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Fwd: Plus definition
In a message dated 5/24/99 2:32:43 PM Central Daylight Time, JustaxPHX writes:
> Subj: Re: Plus definition
> Date: 5/24/99 2:32:43 PM Central Daylight Time
> From: <A HREF="mailto:JustaxPHX">JustaxPHX</A>
> To: <A HREF="mailto:sbigelow@sprint.ca">sbigelow@sprint.ca</A>
> CC: <A HREF="mailto:QSHIPQ">QSHIPQ</A>,
<A HREF="mailto:Dave.Eaton@clear.net.nz">Dave.Eaton@clear.net.nz</A>
>
> >Finally, some one _else_ who gets it...
> >
> >Your car is supported by the air pressure in the tires ..equal and
opposite
> >reactions, and all that, so..
> >
> >A 2800 lbs car with say, 60% of its weight on the front axle, has 1680 lbs
> >on that axle, or 840 lbs per tire.
> >@ 32 lbs per square inch in the tire, how many square inches of patch does
> >it have?
> >
> >840/32 = 26.25 square inches.
> >
> >Wanna change its size? Change vehicle weight, or tire pressure.
> >Wanna change its shape? Change tire size.
> >Wanna change its quality? Change tire brands/models.
>
> If what you're saying is correct, then a nearly flat tire with say, 1psi
of
> pressure, has a contact patch area of 840/1, which works out to 840 in/sq
or
> roughly 5.8 ft/sq ... right? And, of course, a totally flat tire with 0
psi
> of pressure has a contact patch area of 840/0, which works out to ... hmmm.
>
> In reality, there is more than just the air pressure in a tire supporting
> the car. The sidewall stiffness must be considered as well and as you
might
> guess, this is very much affected by the width of the wheel the tire is
> mounted upon hence my original comment to Dave Eaton. The sidewall
stiffness
> provides a minimum "pressure" below which the system cannot fall and thus
> determines the upper limit of contact patch area for any given make/model/
> type of tire. For your formula to be valid, this must be taken into
account
> or else you can end up with some screwy results...
>
> I'm swamped with wrapping up my business this week and moving the office
so
> I unsubscribed to the q-list this morning ... perhaps one of you will
forward
> this to it for me? :^)
>
> JG
>Finally, some one _else_ who gets it...
>
>Your car is supported by the air pressure in the tires ..equal and opposite
>reactions, and all that, so..
>
>A 2800 lbs car with say, 60% of its weight on the front axle, has 1680 lbs
>on that axle, or 840 lbs per tire.
>@ 32 lbs per square inch in the tire, how many square inches of patch does
>it have?
>
>840/32 = 26.25 square inches.
>
>Wanna change its size? Change vehicle weight, or tire pressure.
>Wanna change its shape? Change tire size.
>Wanna change its quality? Change tire brands/models.
If what you're saying is correct, then a nearly flat tire with say, 1psi of
pressure, has a contact patch area of 840/1, which works out to 840 in/sq or
roughly 5.8 ft/sq ... right? And, of course, a totally flat tire with 0 psi
of pressure has a contact patch area of 840/0, which works out to ... hmmm.
In reality, there is more than just the air pressure in a tire supporting the
car. The sidewall stiffness must be considered as well and as you might
guess, this is very much affected by the width of the wheel the tire is
mounted upon hence my original comment to Dave Eaton. The sidewall stiffness
provides a minimum "pressure" below which the system cannot fall and thus
determines the upper limit of contact patch area for any given
make/model/type of tire. For your formula to be valid, this must be taken
into account or else you can end up with some screwy results...
I'm swamped with wrapping up my business this week and moving the office so I
unsubscribed to the q-list this morning ... perhaps one of you will forward
this to it for me? :^)
JG