blooddrive
01-03-2004, 03:18 PM
25655-25679
25655 From: Eric Marble <emarble3@p...>
Date: Tue Jun 19, 2001 9:40pm
Subject: Re: Mathematicians wanted
Remember Joe it's a flexible container not a scuba cylinder so you have to
calculate expansion also..
Padi #56992
p.s. now take it to 60 feet..
Eric
----- Original Message -----
From: "Joe W" <arizonajeep@h...>
To: "'James Towle'" <James.Towle@a...>; "'AZ_VJC'"
<az_vjc@yahoogroups.com>
Sent: Tuesday, June 19, 2001 9:20 PM
Subject: RE: [az_vjc] Mathematicians wanted
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant.
(not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above
for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air can
be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
most
> out of your tires! QuickAIR2 will let you air down for maximum traction
and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at
the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
--------------------------------------------------------------------------------
ADVERTISEMENT
25656 From: linda luik <minihummer@w...>
Date: Tue Jun 19, 2001 9:39pm
Subject: Re: Mathematicians wanted
I am assuming this is with no weight on the tire. That is, the tire is
off the jeep? Or the jeep is up in the air? Or would that make a
difference?
Linda
Joe W wrote:
>
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
> >From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the most
> out of your tires! QuickAIR2 will let you air down for maximum traction and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
> Now... are you sorry you asked?
>
> <grin>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
25657 From: Leah & Geoff <leahlsq@q...>
Date: Tue Jun 19, 2001 8:45pm
Subject: TUCSON - Pima Motor Sports Park needs our help!
I'm back from Alaska and finally at my PC.
This is from Becky Petermann - PLease help out!
I'm leading a run to Rice Peak this Saturday and will
be out at the Park on Sunday.
If you have any questions call me at 546-0501
Leah
----------------------------------------------------------------------------
------------
I have been out at the Pima Motorsports Park for the last three days after
work. We are doing very well, but I need help. And unfortunately I need it
now.
I am hoping that some of you would be willing to give up a Saturday or
Sunday or both to come help me.
We need to put in fencing around our area. We also need to finish a couple
of the obstacles.
When was the last time you got to be the first to drive on a road. Or
actually put in a road? Well here is your chance.
The major problem is that it has to be this weekend.
We need to finish the rock hill.
It is looking pretty good but it needs to be finished.
So please take time out to come help us! I will be going out there
Saturday and Sunday.
Meet at the park at 8:00AM.
This is extremely important. I have done all I can, but now is when I need
help. Please let me know.
25658 From: Clint Ramsey <cwramsey@t...>
Date: Tue Jun 19, 2001 10:13pm
Subject: re: TJ driveline
Sounds like exhaust back pressure making the crap in the cat rattle around
Clint
25659 From: T.J. Nosmo-King <ice626@h...>
Date: Tue Jun 19, 2001 10:39pm
Subject: RE: Mathematicians wanted
For a guy with a low-rider to fix, you sure seem to have an awful lot of
time on your hands...... :-)
__________________________________________________ _______________
Get your FREE download of MSN Explorer at http://explorer.msn.com
25660 From: Joe W <arizonajeep@h...>
Date: Tue Jun 19, 2001 10:46pm
Subject: RE: Mathematicians wanted
Ya... Joyce was gone this evening teaching a SCUBA class and I was all alone
with nothing to do but ponder the proposed question.
<grin>
Besides... he put my name on it so I had to reply... kind of like the triple
dog dare ya from the Christmas Story movie.
Joe West
> -----Original Message-----
> From: T.J. Nosmo-King [mailto:ice626@h...]
> Sent: Tuesday, June 19, 2001 10:40 PM
> To: arizonajeep@h...; James.Towle@a...; az_vjc@yahoogroups.com
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> For a guy with a low-rider to fix, you sure seem to have an
> awful lot of
> time on your hands...... :-)
> __________________________________________________ _______________
> Get your FREE download of MSN Explorer at http://explorer.msn.com
>
>
25661 From: <japdep@a...>
Date: Tue Jun 19, 2001 11:27pm
Subject: FS. D35C rear axle from 90' XJ
complete rear end. Broken pinion gear. Carrier, shafts and housing in
good shape. $50 Email me off list if interested...JIm
25662 From: AzVJC Website <azvjc@a...>
Date: Wed Jun 20, 2001 4:34am
Subject: [TR] Cheap Trick - Saturday, June 23, 2001 - Reminder
Hello All,
Dave O'Boyle is leading a Cheap Trick trail run on Saturday, June 23, 2001.
When: Sat June 23 6am.
Where: Circle K Jackrabbit trail and I10. head south on Jackrabbit
1/2 mile. to Circle K.
Rating: 4.5+
Length: 4 to 6 hrs.
CB Channel: 4
Cheep Trick
It's a nasty one.. Expect body damage.. This one may be impassable. e-mail me if your up to the challange.
davidoboyle@hotmail.com
Very Nasty.
Sign Up List:
Dave O'Boyle Definite
John Manansala Definite
Click Here To Add Your Name to the Sign Up List
Thank You,
- The AzVJC Website
25663 From: AzVJC Website <azvjc@a...>
Date: Wed Jun 20, 2001 4:39am
Subject: [TR] Cheap Trick - Saturday, June 23, 2001 - Reminder
Hello All,
Dave O'Boyle is leading a Cheap Trick trail run on Saturday, June 23, 2001.
When: Sat June 23 6am.
Where: Circle K Jackrabbit trail and I10. head south on Jackrabbit
1/2 mile. to Circle K.
Rating: 4.5+
Length: 4 to 6 hrs.
CB Channel: 4
Cheep Trick
It's a nasty one.. Expect body damage.. This one may be impassable. e-mail me if your up to the challange.
davidoboyle@hotmail.com
I am offline Thursday and Friday so call if you need more info. 623-327-1486
Very Nasty.
Sign Up List:
Dave O'Boyle Definite
John Manansala Definite
Click Here To Add Your Name to the Sign Up List
Thank You,
- The AzVJC Website
25664 From: Jay Eller <jay@t...>
Date: Wed Jun 20, 2001 6:23am
Subject: Re: Mathematicians wanted
"Auntie-M....Auntie-M.....It's a twister!!! It's a twister":-0
Excellent derivation professor! But what about the orifice created by the valve
reducing the flow rate to the tire?
I would guess that with the orifice created by the tire valve, it would take you
just long enough for your toes to start hurting while squatted down filling the
tire........give or take a few comments about "I should have sprung for a damned
CO2 tank!"
Just a little light hearted humour.
--
----------------------------
Jay Eller (http://www.goodnet.com/~eller)
http://www.toyboxoffroad.com
25665 From: <davidoboyle@h...>
Date: Wed Jun 20, 2001 7:04am
Subject: Cheep Trick sat..
Sorry for the double post on the Cheep Trick run.
I added my Phone number to the trip because I am offline Thursday
thru Saturday. Hope to see some Big boys/Girls out this weekend..
This is one SICK trail......... If you don't believe me ask Joey K.
We both walked it and had a hard time even doing that...
See you Sat.
Dave
623-327-1486
25666 From: James Towle <James.Towle@a...>
Date: Wed Jun 20, 2001 7:48am
Subject: RE: Mathematicians wanted
Joe,
Thanks, 23 gallons that's a lot.... I was thinking it would be close to 20,
after looking at some ads, but that seemed a bit much.
I was thinking of getting some air tanks to hook up to my Viair II. And was
wondering if the tanks would make an impact on the air up time. There is a
place in Mesa (www.truckn.com) that sells tanks for a reasonable price.
What do you think, what size (in gallons) tank(s) would make a good impact
on airing up? Or maybe there is not enough room, with this compressor, to
make an impact.
--James
> -----Original Message-----
> From: Joe W [mailto:arizonajeep@h...]
> Sent: Tuesday, June 19, 2001 9:20 PM
> To: 'James Towle'; 'AZ_VJC'
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is
> constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and
> subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere
> above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with
> ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*.
> Get the most
> out of your tires! QuickAIR2 will let you air down for maximum
> traction and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the
> tire at the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
25667 From: Mike Baney <jeepin_in_az@y...>
Date: Wed Jun 20, 2001 8:03am
Subject: Re: TJ Driveline ?
I didn't think you could notice a loss of power in a 4 cyl? <big grin>
Mike
--- "T.J. Nosmo-King" <ice626@h...> wrote:
> I think that if it was the cat, you would not only hear the rattling
> noise
> but you would notice a DEFINITE loss in power during accerlation due to
> the
> loose pieces clogging the system....at least that is what happend to me
> when
> my cat broke up....a little pounding with a steel rod and the emptying
> of
> the ceramic pieces made a great difference in air flow (amazing what an
> empty container can do :-) )
> __________________________________________________ _______________
> Get your FREE download of MSN Explorer at http://explorer.msn.com
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
>
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25668 From: Mike Baney <jeepin_in_az@y...>
Date: Wed Jun 20, 2001 8:11am
Subject: RE: Mathematicians wanted
WOW!!
