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OT? Weigh your car by checking tire pressure?

C

ChairmanOfTheBored

Jan 1, 1970
0


The only thing more retarded than a cross-posting original post author,
is some retarded twit that actually thinks that his filter file edit
session announcements mean a fucking thing to anyone.

You are one such retard.
 
C

ChairmanOfTheBored

Jan 1, 1970
0
I don't know why a tire of larger width would have less rolling
resistance, unless it's just that tires of larger width usually have a
smaller sidewall (proportionally) when used on the same car.


As rolling resistance would be a function of the weight they bear, one
would think that the larger distributed weight area (contact patch) would
result in a lower rolling resistance, as there would be less overall
rubber sidewall distortion.

However, they may have a lower fill pressure as well, which might be a
reason for a claim of higher rolling resistance.
 
M

Matthew T. Russotto

Jan 1, 1970
0
As rolling resistance would be a function of the weight they bear, one
would think that the larger distributed weight area (contact patch) would
result in a lower rolling resistance, as there would be less overall
rubber sidewall distortion.

If the tire pressures are the same, the contact patch is the same
(approximately), regardless of the tire width.
However, they may have a lower fill pressure as well, which might be a
reason for a claim of higher rolling resistance.

Typically lower pressures result in higher rolling resistance, due to
increased sidewall flex.
 
D

Don Klipstein

Jan 1, 1970
0
As rolling resistance would be a function of the weight they bear, one
would think that the larger distributed weight area (contact patch) would
result in a lower rolling resistance, as there would be less overall
rubber sidewall distortion.

However, they may have a lower fill pressure as well, which might be a
reason for a claim of higher rolling resistance.

Increasing contact patch length, whether by increasing weight or
decreasing pressure or decreasing tire width, increases sidewall
distortion.

If tire width is increased but pressure and weight are unchanged, the
contact patch length will decrease. I think that would decrease rolling
resistance.

One factor that appears significant is ratio of radii from axle center
to tire surface, at center of contact patch and at forward/rearward
extremes of the contact patch. The closer this ratio is to 1, the less
rolling resistance will be. One reason I see: Less rubbing of tread
against the road surface upon meeting and leaving the road surface. Also
less deformation counts at least a little, even other than by actual
scraping/rubbing of tread on road surface. It appears to me that 1 minus
this radius ratio strongly influences the rolling resistance coefficent.

And tread duration goes largely as:

* Directly proportionately with contact patch width
* Directly proportionately with depth of rubber to wear away
* Inversely proportionately with pressure
* Inversely proportional to either the above {1 minus radius ratio}, or to
actual rolling resistance coefficient (ratio of rolling resistance to
weight supported by the tire including its own)

As pressure increases, tire life tends to increase as long as product of
rolling resistance coefficient (or the above radius ratio) and pressure
decreases more than contact patch width decreases.

In general, tires for vehicles having 4 or more wheels have contact
patch width varying only slightly inversely with pressure, unless the
contact patch width is narrowed by a large pressure uptick, and then maybe
only after the tire has been "broken in" at the lower pressure. That
problem tends to require either overinflation or proper pressure after
experiencing significant wear while underinflated.

One more thing: Assuming tire design that has contact patch width not
varying much with pressure (or ratio of pressure to weight loading), the
above radius ratio is approximating the cosine of the angle between
straight down and from axle center to either forward or rearward end of
the contact patch.
This (1-minus-radius_ratio) in such case tends to be close to inversely
proportional to square of tire pressure (above atmospheric pressure).
That appears to me to indicate that things tend to get better with such
tires as pressure gets higher, as long as the pressure is not exsceeding
the pressure capability of the tire.

One more thing: Tire pressure is supposed to be measured "cold" - when
the tires are not heated up by using them. If the "cold" pressure does
not exceed the maximum rating for the tire, then the "warmed up pressure"
is supposed to be unable to be "excessive" unless the weigt loading
exceeds the tire rating for that, vehicle speed is excessive, or the
ambient temperature takes a jump big enough to be a considerable factor.

