| Topic: |
Science > Physics |
| User: |
"The Ghost In The Machine" |
| Date: |
27 Oct 2006 07:05:46 PM |
| Object: |
Carnot factoring in Mini drag |
I've just thought of something, and am just
wondering whether I'm doing this right.
http://cgi.ebay.com/ebaymotors/IMMACULATE-LIKE-NEW-2006-MINI-COOPER-S-ONLY-5K-MILES_W0QQitemZ270040376772QQihZ017QQcategoryZ107009QQcmdZViewItem
indicates height of 55.8" or 1.417 m, width of 66.5"
or 1.690 m, and a drag coefficient of 0.36 for a Mini
Cooper S. (This is a small 4-door sedan which is
proving popular in the US.)
The density, in kg / m^3, is equal to 0.029 kg/mol * n/V,
= 0.029 * nRT/(VRT) = 0.029 * PV/(VRT) = 0.029 * P/(RT),
where 0.029 kg/mol is the molar density of air, approximately.
P = 101325 Pascal,
R = 8.314472 J/(mol K),
T = let's say 293 K for a nice day.
Therefore rho = 0.029 kg/mol * 101325 Pa/(8.314472 J/(mol K) * 293K),
or 1.2062 kg/m^3.
If we assume highway travel at 65 mph, that translates into a velocity
of 29.0576 m/s.
Therefore, the theoretical drag force is
F = 0.36 * (1.2062 * 29.076^2)/2 * 1.417 * 1.690
= 439 N
and the theoretical power loss because of this drag is F * d / t
= F * v = 12.756 kW. Therefore, this is the minimum amount of
*mechanical power* that is needed to forestall drag, when one is driving
on an absolutely flat surface in a calm day at about 20 C.
Carnot, however, pushes its way in because gasoline and/or diesel must
be burned, which basically makes for a heat engine (the cool side being
the air, the hot side probably being the ignition temperature of the
flame or spark).
The gasoline equivalent is given as 132 MJ/gallon but it's far from
clear whether that's chemical energy (basically heat related) or
mechanical energy. I suspect the former but one can easily estimate it,
given gasoline's density (http://www.simetric.co.uk/si_liquids.htm
indicates 737.22 kg/m^3) and molar mass (if one assumes pure octane one
gets 0.114 kg/mol; heptane gets 0.100 kg/mol).
The reaction energy for pure octane is of course
C8H18 + 12.5 O2 = 8 CO2 + 9H2O + E
where we're
breaking 7 C-C bonds or -7 * 607 kJ/mol,
breaking 18 C-H bonds or -18 * 338 kJ/mol,
breaking 12.5 O=O bonds or -12.5 * 498.6 kJ/mol
forming 16 C=O bonds or +16 * 1076.5 kJ/mol
forming 18 H-O bonds or +18 * 427.6 kJ/mol,
for a total gain of
8.348 MJ/mol or 73.23 MJ/kg or 53.98 GJ/m^3
or 204 MJ/gallon, chemical.
Gasoline's ignition temperature is about 530K so one
gets a Carnot efficiency of (530 - 293) / 530 or 45%
or 91 MJ/gallon, mechanical. (At least in theory.
Enthalpy calculations are estimates.)
Given 132 MJ/gal, one gets a gallon duration
of 10348 seconds or 186.8 mpg. That's the
absolute best one can do.
Given 91 MJ/gal, one gets 128.8 mpg.
Any claims higher than about 200 mpg, given a Mini Cooper
S form factor (one might do a lot better for example with
a cowled bicycle), I for one have to take with a large
grain of suspicion.
Did I miss anything?
--
#191,
Linux. Because it's not the desktop that's
important, it's the ability to DO something
with it.
--
Posted via a free Usenet account from http://www.teranews.com
.
|
|
| User: "HR" |
|
| Title: Re: Carnot factoring in Mini drag |
28 Oct 2006 03:55:04 AM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:qfja14-kg2.ln1@sirius.tg00suus7038.net...
I've just thought of something, and am just
wondering whether I'm doing this right.
http://cgi.ebay.com/ebaymotors/IMMACULATE-LIKE-NEW-2006-MINI-COOPER-S-ONLY-5K-MILES_W0QQitemZ270040376772QQihZ017QQcategoryZ107009QQcmdZViewItem
indicates height of 55.8" or 1.417 m, width of 66.5"
or 1.690 m, and a drag coefficient of 0.36 for a Mini
Cooper S. (This is a small 4-door sedan which is
proving popular in the US.)
