OT: Nitrogen filled tires

[Jim Thompson]

I have to admit that this silly idea could be done, for example with
the "spinner" weight connected directly to the shaft of a very tiny
single piston compressor similar to a model airplane engine only
smaller, with reed in/out valves in the head and no side ports.
Bearing drag and the compressor shaft torque would spin the weight up
during acceleration, delivering a few strokes of compression, and
deceleration would deliver a few more in the oposite direction
(rotation direction is irrelevant to compressor). You could put all
the air in the tire and let out the excess with a relief valve set at
32 PSIG or whatever is desired.

But I don't like the idea, too cumbersome and delicate, and the
spinners would look lousy on my Yugo. Plus the extra unsprung weight
would spoil its handling.

What is wrong with the usual approach of instaling a compressed air
rotary joint on each axle, connected to the tire through a Schrader
valve in the wheel (so it holds air when removed). The stationary end
of each rotary joint connects to a pressure transducer and fill (from
engine driven compressor) and vent solenoid valves, controlled by the
tire inflation computer. A dashboard control lets the driver adjust
tire inflation from inside, so for instance pressure could be dropped
to 20 PSIG for driving on sand and back to 32 for highway driving, or
upped a bit for driving in rain. This arrangement works pretty well
on my Yugo .

Glen

My thought was a diaphragm type pump built into the wheel itself. As
the wheel rotates it passes a magnet attached to the strut.

...Jim Thompson

 

[Rich Grise]

23:39 -0700, John Larkin
[snip]
No,he's right;those wheel spinners use low friction bearings,and any
crud will stop them.

So, you won't do the math either.

Math without practical experience is useless.

Practical experience without math is amateur guesswork.

John

John, I think you've got yourself out on a limb, on the wrong side of
the saw ;-)

I think it's you who has to put some numbers to it.

I already did. And Guy confirmed it.

Macon?? That's almost as good as having "...bored..." do it ;-)

No, MissingProng can't do math at all. He's entirely number-phobic.

I think you need a pendulum of _substantial weight_ so that it
_doesn't rotate, but acts as a "pinning" point for the piston(s).

That could work, too. But we'd need a small amount of energy per day.
I'm guessing 1 kilojoule would be plenty - and easy to get - based on
the performance of a little cigaret-lighter-plugin compressor I have.
The bling-spinner idea was cute, and would be a nice sales gimmick.

What
weight does it take for a 1/2" piston working against 30PSI, and what
are the respective arm lengths?

Why 1/2"? The amount of air we'd need is tiny.

I'm not going to design it, much less build one to prove something to
rude strangers, but the numbers seem well in the ballpark of
feasibility. That's the first step in engineering, a quick numerical
estimate to see if an idea has a chance of working. This one does.

I've already got the design in my head; a ~ 100:1 or so planetary gear,
and a little eccentric in the middle of a radial-type pump.

30-35 PSI at 0.01 cfm available should keep a tire inflated just
fine. We're not using this for jet propulsion, after all! ;-)

Cheers!
Rich

 

[Joel Koltner]

Gas stations have had pay toilets for decades.

In California? I've never encountered any myself, but I've only spent
significant periods of time in Oregon and Wisconsin.

I doubt if it helps defray
the cost of cleaning/maintenance very much, but it keeps the riff-raff
out, which dramatically cuts those costs.

Yes.

 

[Rich Grise]

Tell that to Faraday.

What would Hero do?
http://en.wikipedia.org/wiki/Hero_of_Alexandria

Cheers!
Rich

 

[Rich Grise]

you just are not going to get any useful energy transfer from the
spinning auto wheel to your pump gearing. You evidently have not
observed a wheel spinner in action,or you'd know what I'm talking about.

We're no longer talking about the bling-bling pimp spinners, but
something that's weighted, possibly hidden inside the hubcap or
embedded in the wheel itself - you can still put the bling on
the outside.

Cheers!
Rich

 

[Richard The Dreaded Libertarian]

Q.E.D.

What's been demonstrated? That you're a member of the cult of
antismokerism?

Thanks,
Rich

 

[ChairmanOfTheBored]

I did some math in my head last night, to lull myself to sleep.

Imagine a reasonable-mass metal ring parallel to the wheel, half the
diameter of the tire, connected by a disk or spokes to the pump shaft
at the center of the wheel, with the pump body fixed to the wheel.

Imagine a car accelerates to, say, 60 k/h and back to zero, maybe 20
times a day. One could get kilojoules of work per day at the pump
shaft, easily enough to keep a tire inflated. If you used a full-size
spinning wheel cover, the increased radius helps even more (squared,
even) and using air drag helps more yet, since that add a continuous
power source that increases greatly with speed.