--- Joe W <arizonajeep@h...> wrote:
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire
> pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant.
> (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
> the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and
> it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above
> for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at
> first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air
> can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according
> to
> quickair's advertisements)...which means that you should be able to air
> up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12
> to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
> The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in
> less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
> most
> out of your tires! QuickAIR2 will let you air down for maximum traction
> and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at
> the
> pressures you are asking and quickair2's numbers show that you can go
> from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
>
__________________________________________________
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Get personalized email addresses from Yahoo! Mail
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25669 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 8:24am
Subject: Jeep @ Dealer
I didn't notice power loss, but the noise was gradually getting
louder till the other day when it became *extremely* noticeable.
With stock gears and 33" swampers, you don't exactly say "loss" of
power in a TJ....more like, change in driving dynamics. =)
The jeep is in the shop right now, but I watched them move it from
the service bay to the back, they drove it in REVERSE all the way
around the building, I'm so pissed they are gonna get a earfull when
I go back!!
When I took it in I told the guy I had front and rear lockers so
drive conservatively if they drive it and the guy said "you have
what?".......
The look on the service guy's faces when I drove up was worth a
million bucks though.
25670 From: Chris Krieg <rv6a@m...>
Date: Wed Jun 20, 2001 8:22am
Subject: Re: Mathematicians wanted
Yeah, thats the same answer I came up with.
Chris K
> --- Joe W <arizonajeep@h...> wrote:
>> James,
>>
>> Let's put down some variables:
>>
>> V1 is the initial volume of the tire to produce the initial tire
>> pressure.
>> V2 is the final volume of the tire to produce the required increase in
>> pressure.
>> P1 is the initial pressure of the tire.
>> P2 is the final pressure of the tire.
>> T1 is the initial temperature of the air in the tire at P1
>> T2 is the final temperature of the air in the tire at P2
>>
>> Assumption; Temperature of the compressed air in the tire is constant.
>> (not
>> really true since the compressed air heats up as it is compressed).
>>
>> Now for the equations, starting with the Natural Gas Law and our initial
>> conditions (initial tire pressure):
>>
>> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>>
>> P1*V1 = K*T1
>>
>> Doing the same for the final conditions (final tire pressure):
>> P2*V2 = K*T2
>>
>> Setting the initial condition equal to the final condition produces the
>> equation:
>>
>> (P1*V1)/K*T1 = (P2*V2)/K*T2
>>
>> But... since T1 = T2 and K is the same constant for both sides of the
>> equation, the equation simplifies to:
>>
>> P1*V1 = P2*V2
>>
>> Since we know the starting volume (V1) and both the starting and ending
>> pressures (P1 and P2), we can solve for the single unknown, V2:
>>
>> EQUATION 1 : (P1*V1)/P2 = V2
>>
>> Now I guess we need to plug in some numbers:
>>
>> The area of a cylinder is pi*R^2*H
>>
>> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
>> the
>> two to find the volume:
>>
>> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
>> Width of the tire is H = 12.5"
>>
>> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>>
>> You didn't state the width of your wheel so I'll assume it is 8" and
>> it's
>> radius is 7.5"
>>
>> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>>
>> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>>
>>
>> From your specifications:
>>
>> P1 = 12psi
>>
>> P2 = 28psi
>>
>> So... now all we have to do is solve EQUATION 1 derived somewhere above
>> for
>> the volume V2
>>
>> V2=P1*V1/P2
>>
>> Plugging in the numbers:
>>
>> V2 = 12psi*9,277in^3/28psi
>>
>> V2 = 3,976 in^3
>>
>> So... in order to increase the pressure of your tire, the volume of air
>> which had to be produced by your pump is:
>>
>> Delta V = V1-V2
>>
>> Delta V = 9,277-3,976
>>
>> Delta V= 5,301in^3
>>
>> Using the handy dandy conversion from cubic inches to gallons of 1gal =
>> 231in^3 give us the delta in gallons of air of:
>>
>> 5,301in^3/231in^3/gal = 22.95 gallons
>>
>> Now... for a sanity check to see if my numbers are reasonable... at
>> first
>> blush... 22.95 gallons seems large:
>>
>>
>> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air
>> can be
>> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according
>> to
>> quickair's advertisements)...which means that you should be able to air
>> up
>> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>>
>> 2.12 minutes!!!
>>
>>
>> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12
>> to
>> 28psi in 2.12 minutes....
>>
>> Let's look at what quickair's advertisement on their web page says:
>>
>> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
>> The
>> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in
>> less
>> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
>> most
>> out of your tires! QuickAIR2 will let you air down for maximum traction
>> and
>> performance and reinflate quickly. Capable of reseating tire on rim."
>>
>>
>> So... I calculate that it will take 2.12 minutes to inflate the tire at
>> the
>> pressures you are asking and quickair2's numbers show that you can go
>> from
>> 15psi to 30psi in "less than 2.5 minutes".
>>
>> I guess the sanity check worked and the numbers are very close given my
>> rough estimates and assumptions.
>>
>> Your answer is 22.95 +- .25 gallons.
>>
>> Kind Regards,
>>
>> Joe West
>>
>>
>> Now... are you sorry you asked?
>>
>> <grin>
>>
>>
>>> -----Original Message-----
>>> From: James Towle [mailto:James.Towle@a...]
>>> Sent: Tuesday, June 19, 2001 5:20 PM
>>> To: AZ_VJC
>>> Subject: [az_vjc] Mathematicians wanted
>>>
>>>
>>>
>>> Okay who is the math people? Here's the question:
>>>
>>> If you have a 33" x 12.5" on a 15" rim, want is the amount in
>>> gallons it
>>> takes to pressurize the tire from 12#'s to 28#'s if you have a
>>> compressor/tank filling at 90PSI? Lets say the CFM is limited
>>> to what a 1/4"
>>> hose can take at the given PSI.
>>>
>>> I know this is some sort of integration problem but, I am not
>>> sure where to
>>> start.
>>>
>>> Joe W. - It isn't thermodynamics but.....
>>>
>>> --James
>>>
>>>
>>>
>>>
>>> Your use of Yahoo! Groups is subject to
>>> http://docs.yahoo.com/info/terms/
>>>
>>>
>>
>>
>>
>>
>> Your use of Yahoo! Groups is subject to
>> http://docs.yahoo.com/info/terms/
>>
>>
>
>
> __________________________________________________
> Do You Yahoo!?
> Get personalized email addresses from Yahoo! Mail
> http://personal.mail.yahoo.com/
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
25671 From: <msa12171@a...>
Date: Wed Jun 20, 2001 8:18am
Subject: Re: Mathematicians wanted
Joe, you such a geek. :)
--- In az_vjc@y..., "Joe W" <arizonajeep@h...> wrote:
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire
pressure.
> V2 is the final volume of the tire to produce the required increase
in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is
constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our
initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces
the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of
the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and
ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and
subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8"
and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic
inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere
above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of
air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of
1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at
first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi
according to
> quickair's advertisements)...which means that you should be able to
air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire
from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with
ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi
in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get
the most
> out of your tires! QuickAIR2 will let you air down for maximum
traction and
> performance and reinflate quickly. Capable of reseating tire on
rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the
tire at the
> pressures you are asking and quickair2's numbers show that you can
go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close
given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
25672 From: tom lafrance <tlafrance@j...>
Date: Wed Jun 20, 2001 8:30am
Subject: Manual CJ7 steering box needed
Hello Group,
I'm in need of a manual steering box for my 78 CJ7. I'm hoping someone
has a spare lurking around the garage left over from a p/s conversion?
Thanks,
Tom
25673 From: Chris R. <my1stjeep@e...>
Date: Wed Jun 20, 2001 8:52am
Subject: Re: Re: Mathematicians wanted(Curve Ball Thrown...)
Well here is a Curve Joe. With the expansion or compression of air among
other things due to surrounding tempatures how would your numbers be
affected if you were in say Alaska vs Death Valley?
Chris
http://www.hotstuff.alloffroad.com/
My1stJeep@e...
'97 TJ
___
[___]
-(O|||||O)-
=======
[||]----O--[||]
----------------------------------------------------
"He who asks is a fool for five minutes, but he who does not ask remains a
fool forever."