The density, in kg / m^3, is equal to 0.029 kg/mol * n/V,
= 0.029 * nRT/(VRT) = 0.029 * PV/(VRT) = 0.029 * P/(RT),
where 0.029 kg/mol is the molar density of air, approximately.
P = 101325 Pascal,
R = 8.314472 J/(mol K),
T = let's say 293 K for a nice day.
Therefore rho = 0.029 kg/mol * 101325 Pa/(8.314472 J/(mol K) * 293K),
or 1.2062 kg/m^3.
If we assume highway travel at 65 mph, that translates into a velocity
of 29.0576 m/s.
Therefore, the theoretical drag force is
F = 0.36 * (1.2062 * 29.076^2)/2 * 1.417 * 1.690
= 439 N
and the theoretical power loss because of this drag is F * d / t
= F * v = 12.756 kW. Therefore, this is the minimum amount of
*mechanical power* that is needed to forestall drag, when one is driving
on an absolutely flat surface in a calm day at about 20 C.
Carnot, however, pushes its way in because gasoline and/or diesel must
be burned, which basically makes for a heat engine (the cool side being
the air, the hot side probably being the ignition temperature of the
flame or spark).
The gasoline equivalent is given as 132 MJ/gallon but it's far from
clear whether that's chemical energy (basically heat related) or
mechanical energy. I suspect the former but one can easily estimate it,
given gasoline's density (http://www.simetric.co.uk/si_liquids.htm
indicates 737.22 kg/m^3) and molar mass (if one assumes pure octane one
gets 0.114 kg/mol; heptane gets 0.100 kg/mol).
The reaction energy for pure octane is of course
C8H18 + 12.5 O2 = 8 CO2 + 9H2O + E
where we're
breaking 7 C-C bonds or -7 * 607 kJ/mol,
breaking 18 C-H bonds or -18 * 338 kJ/mol,
breaking 12.5 O=O bonds or -12.5 * 498.6 kJ/mol
forming 16 C=O bonds or +16 * 1076.5 kJ/mol
forming 18 H-O bonds or +18 * 427.6 kJ/mol,
for a total gain of
8.348 MJ/mol or 73.23 MJ/kg or 53.98 GJ/m^3
or 204 MJ/gallon, chemical.
Gasoline's ignition temperature is about 530K so one
gets a Carnot efficiency of (530 - 293) / 530 or 45%
or 91 MJ/gallon, mechanical. (At least in theory.
Enthalpy calculations are estimates.)
Given 132 MJ/gal, one gets a gallon duration
of 10348 seconds or 186.8 mpg. That's the
absolute best one can do.
Given 91 MJ/gal, one gets 128.8 mpg.
Any claims higher than about 200 mpg, given a Mini Cooper
S form factor (one might do a lot better for example with
a cowled bicycle), I for one have to take with a large
grain of suspicion.
Did I miss anything?
--
#191,
Linux. Because it's not the desktop that's
important, it's the ability to DO something
with it.
--
Power train friction losses?
.
|
|
|
|
| User: "HR" |
|
| Title: Re: Carnot factoring in Mini drag |
28 Oct 2006 05:48:37 AM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:qfja14-kg2.ln1@sirius.tg00suus7038.net...
I've just thought of something, and am just
wondering whether I'm doing this right.
http://cgi.ebay.com/ebaymotors/IMMACULATE-LIKE-NEW-2006-MINI-COOPER-S-ONLY-5K-MILES_W0QQitemZ270040376772QQihZ017QQcategoryZ107009QQcmdZViewItem
indicates height of 55.8" or 1.417 m, width of 66.5"
or 1.690 m, and a drag coefficient of 0.36 for a Mini
Cooper S. (This is a small 4-door sedan which is
proving popular in the US.)
The density, in kg / m^3, is equal to 0.029 kg/mol * n/V,
= 0.029 * nRT/(VRT) = 0.029 * PV/(VRT) = 0.029 * P/(RT),
where 0.029 kg/mol is the molar density of air, approximately.