If you have excessive weight loading on your tires, then they are unsafe
at any pressure - their deformation varies directly with ratio of weight
loading to pressure, and any fatigue effects of that get worse when the
deformation is greater or the when "same deformation" occurs at higher
pressure - excessive weight loading is bad no matter what you do.
However, in my experience of overloading bicycle tires (by being
adventurous as a bicycle messenger by carrying heavier packages or
clusters thereof), I have found "least-worst" (still "living dangerously")
results from having pressure measured-cold at or a little above the
pressure marking on the sidewall of the tire - I don't see tires having
tolerance of excessive weight load improved by lower pressure; lower
pressure with excessive weight increases bigtime flexing of the sidewall,
and fatigue in the sidewall towards the bead.

- Don Klipstein ([email protected])
 
G

Gunner Asch

Jan 1, 1970
0
Responses don't count, dipshit. I have to ensure that my post gets read
by the twit that posted across several groups, many of which he may not
even visit or pull headers from. Get a clue.


so you are admitting to being the DUMB BASTARD that doesnt know a
fucking thing about Usenet who made that stupid cross-posted SHIT.

Thank you for playing

Gunner
 
J

JosephKK

Jan 1, 1970
0
Brian Whatcott [email protected] posted to sci.electronics.design:
Here's a question on an engineering group that has an
engineering answer!

There is a pressure sensitive material available (for gaskets etc.)
that color codes the pressure that it experiences.
That would be one analytical approach.

Brian W

If you use structural strength glass (or other structural transparent
material) with side (edge) lighting the contact area (non-moving)
could be directly observed. Perhaps with a "window" smoothly
embedded in pavement and careful photograph timing the contact patch
could be measured in a variety of circumstances.
 
W

Willem

Jan 1, 1970
0
mpm wrote:
) Take the car and launch it into orbit.
) Then, measure the tire pressure and determine the weight.

Easy:

The contact patch with the ground is zero. Multiply this by
the tire pressure and you get a weight of, you guessed it, zero.


(Determining the mass, however, is an entirely different matter.)
:)


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT
 
M

Michael A. Terrell

Jan 1, 1970
0
Gunner said:
Evidently it was you.

Gunner


His real name is 'dimbulb', and he morphs more often than he bathes.
Thats why all his posts stink. ;-)

Don't waste your tinme on him, he's as big a loser as Hawkie, and
Sloman.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
C

ChairmanOfTheBored

Jan 1, 1970
0
If the tire pressures are the same, the contact patch is the same
(approximately), regardless of the tire width.


Typically lower pressures result in higher rolling resistance, due to
increased sidewall flex.


Absolutely untrue. There are far too many differences (variables) in
tire construction, such as sidewall strength and the cross sectional
thickness of the tire face, and I am not talking about the tread.

If we were talking about a bunch of balloons which all had the same
wall thicknesses, I would agree, but here... no way.

Look at a dragster tire if you want a comparison. They have very weak
sidewalls. It shows up as soon as they accelerate off the line in
expanded tire diameter.
 
C

ChairmanOfTheBored

Jan 1, 1970
0
If you have excessive weight loading on your tires, then they are unsafe
at any pressure - their deformation varies directly with ratio of weight
loading to pressure, and any fatigue effects of that get worse when the
deformation is greater or the when "same deformation" occurs at higher
pressure - excessive weight loading is bad no matter what you do.


Even with normal weight loading, if one is driving around curves like
speed racer, there is going to be excessive distortion introduced,
reducing tire life.
 
C

ChairmanOfTheBored

Jan 1, 1970
0
so you are admitting to being the DUMB BASTARD that doesnt know a
fucking thing about Usenet who made that stupid cross-posted SHIT.

Thank you for playing

No, you retarded ****. I directly referred to the original poster, and
if you weren't a fucking illiterate bastard you'd have seen that, and no,
dipshit, it is not a game, so thank you for being the stupid fucktard I
knew you would be.
 
