI of a tire is to support the weight of the bike and rider off of the tire's pressure patch area that is in contact to the ground, it has nothing to do with the force of the whole thing applied to the ground - therefore reducing the PSI in the tire to 0 is akin to... hovering? A 200 pound bike/rider assembly still has 200# of dead weight (force) applied at all times to the ground, and relatively the same force applied once it gets moving, regardless of the tire pressure. The force of the assembly is still cutting a path in the softer soil.

If you drop the pressure, the tire spreads out, giving the same weight distributed over more area, thus less ground pressure. The inward and outward pressures will more or less equalize, though the stability of the tire will hold up some of the weight as well.

Imagine balancing a dinner plate on top of a balloon - the weight of the plate will raise the pressure inside the balloon to a level sufficient to hold up the plate's weight. Likewise putting more weight on your bike will raise the internal pressure in the tires until the weight is balanced.

If you drop the pressure too much (or put too much weight on the bike), the tire does nothing and your wheels will just be resting on the ground.
-Jason Dorje Short
crossfitter
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### Re: Please stay on the trail.

jdorje wrote:
wildlobo71 wrote:The required PSI of a tire is to support the weight of the bike and rider off of the tire's pressure patch area that is in contact to the ground, it has nothing to do with the force of the whole thing applied to the ground - therefore reducing the PSI in the tire to 0 is akin to... hovering? A 200 pound bike/rider assembly still has 200# of dead weight (force) applied at all times to the ground, and relatively the same force applied once it gets moving, regardless of the tire pressure. The force of the assembly is still cutting a path in the softer soil.

If you drop the pressure, the tire spreads out, giving the same weight distributed over more area, thus less ground pressure. The inward and outward pressures will more or less equalize, though the stability of the tire will hold up some of the weight as well.

Imagine balancing a dinner plate on top of a balloon - the weight of the plate will raise the pressure inside the balloon to a level sufficient to hold up the plate's weight. Likewise putting more weight on your bike will raise the internal pressure in the tires until the weight is balanced.

If you drop the pressure too much (or put too much weight on the bike), the tire does nothing and your wheels will just be resting on the ground.

The internal pressure really won't change very much. It will change a little as the tire deforms, sure, but that deformation is a small fraction of the total volume of the tire. In your balloon example, what really changes is the contact area between the balloon and plate. In the 0 psi case, the entire load is carried by the frame and the ground pressure is extremely high due to the thin edges and rigidity of the rims.
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wildlobo71
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### Re: Please stay on the trail.

In conclusion, I am happy being an architect.

(Stay on the trails, with properly inflated tires, people.)
You've never really stopped even if you feel like you have.
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### Re: Please stay on the trail.

crossfitter wrote:
The internal pressure really won't change very much. It will change a little as the tire deforms, sure, but that deformation is a small fraction of the total volume of the tire. In your balloon example, what really changes is the contact area between the balloon and plate. In the 0 psi case, the entire load is carried by the frame and the ground pressure is extremely high due to the thin edges and rigidity of the rims.

Vids
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### Re: Please stay on the trail.

Here is an example to help things along (using round numbers for simplicity's sake):

200 lb rider + 20 lb bike = 220 lb total weight

10 square inches of each tire contacting ground (assumes 2 inches wide and 5 inches long contact area) = 20 square inches contact area

220 lb / 20 sq in = 11 lb/sq in (or psi) of force on ground at each tire (I'm neglecting the fact that probably more weight is on back tire to make things easy)

The only way the internal tire psi affects the psi exerted on the ground is by changing the contact area (less psi = bigger contact area). But once you get to 0 psi or whatever point when the wheel touches the ground your force increases dramatically since your only contact area is the two metal sides of the wheel, which actually act like point forces rather than distributing the weight.

For the purposes of this debate I recommend ignoring the 0 psi example unless you happen to know people who like to mountain bike without tires on.
MUni Rider
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### Re: Please stay on the trail.

Now what?

shoe-bike-704130.jpg (26.31 KiB) Viewed 283 times
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