On Wed, 20 Jun 2001 15:18:24 -0000, msa12171@a... wrote:
> Joe, you such a geek. :)
>
>
>
>
>
> --- In az_vjc@y..., "Joe W" <arizonajeep@h...> wrote:
> > James,
> >
> > Let's put down some variables:
> >
> > V1 is the initial volume of the tire to produce the initial tire
> pressure.
> > V2 is the final volume of the tire to produce the required increase
> in
> > pressure.
> > P1 is the initial pressure of the tire.
> > P2 is the final pressure of the tire.
> > T1 is the initial temperature of the air in the tire at P1
> > T2 is the final temperature of the air in the tire at P2
> >
> > Assumption; Temperature of the compressed air in the tire is
> constant. (not
> > really true since the compressed air heats up as it is compressed).
> >
> > Now for the equations, starting with the Natural Gas Law and our
> initial
> > conditions (initial tire pressure):
> >
> > P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
> >
> > P1*V1 = K*T1
> >
> > Doing the same for the final conditions (final tire pressure):
> > P2*V2 = K*T2
> >
> > Setting the initial condition equal to the final condition produces
> the
> > equation:
> >
> > (P1*V1)/K*T1 = (P2*V2)/K*T2
> >
> > But... since T1 = T2 and K is the same constant for both sides of
> the
> > equation, the equation simplifies to:
> >
> > P1*V1 = P2*V2
> >
> > Since we know the starting volume (V1) and both the starting and
> ending
> > pressures (P1 and P2), we can solve for the single unknown, V2:
> >
> > EQUATION 1 : (P1*V1)/P2 = V2
> >
> > Now I guess we need to plug in some numbers:
> >
> > The area of a cylinder is pi*R^2*H
> >
> > I'll model the tire as a cylinder, the wheel as a cylinder, and
> subtract the
> > two to find the volume:
> >
> > Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> > Width of the tire is H = 12.5"
> >
> > Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
> >
> > You didn't state the width of your wheel so I'll assume it is 8"
> and it's
> > radius is 7.5"
> >
> > Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
> >
> > The difference in the two volumes is the volume V1 = 9,277 cubic
> inches.
> >
> >
> > From your specifications:
> >
> > P1 = 12psi
> >
> > P2 = 28psi
> >
> > So... now all we have to do is solve EQUATION 1 derived somewhere
> above for
> > the volume V2
> >
> > V2=P1*V1/P2
> >
> > Plugging in the numbers:
> >
> > V2 = 12psi*9,277in^3/28psi
> >
> > V2 = 3,976 in^3
> >
> > So... in order to increase the pressure of your tire, the volume of
> air
> > which had to be produced by your pump is:
> >
> > Delta V = V1-V2
> >
> > Delta V = 9,277-3,976
> >
> > Delta V= 5,301in^3
> >
> > Using the handy dandy conversion from cubic inches to gallons of
> 1gal =
> > 231in^3 give us the delta in gallons of air of:
> >
> > 5,301in^3/231in^3/gal = 22.95 gallons
> >
> > Now... for a sanity check to see if my numbers are reasonable... at
> first
> > blush... 22.95 gallons seems large:
> >
> >
> > 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> air can be
> > pushed out of a quickair II at the rate of 1.45CFM (at 40psi
> according to
> > quickair's advertisements)...which means that you should be able to
> air up
> > your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
> >
> > 2.12 minutes!!!
> >
> >
> > Hmmmm... seems reasonable that a quickair II can inflate a tire
> from 12 to
> > 28psi in 2.12 minutes....
> >
> > Let's look at what quickair's advertisement on their web page says:
> >
> > "The QuickAIR2 was developed to inflate 33" and larger tires with
> ease. The
> > QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi
> in less
> > than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get
> the most
> > out of your tires! QuickAIR2 will let you air down for maximum
> traction and
> > performance and reinflate quickly. Capable of reseating tire on
> rim."
> >
> >
> > So... I calculate that it will take 2.12 minutes to inflate the
> tire at the
> > pressures you are asking and quickair2's numbers show that you can
> go from
> > 15psi to 30psi in "less than 2.5 minutes".
> >
> > I guess the sanity check worked and the numbers are very close
> given my
> > rough estimates and assumptions.
> >
> > Your answer is 22.95 +- .25 gallons.
> >
> > Kind Regards,
> >
> > Joe West
> >
> >
> > Now... are you sorry you asked?
> >
> > <grin>
> >
> >
> > > -----Original Message-----
> > > From: James Towle [mailto:James.Towle@a...]
> > > Sent: Tuesday, June 19, 2001 5:20 PM
> > > To: AZ_VJC
> > > Subject: [az_vjc] Mathematicians wanted
> > >
> > >
> > >
> > > Okay who is the math people? Here's the question:
> > >
> > > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > > gallons it
> > > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > > to what a 1/4"
> > > hose can take at the given PSI.
> > >
> > > I know this is some sort of integration problem but, I am not
> > > sure where to
> > > start.
> > >
> > > Joe W. - It isn't thermodynamics but.....
> > >
> > > --James
> > >
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> > > http://docs.yahoo.com/info/terms/
> > >
> > >
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
__________________________________________________ _____
Send a cool gift with your E-Card
http://www.bluemountain.com/giftcenter/
25674 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 8:57am
Subject: Re: ladies run
I'm sure Tara would lead a run, but I can say that cause she probably
isn't reading this. But to add, I'm not sure I'd want her to lead,
she's gone through stuff before me cause I didn't think we could make
it.
Let's just say she has no fear of water; or breaking things (axles to
be specific). =)
25675 From: Joe W <arizonajeep@h...>
Date: Wed Jun 20, 2001 9:03am
Subject: RE: Mathematicians wanted
The orifice created by the valve reducing the flow rate would not affect the
number of gallons required to fill the tire, all it would do is increase the
time required to fill it.
Joe West
> -----Original Message-----
> From: Jay Eller [mailto:jay@t...]
> Sent: Wednesday, June 20, 2001 6:23 AM
> Cc: 'AZ_VJC'
> Subject: Re: [az_vjc] Mathematicians wanted
>
>
> "Auntie-M....Auntie-M.....It's a twister!!! It's a twister":-0
>
> Excellent derivation professor! But what about the orifice
> created by the valve
> reducing the flow rate to the tire?
>
25676 From: Joe W <arizonajeep@h...>
Date: Wed Jun 20, 2001 9:10am
Subject: RE: Mathematicians wanted
James,
There is a tradeoff in tank size vs. tire filling time. Unless you can
store enough air to fill ALL the tires without turning on the compressor,
I'd stick with a 3 to 6 gallon tank.
Remember... once the compressor turns on while you are filling a tire, it
has to pump enough air to pressurize both the tire AND your air storage
system. Think of it this way... If you had zero air storage tanks, the
compressor would only have to fill the tire being aired up. With storage
tanks, the compressor has to fill both the tire AND the storage tanks...
which would take longer and longer the bigger your storage tank is.
The optimum solution would be to have a 3 to 6 gallon tank, drain the air
out of it to fill your tires and then shut off a valve to the tank so that
when you are using the compressor instead of stored air, the compressor
doesn't have to fill the tank at the same time it is filling the tire.
Then... when you are done filling the tires, you can open the valve to the
tank and fill the tank once you are rolling.
Joe West
> -----Original Message-----
> From: James Towle [mailto:James.Towle@a...]
> Sent: Wednesday, June 20, 2001 7:48 AM
> To: AZ_VJC; arizonajeep@h...
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> Joe,
>
> Thanks, 23 gallons that's a lot.... I was thinking it would
> be close to 20,
> after looking at some ads, but that seemed a bit much.
>
> I was thinking of getting some air tanks to hook up to my
> Viair II. And was
> wondering if the tanks would make an impact on the air up
> time. There is a
> place in Mesa (www.truckn.com) that sells tanks for a
> reasonable price.
>
> What do you think, what size (in gallons) tank(s) would make
> a good impact
> on airing up? Or maybe there is not enough room, with this
> compressor, to
> make an impact.
>
> --James
>
> > -----Original Message-----
> > From: Joe W [mailto:arizonajeep@h...]
> > Sent: Tuesday, June 19, 2001 9:20 PM
> > To: 'James Towle'; 'AZ_VJC'
> > Subject: RE: [az_vjc] Mathematicians wanted
> >
> >
> > James,
> >
> > Let's put down some variables:
> >
> > V1 is the initial volume of the tire to produce the initial
> tire pressure.
> > V2 is the final volume of the tire to produce the required
> increase in
> > pressure.
> > P1 is the initial pressure of the tire.
> > P2 is the final pressure of the tire.