C

ChairmanOfTheBored

Jan 1, 1970
0
His real name is 'dimbulb', and he morphs more often than he bathes.
Thats why all his posts stink. ;-)

Don't waste your tinme on him, he's as big a loser as Hawkie, and
Sloman.


**** you, TerrellTard!
 
M

Michael A. Terrell

Jan 1, 1970
0
ChairmanOfTheBored said:
**** you, TerrellTard!


Once again: I'm not into queers like you are, dimbulb.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida
 
B

Brian Whatcott

Jan 1, 1970
0
If you use structural strength glass (or other structural transparent
material) with side (edge) lighting the contact area (non-moving)
could be directly observed. Perhaps with a "window" smoothly
embedded in pavement and careful photograph timing the contact patch
could be measured in a variety of circumstances.

This approach works well for automating fingerprint acquisition, I
notice. But mapping the pressure would be interesting.

Brian Whatcott Altus OK
 
J

jw

Jan 1, 1970
0
There are two sources of error in that method. The first, which is the one I
think you are referring to is the increase in pressure as you put weight on
the tyre. This will be the inverse of the reduction in volume of the tyre as
the bottom of the tread flattens and will be very small.

The second, which is the significant one, is the amount of the load which is
being supported by the stiffness of the tyre tread and sidewalls. If the
tyre were a balloon, i.e. perfectly elastic, the measurement as above would
be exactly correct.

So the result they should have obtained is a calculated weight somewhat less
than the actual weight of the car.

We did this experiment back in high school in Physics class. Everyone
grouped up in teams of 3 I think and proceeded to drive our cars onto
sheets of paper to get the imprint of the contact area and then took
measurements of the tire pressure. Afterwards each vehicle was taken
to the local grain elevator and scaled(results hidden from the
teams). Considering I was a bit of a gearhead at the time it was
silly because I knew exactly what my car weighed and the weight
distribution, but in any case.....

I argued with our teacher that the experiment was flawed. Assuming
perfectly elastic tires, that yes theoretically it should work, but
that tires were not perfectly elastic. My prime counter example being
an air tank. The pressure in a tank could be 20psia or 200psia and
the contact area would not change. He mumbled something about me
being a trouble maker and told us to finish our calculations.

FWIW: The results did work our "reasonably" close. I dont' remember
exact, but I think most were within 10-15% of actual weight.
Certainly proved there was a strong correlation. This did lead to a
discussion of what errors were present and how we could have gotten
closer to the right answer.

I was always just a little bit of a trouble maker in that class(well
most of my science/math classes really). I challenged the assertions
of the teachers, occasionally getting concessions that things were
presented in greatly simplified means so "the rest" could understand
them.

JW
 
E

Ed Murphy

Jan 1, 1970
0
I'm tempted to come round and let your tyres down, and watch you try
to catch your now weightless car as it drifts away across the sky...

What was the name of that one Donald Duck cartoon in which he
destroys a car that was going to be given to him as a gift?
 
E

Ed Murphy

Jan 1, 1970
0
ChairmanOfTheBored said:
The only thing more retarded than a cross-posting original post author,
is some retarded twit that actually thinks that his filter file edit
session announcements mean a fucking thing to anyone.

You are one such retard.

Fish!
 
P

Paul Hovnanian P.E.

Jan 1, 1970
0
Rich said:
Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:
http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

That's correct. But if you change the tire pressure, the footprint
changes as well. The change in footprint isn't linear, being dependent
on sidewall stiffness and other factors.
Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

It depends on the change in volume of the tire as the tire's contact
patch flattens out and takes up the vehicles load.
Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

Thanks,
Rich

Think about this: I can over or under inflate my tires easily by a
factor of two (or more) without the wheel's bead hitting bottom. The
vehicle's weight didn't change by a factor of two.

Neither their theories nor their tires hold any air.
 
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