P = 101325 Pascal,
R = 8.314472 J/(mol K),
T = let's say 293 K for a nice day.
Therefore rho = 0.029 kg/mol * 101325 Pa/(8.314472 J/(mol K) * 293K),
or 1.2062 kg/m^3.
If we assume highway travel at 65 mph, that translates into a velocity
of 29.0576 m/s.
Therefore, the theoretical drag force is
F = 0.36 * (1.2062 * 29.076^2)/2 * 1.417 * 1.690
= 439 N
and the theoretical power loss because of this drag is F * d / t
= F * v = 12.756 kW. Therefore, this is the minimum amount of
*mechanical power* that is needed to forestall drag, when one is driving
on an absolutely flat surface in a calm day at about 20 C.
Carnot, however, pushes its way in because gasoline and/or diesel must
be burned, which basically makes for a heat engine (the cool side being
the air, the hot side probably being the ignition temperature of the
flame or spark).
The gasoline equivalent is given as 132 MJ/gallon but it's far from
clear whether that's chemical energy (basically heat related) or
mechanical energy. I suspect the former but one can easily estimate it,
given gasoline's density (http://www.simetric.co.uk/si_liquids.htm
indicates 737.22 kg/m^3) and molar mass (if one assumes pure octane one
gets 0.114 kg/mol; heptane gets 0.100 kg/mol).
The reaction energy for pure octane is of course
C8H18 + 12.5 O2 = 8 CO2 + 9H2O + E
where we're
breaking 7 C-C bonds or -7 * 607 kJ/mol,
breaking 18 C-H bonds or -18 * 338 kJ/mol,
breaking 12.5 O=O bonds or -12.5 * 498.6 kJ/mol
forming 16 C=O bonds or +16 * 1076.5 kJ/mol
forming 18 H-O bonds or +18 * 427.6 kJ/mol,
for a total gain of
8.348 MJ/mol or 73.23 MJ/kg or 53.98 GJ/m^3
or 204 MJ/gallon, chemical.
Gasoline's ignition temperature is about 530K so one
gets a Carnot efficiency of (530 - 293) / 530 or 45%
or 91 MJ/gallon, mechanical. (At least in theory.
Enthalpy calculations are estimates.)
Given 132 MJ/gal, one gets a gallon duration
of 10348 seconds or 186.8 mpg. That's the
absolute best one can do.
Given 91 MJ/gal, one gets 128.8 mpg.
Any claims higher than about 200 mpg, given a Mini Cooper
S form factor (one might do a lot better for example with
a cowled bicycle), I for one have to take with a large
grain of suspicion.
Did I miss anything?
Drive train friction?
.
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: Carnot factoring in Mini drag |
28 Oct 2006 11:31:36 AM |
|
|
In sci.physics, HR
<Dracos@spitenet.com>
wrote
on Sat, 28 Oct 2006 03:48:37 -0700
<12k6dc6q68b6p92@corp.supernews.com>:
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:qfja14-kg2.ln1@sirius.tg00suus7038.net...
I've just thought of something, and am just
wondering whether I'm doing this right.
http://cgi.ebay.com/ebaymotors/IMMACULATE-LIKE-NEW-2006-MINI-COOPER-S-ONLY-5K-MILES_W0QQitemZ270040376772QQihZ017QQcategoryZ107009QQcmdZViewItem
indicates height of 55.8" or 1.417 m, width of 66.5"
or 1.690 m, and a drag coefficient of 0.36 for a Mini
Cooper S. (This is a small 4-door sedan which is
proving popular in the US.)
The density, in kg / m^3, is equal to 0.029 kg/mol * n/V,
= 0.029 * nRT/(VRT) = 0.029 * PV/(VRT) = 0.029 * P/(RT),
where 0.029 kg/mol is the molar density of air, approximately.
P = 101325 Pascal,
R = 8.314472 J/(mol K),
T = let's say 293 K for a nice day.
Therefore rho = 0.029 kg/mol * 101325 Pa/(8.314472 J/(mol K) * 293K),
or 1.2062 kg/m^3.
If we assume highway travel at 65 mph, that translates into a velocity
of 29.0576 m/s.
