A simplistic approach to hydrogen powered cars
Max Goodman
I cannot pretend to understand the issues involved in the creation of a
hydrogen powered car but, for little more than my own satisfaction,
following are my thoughts on the matter:
Hydrogen to be stored as gas, on-board, in a removable gas bottle. I
imagine that this would be mounted amidships between the drivers seat and
the front passenger seat and run within a fixed column from the floor to
the roof.
The disadvantage of this system placement is that it would ruin the view
from the rear-view mirror, the advantage is that it would place the unit in
the safest possible place, namely within the passenger cell itself. (more
below on safety)
The first issue is that re-fuelling would be done via replacement of the
entire bottle, not by refilling in-situ. This would have to be done from
underneath vand would require the develpment of a type of conveyer belt to
replace the bottle (in appearance it would resemble an automated car wash
except that underneath the car being pulled along would be the equipment to
remove, replace and presumably refill the bottles.
This therefore is another potential disadvantage in that the cost of setup
for this type of system is relatively high.
The net result of this switch, away from petrol powered vehicles would be
that the whole process would be far more pleasant than the current
self-service gas station which requires the driver to leave the vehicle,
inhale the benzene and other noxious chemicals whilst re-fuelling and then
finally to go and pay.
I believe (hope) that existing engines could be modified to run on a
hydrogen replacement fuel and that further, any front wheel drive car would
be adapted likewise (rear wheel drive would present a problem as the drive
shaft is in the way).
Refuelling of the gas bottles themselves would be done at the gas-station.
This would mean holding a stock of ready filled bottles in addition to
bottles being filled at the time. The amount of stock would depend on the
rate and quantity of bottle filling achievable.
As far as safety is concerned the extending of the bottle from floor to
roof of te car allows a built in weak point to be engineered into the roof
line. In the event of a head on collision this would be desgned to rupture
allowing rapid excape of the hydrogen gas out of harms way. Further within
the tank would be a greased ball bearing, held magnetically in place in a
cup on the base. In the event that the car overturned the ball bearing
would fall to the neck of the bottle blocking the escape of the gas and
preventing a fire hazard.
Told you it was a simple approach.
Anyone got any ideas on what's wrong with this, doubtless there's a lot!!
William Morse
In article <01bc01f2$2a27dc20$27cd...@maxg.hol.gr>,
"Max Goodman" <ma...@hol.gr> wrote:
(snip)
>Hydrogen to be stored as gas, on-board, in a removable gas bottle. I
>imagine that this would be mounted amidships between the drivers seat and
>the front passenger seat and run within a fixed column from the floor to
>the roof.
>
>The first issue is that re-fuelling would be done via replacement of the
>entire bottle, not by refilling in-situ. This would have to be done from
>underneath vand would require the develpment of a type of conveyer belt to
>replace the bottle (in appearance it would resemble an automated car wash
>except that underneath the car being pulled along would be the equipment to
>remove, replace and presumably refill the bottles.
>
>This therefore is another potential disadvantage in that the cost of setup
>for this type of system is relatively high.
>
safeguards there is still a danger of leakage and burning in the passenger
compartment, which can be eliminated by placement outside of the compartment.
However, the idea of a separate container which can be quickly replaced in the
vehicle and then more slowly filled at the station makes a lot of sense.
Yours,
Bill Morse
Peter M.T.
William Morse wrote:
>
> In article <01bc01f2$2a27dc20$27cd...@maxg.hol.gr>,
> "Max Goodman" <ma...@hol.gr> wrote:
> (snip)
> safeguards there is still a danger of leakage and burning in the passenger
> compartment, which can be eliminated by placement outside of the compartment.
> However, the idea of a separate container which can be quickly replaced in the
> vehicle and then more slowly filled at the station makes a lot of sense.
>
> Yours,
>
in other programs}
Anyway now for my two bits:
Look at the way LPG cars have their tanks, Hydrogen tanks are no different.
also I remember hearing about a catalyst that can be put in the tank to store
Hudrogen as Hydrazane. The extraction of Hydrogen from this catalyst is leached
slowly, thus removing the danger of an explosion.
Cheers
Peter T.
Nihar Kirtidev Bhatt
William Morse (wdm...@dreamscape.com) wrote:
: In article <01bc01f2$2a27dc20$27cd...@maxg.hol.gr>,
: "Max Goodman" <ma...@hol.gr> wrote:
: (snip)
: >imagine that this would be mounted amidships between the drivers seat and
: >the front passenger seat and run within a fixed column from the floor to
: >the roof.
: I don't aqree with you on the best placement of the bottle - even with your
: compartment, which can be eliminated by placement outside of the compartment.
: However, the idea of a separate container which can be quickly replaced in the
: vehicle and then more slowly filled at the station makes a lot of sense.
: Bill Morse
------------------------------------------------------------------------------
Before trying to store Hydrogen gas in cars one needs to consider
the volume efficiency and safety aspects in terms of BTU stored/Vs/volume
required. Assuming storage at 2500PSI a large bottle 5ft tall by 9" dia
will store as many BTUs as less than a gallon of gasoline.
Kirtidev Bhatt
Michael Hannon
Storage of H2 gas is totally unnecessary when water can be stored and
separated as needed onboard, and there are methods available, some
basically untried, for doing this. We are on the verge of having these
methods come before us in a general fashion, but I've gotten mail from
someone who has been using H on-demand for years for himself, so there's
no question that it can be done, and in many ways, efficiently enough to
be feasible right now, and some of which give more energy than they take
- over-unity, using natural forces on earth which are present, which if
done in space, would not work, because the conditions would not create
over-unity, such as atmospheric pressure, and implosion. 15 psi
(atmospheric pressure) can run one hell of an implosion engine,
generating more power than is required to produce the Brown's gas needed
to cause it.
Michael
Harry H Conover
Michael Hannon (oha...@worldnet.att.net) wrote:
:
: Storage of H2 gas is totally unnecessary when water can be stored and
: separated as needed onboard...
Er... Using precisely what source of energy?
: basically untried, for doing this. We are on the verge of having these
: methods come before us in a general fashion, but I've gotten mail from
: someone who has been using H on-demand for years for himself,
Please post his name and address so we may verify such a fantastic claim
and, failing that, laugh at him in public.
: so there's
: no question that it can be done,
Sorry, wrong! There's every reason to believe that is cannot be done.
: and in many ways, efficiently enough to
: be feasible right now, and some of which give more energy than they take
: - over-unity, using natural forces on earth which are present, which if
: done in space, would not work, because the conditions would not create
: over-unity, such as atmospheric pressure, and implosion. 15 psi
: (atmospheric pressure) can run one hell of an implosion engine,
: generating more power than is required to produce the Brown's gas needed
: to cause it.
And if frogs had wings, then they could fly!
Michael, you truly are a 'piece of work!' (You make us smile.)
Harry C.
Kevin Jones
Michael Hannon wrote:
<SNIP>
> because the conditions would not create
> over-unity, such as atmospheric pressure, and implosion. 15 psi
> (atmospheric pressure) can run one hell of an implosion engine,
> generating more power than is required to produce the Brown's gas needed
> to cause it.
>
> Michael
Try running a car engine on 15psig with compressed air, and see if you
get any usable power out of it. 15psig will almost exactly imitate the
pressure difference you've inferred by an "implosion engine" running via
atmospheric pressure. See if you can overcome frictional losses,valve
spring forces, etc and even keep the rotating assembly in motion.
K. Jones
The opinions expressed are my own, not those of my employer, or anyone
else.
Michael Hannon
Kevin Jones wrote:
>
> Michael Hannon wrote:
>
> <SNIP>
>
> > because the conditions would not create
> > over-unity, such as atmospheric pressure, and implosion. 15 psi
> > (atmospheric pressure) can run one hell of an implosion engine,
> > generating more power than is required to produce the Brown's gas needed
> > to cause it.
> >
> > Michael
>
> Try running a car engine on 15psig with compressed air, and see if you
> get any usable power out of it. 15psig will almost exactly imitate the
> pressure difference you've inferred by an "implosion engine" running via
> atmospheric pressure. See if you can overcome frictional losses,valve
> spring forces, etc and even keep the rotating assembly in motion.
Let's just do a little math: a 4" diam. pistom has an area of 12.56 sq.
in. 15 lbs./sq. in. X 12.56 sq.in. = 188.4 lbs. of force on that piston,
through the entire stroke (ON ONE PISTON). Think, Kevin, before you
speak. Differential pressure in the atmosphere isn't the same as
atmosphere/vacuum as well. I could go on, and on, but what's the use.
Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
288 ft-lbs of torque.
Think, Kevin - don't attack and obfuscate.
Ian Johnston
Michael Hannon (oha...@worldnet.att.net) wrote:
: Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
: at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
: 288 ft-lbs of torque.
: Think, Kevin - don't attack and obfuscate.
So why do silly car manufacturers insist on using hundreds of psi to do the
same thing?
Ian
Harry H Conover
Michael Hannon (oha...@worldnet.att.net) wrote:
: Differential pressure in the atmosphere isn't the same as
: atmosphere/vacuum as well.
Perhaps you would care to expound on the differences, since most of
us poor, misguided technical types tend to believe that the differential
pressure in each situation causes identical forces on the piston.