Do the math... it's easy.

You STILL don't get it. When the car brakes, the spinner assembly you
are on about is ALSO decelerated. Near ZERO net energy.

You would need an off center weight, and that breaks the rule of the
wheel being a balanced assembly.

 

[Don Klipstein]

On Mon, 15 Oct 2007 16:23:39 -0700, John Larkin



[snip]

No,he's right;those wheel spinners use low friction bearings,and any
crud will stop them.

So, you won't do the math either.

John





Math without practical experience is useless.

Practical experience without math is amateur guesswork.

John

John, I think you've got yourself out on a limb, on the wrong side of
the saw ;-)

I think it's you who has to put some numbers to it.

...Jim Thompson

I already did. And Guy confirmed it.

John

Macon?? That's almost as good as having "...bored..." do it ;-)

I think you need a pendulum of _substantial weight_ so that it
_doesn't rotate, but acts as a "pinning" point for the piston(s). What
weight does it take for a 1/2" piston working against 30PSI, and what
are the respective arm lengths?

If someone is going to get a bit creative and make a spinner-powered
tire pump, why not make the piston diameter 1/8 inch? That takes .37
pound. For that matter, how about 1/16 inch? That takes a bit over .09
pound of force.

Now, let's say you have .3 G braking and a ring-shped flywheel "spinner"
of diameter 60% of that of the tire. More like a big ring than "bling",
but this is for function and angineering calculations - it could
concievably be "blinged up" afterwards.

Ring with points accelerating (even if negatively) by .18 G only needs
to weigh a little over half a pound to have such
acceleration/decelleration provide enough force to push a 1/16 inch
diameter column of air 30 PSI above atmosphere.
For that matter, the flywheel could have a half pound outer rig welded
to it but have some point closer to the axle drive the pump. And there is
such a thing as gears.

Let's say the pump has a stroke of 1 inch, a bore of 1/16 inch diameter,
meaning volume of about 1/325 cubic inch. Let's say that a 35 to zero
braking at .3 G occurs 4 times a day, a 25 to zero braking occurs 6 times
a day and a 55 to 20 MPH braking occurs twice a day, 6 days a week. (Of
course I am pulling numbers out of a hat, probably on the conservative
side, for "typical driving".)
This adds up to 55 seconds a day if braking at .3 G, with average speed
reduction of 30 MPH per braking. I would consider the available kinetic
energy from the flywheel to be 1/2 times its mass (if in form of a ring)
times square of speed reduction of braking. So I would use half this 30
MPH braking figure for effective flywheel speed for calculating average
RPM of a pump shaft during braking.

I am guessing a fairly typical tire has a circumference of 6 feet - I am
feeling a little too lazy just now to go to the parking lot with a tape
measure. But 15 MPH and 6 feet means 3.67 revolutions per second. Times
55 seconds of braking a day, is 201-202 revolutions. Even if the pump has
only one cycle per cycle of flywheel outrunning the tire, that is 201-202
strokes. This is roughly 5/8 cubic inch of air for 1/16 inch diameter
bore, 1 inch stroke.
If the flywheel output to the pump is only 1.5 or so inches from the
centerline of the axle, a half pound ring flywheel, it will deliver enough
force to push a 1/8 inch diameter column of air 30 PSI above atmosphere.
That means 2.5 cubic inches a day. If the pump has so much drag with a
small bore to need force double or triple that required just to push the
air, make the flywheel a ring of 1 to 1.6 pounds.

- Don Klipstein ([email protected])

 

[krw]

"most" gas stations have free air? Not in Orlando.
Most have changed to coin-operated compressors,and often have been
vandalized,hoses cut.Hess is the only one left with free air.

In NY, any station that has air for their use must supply it free for
customers. This gets the quicky-mart off the hook for free air.
Service stations with mechanics (that change oil, for example) must
also accept small amounts of oil from anyone who comes in off the
street. Both are reasonable.

I ended up buying my own portable 12V compressor,a nice one. Cost me $21
USD.

My wife bought one because the couldn't figure out how to turn my
compressor on (it's in the basement) and burned up the 120V tire
pump. The thing is pretty much useless.

 

[Herbert John \Jackie\ Gleason]

The oxygen on the outside of the tire isn't at 150-200 PSIG, more like
3 PSIA. (approx 20% of 1 atmosphere).

And what the hell do I know? I didn't design the damn things!

WRONG! It is like series current. The current is the same everywhere.

HERE, at sea level, ALL the gasses in the atmosphere HERE is at the
same pressure!

How could you possibly think that the air, since it is only 20% of what
we breathe here, only has one fifth of the pressure on it?

The entire atmosphere weighs in on us all! Same with sea level, and
pressure.