Therefore, the theoretical drag force is
F = 0.36 * (1.2062 * 29.076^2)/2 * 1.417 * 1.690
= 439 N
and the theoretical power loss because of this drag is F * d / t
= F * v = 12.756 kW. Therefore, this is the minimum amount of
*mechanical power* that is needed to forestall drag, when one is driving
on an absolutely flat surface in a calm day at about 20 C.
Carnot, however, pushes its way in because gasoline and/or diesel must
be burned, which basically makes for a heat engine (the cool side being
the air, the hot side probably being the ignition temperature of the
flame or spark).
The gasoline equivalent is given as 132 MJ/gallon but it's far from
clear whether that's chemical energy (basically heat related) or
mechanical energy. I suspect the former but one can easily estimate it,
given gasoline's density (http://www.simetric.co.uk/si_liquids.htm
indicates 737.22 kg/m^3) and molar mass (if one assumes pure octane one
gets 0.114 kg/mol; heptane gets 0.100 kg/mol).
The reaction energy for pure octane is of course
C8H18 + 12.5 O2 = 8 CO2 + 9H2O + E
where we're
breaking 7 C-C bonds or -7 * 607 kJ/mol,
breaking 18 C-H bonds or -18 * 338 kJ/mol,
breaking 12.5 O=O bonds or -12.5 * 498.6 kJ/mol
forming 16 C=O bonds or +16 * 1076.5 kJ/mol
forming 18 H-O bonds or +18 * 427.6 kJ/mol,
for a total gain of
8.348 MJ/mol or 73.23 MJ/kg or 53.98 GJ/m^3
or 204 MJ/gallon, chemical.
Gasoline's ignition temperature is about 530K so one
gets a Carnot efficiency of (530 - 293) / 530 or 45%
or 91 MJ/gallon, mechanical. (At least in theory.
Enthalpy calculations are estimates.)
Given 132 MJ/gal, one gets a gallon duration
of 10348 seconds or 186.8 mpg. That's the
absolute best one can do.
Given 91 MJ/gal, one gets 128.8 mpg.
Any claims higher than about 200 mpg, given a Mini Cooper
S form factor (one might do a lot better for example with
a cowled bicycle), I for one have to take with a large
grain of suspicion.
Did I miss anything?
Drive train friction?
Good point; also steering power (if the Mini Cooper S has
power steering), tire friction, headlights, miscellany
such as fuel pump, fans, and heater core. I don't know
how to calculate those but they will require additional
power and thus reduce gas mileage even further.
Admittedly, the heater core might increase efficiency by
a small amount since it allows heat to be dumped into
the passenger compartment. However, efficiency would
be increased more were the heater core's output simply
dumped into the outside air.
--
#191,
Is it cheaper to learn Linux, or to hire someone
to fix your Windows problems?
--
Posted via a free Usenet account from http://www.teranews.com
.
|
|
|
|
|
| User: "Androcles" |
|
| Title: Re: Carnot factoring in Mini drag |
27 Oct 2006 08:58:21 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:qfja14-kg2.ln1@sirius.tg00suus7038.net...
| I've just thought of something, and am just
| wondering whether I'm doing this right.
Good grief. Be careful with that thinking.
|
|
http://cgi.ebay.com/ebaymotors/IMMACULATE-LIKE-NEW-2006-MINI-COOPER-S-ONLY-5K-MILES_W0QQitemZ270040376772QQihZ017QQcategoryZ107009QQcmdZViewItem
|
| indicates height of 55.8" or 1.417 m, width of 66.5"
| or 1.690 m, and a drag coefficient of 0.36 for a Mini
| Cooper S. (This is a small 4-door sedan which is
| proving popular in the US.)
| The density, in kg / m^3, is equal to 0.029 kg/mol * n/V,
| = 0.029 * nRT/(VRT) = 0.029 * PV/(VRT) = 0.029 * P/(RT),
| where 0.029 kg/mol is the molar density of air, approximately.
|
| P = 101325 Pascal,
| R = 8.314472 J/(mol K),
| T = let's say 293 K for a nice day.
|
| Therefore rho = 0.029 kg/mol * 101325 Pa/(8.314472 J/(mol K) * 293K),
| or 1.2062 kg/m^3.
|
| If we assume highway travel at 65 mph, that translates into a velocity
| of 29.0576 m/s.