Please explain our mistake.
Harry C.
Ian Johnston
Michael Hannon (oha...@worldnet.att.net) wrote:
: Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
: at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
: 288 ft-lbs of torque.
OK, here's some thermodynamics. The work done is the integral of p dv. Since
the pressure is constant, that gives the work done by a volume v as
p v
If this happens n times a second, the rate of working is
P = n p v
In this case, p = 15psi = 10^5 Pa, n = 1000/60 = 20 (for simplicity) and
v = 150 cu in = 2 litres (again for simplicity). Note that the volume is
half the engine swept volume because in a four stroke cycle only half the
cylinders produce power each cycle. Anyway, that gives a power output of
P = 20 * 10^5 * 2 * 10^-3
= 4000 W
And since torque = power / angular velocity, and angular velocity = 2 Pi *
frequency,
T = 4000 / (20 * 2 * Pi)
= 4000 / 125
= 32 Nm
So at 100% efficiency you will get 4kW (which is about 6hp) and 32 Nm torque
(which is about 20 foot pounds).
At the 30% efficiency you accepted you would have 2hp and 6 ft lbs.
Perhaps you could post your working?
Ian
Michael Hannon
TADedek
> So why do silly car manufacturers insist on using hundreds of psi to do the
> same thing?
>
> Ian
Ummmmm, let me guess. They are using hundreds of psi to do the same
thing because the petroleum companies (including opec members) are
strong arming them into compliance. To go against them would mean the
automobile would be squashed into a coverup and subjects of tabloids.
Do I get the pretty prize??? PLEASE???
Sheldon
Don't be pointin your flame throwers at me---this thread needed an
injection of humor and you know it---even if the humor is bad.
Michael Hannon
Wah, no, massuhs, iz cuz y'all don' wan' feedum', special cuz then we'd
awl be ekul, lak God sez, when y'all reale wonna run evethin' an' have
folks lak us'n talkin' lak dis t'yall cuzn' i' make y'all feelz betta
'bout yo empty lavs, ah rek'n, bu', az y'no weinz jus dumm, don no
nuthin' bou nuthin', sep wha y'all sez iz tru, and y'all noze evethin.'
Wa's mah fishin' pol' - don' wanna miss them hone pou'!
Ain' got no munneh,
Ain' got no station, hone',
Bu' ah gus you,
Yeh, bebe, ah gus you.
Ain' gu' no cah, bebe,
Don' need one nah, so mebe,
We git lucke' nex' laf hone',
Mebe den we gess some munneh....
Den we's kin be hah an' mateh,
Makem pay th' weh we duz, sweeteh',
Makem peh' reel goo', now bebe,
Wen' da Lo' cum down fuh usn', bebe,
Wen' da Lo' cum down.......
Harry H Conover
Kevin Jones
Michael Hannon wrote:
>
> Kevin Jones wrote:
> >
> > Michael Hannon wrote:
> >
> > <SNIP>
> >
> > > because the conditions would not create
> > > over-unity, such as atmospheric pressure, and implosion. 15 psi
> > > (atmospheric pressure) can run one hell of an implosion engine,
> > > generating more power than is required to produce the Brown's gas needed
> > > to cause it.
> > >
> > > Michael
> >
> > Try running a car engine on 15psig with compressed air, and see if you
> > get any usable power out of it. 15psig will almost exactly imitate the
> > pressure difference you've inferred by an "implosion engine" running via
> > atmospheric pressure. See if you can overcome frictional losses,valve
> > spring forces, etc and even keep the rotating assembly in motion.
> There is SO much about what you just said that needs addressing, Kevin.
> Let's just do a little math: a 4" diam. pistom has an area of 12.56 sq.
> in. 15 lbs./sq. in. X 12.56 sq.in. = 188.4 lbs. of force on that piston,
> through the entire stroke (ON ONE PISTON). Think, Kevin, before you
> speak. Differential pressure in the atmosphere isn't the same as
> atmosphere/vacuum as well. I could go on, and on, but what's the use.
> Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
> at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
> 288 ft-lbs of torque.
> Think, Kevin - don't attack and obfuscate.
Ok, I'm thinking. I'm thinking this has gotta be a troll, because
(quote)"there is SO much about what you just said that needs
addressing," Mike, I'll bite.
Before we "do a little math" howabout a plain old real-life example? An
average internal combustion engine compresses the air/fuel charge to
anywhere from 120psig to 150psig before it even ignites the mixture,
then drives the piston down. Using these tremendous forces, a plain old
small block, 302cu inch, V8 develops only around 160 Horsepower. You
can go out and spend thousands and thousands of dollars on parts just to
build up a small block capable of producing your "400" hp. Why bother
if only 15psig will produce hundreds of horses instead??
Also, before we "do a little math" we have to understand the concept of
"atmosphere/vacuum" as well. What is usually called "atmospheric
pressure" is equal to 0psig (g for "gauge"),about 14.7psia (a for
"absolute", important, remember this), or about 101.3 kPa in metric. A
"perfect vacuum" is considered 0psia, atmospheric pressure is about
15psia. The difference between the two is 15psi, vacuum or no vacuum.
Think of it like a thermometer, and atmospheric pressure (0psig) as
0degF.
The pressure difference between a perfect vacuum of -15psig and
atmospheric pressure at 0psig is no different than between 0psig and
15psig, it's still only 15psi. Just as the difference in temperature
between -15degF and 0degF or the difference between 0degF and 15degF, is
15degF. This is why I made the suggestion of trying it with
15psi(gauge) air pressure, it would replicate the maximum pressure
difference obtainable with an "implosion" engine.
An "implosion" engine would "work" by the same principal that allows a
piston pump to "suck" water up from a well. A vacuum is created within
the cylinder of the pump, and "atmospheric pressure" bearing down on the
well space "pushes" the water into the cylinder. Your "implosion" would
create a "negative pressure" inside the cylinder, and atmospheric
pressure on the underside of the piston, would drive the piston up.
Maximum pressure difference attainable is 15psi, vacuum or not, you
could use 15psig compressed air to obtain the same results.(quote)
"Think Mike, before you speak"
Now for the math:
For the sake of simplicity we'll assume a constant pressure on the
(flat) piston head, zero clearance volume, and no frictional losses at
all, athough these are, in a practical sense, unachievable ideal
conditions.
Power developed by reciprocating engines is calculated by PLAN
(mean effective Pressure x effective Area of piston x Length of stroke x
Number of power strokes per second)
You've indicated a 300cuin V6 with a piston diameter of 4 inches,
which gives the stroke to be about 4 inches
Pressure = atmospheric = 101.3kPa
Length = 4inches = 0.1016 meters
Area = 12.56 sqin = 0.0081 sqmeters (0.1016 x 0.1016 x 0.7854)
Number strokes = 8.3333 (see below)
A single acting four-stroke cycle has 1 power stroke/2 revolutions
1000 Rpm = 500 power strokes per minute, or 8.3333 power strokes per
second
Therfore, power per cylinder = 101.3 x 0.1016 x 0.0081 x 8.3333
= 0.6947kW
Total indicated power = 0.6947 x 6 (6 cyl engine)
= 4.1682kW
Horsepower at 100% eff = 5.477 (kW x 1.341)
About 5 1/2 Hp on a "perfect" machine Mike, at 30% eff (you quoted)
it would develop 1.6 HP.
(quote) "Think Mike, - don't attack and obfuscate"
If you can show your work, where you develop 400HP at 30% eff, please
show us, as I could make no sense of your earlier incomplete
calculation.
K. Jones
The opinions expressed are my own, and not those of my employer, or
anyone else.
Eric Poulson
Michael Hannon wrote:
> Let's just do a little math: a 4" diam. pistom has an area of 12.56 sq.
> in. 15 lbs./sq. in. X 12.56 sq.in. = 188.4 lbs. of force on that piston,
> through the entire stroke (ON ONE PISTON). Think, Kevin, before you
> speak. Differential pressure in the atmosphere isn't the same as
> atmosphere/vacuum as well. I could go on, and on, but what's the use.
> Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
> at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
> 288 ft-lbs of torque.
Using the math I learned:
300 in^3 * 1 atm = .00492 m^3 * 101000 Pa = 498 Joules of work done each
revolution assuming a two stroke design (in a four stroke, this value would
be half that).
Power = work/time = (498 J/rev) / (60 sec/1000 RPM) = 8302 Watts = 11 HP
Torque = work/radian = 498 J / 2*pi rad = 79 Newton meters = 58 ft lbs
These numbers include NO friction losses.
In a 300cu V6 with 4" dia. pistons, the stroke is 3.98", indicating the
crank radius to be 1.99" (about 2"). The best we can do is to have the
crank angles spaced 30 degrees: then, the peak torque is
12.56"*15 psi * 2" + 2*(12.56*15) * 2"*sin(30) = 63 ft-lbs.
Note that the 58 ft-lbs above is average torque, while the 63 ft-lbs is
peak torque using your piston/stroke selection. Michael, this is not a
personal attack, just accurate arithmetic.
Eric Poulson sl...@cc.usu.edu
John Jacob
>
> Michael Hannon wrote:
> > Let's just do a little math: a 4" diam. pistom has an area of 12.56 sq.
