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Please stay on the trail.

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

Postby wildlobo71 » Tue Oct 02, 2012 12:06 pm

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.)
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Re: Please stay on the trail.

Postby ajkagy » Tue Oct 02, 2012 12:07 pm

P = F / A
Pressure = Force / Area

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

Postby jdorje » Tue Oct 02, 2012 12:23 pm

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.
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Re: Please stay on the trail.

Postby crossfitter » Tue Oct 02, 2012 12:31 pm

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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Re: Please stay on the trail.

Postby wildlobo71 » Tue Oct 02, 2012 12:38 pm

In conclusion, I am happy being an architect.

(Stay on the trails, with properly inflated tires, people.)
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Re: Please stay on the trail.

Postby madbuck » Tue Oct 02, 2012 12:46 pm

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.


Crossfitter's comments on this thread remain correct.

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Re: Please stay on the trail.

Postby Vids » Tue Oct 02, 2012 12:57 pm

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. :?

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Re: Please stay on the trail.

Postby MUni Rider » Tue Oct 02, 2012 1:23 pm

Now what? :-k

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shoe-bike-704130.jpg (26.31 KiB) Viewed 205 times
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Re: Please stay on the trail.

Postby madbuck » Tue Oct 02, 2012 1:31 pm

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


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 the support of a properly inflated tire is precisely the dominant mechanism for supporting the weight of the bike and rider (with a much lesser, but nonzero, influence of the tire's rigidity. Consider a tire closer to a balloon (almost no rigidity) and a steel tank (very much), to use examples earlier in this thread; also consider how a 20lb bike sinks to the ground on a flat tire, to get an idea of the degree of support from the sidewalls alone).

A cushion of air in a tire -- if sufficient pressure exists to keep the rim off the ground -- is responsible for nearly all of the support of the weight above it.
A tire is a container for holding sufficient air to support the weight above it. (Bridgestone).
The volume of the tire remains nearly constant (sinks lower and stretches fatter); the mass of the air in the tire remains constant; the pressure remains nearly constant.

The issue with your example is that the numbers are both simplifed and made up :-D
But if those were real weight and contact patch sizes, the pressure in the tire to yield those results would be close to 11psi.

See mtnfiend's actual experiment earlier in the thread. It was a great match of empirical observation to theory -- and he began the experiment being dubious of the theory!

In very low psi examples, when the pressure isn't enough to hold up the entire weight, gravitational forces are sufficient to overcome all of the air pressure in the tires, and any remaining unsupported weight is supported by the structure of the bike. Maybe this is the disconnect between people saying "internal psi != external psi" -- because it's an example when not all of the support is given by the air pressure.

Perhaps it helps to think of an inflated tire as a crude scale: the more weight pushing down on it, the larger the contact patch is, up until the internal pressure of the tire is overcome. In that case, the analogy would be a fat person standing on a spring-loaded scale, and bottoming out the scale! All we could conclude is that the weight was at least heavy enough to overcome the support of the air (or spring).

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Re: Please stay on the trail.

Postby TallGrass » Tue Oct 02, 2012 1:34 pm

w = mg; weight equals mass times acceleration of gravity
f = ma; force equals mass times acceleration
Gravity is often treated like a constant (9.8m/s^2, 32.2'/s^2), but it varies by altitude, crust (oceanic, continental), as well as ground composition (e.g. if standing above a large ore deposit) hence the need for different equations, especially for the Mars rovers.
2.2lbs does not equal 1kg as one is force the other mass. In space it would basically be 0lb but still 1kg. Heard of the 14er weight loss program? Unfortunately you get it all back by the trailhead!

"Outside PSI = inside PSI" doesn't apply if the vessel also exerts an inward force. E.g., inflated to 3atm, a wheel's tire/tube/rim only needs to exert an inward force of 2atm which coupled with the 1atm outside yields equilibrium.

"Tire contact patch(es) [area] X tire PSI [pneumatic] = rider+bike weight" may be close in some circumstances but clearly is non-determinant as evidenced by a bike with both tires in a rounded muddy rut will have a contact area factors larger, and one with both tires rolling over angle iron factors less, but both will have the same weight on the ground and PSI in the tires. If it is fuzzy with bicycles, it sure isn't with 700lb of rider+motorcycle+gear (my 14er approach vehicle :-D ).


And who's to say the elephant isn't wearing floral pumps or a nice patent leather stilettos? (cue Pampilion)
Let's just hope they wear something more appropriate on a 14er trail. :)

MUni Rider wrote:Now what? :-k
Image


Jove help you if you go barreling down Elbert on that and blow a lace!
With right on front and left on rear, it could induce some interesting crabbing tendencies if it pro- or supinates!
Not sure if I'll do more 14ers. The trip reports are too tiring. :wink:

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Re: Please stay on the trail.

Postby tlongpine » Tue Oct 02, 2012 2:00 pm

Now that this matter has been resolved...

What ground force (expressed as PSI) is required to damage tundra?
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Re: Please stay on the trail.

Postby Greenhouseguy » Tue Oct 02, 2012 2:26 pm

tlongpine wrote:Now that this matter has been resolved...

What ground force (expressed as PSI) is required to damage tundra?


Wet tundra, dry tundra, frozen tundra, or snow-covered tundra? It makes a difference.
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