|
| Therefore, the theoretical drag force is
|
| F = 0.36 * (1.2062 * 29.076^2)/2 * 1.417 * 1.690
| = 439 N
|
| and the theoretical power loss because of this drag is F * d / t
| = F * v = 12.756 kW. Therefore, this is the minimum amount of
| *mechanical power* that is needed to forestall drag, when one is driving
| on an absolutely flat surface in a calm day at about 20 C.
|
| Carnot, however, pushes its way in because gasoline and/or diesel must
| be burned, which basically makes for a heat engine (the cool side being
| the air, the hot side probably being the ignition temperature of the
| flame or spark).
Start/Finish: Llyn Mawr Nature Reserve, Carno
The hot side probably being the coolant temperature when one is driving
on an absolutely flat surface going uphill on a calm day at about 0 C
and the interior of the vehicle is 20 C
| The gasoline equivalent is given as 132 MJ/gallon but it's far from
| clear whether that's chemical energy (basically heat related) or
| mechanical energy. I suspect the former but one can easily estimate it,
| given gasoline's density (http://www.simetric.co.uk/si_liquids.htm
| indicates 737.22 kg/m^3) and molar mass (if one assumes pure octane one
| gets 0.114 kg/mol; heptane gets 0.100 kg/mol).
|
| The reaction energy for pure octane is of course
|
| C8H18 + 12.5 O2 = 8 CO2 + 9H2O + E
|
| where we're
|
| breaking 7 C-C bonds or -7 * 607 kJ/mol,
| breaking 18 C-H bonds or -18 * 338 kJ/mol,
| breaking 12.5 O=O bonds or -12.5 * 498.6 kJ/mol
| forming 16 C=O bonds or +16 * 1076.5 kJ/mol
| forming 18 H-O bonds or +18 * 427.6 kJ/mol,
| for a total gain of
| 8.348 MJ/mol or 73.23 MJ/kg or 53.98 GJ/m^3
| or 204 MJ/gallon, chemical.
|
| Gasoline's ignition temperature is about 530K so one
| gets a Carnot efficiency of (530 - 293) / 530 or 45%
| or 91 MJ/gallon, mechanical. (At least in theory.
| Enthalpy calculations are estimates.)
|
| Given 132 MJ/gal, one gets a gallon duration
| of 10348 seconds or 186.8 mpg. That's the
| absolute best one can do.
Is that a US gallon or an imperial gallon?
| Given 91 MJ/gal, one gets 128.8 mpg.
|
| Any claims higher than about 200 mpg, given a Mini Cooper
| S form factor (one might do a lot better for example with
| a cowled bicycle), I for one have to take with a large
| grain of suspicion.
|
| Did I miss anything?
|
A brain if you think a Mini has 4 doors.
In Britain we don't count the sun roof and hatchback as doors. Probably.
Androcles
.
|
|
|
|
| User: "=?UTF-8?Q?Jeff=E2=80=A6Relf?=" |
|
| Title: Win_XP, fuels, engines, billionaires... and a girl in heat. |
04 Nov 2006 06:52:43 AM |
|
|
Hi The_Ghost_In_The_Machine, You wrote:
Linux. Because it's not the desktop that's
important, it's the ability to DO something
Win_XP does more than run a desktop; in fact, you can dump the desktop.
Win_XP runs shitloads of software and hardware that Linux can't touch,
not the least of which is Visual_Studio_2005.
Besides... Why not have _Both_ ? Novell and MicroSoft have joined forces.
As for the best miles per gallon one could get...
As I told T.J., " Math On Everything " is not enough.
You can't know the _Real_ miles per gallon without doing actual road tests;
i.e. on congested, hot highways, commuting to work 5 days a week,
and taking trips to the mountains on the weekends, pulling gear.
Teasing us, T.J. said he did all that,
but wouldn't provide a web site with the details,
saying he didn't want others to beat him to the market.
Further, he doesn't believe his own " end of the oil age " rhetoric because
he's heavily investing in commodities which'd plunge in value, if it happened.
Personally, I believe he's growing sugar cane in Brazil,
which, when converted to alcohol, is a semi-reasonable alternative fuel.
Also, from what little I know, oxygen-boosted engines make sense.