> > in. 15 lbs./sq. in. X 12.56 sq.in. = 188.4 lbs. of force on that piston,
> > through the entire stroke (ON ONE PISTON). Think, Kevin, before you
> > speak. Differential pressure in the atmosphere isn't the same as
> > atmosphere/vacuum as well. I could go on, and on, but what's the use.
> > Let's just say that if you calculate out a 300 cu in V6 at 1,000 rpm's,
> > at a constant 15 psi, at an efficiency of 30%, you get over 400 hp, and
> > 288 ft-lbs of torque.
>
I've verified Eric's numbers: they are accurate. A car engine running on
15 psig compressed air would deliver no more than 11 HP at 100% efficiency,
nowhere near the 400 HP mentioned above. And, in fact, differential
pressure in the atmosphere is exactly the same as atmosphere/vacuum
pressure, as far as concerns the force on the piston and amount of P-V work
that can be done.
J.S. Jacob jja...@valkyry.ece.usu.edu
DGoncz
I have always wanted to build an air compressor from two V-8 engines tied
together at the shaft. But I know you can't get power from nowhere. I read
a little about the 400HP calculation and can only add that to compare the
performance of two engines one must use the same calculation for both,
using different numbers for starting pressure and even possibly
efficiency. But the sequence of calculations and the assumptions made must
be the same for both engines.
Using a figure of 600 psi for an internal combustion engine running on
gasoline, that 300 cubic inch V8 would generate 16,000 HP! The 400 HP
figure and the 16,000 HP figure are derived the same way and can be used
to make a valide comparison. It is important to compare model results to
model results and built engines to built engines. Tellling the proponents
of implosion engines that the model is way off doesn't help. Comparing the
theoretical 400 HP figure with a real world figure of 160 HP in a running
engine takes us way back to apples and oranges from third grade math.
Theoretical calculations made the same way using reasonable assumptions
will provide a true comparison of various engine's potential for
performance. This is how engineers choose the type of turbine for a
particular aircraft, how many stages, type of turbine, burner heat input,
etc. All theoretical. Then there is the risk the built engine won't
perform as calculated. I once calculated that a single 14 ounce tank of
propane could propel me and my bicycle at 20 mph for three hours using jet
propulsion. I calculated l00 pounds of thrust for the engine. On test, it
made about three. I revised my calculations. Please stop the name calling.
And by the way, what the heck is Brown's Gas? Why do people say it
implodes? Doesn't the water formed turn immediately into high pressure
steam due to the heat of combustion of the hydrogen burning with oxygen? I
would like to try a Brown's Gas welder some time. I tried a jewler's
hydroxy torch at a gem show. It was neat. You can convert foot-pounds to
watts accurately beacause they are both units of energy, but calculating
the performace of an engine by converting psi to HP is too damn simple. It
is more accurate to calculate the energy available in the fuel and figure
out how much fuel you'll need to burn. It's not simply because there are
efficiency considerations in getting the machine to convert pressure to
rotation, it is because the assumptions in the model have to cover the
conversion from force to energy to power, and the assumptions are too
simple to provide accurate answers. Still, if you want to compare two
technologies, it makes sense to use the same method to compare them, not
to compare a completely theoretical power figure to the performance of a
real, built engine. You have to compare the theoretical performance of the
new technology with the theoretical performance of the old to have a valid
comparison.
Look at it this way: it takes two minutes to do the simple math. But it
takes hours to build an engine. If we could build engines in two minutes,
we would all be trying all kinds of fascinating and possibly revolutionary
engine ides every day, and the field I work in, automatic manufacturing,
promises to make this a reality someday. But for today, it makes sense to
spend more than two minutes on the math.
DGo...@aol.com
A.A.S. Mech.Engr.Tech., CAD/CAM, NVCC, 1990
"Anger is like beer: when you have too much, you have to let it go."
Trying to build a self-reproducing universal machine tool in inexpensive kit form.
http://home.aol.com/DGoncz (Just a couple
Michael Hannon
Now try adding the vacuum figures you yourself came up with to your
results. 400 hp, or 6 hp? -You tell me. It implodes (not exploding,
creating heat and pressure, steam and vapor, but imploding, creating a
high vacuum, with no steam or vapor, immediately after the implosion),
at a rate of ~1860/1 - add that to the 15psi on the other side of the
piston, and see what you get.
You won't get anything like what this gas mixture can do from a
jeweler's hydroxy torch. It's a whole 'nother animal. Do a search on it.
Michael
Michael Hannon
Kevin Jones wrote:
>
> DGoncz wrote:
>
> <snip>
>
> > Using a figure of 600 psi for an internal combustion engine running on
> > gasoline, that 300 cubic inch V8 would generate 16,000 HP! The 400 HP
> > figure and the 16,000 HP figure are derived the same way and can be used
> > to make a valide comparison.
>
> running solely on the difference in pressure between a near "perfect"
> vacuum, and atmospheric pressure. Browns's gas was to be "imploded" in
> the cylinder to create the vacuum.
> How did you come up with 16,000HP?!?? The math is wrong somewhere! A
> 300 cu inch reciprocating engine, powered by a pressure differential of
> 600psi, could not produce more than 230HP. This would be a "perfect"
> engine with no losses whatsoever. If you believe this to be incorrect,
> please correct our maths.
> ((600psi x 300cu)/12 X 500pwrstks/min)/33,000 = Max HP (im)possible
>
> > It is important to compare model results to
> > model results and built engines to built engines. Tellling the proponents
> > of implosion engines that the model is way off doesn't help.
>
> if only to demonstrate the impracticality of the device.
>
> <snip,snip>
>
> > Look at it this way: it takes two minutes to do the simple math. But it
> > takes hours to build an engine. If we could build engines in two minutes,
> > we would all be trying all kinds of fascinating and possibly revolutionary
> > engine ides every day, and the field I work in, automatic manufacturing,
> > promises to make this a reality someday. But for today, it makes sense to
> > spend more than two minutes on the math.
>
> building is a complete waste of time. If a very simple calculation will
> show that a "perfect" machine is only capable of a few HP instead of the
> hundreds of HP previously guessed/wanted/needed there is no reason to
> spend anymore time proving a less efficient machine will produce even
> less HP.
>
> K. Jones
> The opinions expressed are my own, and not those of my employer, or
> anyone else.
Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
someone mentioned as the result of a weaker implosion) on the other side
of 15 psi isn't anything like you think it is. It IS NOT a pressure
differential of 15 psi, or 30, or 600 psi. AND, there is absolutely no
reason to assume that a vacuum engine has to run on 4 strokes, which it
doesn't.
Michael
Kevin Jones
vgo...@ibm.net
TADedek wrote:
>
> > So why do silly car manufacturers insist on using hundreds of psi to do the
> > same thing?
> >
> > Ian
>
> Ummmmm, let me guess. They are using hundreds of psi to do the same
> thing because the petroleum companies (including opec members) are
> strong arming them into compliance. To go against them would mean the
> automobile would be squashed into a coverup and subjects of tabloids.
>
> Do I get the pretty prize??? PLEASE???
>
> Sheldon
>
Sorry, no prize. The reason car manufactures "insist on using hundreds
of psi" is because fossil fuel is still cheap. It is unfortunate that we
are addicted to this form of energy to the point where we are disturbing
our atmosphere, and poisoning our air and habitat.
The sooner we learn to depend on less psi, and hydrogen, the longer we
will survive.
Victor
--
________________
vgo...@ibm.net
________________
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
>someone mentioned as the result of a weaker implosion) on the other side
>of 15 psi isn't anything like you think it is. It IS NOT a pressure
>differential of 15 psi, or 30, or 600 psi.
OK, Mike, I'll bite. If you don't think it's a pressure difference,
what do you think it is?
Bill Ward (just knowing I'll regret this)
Ian Johnston
Michael Hannon (oha...@mailroom.worldnet.att.net) wrote:
: Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
: someone mentioned as the result of a weaker implosion) on the other side
: of 15 psi isn't anything like you think it is. It IS NOT a pressure
: differential of 15 psi, or 30, or 600 psi.
Are you trying to tell us that a piston with 15psi on one side and a high
vacuum on the other behaves differently from one with 30psi and 15psi on
the two sides?
: AND, there is absolutely no
: reason to assume that a vacuum engine has to run on 4 strokes, which it
: doesn't.
A fair point - it might well be two stroke. Which takes my estimate of the
power available from your 300cu in engine at 1000rpm up to 12hp or so.
Why won't you post the working that gives you 400hp?
Ian
DGoncz
The figure of 16,000 HP means that using a regular engine would outperform
a vacuum engine, even using math that wildly inflates estimates of power
output. I was trying to show that the math being used was too simple.
