hydroplaning

[Livolsi, Stephane]

Ok, your calculation makes sense but there are other factors.  say you
throw a 250lb rider on and according to your calculations you would have


250 + 25 = 275/2 = 137.5 lbs on each tire.

The pressure in the tire is still 90 psi so to calculate the contact
patch:

90 = 137.5/x

90x = 137.5

X = 1.52 inches per tire.

If I understand your explanation, you are saying that in order to reduce
137.5 lbs pressure down to 90 psi, you would have to spread it out over
1.52 inches.  I can see the logic in that AND I can see it's relation to
the internal pressure of the tire.  

Dammit I think you're right...but...

What I don't understand is the ignoring of external factors such as tire
compound.. 

.. change the tires to a stiffer or softer compound and you are saying
the contact patch will remain the same 1.52 inches? 




> -----Original Message-----
> From: Huw Powell [mailto:audi at humanspeakers.com] 
> Sent: May 28, 2004 12:47 PM
> To: Livolsi, Stephane
> Cc: Audi Quattro List
> Subject: Re: hydroplaning
> 
> 
> 
> > This discussion is interesting - NAC - but interesting and 
> so far no 
> > one has insulted anyone else...so far....:)
> > 
> > Looking at a different example...how does a bike weighing 
> say, 25 lbs, 
> > having 2 tires pumped up to 90psi exert 180psi on 2 little contact 
> > patches where the rubber meets the ground?  At 90 psi those 
> tires are 
> > gonna be firm with very little contact patch.
> 
> I assume you mean when no one is on the bike?
> 
> 25 pounds/2 (assuming equal weight distr.) = 12.5 per tire
> 
> 90 p/si = 12.5 p / "x" si
> 
> 90 x = 12.5
> 
> x = 12.5/90 = .138 approx sq in.
> 
> just over an eighth of a square inch per tire, yes, I'd call 
> that small.
> 
> -- 
> Huw Powell
> 
http://www.humanspeakers.com/audi

http://www.humanthoughts.org/