Moving at 10 knots, Fearless_Johnny, the tug he's thinking of giving me,
weights 400 tons ( unloaded ) and burns a ton ( about 300 gallons )
of diesel per day. Loaded, it burns 4 tons a day.
So it only costs 2,500 dollars per day to operate ( I think he said ),
well within anyone's budget ( ha ha ).
The tank holds 25 thousand gallons ( or 85 tons given 6.8 lbs per gallon ).
As far as I can tell, T.J.'s life is very different from ours,
with our tiny budgets; he runs a floating city... thinking at that scale.
P.S. You should see what Patinha isn't wearing right now... holy *****.
She's 19, I'm 46. She's 90 pounds, I'm 180. She wants kids, I don't.
She wants a house, I don't. She likes drugs, I don't. The list goes on.
She claims she'd quit drugs if she had kids... but I wonder.
I got her food stamps ( no small feat ) and she's actually eating now,
which means her boyfriend might get her pregnant ( before he goes to jail ).
Who'll wind up with the kids ? me ? the state ? her mom ?
.
|
|
|
|
| User: "MathFreak NoMore" |
|
| Title: Re: Carnot factoring in Mini drag |
29 Oct 2006 01:45:54 PM |
|
|
On Fri, 27 Oct 2006 17:05:46 -0700, The Ghost In The
Machine wrote:
(This is a small 4-door sedan which is
proving popular in the US.)
No less than forty years behind Tehranese.
--
"When the question is if he was a fag or a god,
you hide the facts like Baha'is do."
- Maleki
.
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: Carnot factoring in Mini drag |
29 Oct 2006 03:56:25 PM |
|
|
In sci.physics, MathFreak NoMore
<MathFreakNoMore@FakeAddress.com>
wrote
on Sun, 29 Oct 2006 13:45:54 -0600
<1wn3kv1cvhjuc$.1qdue5kyp95e1$.dlg@40tude.net>:
On Fri, 27 Oct 2006 17:05:46 -0700, The Ghost In The
Machine wrote:
(This is a small 4-door sedan which is
proving popular in the US.)
No less than forty years behind Tehranese.
I'm sorry; that particular sentence did not make any sense whatsoever,
nor were we discussing Iranian cars (Google indicates there are in fact
such: Khodro, Samand, Peykan X7, New Peykan; stats for such are
available at http://www.farhangsara.com/cars/paykanx7.htm).
The Mini Cooper S is but one of many automobiles; if you prefer I can
redo the computations using a Ford Excursion (which is the biggest SUV
on the planet). Or perhaps you'd prefer I use a cowled Segway? :-)
--
#191,
Useless C++ Programming Idea #1123133:
void f(FILE * fptr, char *p) { fgets(p, sizeof(p), fptr); }
--
Posted via a free Usenet account from http://www.teranews.com
.
|
|
|
| User: "Androcles" |
|
| Title: Re: Carnot factoring in Mini drag |
29 Oct 2006 06:00:04 PM |
|
|
"The Ghost In The Machine" <ewill@sirius.tg00suus7038.net> wrote in message
news:9lkf14-afe.ln1@sirius.tg00suus7038.net...
| In sci.physics, MathFreak NoMore
| <MathFreakNoMore@FakeAddress.com>
| wrote
| on Sun, 29 Oct 2006 13:45:54 -0600
| <1wn3kv1cvhjuc$.1qdue5kyp95e1$.dlg@40tude.net>:
| > On Fri, 27 Oct 2006 17:05:46 -0700, The Ghost In The
| > Machine wrote:
| >
| >> (This is a small 4-door sedan which is
| >> proving popular in the US.)
| >
| > No less than forty years behind Tehranese.
| >
|
| I'm sorry; that particular sentence did not make any sense whatsoever,
Yes it did.
Tehran is in Iran where Tehranese, a dialect of Farsi, is spoken.
| The Mini Cooper S is but one of many automobiles; if you prefer I can
| redo the computations using a Ford Excursion (which is the biggest SUV
| on the planet). Or perhaps you'd prefer I use a cowled Segway? :-)
You are sorry; that particular sentence did not make any sense whatsoever
(but then, you never do).
Androcles
.
|
|
|
|
|
|
|
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|
|