Yes, a two minute calculation can show that the proposed vacuum engine
should not be built, or at least that it won't provide enough power to
make a profit in the marketplace. What bothered me was that the short
calculation was compared to a real working device. Showing that the short
calculation produces a result totally different from reality still allows
the argument that the proposed engine is totally different from current
engines. Using the same math allows the math itself to be tested and
tried, accepted or rejected. My argument is that since the same short
calculation says a regular engine would outperform a vacuum engine, that a
built real engine would still outperform a built implosion engine. So
there is no reason to build an implosion engine. Unless we want further
proof. Beliefs cannot be effectively challenged outside the belief system
of the believer. That is where all the name-calling comes from. I try to
respect the belief systems of people I meet on the various NGs. I think
that instead of trying to directly attack the belief that Brown's Gas
implodes, it is more reasonable and fair to work within the belief system
of the believers and let them make up their own minds. Thinking that a
simple calculation proves that a new engine type is better is arrogant. So
is thinking that someone will change their belief on the basis of a few
words, no matter how well supported the words are. I am arguring for more
tolerance and depth, not more words. Extensive quotes and a two line reply
are not the way to spread tolerance and respect. In depth replies help
more. I have a lot of respect for belief systems, scientific or not. It's
like religious tolerance, a fundamental American personal right. Just keep
trying to work from the known to the unknown.
Michael Hannon
Bill Ward wrote:
>
> Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>
> >Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
> >someone mentioned as the result of a weaker implosion) on the other side
> >of 15 psi isn't anything like you think it is. It IS NOT a pressure
> >differential of 15 psi, or 30, or 600 psi.
>
> what do you think it is?
>
> Bill Ward (just knowing I'll regret this)
>
> >reason to assume that a vacuum engine has to run on 4 strokes, which it
> >doesn't.
Keep biting, Bill. If you think that the pressure differential between
15psi and zero psi is the same as the difference between 15psi and
0.42 Torr, you are very sadly mistaken. What is the difference between
0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
reduction in pressure by implosion)? Do you think that there is no
difference, that 0 psi=.42 Torr?
Michael
Michael Hannon
I disagree about you conclusion of a hydrogen implosion vacuum engine,
DGonzc, but only on the grounds that it isn't economically feasible now,
especially with the opposition it obviously has. I STILL haven't seen
what I would call accurate descriptions of what is taking place.
I have a question about what you think the vacuum produced by the 1830/1
implosion, starting from 1 atm, is, in negative atmosheres, if there is
no vapor or steam produced, but just vacuum and water, and how you would
translate that into comparable psi in terms of reverse pressure? How
would you describe the vacuum in the cylinder at halfway through the
suction stroke? Would there be a decrease in the vacuum by 50%?
Michael
Michael Hannon
My sentiments exactly, Victor, and if it takes the rest of my life being
called every name in the book for trying to pursue just that in new ways
for that sake, so be it. If it were left up to these people who think
this is some kind of smug joke, our children, and theirs, until there
are none left, will definitely pay the price.
Michael
Robert Erck
Shame on you! Mike is correct!!
The difference between 15 psi and zero is 15 psi.
But the difference between 15 psi and 0.42 torr is 14.999447 psi.
Why, with only a few tens of thousands of dollars of ultra-precise
pressure equipment, I'll bet even my grandma could tell the difference.
--------------
In article <332EEE...@mailroom.worldnet.att.net>,
Robert Erck
In article <332EEE...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
>
> Keep biting, Bill. If you think that the pressure differential between
> 15psi and zero psi is the same as the difference between 15psi and
> 0.42 Torr, you are very sadly mistaken. What is the difference between
> 0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
> reduction in pressure by implosion)? Do you think that there is no
> difference, that 0 psi=.42 Torr?
> Michael
------------
The ratio of the rest mass of the proton to the rest mass of the electron
is 1836/1. This is suspiciously close to the 1830/1 ratio that Hannon is
getting.
By golly, I think that Hannon is working on an antimatter engine!
Zefram Cochran, move over, Hannon is developing WARP DRIVE!
Kevin Jones
Michael Hannon wrote:
<lengthy multi-replies snipped>
> Too bad it isn't that simple, Kevin.
Why not? We're still waiting for the math that shows 400HP and 288
ft.lbs. of torque at 30% efficiency, out of a 300 cu.in. V6 driven by
"atmospheric pressure/vacuum", and at 1000rpm, Mike. You've gone
un-characteristally quiet when asked to show a simple calculation,
Mike. Show us how you came up with those numbers, and perhaps I'll
understand why it isn't that simple, as words may confuse and
"obfuscate", numbers do not. This isn't "proprietary information", or
anything else Mike, just simple ol' math. You supplied the numbers
Mike, show us where you came up with those numbers, if not out of thin
(hot) air.
Think Mike (quote),don't attack and obfuscate , simply show us the math.
A vacuum, even of .42 Torr (which
> someone mentioned as the result of a weaker implosion) on the other side
> of 15 psi isn't anything like you think it is. It IS NOT a pressure
> differential of 15 psi, or 30, or 600 psi.
Oh really?!? What IS it then??!?
You started a thread the *SAME* day you wrote this, in this *SAME*
newgroup *ASKING* "WHAT IS A PERFECT VACUUM?", *SELF ADMITTING* you
don't know what the concept IS, or have a clue what you are talking
about with regards to the above post, but somehow you feel in a position
qualified to make "factual" statements about it and tell me it "isn't
anything like I think it is". I'm rolling on the floor laughing Mike,
because you've just proven you like to argue with everybody, just for
the sake of arguing, even if you don't know what the topic is you're
arguing about!
Dont think Mike, just rant like an idiot, and show the world how a 14
year old throws a tempertantrum.
What I think a perfect vacuum is measured at Mike, is *zero pounds per
square inch absolute*.
But the real question is Mike, What do YOU think it is??!?
Think Mike (quote), don't attack and obfuscate
What I think (standard) atmospheric pressure is Mike, is *14.69 pounds
per square inch absolute*
But the real question is Mike, What do YOU think it is??!?
Think Mike (quote), don't attack and obfuscate
The difference between the two is slightly less than 15 pounds per
square inch. No mystery at all Mike. In the real world, "atmospheric
pressure" obviously varies depending on height above/below sea level,
temperature, etc, but is suffice for our calculations to demonstrate the
difference between 2HP and 400HP
> and there is absolutely no reason to assume that a vacuum engine has to run on 4 strokes, which it doesn't.
> Michael
This is true, and I'll agree with you Mike, you are right. It was
assumed because you mentioned a 300 cu.in. V6, most engines of that size
would be 4 strokes (this is a large V6), my error.
Fine, we'll make it a 2 stroke. 11.37HP x 30% = 3.4HP Mike, not "over
400 HP"
Tell you what, we'll make it a double acting 2 stroke, with 2
powerstrokes per revolution....... 22.7HP x 30% = 6.8HP Mike.
Put up (your calculations) or shut up, Mike.
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>Bill Ward wrote:
>>
>> Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>>
>> >Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
>> >someone mentioned as the result of a weaker implosion) on the other side
>> >of 15 psi isn't anything like you think it is. It IS NOT a pressure
>> >differential of 15 psi, or 30, or 600 psi.
>>
>> OK, Mike, I'll bite. If you don't think it's a pressure difference,
>> what do you think it is?
>>
>> Bill Ward (just knowing I'll regret this)
>>
>> > AND, there is absolutely no
>> >reason to assume that a vacuum engine has to run on 4 strokes, which it
>> >doesn't.
>> >Michael
>Keep biting, Bill. If you think that the pressure differential between
>15psi and zero psi is the same as the difference between 15psi and
>0.42 Torr, you are very sadly mistaken. What is the difference between
>0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
>reduction in pressure by implosion)? Do you think that there is no
>difference, that 0 psi=.42 Torr?
>Michael
Mike, you misunderstood my question. I know what _I_ think, I'm asking
what _you_ think it is if "It IS NOT a pressure differential..."
Bill Ward
Michael Hannon
Ian Johnston wrote:
>
> Michael Hannon (oha...@mailroom.worldnet.att.net) wrote:
>
> : Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
> : someone mentioned as the result of a weaker implosion) on the other side
> : of 15 psi isn't anything like you think it is. It IS NOT a pressure
> : differential of 15 psi, or 30, or 600 psi.
>
> vacuum on the other behaves differently from one with 30psi and 15psi on
> the two sides?
>
> : reason to assume that a vacuum engine has to run on 4 strokes, which it
> : doesn't.
>
> power available from your 300cu in engine at 1000rpm up to 12hp or so.
>
> Why won't you post the working that gives you 400hp?
>
> Ian
Why? Because I want you to keep showing us all how you think, Ian. You
have no concept of what a vacuum is, or is capable of, and you should
learn, because maybe it will teach you a little wisdom about attacking
things you know little about, and that includes the whole Brown's gas
community. 400hp is CONSERVATIVE, Ian. A good vacuum engine can run on
.01 psi external pressure, never mind 15 psi. Suss it out.
Michael
Michael Hannon
Kevin Jones wrote:
Blah, Blah, Blah.......or shut up, Mike.
>
> K. Jones
Sorry, Kevin....no time to waste on you this century.
Michael Hannon
Bill Ward wrote:
>
> Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>
> >Bill Ward wrote:
> >>
> >>
> >> >Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
> >> >someone mentioned as the result of a weaker implosion) on the other side
> >> >of 15 psi isn't anything like you think it is. It IS NOT a pressure
> >> >differential of 15 psi, or 30, or 600 psi.
> >>
> >> what do you think it is?
> >>
> >> Bill Ward (just knowing I'll regret this)
> >>
> >> >reason to assume that a vacuum engine has to run on 4 strokes, which it
> >> >doesn't.