> > T1 is the initial temperature of the air in the tire at P1
> > T2 is the final temperature of the air in the tire at P2
> >
> > Assumption; Temperature of the compressed air in the tire is
> > constant. (not
> > really true since the compressed air heats up as it is compressed).
> >
> > Now for the equations, starting with the Natural Gas Law
> and our initial
> > conditions (initial tire pressure):
> >
> > P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
> >
> > P1*V1 = K*T1
> >
> > Doing the same for the final conditions (final tire pressure):
> > P2*V2 = K*T2
> >
> > Setting the initial condition equal to the final condition
> produces the
> > equation:
> >
> > (P1*V1)/K*T1 = (P2*V2)/K*T2
> >
> > But... since T1 = T2 and K is the same constant for both
> sides of the
> > equation, the equation simplifies to:
> >
> > P1*V1 = P2*V2
> >
> > Since we know the starting volume (V1) and both the
> starting and ending
> > pressures (P1 and P2), we can solve for the single unknown, V2:
> >
> > EQUATION 1 : (P1*V1)/P2 = V2
> >
> > Now I guess we need to plug in some numbers:
> >
> > The area of a cylinder is pi*R^2*H
> >
> > I'll model the tire as a cylinder, the wheel as a cylinder, and
> > subtract the
> > two to find the volume:
> >
> > Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> > Width of the tire is H = 12.5"
> >
> > Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
> >
> > You didn't state the width of your wheel so I'll assume it
> is 8" and it's
> > radius is 7.5"
> >
> > Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
> >
> > The difference in the two volumes is the volume V1 = 9,277
> cubic inches.
> >
> >
> > From your specifications:
> >
> > P1 = 12psi
> >
> > P2 = 28psi
> >
> > So... now all we have to do is solve EQUATION 1 derived somewhere
> > above for
> > the volume V2
> >
> > V2=P1*V1/P2
> >
> > Plugging in the numbers:
> >
> > V2 = 12psi*9,277in^3/28psi
> >
> > V2 = 3,976 in^3
> >
> > So... in order to increase the pressure of your tire, the
> volume of air
> > which had to be produced by your pump is:
> >
> > Delta V = V1-V2
> >
> > Delta V = 9,277-3,976
> >
> > Delta V= 5,301in^3
> >
> > Using the handy dandy conversion from cubic inches to
> gallons of 1gal =
> > 231in^3 give us the delta in gallons of air of:
> >
> > 5,301in^3/231in^3/gal = 22.95 gallons
> >
> > Now... for a sanity check to see if my numbers are
> reasonable... at first
> > blush... 22.95 gallons seems large:
> >
> >
> > 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> > air can be
> > pushed out of a quickair II at the rate of 1.45CFM (at
> 40psi according to
> > quickair's advertisements)...which means that you should be
> able to air up
> > your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
> >
> > 2.12 minutes!!!
> >
> >
> > Hmmmm... seems reasonable that a quickair II can inflate a
> tire from 12 to
> > 28psi in 2.12 minutes....
> >
> > Let's look at what quickair's advertisement on their web page says:
> >
> > "The QuickAIR2 was developed to inflate 33" and larger tires with
> > ease. The
> > QuickAIR2 will inflate a 33x12.5 tire from totally flat to
> 30 psi in less
> > than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*.
> > Get the most
> > out of your tires! QuickAIR2 will let you air down for maximum
> > traction and
> > performance and reinflate quickly. Capable of reseating
> tire on rim."
> >
> >
> > So... I calculate that it will take 2.12 minutes to inflate the
> > tire at the
> > pressures you are asking and quickair2's numbers show that
> you can go from
> > 15psi to 30psi in "less than 2.5 minutes".
> >
> > I guess the sanity check worked and the numbers are very
> close given my
> > rough estimates and assumptions.
> >
> > Your answer is 22.95 +- .25 gallons.
> >
> > Kind Regards,
> >
> > Joe West
> >
> >
> > Now... are you sorry you asked?
> >
> > <grin>
> >
> >
> > > -----Original Message-----
> > > From: James Towle [mailto:James.Towle@a...]
> > > Sent: Tuesday, June 19, 2001 5:20 PM
> > > To: AZ_VJC
> > > Subject: [az_vjc] Mathematicians wanted
> > >
> > >
> > >
> > > Okay who is the math people? Here's the question:
> > >
> > > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > > gallons it
> > > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > > to what a 1/4"
> > > hose can take at the given PSI.
> > >
> > > I know this is some sort of integration problem but, I am not
> > > sure where to
> > > start.
> > >
> > > Joe W. - It isn't thermodynamics but.....
> > >
> > > --James
> > >
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> > > http://docs.yahoo.com/info/terms/
> > >
> > >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
25677 From: Roger Tomas <tomasr@a...>
Date: Wed Jun 20, 2001 9:42am
Subject: Re: Mathematicians wanted
James Towle wrote:
>
> <snip>
> What do you think, what size (in gallons) tank(s) would make a good impact
> on airing up? Or maybe there is not enough room, with this compressor, to
> make an impact.
My opinion is that you should get a compressor that is capable of effectively
airing up tires without a tank and then add a small tank for the sole purpose
of assisting in reseating a popped bead.
A York and a small tank is a great set-up.
-Roger
25678 From: <hunteroffroad@a...>
Date: Wed Jun 20, 2001 5:44am
Subject: Re: Mathematicians wanted
My God Joe.. I am impressed but you need to get out just a bit more. : )
John
25679 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 10:06am
Subject: Jeep
Darner says there is nothing wrong with my jeep, the cat is fine.
However, I still have a very odd noise coming from somewhere under
the jeep mostly under deceleration. If is is a vibration of some
sort but it is not physical enough to allow you to feel it, only hear
it. It's not a squeaking noise, so I thought I could rule out a
bearing of some sort. I haven't lost any power, and the jeep drives
fine. It is not a rotational sound, but very constant. It doesn't
sound like gears, at least it doesn't sound like the noise I heard
when Tara stripped the ring/pinion on her TJ a while back.
I'm at a total loss.
Plus I'm pissed that I paid them $40 to tell me there isn't anything
wrong with the jeep.
Dan
25655 From: Eric Marble <emarble3@p...>
Date: Tue Jun 19, 2001 9:40pm
Subject: Re: Mathematicians wanted
Remember Joe it's a flexible container not a scuba cylinder so you have to
calculate expansion also..
Padi #56992
p.s. now take it to 60 feet..
Eric
----- Original Message -----
From: "Joe W" <arizonajeep@h...>
To: "'James Towle'" <James.Towle@a...>; "'AZ_VJC'"
<az_vjc@yahoogroups.com>
Sent: Tuesday, June 19, 2001 9:20 PM
Subject: RE: [az_vjc] Mathematicians wanted
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant.
(not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above
for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air can
be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
most
> out of your tires! QuickAIR2 will let you air down for maximum traction
and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at
the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
--------------------------------------------------------------------------------
ADVERTISEMENT
25656 From: linda luik <minihummer@w...>
Date: Tue Jun 19, 2001 9:39pm
Subject: Re: Mathematicians wanted
I am assuming this is with no weight on the tire. That is, the tire is
off the jeep? Or the jeep is up in the air? Or would that make a
difference?
Linda
Joe W wrote:
>
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
> >From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the most
> out of your tires! QuickAIR2 will let you air down for maximum traction and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
> Now... are you sorry you asked?
>
> <grin>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
25657 From: Leah & Geoff <leahlsq@q...>
Date: Tue Jun 19, 2001 8:45pm
Subject: TUCSON - Pima Motor Sports Park needs our help!
I'm back from Alaska and finally at my PC.
This is from Becky Petermann - PLease help out!
I'm leading a run to Rice Peak this Saturday and will
be out at the Park on Sunday.
If you have any questions call me at 546-0501
Leah
----------------------------------------------------------------------------
------------
I have been out at the Pima Motorsports Park for the last three days after
work. We are doing very well, but I need help. And unfortunately I need it
now.
I am hoping that some of you would be willing to give up a Saturday or
Sunday or both to come help me.
We need to put in fencing around our area. We also need to finish a couple
of the obstacles.
When was the last time you got to be the first to drive on a road. Or
actually put in a road? Well here is your chance.
The major problem is that it has to be this weekend.
We need to finish the rock hill.
It is looking pretty good but it needs to be finished.
So please take time out to come help us! I will be going out there
Saturday and Sunday.
Meet at the park at 8:00AM.
This is extremely important. I have done all I can, but now is when I need
help. Please let me know.