Where did you pull that one from?

 

[Herbert John \Jackie\ Gleason]

Free air? Do they also have a guy who comes out and asks, "fill 'er up?"

Where do I find this time warp? ;-)

Cheers!
Rich

At your state legislature.

 

[ChairmanOfTheBored]

Mind the wrap:
http://patimg1.uspto.gov/.piw?Docid...&SectionNum=&idkey=NONE&Input=View+first+page
http://patimg1.uspto.gov/.piw?Docid...&SectionNum=&idkey=NONE&Input=View+first+page
http://patimg2.uspto.gov/.piw?Docid...&SectionNum=&idkey=NONE&Input=View+first+page

If you can turn a counter, you can gear it down and turn a pump.

Hope This Helps!
Rich

Blank, blank, and blank.

Learn how to post a link.

Also, a counter is a miniscule amount of requisite force needed to turn
it.

A compressor would need one or more orders of magnitude more force
applied, and would spin with the wheel, instead of doing the job needed.

 

[ChairmanOfTheBored]

So, you won't do the math either.

OK, dope ass. The OTHER REASON you will not find it IN the wheel
assembly is because they have spent YEARS REDUCING the unsprung weight of
the cars suspension, only to have you come along and **** it all up
again.

NO! Absolutely not. That is aside from the fact that you are just
asking for a design that would need to be fixed so often that it would
HAVE to bear the FORD (Found On Road Dead) moniker!

The right place to put it is in the Up/Down suspension travel, with the
ENTIRE weight of that corner of the car upon it! It could be made to
replace the dampening effector (shock absorber), so nothing about the
mass of things would need to be changed. And ONE HELL of a lot MORE work
would be getting done!

 

[John Larkin]

32:10 -0700, Jim Thompson
On Mon, 15 Oct 2007 16:23:39 -0700, John Larkin
[snip]
No,he's right;those wheel spinners use low friction bearings,and any
crud will stop them.

So, you won't do the math either.

Math without practical experience is useless.

Practical experience without math is amateur guesswork.

John

John, I think you've got yourself out on a limb, on the wrong side of
the saw ;-)

I think it's you who has to put some numbers to it.

I already did. And Guy confirmed it.

Macon?? That's almost as good as having "...bored..." do it ;-)

No, MissingProng can't do math at all. He's entirely number-phobic.

I think you need a pendulum of _substantial weight_ so that it
_doesn't rotate, but acts as a "pinning" point for the piston(s).

That could work, too. But we'd need a small amount of energy per day.
I'm guessing 1 kilojoule would be plenty - and easy to get - based on
the performance of a little cigaret-lighter-plugin compressor I have.
The bling-spinner idea was cute, and would be a nice sales gimmick.

What
weight does it take for a 1/2" piston working against 30PSI, and what
are the respective arm lengths?

Why 1/2"? The amount of air we'd need is tiny.

I'm not going to design it, much less build one to prove something to
rude strangers, but the numbers seem well in the ballpark of
feasibility. That's the first step in engineering, a quick numerical
estimate to see if an idea has a chance of working. This one does.

I've already got the design in my head; a ~ 100:1 or so planetary gear,
and a little eccentric in the middle of a radial-type pump.

I don't think you need to gear down; just make the pump displacement
smaller. A really tiny piston pump would work, but there's probably
something cheaper and more clever, like a diaphragm or a peristaltic
pump or something.

30-35 PSI at 0.01 cfm available should keep a tire inflated just
fine. We're not using this for jet propulsion, after all! ;-)

Numbers like that would be plenty.

John

 

[ChairmanOfTheBored]

The (still in preliminary stage, not a completed design) tire
inflator/spinner design. He says (see above) that it won't
work, using a sophisticated analysis technique that consists
of asserting that various folks are "obviously too stupid to
see that it would not work" rather than doing some simple
calculations. I did the calculations and I say that it is
workable -- or to be more precise, if it isn't workable it
won't be because a spinner has too little energy available
to keep a tire from slowly losing pressure.

Leave it to an overzealous would be "engineer" to leave out the ONVIOUS
reasons for why such things are not around already.

What? You think you came up with that CRAP?

The word/term for today is UNSPRUNG WEIGHT.

You know, that thing that AUTO DESIGNERS have spent decades removing,
and improving on, only to have a twit like you come along and say that
you want to throw a couple pounds back into each wheel.

Smooth move, X-lax!

 

[John Larkin]

On Mon, 15 Oct 2007 16:32:10 -0700, Jim Thompson

On Mon, 15 Oct 2007 16:23:39 -0700, John Larkin



[snip]

No,he's right;those wheel spinners use low friction bearings,and any
crud will stop them.