>
> >Keep biting, Bill. If you think that the pressure differential between
> >15psi and zero psi is the same as the difference between 15psi and
> >0.42 Torr, you are very sadly mistaken. What is the difference between
> >0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
> >reduction in pressure by implosion)? Do you think that there is no
> >difference, that 0 psi=.42 Torr?
> >Michael
>
> Mike, you misunderstood my question. I know what _I_ think, I'm asking
>
> Bill Ward
Dear Bill,
Here goes:
In a cylindrical container as illustrated below,
whose left end (S) is completely closed and sealed,
while its right end (E) is open to the atmosphere and
which has vents to the atmosphere, as shown, between
piston (A) and a middle immovable wall which (W) is
well-sealed at the point where the moveable rod of the
solid rod-piston assembly (pistons A and B) passes through it,
and there is a vacuum to the left of piston A (pulling it to
the left) and pressure left of piston B, (pushing it to the
right) acting in opposition to that vacuum. (The pistons,
rigidly fixed to the rod, can slide freely to the left or
right, in unison, and are well-sealed at the points where
they contact the chamber)
If those pistons (whose sizes are identical) are to remain
in exactly the positions they are, how much Pressure (P)
is required to counteract a Vacuum (V) of (777/1830) 0.42 Torr?
(S) Open air (W)
|_____________________\|||______________________
| |-------| ||| >>>P>>> |-------|
| Vacuum | |Vent||| Pressure | |
| <<<<<< | | ||| | |
| V | Piston|----|||---Rod----| Piston| Open (E)
| V | A |----|||----------| B | air
| <<<<<< | | ||| | |
| Vacuum | | ||| Pressure | |
| |_______|Vent||| >>>P>>> |_______|
|--------------------- |||----------------------
(S) Open air/ (W)
Michael
Michael Hannon
Harry H Conover wrote:
>
> vgo...@ibm.net wrote:
> :
> : Sorry, no prize. The reason car manufactures "insist on using hundreds
> : of psi" is because fossil fuel is still cheap. It is unfortunate that we
> : are addicted to this form of energy to the point where we are disturbing
> : our atmosphere, and poisoning our air and habitat.
> :
> : The sooner we learn to depend on less psi, and hydrogen, the longer we
> : will survive.
>
> which of course it isn't.
>
> Please do not be discouraged by the fact the it takes more fossil or
> nuclear fuels to produce the hyrdrogen, than use of the hydrogen can
> possibly ever yield.
>
> The Conservation of Energy is not just a good idea, it's nature's law!
>
> Sorry, just the facts.
>
> Harry C.
Funny, Harry - didn't you see the publications reffered to in either
this NG, or oner of the other energy-related NG's from Lawrence
Livermore directly contradicting what you have just said about H not
being any energy source? You spend way too much time not seeing what is
right in front of you. They have published whole studies, including
charts,
on the viability of H in cars, while you're sitting here playing with
yourself. Tsk, Tsk, Tsk.
It goes like this - you take nuclear sources electricity, generate hH in
one of the new super-efficient generators, put it in a tank, and run
your car's engine on it. Where are the fossil fuels in that?
I saw a Taurus in the VA hospital parking lot the other day that had a
small logo on the side of a leaf on a branch, with the letters FFV under
it. What does that stand for?
Michael
Michael Hannon
Robert Erck wrote:
>
> In article <332EEE...@mailroom.worldnet.att.net>,
> oha...@mailroom.worldnet.att.net wrote:
> >
> > 15psi and zero psi is the same as the difference between 15psi and
> > 0.42 Torr, you are very sadly mistaken. What is the difference between
> > 0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
> > reduction in pressure by implosion)? Do you think that there is no
> > difference, that 0 psi=.42 Torr?
> > Michael
>
>
> The ratio of the rest mass of the proton to the rest mass of the electron
> is 1836/1. This is suspiciously close to the 1830/1 ratio that Hannon is
> getting.
>
> By golly, I think that Hannon is working on an antimatter engine!
>
> Zefram Cochran, move over, Hannon is developing WARP DRIVE!
You caught me with my pants down Zefram - I confess.
Michael Hannon
Bill Ward wrote:
>
> Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
>
> >Too bad it isn't that simple, Kevin. A vacuum, even of .42 Torr (which
> >someone mentioned as the result of a weaker implosion) on the other side
> >of 15 psi isn't anything like you think it is. It IS NOT a pressure
> >differential of 15 psi, or 30, or 600 psi.
>
> OK, Mike, I'll bite. If you don't think it's a pressure difference,
> what do you think it is?
>
> Bill Ward (just knowing I'll regret this)
>
> > AND, there is absolutely no
> >reason to assume that a vacuum engine has to run on 4 strokes, which it
> >doesn't.
> >Michael
Bill, I wouldn't flame you for anything - here's round 2 of
"physics quiz:"
Taking a situation similar to the first "physics quiz," the closed
left end of the cylindrical container has been removed and replaced by
another piston (C), which is rigidly attached to its own rod, which
extends through a vented wall to the open air. This rod is attached at
its other end to a powerful winch capable of pulling it and the piston
backwards (to the left), and holding the rod and piston absolutely
stationary, which it is doing at the beginning of this test. Pistons A
and C are touching, with no space between them, while piston B also has
no space between it and wall W (they are touching). Air is pumped
between piston B and wall W, moving piston B to the right, and thus,
also piston A to the right as well, creating a partial vacuum between
pistons A and C. When the pressure between piston B and wall W reaches 1
atmosphere (15 psi), everything is stopped. What is the pressure
difference between the partial vacuum in the volume between pistons A
and C, when the pressure between piston B and wall W is 15psi (equal to
the pressure of the open air outside)? 15psi, or what? If it is 15 psi,
what is the pressure in the volume between A and C?
15psi (open air) minus 15psi (pressure difference) = 0 psi, or what?
During this operation, piston C has been held firmly stationary,
while piston A has moved a distance(J). If we now lock pistons A and B
in their positions, maintaining the partial vacuum between A and C, and
the pressure between B and W, turn on the winch, and pull piston C
backwards a distance equal to 1830 X the distance J (which piston A has
traveled from piston C previously), what is the pressure between A and C
now? Still 0, or what? How many psi pumped between piston B and wall W
will it take to keep piston B in its same position after pistons A and B
are unlocked, and are free to respond to the vacuum created between
pistons A and C after C has been moved that distance? Is the vacuum
between A and C equal to 1 Atm (777 Torr) divided by 1830, which equals
0.42 Torr, 0 psi, or what?
Open air (W)
(Q)__________________________________________\|||______________________
Open Air |------| |------| ||| >>>P>>>
|------| | | | | |Vent||| Pressure
| |
W| | | | | ||| | |
I|-----Rod--------|Piston| |Piston|----|||---Rod---- |Piston|Open
N|----------------| C | (?) | A |----|||---------- | B |air
C| | | | | ||| | |
H| | | | | ||| Pressure | |
Open Air |______|---J----|______|Vent||| >>>P>>> |______|
(Q)------------------------------------------ |||----------------------
Open air/ (W)
Harry H Conover
Harry H Conover
Bill Ward (bward*remove_this*@ix.netcom.com) wrote:
:
: Mike, you misunderstood my question. I know what _I_ think, I'm asking
: what _you_ think it is if "It IS NOT a pressure differential..."
:
Bill, don't get sucked in too far. Michael clearly has no understanding
of the pneumatic significance of differential pressure, has no idea of
what a Torr signifies, plus no interest in learning about anything that
flies-in-the-face of his dreams.
He is not interested in science. He wants to forge a 'sampo' and no
amount of technical fact will deter his pursuit of inventing a machine
that produces 'something for nothing.'
It's a trait that he, sadly, shares with many, many other empty-headed
dreamers that waste their lives chasing nonsense!
Harry C.
DGoncz
Michael posted:
I disagree about you conclusion of a hydrogen implosion vacuum engine,
DGonzc, but only on the grounds that it isn't economically feasible now,
especially with the opposition it obviously has. I STILL haven't seen
what I would call accurate descriptions of what is taking place.
I have a question about what you think the vacuum produced by the 1830/1
implosion, starting from 1 atm, is, in negative atmosheres, if there is
no vapor or steam produced, but just vacuum and water, and how you would
translate that into comparable psi in terms of reverse pressure? How
would you describe the vacuum in the cylinder at halfway through the
suction stroke? Would there be a decrease in the vacuum by 50%?
Michael
And my reply is:
I don't think any mixture of hydrogen and oxygen will implode, and I don't
think the Brown's Gas generator generates monatomic H and O. I think an
estimate of 400 HP for a 300 cid implosion engine is approximately as
accurate as my estimate of 16,000 HP for a 300 cid gasoline engine. And I
think both estimates are way off. I am trying to meet the proponents of
implosion power on their own terms by using the same math sequence with
one critical initial difference. The earlier post (I am still learning
your names here, and I try to be respectful.) started with atmospheric
pressure, about 15 psi, and ended with 400 HP for an implosion engine. I
started with 600 psi (a conservative value) and ended with 16,000 HP for a
gasoline engine. Using the exact same steps between start and finish for
the two figures.