25658 From: Clint Ramsey <cwramsey@t...>
Date: Tue Jun 19, 2001 10:13pm
Subject: re: TJ driveline
Sounds like exhaust back pressure making the crap in the cat rattle around
Clint
25659 From: T.J. Nosmo-King <ice626@h...>
Date: Tue Jun 19, 2001 10:39pm
Subject: RE: Mathematicians wanted
For a guy with a low-rider to fix, you sure seem to have an awful lot of
time on your hands...... :-)
__________________________________________________ _______________
Get your FREE download of MSN Explorer at http://explorer.msn.com
25660 From: Joe W <arizonajeep@h...>
Date: Tue Jun 19, 2001 10:46pm
Subject: RE: Mathematicians wanted
Ya... Joyce was gone this evening teaching a SCUBA class and I was all alone
with nothing to do but ponder the proposed question.
<grin>
Besides... he put my name on it so I had to reply... kind of like the triple
dog dare ya from the Christmas Story movie.
Joe West
> -----Original Message-----
> From: T.J. Nosmo-King [mailto:ice626@h...]
> Sent: Tuesday, June 19, 2001 10:40 PM
> To: arizonajeep@h...; James.Towle@a...; az_vjc@yahoogroups.com
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> For a guy with a low-rider to fix, you sure seem to have an
> awful lot of
> time on your hands...... :-)
> __________________________________________________ _______________
> Get your FREE download of MSN Explorer at http://explorer.msn.com
>
>
25661 From: <japdep@a...>
Date: Tue Jun 19, 2001 11:27pm
Subject: FS. D35C rear axle from 90' XJ
complete rear end. Broken pinion gear. Carrier, shafts and housing in
good shape. $50 Email me off list if interested...JIm
25662 From: AzVJC Website <azvjc@a...>
Date: Wed Jun 20, 2001 4:34am
Subject: [TR] Cheap Trick - Saturday, June 23, 2001 - Reminder
Hello All,
Dave O'Boyle is leading a Cheap Trick trail run on Saturday, June 23, 2001.
When: Sat June 23 6am.
Where: Circle K Jackrabbit trail and I10. head south on Jackrabbit
1/2 mile. to Circle K.
Rating: 4.5+
Length: 4 to 6 hrs.
CB Channel: 4
Cheep Trick
It's a nasty one.. Expect body damage.. This one may be impassable. e-mail me if your up to the challange.
davidoboyle@hotmail.com
Very Nasty.
Sign Up List:
Dave O'Boyle Definite
John Manansala Definite
Click Here To Add Your Name to the Sign Up List
Thank You,
- The AzVJC Website
25663 From: AzVJC Website <azvjc@a...>
Date: Wed Jun 20, 2001 4:39am
Subject: [TR] Cheap Trick - Saturday, June 23, 2001 - Reminder
Hello All,
Dave O'Boyle is leading a Cheap Trick trail run on Saturday, June 23, 2001.
When: Sat June 23 6am.
Where: Circle K Jackrabbit trail and I10. head south on Jackrabbit
1/2 mile. to Circle K.
Rating: 4.5+
Length: 4 to 6 hrs.
CB Channel: 4
Cheep Trick
It's a nasty one.. Expect body damage.. This one may be impassable. e-mail me if your up to the challange.
davidoboyle@hotmail.com
I am offline Thursday and Friday so call if you need more info. 623-327-1486
Very Nasty.
Sign Up List:
Dave O'Boyle Definite
John Manansala Definite
Click Here To Add Your Name to the Sign Up List
Thank You,
- The AzVJC Website
25664 From: Jay Eller <jay@t...>
Date: Wed Jun 20, 2001 6:23am
Subject: Re: Mathematicians wanted
"Auntie-M....Auntie-M.....It's a twister!!! It's a twister":-0
Excellent derivation professor! But what about the orifice created by the valve
reducing the flow rate to the tire?
I would guess that with the orifice created by the tire valve, it would take you
just long enough for your toes to start hurting while squatted down filling the
tire........give or take a few comments about "I should have sprung for a damned
CO2 tank!"
Just a little light hearted humour.
--
----------------------------
Jay Eller (http://www.goodnet.com/~eller)
http://www.toyboxoffroad.com
25665 From: <davidoboyle@h...>
Date: Wed Jun 20, 2001 7:04am
Subject: Cheep Trick sat..
Sorry for the double post on the Cheep Trick run.
I added my Phone number to the trip because I am offline Thursday
thru Saturday. Hope to see some Big boys/Girls out this weekend..
This is one SICK trail......... If you don't believe me ask Joey K.
We both walked it and had a hard time even doing that...
See you Sat.
Dave
623-327-1486
25666 From: James Towle <James.Towle@a...>
Date: Wed Jun 20, 2001 7:48am
Subject: RE: Mathematicians wanted
Joe,
Thanks, 23 gallons that's a lot.... I was thinking it would be close to 20,
after looking at some ads, but that seemed a bit much.
I was thinking of getting some air tanks to hook up to my Viair II. And was
wondering if the tanks would make an impact on the air up time. There is a
place in Mesa (www.truckn.com) that sells tanks for a reasonable price.
What do you think, what size (in gallons) tank(s) would make a good impact
on airing up? Or maybe there is not enough room, with this compressor, to
make an impact.
--James
> -----Original Message-----
> From: Joe W [mailto:arizonajeep@h...]
> Sent: Tuesday, June 19, 2001 9:20 PM
> To: 'James Towle'; 'AZ_VJC'
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is
> constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and
> subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere
> above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according to
> quickair's advertisements)...which means that you should be able to air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with
> ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*.
> Get the most
> out of your tires! QuickAIR2 will let you air down for maximum
> traction and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the
> tire at the
> pressures you are asking and quickair2's numbers show that you can go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
25667 From: Mike Baney <jeepin_in_az@y...>
Date: Wed Jun 20, 2001 8:03am
Subject: Re: TJ Driveline ?
I didn't think you could notice a loss of power in a 4 cyl? <big grin>
Mike
--- "T.J. Nosmo-King" <ice626@h...> wrote:
> I think that if it was the cat, you would not only hear the rattling
> noise
> but you would notice a DEFINITE loss in power during accerlation due to
> the
> loose pieces clogging the system....at least that is what happend to me
> when
> my cat broke up....a little pounding with a steel rod and the emptying
> of
> the ceramic pieces made a great difference in air flow (amazing what an
> empty container can do :-) )
> __________________________________________________ _______________
> Get your FREE download of MSN Explorer at http://explorer.msn.com
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
>
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25668 From: Mike Baney <jeepin_in_az@y...>
Date: Wed Jun 20, 2001 8:11am
Subject: RE: Mathematicians wanted
WOW!!
--- Joe W <arizonajeep@h...> wrote:
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire
> pressure.
> V2 is the final volume of the tire to produce the required increase in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is constant.
> (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
> the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8" and
> it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere above
> for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of 1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at
> first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air
> can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according
> to
> quickair's advertisements)...which means that you should be able to air
> up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12
> to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
> The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in
> less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
> most
> out of your tires! QuickAIR2 will let you air down for maximum traction
> and
> performance and reinflate quickly. Capable of reseating tire on rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the tire at
> the
> pressures you are asking and quickair2's numbers show that you can go
> from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>
>
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25669 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 8:24am
Subject: Jeep @ Dealer
I didn't notice power loss, but the noise was gradually getting
louder till the other day when it became *extremely* noticeable.
With stock gears and 33" swampers, you don't exactly say "loss" of
power in a TJ....more like, change in driving dynamics. =)
The jeep is in the shop right now, but I watched them move it from
the service bay to the back, they drove it in REVERSE all the way
around the building, I'm so pissed they are gonna get a earfull when
I go back!!
When I took it in I told the guy I had front and rear lockers so
drive conservatively if they drive it and the guy said "you have
what?".......
The look on the service guy's faces when I drove up was worth a
million bucks though.
25670 From: Chris Krieg <rv6a@m...>
Date: Wed Jun 20, 2001 8:22am
Subject: Re: Mathematicians wanted
Yeah, thats the same answer I came up with.
Chris K
> --- Joe W <arizonajeep@h...> wrote:
>> James,
>>
>> Let's put down some variables:
>>
>> V1 is the initial volume of the tire to produce the initial tire
>> pressure.
>> V2 is the final volume of the tire to produce the required increase in
>> pressure.
>> P1 is the initial pressure of the tire.
>> P2 is the final pressure of the tire.
>> T1 is the initial temperature of the air in the tire at P1
>> T2 is the final temperature of the air in the tire at P2
>>
>> Assumption; Temperature of the compressed air in the tire is constant.
>> (not
>> really true since the compressed air heats up as it is compressed).