So, you won't do the math either.

John





Math without practical experience is useless.

Practical experience without math is amateur guesswork.

John

John, I think you've got yourself out on a limb, on the wrong side of
the saw ;-)

I think it's you who has to put some numbers to it.

...Jim Thompson

I already did. And Guy confirmed it.

John

Macon?? That's almost as good as having "...bored..." do it ;-)

I think you need a pendulum of _substantial weight_ so that it
_doesn't rotate, but acts as a "pinning" point for the piston(s). What
weight does it take for a 1/2" piston working against 30PSI, and what
are the respective arm lengths?

If someone is going to get a bit creative and make a spinner-powered
tire pump, why not make the piston diameter 1/8 inch? That takes .37
pound. For that matter, how about 1/16 inch? That takes a bit over .09
pound of force.

Now, let's say you have .3 G braking and a ring-shped flywheel "spinner"
of diameter 60% of that of the tire. More like a big ring than "bling",
but this is for function and angineering calculations - it could
concievably be "blinged up" afterwards.

Ring with points accelerating (even if negatively) by .18 G only needs
to weigh a little over half a pound to have such
acceleration/decelleration provide enough force to push a 1/16 inch
diameter column of air 30 PSI above atmosphere.
For that matter, the flywheel could have a half pound outer rig welded
to it but have some point closer to the axle drive the pump. And there is
such a thing as gears.

Let's say the pump has a stroke of 1 inch, a bore of 1/16 inch diameter,
meaning volume of about 1/325 cubic inch. Let's say that a 35 to zero
braking at .3 G occurs 4 times a day, a 25 to zero braking occurs 6 times
a day and a 55 to 20 MPH braking occurs twice a day, 6 days a week. (Of
course I am pulling numbers out of a hat, probably on the conservative
side, for "typical driving".)
This adds up to 55 seconds a day if braking at .3 G, with average speed
reduction of 30 MPH per braking. I would consider the available kinetic
energy from the flywheel to be 1/2 times its mass (if in form of a ring)
times square of speed reduction of braking. So I would use half this 30
MPH braking figure for effective flywheel speed for calculating average
RPM of a pump shaft during braking.

I am guessing a fairly typical tire has a circumference of 6 feet - I am
feeling a little too lazy just now to go to the parking lot with a tape
measure. But 15 MPH and 6 feet means 3.67 revolutions per second. Times
55 seconds of braking a day, is 201-202 revolutions. Even if the pump has
only one cycle per cycle of flywheel outrunning the tire, that is 201-202
strokes. This is roughly 5/8 cubic inch of air for 1/16 inch diameter
bore, 1 inch stroke.
If the flywheel output to the pump is only 1.5 or so inches from the
centerline of the axle, a half pound ring flywheel, it will deliver enough
force to push a 1/8 inch diameter column of air 30 PSI above atmosphere.
That means 2.5 cubic inches a day. If the pump has so much drag with a
small bore to need force double or triple that required just to push the
air, make the flywheel a ring of 1 to 1.6 pounds.

Yup. Even with your ultra-conservative numbers, it works.

John

 

[ChairmanOfTheBored]

...which killed the small copmanies that were providing coin-operated
air, and led to a shortage of stations that have air hoses. Someone
has to pay for the expense of purchasing the air pumps, maintaining
them, running them with electricity, etc.

Thank god they didn't require grocery stores to provide "free" food...

You bring one term to mind...

"Big Dope".

A line from the film "All of Me" with Lilly Tomlin and Steve Martin.
The "Indian" guy says it.

Great flic.

You: Big Dope.

 

[Herbert John \Jackie\ Gleason]

The operative word here being "customers" - that's somebody who
buys something. They shouldn't have to provide free air for somdbody
who just pops in off the street and uses their air without buying
anything, ergo the coin-op air makes them a "customer".

Naah - that's way too sane for California bureaucrats.

You're a goddamned idiot.

 

[ChairmanOfTheBored]

Practical experience without math is amateur guesswork.

John

Yes, and fucked up math spewing forth from an idiot that doesn't know
what he is talking about, like you, proves the level of retardedness
which you possess.

 

[ChairmanOfTheBored]

[snip]

No,he's right;those wheel spinners use low friction bearings,and any
crud will stop them.

So, you won't do the math either.

John





Math without practical experience is useless.

Practical experience without math is amateur guesswork.

John

John, I think you've got yourself out on a limb, on the wrong side of
the saw ;-)

I think it's you who has to put some numbers to it.

...Jim Thompson

I already did. And Guy confirmed it.

And you are both being horse with blinders on idiots about it.

Let alone the fact that nobody in their right mind is going to want to
ADD weight to their wheel.