The demo of a water filled vessel, filled with gas mix, then ignited,
causing water to rush in, probably allows the heat of combustion to
dissipate through the clear walls of the vessel by infrared radiation, and
the water of combustion is most likely soon dissolved in the inrushing
water. I think in a mirrored vessel, or in a metal cylinder with a metal
piston, the results would be different. I would like to see the demo with
the transparent vessel and water rushing in, and I am available to build
experimental engines for anyone at the lowest possible price consistent
with safety. I am ready to make new friends, no matter what they believe
or say. I have never ignited a balloon full of H2/O2 with an internal
spark, and this might be a good thing to try, as the cigarette lighter
flame I used in my trials caused the balloon to break. For all I know it
would deflate or implode. The small balloon full I ignited certainly made
a loud bang.
The force on the piston of an implosion engine comes from atoms of
atmosphere banging into the crankcase side of the piston, at various
angles, not always directly in line with the piston's travel. The
atmospheric atoms do this because they have kinetic energy due to the
ambient heat. The kinetic energy of the atoms varies widely. They do not
all have the same energy. Since they are small, the impacts average out.
The accepted figure for this averaged force is about 15 pounds of force
per square inch of surface area. If you want a compact high power engine
that wastes lots of fuel value, which someone might want if hydrogen power
has all it claims to, I'd recommend a vane motor. I have a die grinder,
about a horsepower, that reaches 20,000 rpm easily. It runs on compressed
air from about a one horsepower compressor. It is a little larger than my
fist. Operated on a continously flowing stream of mixed fuel and air or
stoichiometric H2 and O2, and somehow igniting the fuel (kind of like a
Wankel engine) it would go faster. How fast, I have no idea. And it would
heat up. The neat thing about compressed air tools is they get colder the
more you use them. This is because the air expands through the vane motor,
doing work on the way. It's a little counterintuitive until you feel the
heat coming from the compressor, and from the electric motor that drives
it. It's not magic. I guess it would run backwards with an imploding gas
mix stream, but since the velocity of combusiton in H/O is very high, it
might produce useable power in a small package.Come to think of it, this
would be neat to try. But not in my apartment.
Requoting:
How would you describe the vacuum in the cylinder at halfway through the
suction stroke? Would there be a decrease in the vacuum by 50%?
My answer:
Amazingly, NO! The pressure on the piston of an implosion engine is nearly
constant during its stroke because the vacuum doesn't change much. It's
the atmosphere pushing on the other side that does the work, and as long
as that side is open to the air, the force is approximately constant. With
a perfect vacuum the force is constant right to the end of the stroke. I
must repeat, there is no such thing as a perfect vacuum. With a 1/1800
expansion ratio, the piston would get to within 1/1800 of its stroke
before the force balanced. At 2/1800 of the stroke from the end, the force
would be half. At 3/1800 from top dead center the force would be 1/3. But
for most of the stroke the force would be nearly the same. For instance,
at 900/1800 of the stroke, or halfway, the force would be within 1/900 of
100%, or about 99.9% of the original force. Pretty neat, still not magic.
I have been working on a list of 63 invention/ideas for the last few
weeks, just a few minutes every few days, and I forgot about this one:
Install a piston in the down tube of a bicycle, near the head tube.
Evacuate and seal the area between the head tube and the piston. Connect a
thin, strong, (titanium) tape to the piston. Connect the tape to a one way
roller clutch bearing on the crankset. In normal operation, the bearing
would present little drag to the user, but, while coming to a stop, the
user could back pedal, drawing the piston toward the crankset. The user
would have to stand on the pedal to keep the mechanism "cocked", but when
the light changed, the atmosphere would push the piston back to its
original position. For a 1.25 inch ID, 1.375 inch OD tube, the piston
would pull on the tape with 15 psi times pi times 1.25 inches divided by
4, or about twenty pounds of force. If the OD on the bearing was an inch,
this would be about ten inch-pounds of torque. That's the same torque as
pressing on the pedals (6 inches from the crank) with about one pound of
force. (Just enough to get started again.) The one pound pedaling force
would be present for twenty inches divided by pi, or about six full
revolutions of the cranks. You may have this invention if you wish. I
will add it to my list. The down tube would probably crumple from the
external pressure anyway.
I have some old CRT's if you would like to implode them for further proof.
That would make a neat Shuttle experiment. It is the force of the
atmosphere pushing in that causes the fragments to accelerate inwards.
I would like to hear more about sunlight, bromine, and the generation of
hydrogen.
Interesting, my post is still not readable through aol's newsreader. But
your quotes of my posts are. Aol sucks. But a vacuum doesn't.
Bill Ward
con...@tiac.net (Harry H Conover) wrote:
>Bill, don't get sucked in too far. Michael clearly has no understanding
>of the pneumatic significance of differential pressure, has no idea of
>what a Torr signifies, plus no interest in learning about anything that
>flies-in-the-face of his dreams.
>He is not interested in science. He wants to forge a 'sampo' and no
>amount of technical fact will deter his pursuit of inventing a machine
>that produces 'something for nothing.'
>It's a trait that he, sadly, shares with many, many other empty-headed
>dreamers that waste their lives chasing nonsense!
> Harry C.
Yeah, Harry, you're right of course, but I do hate to see such
imagination and dedication go to waste for lack of discipline. I keep
watching and hoping to see him get a clue, but no dice so far. At first
his antics seemed kind of humorous, but now they are getting a bit
pathetic. He just doesn't get that science is about measurements, not
beliefs. Maybe someday...
BTW, what the heck's a "sampo"?
Bill Ward
>
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
<snip previous posts>
>Michael
Dear Mike,
Thanks for a clear, well-described question. I'm going to take it
seriously and answer it as clearly as I can.
First off, the quick answer is that pressure P is twice atmospheric
minus the .42 torr "vacuum" pressure, or about 1519.58 torr.
The term "Torr" is in fact a unit of _pressure_ equal to 1mm of
mercury, or 1/760 atmosphere. The pressure inside a "vacuum" chamber is
often measured in Torr, but it's still a pressure, just like
atmospheric pressure. When you start pumping down a vacuum rig, you can
watch the absolute pressure drop from 760 (or whatever the barometric
pressure is that day) down to however low your equipment will go.
What may be throwing you off a bit is the difference between "gauge" and
absolute pressures. Absolute pressure refers to the actual pressure, not
relative to any other pressure, while gauge pressure is the difference
between atmospheric and the inside of the gauge. Gauge pressure will be
negative (a "vacuum") if the absolute pressure inside the gauge is less
than the outside atmospheric pressure. Absolute pressure can never be
negative.
In other words, vacuums don't suck, pressures push. When you suck on a
straw and reduce the pressure in your mouth and the straw, atmospheric
pressure pushes down on the soda, which pushes the liquid up into the
lower pressure area in your mouth.
So on your drawing, the difference in pressure across piston A is
(760-.42) = 759.58 torr, which pulls on the rod in the same direction
that atmospheric pressure E of 760 torr pushes. To remain in
equilibrium, the force from P must equal the sum of those, or 1519.58
torr times the area.
I initially misread your diagram to be balancing the two forces, and
caught the mistake right after posting. (of course :-) I apologize in
advance for the confusion if I didn't cancel the bad post quick enough.
I hope this helps.
Bill Ward
Michael Hannon
phenomena that opposes all conventional rules and beliefs. I hold no bad
sentiments about your feelings. But what will happen if you look at the
"quizes" and discover something, and what will happen when I build a
Vacuum powered Tesla turbine, and it works? What will any of these
people do when they are proven wrong. Will they be the first to admit
their errors? No. Remember, Harry is one guy who said that water is a
great dielectric, when that is true only in special circumstances.
Otherwise, where are all those water capacitors. Do you call that
disciplined, (he runs a test for a few minutes that, if left to run for
weeks would prove him wrong), well-conducted testing and research, or a
staged performance with limited variables, particularly designed to
prove a point that can easily be disproven by anyone who knows about the
weaknesses of that test. Is THAT good science? Check out the "quizes,"
Bill. I'd really like to hear an objective analysis based on sound
reasoning.
Michael
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
<snip previous posts>
First off, the quick answer is that the Pressure P is the same as the
"vacuum" V, or .42 Torr.
The term "Torr" is in fact a unit of _pressure_ equal to 1mm of
mercury, or 1/760 atmosphere. The pressure inside a "vacuum" chamber is
often measured in Torr, but it's still a pressure, just like
atmospheric pressure. When you start pumping down a vacuum rig, you can
watch the absolute pressure drop from 760 (or whatever the barometric
pressure is that day) down to however low your equipment will go.
What may be throwing you off a bit is the difference between "gauge" and
absolute pressures. Absolute pressure refers to the actual pressure, not
relative to any other pressure, while gauge pressure is the difference
between atmospheric and the inside of the gauge. Gauge pressure will be
negative (a "vacuum") if the absolute pressure inside the gauge is less
than the outside atmospheric pressure. Absolute pressure can never be
negative.
In other words, vacuums don't suck, pressures push. When you suck on a
straw and reduce the pressure in your mouth and the straw, atmospheric
pressure pushes down on the soda, which pushes the liquid up into the
lower pressure area in your mouth.
So on your drawing, the difference in pressure across piston A is
(760-.42) = 759.58 torr, which requires an equal pressure difference
across piston B to remain in equilibrium, since the areas are equal.