>>
>> Now for the equations, starting with the Natural Gas Law and our initial
>> conditions (initial tire pressure):
>>
>> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>>
>> P1*V1 = K*T1
>>
>> Doing the same for the final conditions (final tire pressure):
>> P2*V2 = K*T2
>>
>> Setting the initial condition equal to the final condition produces the
>> equation:
>>
>> (P1*V1)/K*T1 = (P2*V2)/K*T2
>>
>> But... since T1 = T2 and K is the same constant for both sides of the
>> equation, the equation simplifies to:
>>
>> P1*V1 = P2*V2
>>
>> Since we know the starting volume (V1) and both the starting and ending
>> pressures (P1 and P2), we can solve for the single unknown, V2:
>>
>> EQUATION 1 : (P1*V1)/P2 = V2
>>
>> Now I guess we need to plug in some numbers:
>>
>> The area of a cylinder is pi*R^2*H
>>
>> I'll model the tire as a cylinder, the wheel as a cylinder, and subtract
>> the
>> two to find the volume:
>>
>> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
>> Width of the tire is H = 12.5"
>>
>> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>>
>> You didn't state the width of your wheel so I'll assume it is 8" and
>> it's
>> radius is 7.5"
>>
>> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>>
>> The difference in the two volumes is the volume V1 = 9,277 cubic inches.
>>
>>
>> From your specifications:
>>
>> P1 = 12psi
>>
>> P2 = 28psi
>>
>> So... now all we have to do is solve EQUATION 1 derived somewhere above
>> for
>> the volume V2
>>
>> V2=P1*V1/P2
>>
>> Plugging in the numbers:
>>
>> V2 = 12psi*9,277in^3/28psi
>>
>> V2 = 3,976 in^3
>>
>> So... in order to increase the pressure of your tire, the volume of air
>> which had to be produced by your pump is:
>>
>> Delta V = V1-V2
>>
>> Delta V = 9,277-3,976
>>
>> Delta V= 5,301in^3
>>
>> Using the handy dandy conversion from cubic inches to gallons of 1gal =
>> 231in^3 give us the delta in gallons of air of:
>>
>> 5,301in^3/231in^3/gal = 22.95 gallons
>>
>> Now... for a sanity check to see if my numbers are reasonable... at
>> first
>> blush... 22.95 gallons seems large:
>>
>>
>> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of air
>> can be
>> pushed out of a quickair II at the rate of 1.45CFM (at 40psi according
>> to
>> quickair's advertisements)...which means that you should be able to air
>> up
>> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>>
>> 2.12 minutes!!!
>>
>>
>> Hmmmm... seems reasonable that a quickair II can inflate a tire from 12
>> to
>> 28psi in 2.12 minutes....
>>
>> Let's look at what quickair's advertisement on their web page says:
>>
>> "The QuickAIR2 was developed to inflate 33" and larger tires with ease.
>> The
>> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi in
>> less
>> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get the
>> most
>> out of your tires! QuickAIR2 will let you air down for maximum traction
>> and
>> performance and reinflate quickly. Capable of reseating tire on rim."
>>
>>
>> So... I calculate that it will take 2.12 minutes to inflate the tire at
>> the
>> pressures you are asking and quickair2's numbers show that you can go
>> from
>> 15psi to 30psi in "less than 2.5 minutes".
>>
>> I guess the sanity check worked and the numbers are very close given my
>> rough estimates and assumptions.
>>
>> Your answer is 22.95 +- .25 gallons.
>>
>> Kind Regards,
>>
>> Joe West
>>
>>
>> Now... are you sorry you asked?
>>
>> <grin>
>>
>>
>>> -----Original Message-----
>>> From: James Towle [mailto:James.Towle@a...]
>>> Sent: Tuesday, June 19, 2001 5:20 PM
>>> To: AZ_VJC
>>> Subject: [az_vjc] Mathematicians wanted
>>>
>>>
>>>
>>> Okay who is the math people? Here's the question:
>>>
>>> If you have a 33" x 12.5" on a 15" rim, want is the amount in
>>> gallons it
>>> takes to pressurize the tire from 12#'s to 28#'s if you have a
>>> compressor/tank filling at 90PSI? Lets say the CFM is limited
>>> to what a 1/4"
>>> hose can take at the given PSI.
>>>
>>> I know this is some sort of integration problem but, I am not
>>> sure where to
>>> start.
>>>
>>> Joe W. - It isn't thermodynamics but.....
>>>
>>> --James
>>>
>>>
>>>
>>>
>>> Your use of Yahoo! Groups is subject to
>>> http://docs.yahoo.com/info/terms/
>>>
>>>
>>
>>
>>
>>
>> Your use of Yahoo! Groups is subject to
>> http://docs.yahoo.com/info/terms/
>>
>>
>
>
> __________________________________________________
> Do You Yahoo!?
> Get personalized email addresses from Yahoo! Mail
> http://personal.mail.yahoo.com/
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
25671 From: <msa12171@a...>
Date: Wed Jun 20, 2001 8:18am
Subject: Re: Mathematicians wanted
Joe, you such a geek. :)
--- In az_vjc@y..., "Joe W" <arizonajeep@h...> wrote:
> James,
>
> Let's put down some variables:
>
> V1 is the initial volume of the tire to produce the initial tire
pressure.
> V2 is the final volume of the tire to produce the required increase
in
> pressure.
> P1 is the initial pressure of the tire.
> P2 is the final pressure of the tire.
> T1 is the initial temperature of the air in the tire at P1
> T2 is the final temperature of the air in the tire at P2
>
> Assumption; Temperature of the compressed air in the tire is
constant. (not
> really true since the compressed air heats up as it is compressed).
>
> Now for the equations, starting with the Natural Gas Law and our
initial
> conditions (initial tire pressure):
>
> P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
>
> P1*V1 = K*T1
>
> Doing the same for the final conditions (final tire pressure):
> P2*V2 = K*T2
>
> Setting the initial condition equal to the final condition produces
the
> equation:
>
> (P1*V1)/K*T1 = (P2*V2)/K*T2
>
> But... since T1 = T2 and K is the same constant for both sides of
the
> equation, the equation simplifies to:
>
> P1*V1 = P2*V2
>
> Since we know the starting volume (V1) and both the starting and
ending
> pressures (P1 and P2), we can solve for the single unknown, V2:
>
> EQUATION 1 : (P1*V1)/P2 = V2
>
> Now I guess we need to plug in some numbers:
>
> The area of a cylinder is pi*R^2*H
>
> I'll model the tire as a cylinder, the wheel as a cylinder, and
subtract the
> two to find the volume:
>
> Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> Width of the tire is H = 12.5"
>
> Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
>
> You didn't state the width of your wheel so I'll assume it is 8"
and it's
> radius is 7.5"
>
> Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
>
> The difference in the two volumes is the volume V1 = 9,277 cubic
inches.
>
>
> From your specifications:
>
> P1 = 12psi
>
> P2 = 28psi
>
> So... now all we have to do is solve EQUATION 1 derived somewhere
above for
> the volume V2
>
> V2=P1*V1/P2
>
> Plugging in the numbers:
>
> V2 = 12psi*9,277in^3/28psi
>
> V2 = 3,976 in^3
>
> So... in order to increase the pressure of your tire, the volume of
air
> which had to be produced by your pump is:
>
> Delta V = V1-V2
>
> Delta V = 9,277-3,976
>
> Delta V= 5,301in^3
>
> Using the handy dandy conversion from cubic inches to gallons of
1gal =
> 231in^3 give us the delta in gallons of air of:
>
> 5,301in^3/231in^3/gal = 22.95 gallons
>
> Now... for a sanity check to see if my numbers are reasonable... at
first
> blush... 22.95 gallons seems large:
>
>
> 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
air can be
> pushed out of a quickair II at the rate of 1.45CFM (at 40psi
according to
> quickair's advertisements)...which means that you should be able to
air up
> your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
>
> 2.12 minutes!!!
>
>
> Hmmmm... seems reasonable that a quickair II can inflate a tire
from 12 to
> 28psi in 2.12 minutes....
>
> Let's look at what quickair's advertisement on their web page says:
>
> "The QuickAIR2 was developed to inflate 33" and larger tires with
ease. The
> QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi
in less
> than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get
the most
> out of your tires! QuickAIR2 will let you air down for maximum
traction and
> performance and reinflate quickly. Capable of reseating tire on
rim."
>
>
> So... I calculate that it will take 2.12 minutes to inflate the
tire at the
> pressures you are asking and quickair2's numbers show that you can
go from
> 15psi to 30psi in "less than 2.5 minutes".
>
> I guess the sanity check worked and the numbers are very close
given my
> rough estimates and assumptions.
>
> Your answer is 22.95 +- .25 gallons.
>
> Kind Regards,
>
> Joe West
>
>
> Now... are you sorry you asked?
>
> <grin>
>
>
> > -----Original Message-----
> > From: James Towle [mailto:James.Towle@a...]
> > Sent: Tuesday, June 19, 2001 5:20 PM
> > To: AZ_VJC
> > Subject: [az_vjc] Mathematicians wanted
> >
> >
> >
> > Okay who is the math people? Here's the question:
> >
> > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > gallons it
> > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > to what a 1/4"
> > hose can take at the given PSI.
> >
> > I know this is some sort of integration problem but, I am not
> > sure where to
> > start.
> >
> > Joe W. - It isn't thermodynamics but.....
> >
> > --James
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
25672 From: tom lafrance <tlafrance@j...>
Date: Wed Jun 20, 2001 8:30am
Subject: Manual CJ7 steering box needed
Hello Group,
I'm in need of a manual steering box for my 78 CJ7. I'm hoping someone
has a spare lurking around the garage left over from a p/s conversion?
Thanks,
Tom
25673 From: Chris R. <my1stjeep@e...>
Date: Wed Jun 20, 2001 8:52am
Subject: Re: Re: Mathematicians wanted(Curve Ball Thrown...)
Well here is a Curve Joe. With the expansion or compression of air among
other things due to surrounding tempatures how would your numbers be
affected if you were in say Alaska vs Death Valley?
Chris
http://www.hotstuff.alloffroad.com/
My1stJeep@e...
'97 TJ
___
[___]
-(O|||||O)-
=======
[||]----O--[||]
----------------------------------------------------
"He who asks is a fool for five minutes, but he who does not ask remains a
fool forever."
On Wed, 20 Jun 2001 15:18:24 -0000, msa12171@a... wrote:
> Joe, you such a geek. :)
>
>
>
>
>
> --- In az_vjc@y..., "Joe W" <arizonajeep@h...> wrote:
> > James,
> >
> > Let's put down some variables:
> >
> > V1 is the initial volume of the tire to produce the initial tire
> pressure.
> > V2 is the final volume of the tire to produce the required increase
> in
> > pressure.
> > P1 is the initial pressure of the tire.
> > P2 is the final pressure of the tire.
> > T1 is the initial temperature of the air in the tire at P1
> > T2 is the final temperature of the air in the tire at P2
> >
> > Assumption; Temperature of the compressed air in the tire is
> constant. (not
> > really true since the compressed air heats up as it is compressed).
> >
> > Now for the equations, starting with the Natural Gas Law and our
> initial
> > conditions (initial tire pressure):
> >
> > P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
> >
> > P1*V1 = K*T1
> >
> > Doing the same for the final conditions (final tire pressure):
> > P2*V2 = K*T2
> >
> > Setting the initial condition equal to the final condition produces
> the
> > equation:
> >
> > (P1*V1)/K*T1 = (P2*V2)/K*T2
> >
> > But... since T1 = T2 and K is the same constant for both sides of
> the
> > equation, the equation simplifies to:
> >
> > P1*V1 = P2*V2
> >
> > Since we know the starting volume (V1) and both the starting and
> ending
> > pressures (P1 and P2), we can solve for the single unknown, V2:
> >
> > EQUATION 1 : (P1*V1)/P2 = V2
> >
> > Now I guess we need to plug in some numbers:
> >
> > The area of a cylinder is pi*R^2*H
> >
> > I'll model the tire as a cylinder, the wheel as a cylinder, and
> subtract the
> > two to find the volume:
> >
> > Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> > Width of the tire is H = 12.5"
> >
> > Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
> >
> > You didn't state the width of your wheel so I'll assume it is 8"
> and it's
> > radius is 7.5"
> >
> > Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
> >
> > The difference in the two volumes is the volume V1 = 9,277 cubic
> inches.
> >
> >
> > From your specifications:
> >
> > P1 = 12psi
> >
> > P2 = 28psi
> >
> > So... now all we have to do is solve EQUATION 1 derived somewhere
> above for
> > the volume V2
> >
> > V2=P1*V1/P2
> >
> > Plugging in the numbers:
> >
> > V2 = 12psi*9,277in^3/28psi
> >
> > V2 = 3,976 in^3
> >
> > So... in order to increase the pressure of your tire, the volume of
> air
> > which had to be produced by your pump is:
> >
> > Delta V = V1-V2
> >
> > Delta V = 9,277-3,976
> >
> > Delta V= 5,301in^3
> >
> > Using the handy dandy conversion from cubic inches to gallons of
> 1gal =
> > 231in^3 give us the delta in gallons of air of:
> >
> > 5,301in^3/231in^3/gal = 22.95 gallons
> >
> > Now... for a sanity check to see if my numbers are reasonable... at
> first
> > blush... 22.95 gallons seems large:
> >
> >
> > 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> air can be
> > pushed out of a quickair II at the rate of 1.45CFM (at 40psi
> according to
> > quickair's advertisements)...which means that you should be able to
> air up
> > your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
> >
> > 2.12 minutes!!!
> >
> >
> > Hmmmm... seems reasonable that a quickair II can inflate a tire
> from 12 to
> > 28psi in 2.12 minutes....
> >
> > Let's look at what quickair's advertisement on their web page says:
> >
> > "The QuickAIR2 was developed to inflate 33" and larger tires with
> ease. The
> > QuickAIR2 will inflate a 33x12.5 tire from totally flat to 30 psi
> in less
> > than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*. Get
> the most
> > out of your tires! QuickAIR2 will let you air down for maximum
> traction and
> > performance and reinflate quickly. Capable of reseating tire on
> rim."
> >
> >
> > So... I calculate that it will take 2.12 minutes to inflate the
> tire at the
> > pressures you are asking and quickair2's numbers show that you can
> go from
> > 15psi to 30psi in "less than 2.5 minutes".
> >
> > I guess the sanity check worked and the numbers are very close
> given my
> > rough estimates and assumptions.
> >
> > Your answer is 22.95 +- .25 gallons.
> >
> > Kind Regards,
> >
> > Joe West
> >
> >
> > Now... are you sorry you asked?
> >
> > <grin>
> >
> >
> > > -----Original Message-----
> > > From: James Towle [mailto:James.Towle@a...]
> > > Sent: Tuesday, June 19, 2001 5:20 PM
> > > To: AZ_VJC
> > > Subject: [az_vjc] Mathematicians wanted
> > >
> > >
> > >
> > > Okay who is the math people? Here's the question:
> > >
> > > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > > gallons it
> > > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > > to what a 1/4"
> > > hose can take at the given PSI.
> > >
> > > I know this is some sort of integration problem but, I am not
> > > sure where to
> > > start.
> > >
> > > Joe W. - It isn't thermodynamics but.....
> > >
> > > --James
> > >
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> > > http://docs.yahoo.com/info/terms/
> > >
> > >
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
__________________________________________________ _____
Send a cool gift with your E-Card
http://www.bluemountain.com/giftcenter/
25674 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 8:57am
Subject: Re: ladies run
I'm sure Tara would lead a run, but I can say that cause she probably
isn't reading this. But to add, I'm not sure I'd want her to lead,
she's gone through stuff before me cause I didn't think we could make
it.
Let's just say she has no fear of water; or breaking things (axles to
be specific). =)
25675 From: Joe W <arizonajeep@h...>
Date: Wed Jun 20, 2001 9:03am
Subject: RE: Mathematicians wanted
The orifice created by the valve reducing the flow rate would not affect the
number of gallons required to fill the tire, all it would do is increase the
time required to fill it.
Joe West
> -----Original Message-----
> From: Jay Eller [mailto:jay@t...]
> Sent: Wednesday, June 20, 2001 6:23 AM
> Cc: 'AZ_VJC'
> Subject: Re: [az_vjc] Mathematicians wanted
>
>
> "Auntie-M....Auntie-M.....It's a twister!!! It's a twister":-0
>
> Excellent derivation professor! But what about the orifice
> created by the valve
> reducing the flow rate to the tire?
>
25676 From: Joe W <arizonajeep@h...>
Date: Wed Jun 20, 2001 9:10am
Subject: RE: Mathematicians wanted
James,
There is a tradeoff in tank size vs. tire filling time. Unless you can
store enough air to fill ALL the tires without turning on the compressor,
I'd stick with a 3 to 6 gallon tank.
Remember... once the compressor turns on while you are filling a tire, it
has to pump enough air to pressurize both the tire AND your air storage
system. Think of it this way... If you had zero air storage tanks, the
compressor would only have to fill the tire being aired up. With storage
tanks, the compressor has to fill both the tire AND the storage tanks...
which would take longer and longer the bigger your storage tank is.