Pressure P must therefore be .42 torr, the same as pressure ("vacuum")
V.
Michael Hannon
Dear DGoncz,
Thanks for your comments - I feel in no way affronted by them,
and actually welcome your considered commentary. This mixture of H and O
does, in fact implode, however, and there are enough people who have
built their own generators who say so to prove that out. The better they
built it, the better their results, but it is a fact.
Just for the heck of it, consider checking out my posts titled "physics
quiz," - I'd like to hear your conclusions. I'm building a vacuum
turbine based on another ridiculed design, for a 55lb., 110 hp "fraud"
that should prove very interesting run on the figures you'll see from
these posts.
Regards,
Michael
Ian Johnston
Michael Hannon (oha...@mailroom.worldnet.att.net) wrote:
: Keep biting, Bill. If you think that the pressure differential between
: 15psi and zero psi is the same as the difference between 15psi and
: 0.42 Torr, you are very sadly mistaken. What is the difference between
: 0(zero)psi and .42 Torr (.42 Torr is the vacuum created by an 1830/1
: reduction in pressure by implosion)? Do you think that there is no
: difference, that 0 psi=.42 Torr?
What do you think the difference is, Michael? Is the force on a piston with
15psi on one side and 0psi on the other more or less than the force on
a piston with 15psi on one side and 0.42 Torr on the other?
By how much, in Torr, do you think atmospheric pressure differs from 15 psi?
Ian
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
<snip earllier posts>
>Bill, I wouldn't flame you for anything - here's round 2 of
>"physics quiz:"
>
Mike, this is a little harder to explain, but I'll give it a shot. BTW,
I see now why you didn't balance the forces in your first version.
> Taking a situation similar to the first "physics quiz," the closed
>left end of the cylindrical container has been removed and replaced by
>another piston (C), which is rigidly attached to its own rod, which
>extends through a vented wall to the open air. This rod is attached at
>its other end to a powerful winch capable of pulling it and the piston
>backwards (to the left), and holding the rod and piston absolutely
>stationary, which it is doing at the beginning of this test. Pistons A
>and C are touching, with no space between them, while piston B also has
>no space between it and wall W (they are touching). Air is pumped
>between piston B and wall W, moving piston B to the right, and thus,
>also piston A to the right as well, creating a partial vacuum between
>pistons A and C. When the pressure between piston B and wall W reaches 1
>atmosphere (15 psi), everything is stopped.
Piston B won't move to the right at all until the pressure between it
and wall W is greater than atmospheric. It needs a pressure differential
to move, and the right side is already at atmospheric.
> What is the pressure
>difference between the partial vacuum in the volume between pistons A
>and C, when the pressure between piston B and wall W is 15psi (equal to
>the pressure of the open air outside)? 15psi, or what?
There is no volume between A and C, because B (and A) couldn't move with
only 15 psi between B and W.
> If it is 15 psi,
>what is the pressure in the volume between A and C?
No volume, no pressure.
>15psi (open air) minus 15psi (pressure difference) = 0 psi, or what?
> During this operation, piston C has been held firmly stationary,
>while piston A has moved a distance(J). If we now lock pistons A and B
>in their positions, maintaining the partial vacuum between A and C, and
>the pressure between B and W, turn on the winch, and pull piston C
>backwards a distance equal to 1830 X the distance J (which piston A has
>traveled from piston C previously)
J=0, 1830 x 0 = 0
>, what is the pressure between A and C
>now?
No volume - no pressure. If you let the drive pressure (B to W) go
above atmospheric, you could move A and C apart, but trust me - you
won't get more energy out than you put in.
>.Still 0, or what? How many psi pumped between piston B and wall W
>will it take to keep piston B in its same position after pistons A and B
>are unlocked, and are free to respond to the vacuum created between
>pistons A and C after C has been moved that distance? Is the vacuum
>between A and C equal to 1 Atm (777 Torr) divided by 1830, which equals
>0.42 Torr, 0 psi, or what?
> Open air (W)
>(Q)__________________________________________\|||______________________
>Open Air |------| |------| ||| >>>P>>>
>|------| | | | | |Vent||| Pressure
>| |
>W| | | | | ||| | |
>I|-----Rod--------|Piston| |Piston|----|||---Rod---- |Piston|Open
>N|----------------| C | (?) | A |----|||---------- | B |air
>C| | | | | ||| | |
>H| | | | | ||| Pressure | |
> Open Air |______|---J----|______|Vent||| >>>P>>> |______|
>(Q)------------------------------------------ |||----------------------
> Open air/ (W)
Mike, I'm not exactly sure where you're trying to go with this, but it
would help a lot if you would just forget the term "vacuum" and
substitute absolute pressure in your thinking. No offense, but gas laws
are very basic physics, and you really ought to get a good book and
learn for yourself rather than taking someone elses word for it
(especially mine). I would suggest some of the Issac Asimov science
series as a place to start.
I'd like to help you more, but I just don't have time to study your
ideas in the detail needed to adequately critique them.
Read the books - physics is fun.
Bill Ward
Ian Johnston
DGoncz (dgo...@aol.com) wrote:
: Yes, a two minute calculation can show that the proposed vacuum engine
: should not be built, or at least that it won't provide enough power to
: make a profit in the marketplace. What bothered me was that the short
: calculation was compared to a real working device.
All the short calculations I have seen (and that includes my own) were using
fundamental laws of physics, maths and thermodynamics. "Real working
devices" didn't come into it, although it's useful to show that the behaviour
of the real devices is very lose to what's predicted.
If a "Brown's Gas" engine is so revolutionary, what assertion do you want to
change -
* that work done = $\int p dv$?
* that force = pressure difference * area?
* that rate of working = torque * angular velocity?
For Hannon's claims to be true, one of the above must be wrong by a factor of
at least ten.
Ian
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
<snip prior posts>
>> >
> Do the physics quizes, Bill.
OK, Mike, I just did.
>Science is also about discovery of
>phenomena that opposes all conventional rules and beliefs. I hold no bad
>sentiments about your feelings. But what will happen if you look at the
>"quizes" and discover something,
I think I'd be more likely to discover something by reading up on the
present state of the art, then looking for holes in the logic. No
offense intended, but you don't come across as knowing very much about
the existing theories. Don't you think you'd be in a better position to
judge if something is new and revolutionary if you know the basic
principles first? You don't have to _believe_ them, just be aware of
them and know how to use them. They're basic because they very
successfully predict how the world works, not because of some mandate
from above. If you find one that's wrong, there's a king in Stockholm
waiting to hang a medal around your neck.
>and what will happen when I build a
>Vacuum powered Tesla turbine, and it works? What will any of these
>people do when they are proven wrong. Will they be the first to admit
>their errors? No.
Actually, Mike, yes. Science is all about proving things wrong. The
catch is that you have to know all the basics and much more to do it.
Ignorance of basics is not a recipe for success.
> Remember, Harry is one guy who said that water is a
>great dielectric, when that is true only in special circumstances.
>Otherwise, where are all those water capacitors.
Harry is right - water is a great dielectric. Water plus ions is a
conductor. He knows that, as does everyone else who's taken basic
chemistry. BTW, I personally have seen a power RF transmitter running
submerged in DI water as the dielectric and cooling medium.. (Mount
Wilson, ca 1958). Whether you believe me (or Harry) or not is
immaterial.
> Do you call that
>disciplined, (he runs a test for a few minutes that, if left to run for
>weeks would prove him wrong), well-conducted testing and research, or a
>staged performance with limited variables, particularly designed to
>prove a point that can easily be disproven by anyone who knows about the
>weaknesses of that test. Is THAT good science?
Mike, that science is good enough to prove his point. He knows, I know,
and just about everybody else in the NG knows that water is in fact a
dielectric. We know because we've had direct experience with it, done
calculations that worked (made accurate predictions), and some of us
have actually measured the conductivity and dielectric constant in a lab
exercise, He was trying to be nice to you and show you how to convince
yourself that pure water is a dielectric. Whether you believe it or not
is entirely your option.
> Check out the "quizes,"
>Bill. I'd really like to hear an objective analysis based on sound
>reasoning.
>Michael
I hope you don't consider the above comments a flame. I think you're
trying to get ahead of yourself and discover some revolutionary new
principle before you have learned the fundamental laws of science. I
admire your tenacity, but I think you'lll have better luck after you
find out exactly what the rules are that you're trying to disprove.
Bill Ward
Harry H Conover
Bill Ward (bward*remove_this*@ix.netcom.com) wrote:
:
: Yeah, Harry, you're right of course, but I do hate to see such
: imagination and dedication go to waste for lack of discipline. I keep
: watching and hoping to see him get a clue, but no dice so far. At first
: his antics seemed kind of humorous, but now they are getting a bit
: pathetic. He just doesn't get that science is about measurements, not
: beliefs. Maybe someday...
:
: BTW, what the heck's a "sampo"?
The 'Sampo' is a magical device appearing in classical Finish myth --
by tradition forged under mystical conditions. Evidently a machine
of some sort, it produces gold, salt, and flour without requiring any
input.
Harry C.