The optimum solution would be to have a 3 to 6 gallon tank, drain the air
out of it to fill your tires and then shut off a valve to the tank so that
when you are using the compressor instead of stored air, the compressor
doesn't have to fill the tank at the same time it is filling the tire.
Then... when you are done filling the tires, you can open the valve to the
tank and fill the tank once you are rolling.
Joe West
> -----Original Message-----
> From: James Towle [mailto:James.Towle@a...]
> Sent: Wednesday, June 20, 2001 7:48 AM
> To: AZ_VJC; arizonajeep@h...
> Subject: RE: [az_vjc] Mathematicians wanted
>
>
> Joe,
>
> Thanks, 23 gallons that's a lot.... I was thinking it would
> be close to 20,
> after looking at some ads, but that seemed a bit much.
>
> I was thinking of getting some air tanks to hook up to my
> Viair II. And was
> wondering if the tanks would make an impact on the air up
> time. There is a
> place in Mesa (www.truckn.com) that sells tanks for a
> reasonable price.
>
> What do you think, what size (in gallons) tank(s) would make
> a good impact
> on airing up? Or maybe there is not enough room, with this
> compressor, to
> make an impact.
>
> --James
>
> > -----Original Message-----
> > From: Joe W [mailto:arizonajeep@h...]
> > Sent: Tuesday, June 19, 2001 9:20 PM
> > To: 'James Towle'; 'AZ_VJC'
> > Subject: RE: [az_vjc] Mathematicians wanted
> >
> >
> > James,
> >
> > Let's put down some variables:
> >
> > V1 is the initial volume of the tire to produce the initial
> tire pressure.
> > V2 is the final volume of the tire to produce the required
> increase in
> > pressure.
> > P1 is the initial pressure of the tire.
> > P2 is the final pressure of the tire.
> > T1 is the initial temperature of the air in the tire at P1
> > T2 is the final temperature of the air in the tire at P2
> >
> > Assumption; Temperature of the compressed air in the tire is
> > constant. (not
> > really true since the compressed air heats up as it is compressed).
> >
> > Now for the equations, starting with the Natural Gas Law
> and our initial
> > conditions (initial tire pressure):
> >
> > P1*V1 = N*R*T1 (Let N*R= constant K) which reduces the equation to:
> >
> > P1*V1 = K*T1
> >
> > Doing the same for the final conditions (final tire pressure):
> > P2*V2 = K*T2
> >
> > Setting the initial condition equal to the final condition
> produces the
> > equation:
> >
> > (P1*V1)/K*T1 = (P2*V2)/K*T2
> >
> > But... since T1 = T2 and K is the same constant for both
> sides of the
> > equation, the equation simplifies to:
> >
> > P1*V1 = P2*V2
> >
> > Since we know the starting volume (V1) and both the
> starting and ending
> > pressures (P1 and P2), we can solve for the single unknown, V2:
> >
> > EQUATION 1 : (P1*V1)/P2 = V2
> >
> > Now I guess we need to plug in some numbers:
> >
> > The area of a cylinder is pi*R^2*H
> >
> > I'll model the tire as a cylinder, the wheel as a cylinder, and
> > subtract the
> > two to find the volume:
> >
> > Radius of the tire is Diameter/2 so the radius of your tire is 16.5"
> > Width of the tire is H = 12.5"
> >
> > Volume of the tire= pi*16.5^2*12.5 = 10,691 cubic inches
> >
> > You didn't state the width of your wheel so I'll assume it
> is 8" and it's
> > radius is 7.5"
> >
> > Volume of the wheel is pi*7.5^2*8 = 1,413 cubic inches
> >
> > The difference in the two volumes is the volume V1 = 9,277
> cubic inches.
> >
> >
> > From your specifications:
> >
> > P1 = 12psi
> >
> > P2 = 28psi
> >
> > So... now all we have to do is solve EQUATION 1 derived somewhere
> > above for
> > the volume V2
> >
> > V2=P1*V1/P2
> >
> > Plugging in the numbers:
> >
> > V2 = 12psi*9,277in^3/28psi
> >
> > V2 = 3,976 in^3
> >
> > So... in order to increase the pressure of your tire, the
> volume of air
> > which had to be produced by your pump is:
> >
> > Delta V = V1-V2
> >
> > Delta V = 9,277-3,976
> >
> > Delta V= 5,301in^3
> >
> > Using the handy dandy conversion from cubic inches to
> gallons of 1gal =
> > 231in^3 give us the delta in gallons of air of:
> >
> > 5,301in^3/231in^3/gal = 22.95 gallons
> >
> > Now... for a sanity check to see if my numbers are
> reasonable... at first
> > blush... 22.95 gallons seems large:
> >
> >
> > 5,301in^3 is equivalent to 3.067ft^3 of air. 3.067 cubic foot of
> > air can be
> > pushed out of a quickair II at the rate of 1.45CFM (at
> 40psi according to
> > quickair's advertisements)...which means that you should be
> able to air up
> > your tire from 12psi to 28psi in just about 3.067/1.45 minutes =
> >
> > 2.12 minutes!!!
> >
> >
> > Hmmmm... seems reasonable that a quickair II can inflate a
> tire from 12 to
> > 28psi in 2.12 minutes....
> >
> > Let's look at what quickair's advertisement on their web page says:
> >
> > "The QuickAIR2 was developed to inflate 33" and larger tires with
> > ease. The
> > QuickAIR2 will inflate a 33x12.5 tire from totally flat to
> 30 psi in less
> > than 5 minutes, or from 15 to 30 psi in less than 2.5 minutes*.
> > Get the most
> > out of your tires! QuickAIR2 will let you air down for maximum
> > traction and
> > performance and reinflate quickly. Capable of reseating
> tire on rim."
> >
> >
> > So... I calculate that it will take 2.12 minutes to inflate the
> > tire at the
> > pressures you are asking and quickair2's numbers show that
> you can go from
> > 15psi to 30psi in "less than 2.5 minutes".
> >
> > I guess the sanity check worked and the numbers are very
> close given my
> > rough estimates and assumptions.
> >
> > Your answer is 22.95 +- .25 gallons.
> >
> > Kind Regards,
> >
> > Joe West
> >
> >
> > Now... are you sorry you asked?
> >
> > <grin>
> >
> >
> > > -----Original Message-----
> > > From: James Towle [mailto:James.Towle@a...]
> > > Sent: Tuesday, June 19, 2001 5:20 PM
> > > To: AZ_VJC
> > > Subject: [az_vjc] Mathematicians wanted
> > >
> > >
> > >
> > > Okay who is the math people? Here's the question:
> > >
> > > If you have a 33" x 12.5" on a 15" rim, want is the amount in
> > > gallons it
> > > takes to pressurize the tire from 12#'s to 28#'s if you have a
> > > compressor/tank filling at 90PSI? Lets say the CFM is limited
> > > to what a 1/4"
> > > hose can take at the given PSI.
> > >
> > > I know this is some sort of integration problem but, I am not
> > > sure where to
> > > start.
> > >
> > > Joe W. - It isn't thermodynamics but.....
> > >
> > > --James
> > >
> > >
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> > > http://docs.yahoo.com/info/terms/
> > >
> > >
> >
>
>
>
>
> Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
25677 From: Roger Tomas <tomasr@a...>
Date: Wed Jun 20, 2001 9:42am
Subject: Re: Mathematicians wanted
James Towle wrote:
>
> <snip>
> What do you think, what size (in gallons) tank(s) would make a good impact
> on airing up? Or maybe there is not enough room, with this compressor, to
> make an impact.
My opinion is that you should get a compressor that is capable of effectively
airing up tires without a tank and then add a small tank for the sole purpose
of assisting in reseating a popped bead.
A York and a small tank is a great set-up.
-Roger
25678 From: <hunteroffroad@a...>
Date: Wed Jun 20, 2001 5:44am
Subject: Re: Mathematicians wanted
My God Joe.. I am impressed but you need to get out just a bit more. : )
John
25679 From: Dan Coley <mt_b@y...>
Date: Wed Jun 20, 2001 10:06am
Subject: Jeep
Darner says there is nothing wrong with my jeep, the cat is fine.
However, I still have a very odd noise coming from somewhere under
the jeep mostly under deceleration. If is is a vibration of some
sort but it is not physical enough to allow you to feel it, only hear
it. It's not a squeaking noise, so I thought I could rule out a
bearing of some sort. I haven't lost any power, and the jeep drives
fine. It is not a rotational sound, but very constant. It doesn't
sound like gears, at least it doesn't sound like the noise I heard
when Tara stripped the ring/pinion on her TJ a while back.
I'm at a total loss.
Plus I'm pissed that I paid them $40 to tell me there isn't anything
wrong with the jeep.
Dan