DW
Michael Hannon wrote:
> Dear Bill,
> Here goes:
> In a cylindrical container as illustrated below,
> whose left end (S) is completely closed and sealed,
> while its right end (E) is open to the atmosphere and
> which has vents to the atmosphere, as shown, between
> piston (A) and a middle immovable wall which (W) is
> well-sealed at the point where the moveable rod of the
> solid rod-piston assembly (pistons A and B) passes through it,
> and there is a vacuum to the left of piston A (pulling it to
> the left) and pressure left of piston B, (pushing it to the
> right) acting in opposition to that vacuum. (The pistons,
> rigidly fixed to the rod, can slide freely to the left or
> right, in unison, and are well-sealed at the points where
> they contact the chamber)
>
> If those pistons (whose sizes are identical) are to remain
> in exactly the positions they are, how much Pressure (P)
> is required to counteract a Vacuum (V) of (777/1830) 0.42 Torr?
>
> (S) Open air (W)
> |_____________________\|||______________________
> | |-------| ||| >>>P>>> |-------|
> | Vacuum | |Vent||| Pressure | |
> | <<<<<< | | ||| | |
> | V | Piston|----|||---Rod----| Piston| Open (E)
> | V | A |----|||----------| B | air
> | <<<<<< | | ||| | |
> | Vacuum | | ||| Pressure | |
> | |_______|Vent||| >>>P>>> |_______|
> |--------------------- |||----------------------
> (S) Open air/ (W)
>
> Michael
Michael,
This was a fun problem. First a few assumptions about the size of this
beast.
They may not matter but it made thinking about it easier:
Assume both pistons are square and are 0.02 m on a side:
Assume each chamber is a cube (to start with) and are 0.02 m on a side.
The initial conditions are is that everything is at 1 atm and at
equilibrium.
That means that each side of each piston sees 1.013x10^5 Pa or 40.52N.
Now if you take one chamber and reduce the pressure to 0.42 Torr or 56
Pa
then the insides of the pistons feel 0.0224N+40.52N = 40.5424N and the
outsides of the pistons still see 40.52N each or 81.04 N. Clearly, this
system is out of equilibrium and the pistons will move. The question is
how far.
To figure this out we use Boyle’s Law PV = c (Assuming a constant Temp.
which
I am, also assuming ideal gas behavior, which I am). This also assumes
that
the piston rod is infinitely small (I don't think that will matter
much).
So for our low pressure chamber c(l) = 56 Pa * 8.0x10^-6 m^3 =
4.48x10^-4 N/m^5
for our high pressure chamber c(h) = 1.013 Pa * 8.0x10^-6 m^3 = 0.8104
N/m^5
Since the volume only has one degree of freedom V = 0.0004*Y. Y is the
distance from the piston to the end of the chamber(in meters) and since
I made the chambers identical Y must be the same for each chamber. So,
(Area of Piston)*c(l)/V + (Area of Piston)*c(h)/V = 81.04 N
Area of Piston = 0.0004m^2 and is the same for both pistons
Volume = 0.0004*Y m^3 and is the same for both chambers.
{0.0004 * c(l)}/(0.0004*Y) + {0.0004* c(h)}/(0.0004*Y) = 81.04
Solving for Y = 1.000552x10^-2 m. Which means the pistons moved in
very close to ½ of their original distance from the end. Which of
course means
the pressure on the high side just about doubled to 202488.05 Pa or 1519
Torr
and the pressure on the low side just about doubled to 111.94 Pa or
0.8398 Torr.
Looking back on this I went duh! It should have been obvious. But
there it is.
I hope this helps,
Donald
DGoncz
I can't read the ASCII art! My newsreader uses proportional spacing!
Interesting question. It seems that pressure on B could balance vacuum on
A, but only if the atmospheric pressure on the other side of A is given.
Since the piston sizes are equal we can balance the pressures.
Assuming atomospheric pressure to the right of A (and to the right of B)
is 777 Torr,
and assuming an infinitely thin connecting rod:
Pressure on A: 777-.42=776.58 Torr, acting left.
Pressure on B must equal 776.58 Torr, acting right:
Pressure on right of B, acting left, 777 Torr.
Pressure on left of B, acting right, 777+776.58 Torr, or P= 1553.58
Torr, about double atmospheric. Close to Bill's answer, I used 777 instead
of 760 Torr as atomospheric.
With a substantial connecting rod, the answer changes. There is an engine
type, a Stelzer (German, sp?) engine that pushes an internal piston back
and forth with explosions at both ends.
They make nice linear generators or air compressors. Designed very much
like this problem.
Let me try, with a connecting rod of 1 x 1 inches, one square inch.
Assuming 16 square inch piston (D = 16 /pi, about 5 inch OD or a 4 x 4
inch square piston),
Assuming atmospheric 777 Torr:
Force acting right on A from .42 Torr = .42 * 16 = 6.72 hybrid units of
force
Force acting left on A from 777 Torr = 777 * 15 = 11655 hybrid units
Force acting left on B from 777 Torr = 777 * 16 = 12432 hybrid units
Net force on A, acting left 11655-6.72 =11648.28 hybrid units (In my head!
No calculator!)
Net force on B must balance 11648.28 hybrid units
Force on B acting right must be 12432+11648.28 hybrid units or 24080.28
Area on left of B for force acting right, 15 sq in
Pressure P = 24080.28 / 15 = 1605.352 Torr
Previous P = 1553.58
The two answers are different enough to make the problem interesting.
Would someone check my work, please? What fun! What have we shown here?
We see that with a thicker connecting rod, the balancing pressure will be
higher.
Michael Hannon
Gee, Bill - I thought this stuff was so simple. Also, since basically
ANY pressure
above 15psi (1 atm), let's say 15.1 psi, will move the piston, you're
nit-picking
to stall the quiz. I'll, go back and critique what you did provide later
- got to
fix my car. It's becoming apparent, Bill, that you have little, if no,
intention of
acting in the spirit of this quiz, attempting to stall it at the getout,
telling me
to go study Azimov (it's been in my library for years, by the way),
instead of con-
tributing to its fruitful end.
Michael
Bill Ward
con...@tiac.net (Harry H Conover) wrote:
>Bill Ward (bward*remove_this*@ix.netcom.com) wrote:
<snip>
>: BTW, what the heck's a "sampo"?
>The 'Sampo' is a magical device appearing in classical Finish myth --
>by tradition forged under mystical conditions. Evidently a machine
>of some sort, it produces gold, salt, and flour without requiring any
>input.
> Harry C.
Thanks, Harry, That's a new one for me. Can't wait to use it on
someone.
Could it be considered a perpetual merchandise machine?
Bill Ward
Michael Hannon
Kevin's types most of my life - they are a clique, depending on
self-congratulation and protection. Have you EVER seen one of them argue
with another on this NG about ANYTHING? Is this natural?
> > Remember, Harry is one guy who said that water is a
> >great dielectric, when that is true only in special circumstances.
> >Otherwise, where are all those water capacitors.
>
> Harry is right - water is a great dielectric. Water plus ions is a
> conductor. He knows that, as does everyone else who's taken basic
> chemistry. BTW, I personally have seen a power RF transmitter running
> submerged in DI water as the dielectric and cooling medium.. (Mount
> Wilson, ca 1958). Whether you believe me (or Harry) or not is
> immaterial.
>
How much of water in that state exists in the world, percentage-wise,
and how is that water you're talking about prepared and maintained? I'm
fully aware of absolutely pure water's dielectric properties, and have
been since prep school chemistry at Mount Hermon, at then Syracuse, and
I believe you, Bill, but that has nothing to do with the fact that
99.99999999999(add as many zero's as you like)% of the water that we
know of is NOT a good dielectric, and the instance you give is basically
just as much an anomoly as Brown's gas. As I have said before, if I
prepare for weeks ahead of time, exercise my throat, drink certain
liquids, and practice a lot, there's one song I've sung all my life that
I could sing very well, once a year. Does that make me a great singer?
Well, my mom thinks so anyway. Now, what were we saying about monatomic
hydrogen?
> > Do you call that
> >disciplined, (he runs a test for a few minutes that, if left to run for
> >weeks would prove him wrong), well-conducted testing and research, or a
> >staged performance with limited variables, particularly designed to
> >prove a point that can easily be disproven by anyone who knows about the
> >weaknesses of that test. Is THAT good science?
>
> Mike, that science is good enough to prove his point. He knows, I know,
> and just about everybody else in the NG knows that water is in fact a
> dielectric. We know because we've had direct experience with it, done
> calculations that worked (made accurate predictions), and some of us
> have actually measured the conductivity and dielectric constant in a lab
> exercise, He was trying to be nice to you and show you how to convince
> yourself that pure water is a dielectric. Whether you believe it or not
> is entirely your option.
>
No, it isn't good enough to prove his point - it is misleading. Put two
iron, or aluminum plates, not wires (this is a test for dielectric
properties - when have you ever seen a capacitor made from two short
pieces of wire?), and leave them conducting for days so that the decay
of dielectric properties is shown, and it would be a good test. It is a
contrived test to prove a point which is based on momentary results
about a dielectric which basically no one in this world uses except in
particular circumstances, where the facilities to maintain a basically
anomolous situation (chemically pure water in a non-reactive
environment)
are maintained at high cost (once again, sounds like monatomic H and O
production).
> > Check out the "quizes,"
> >Bill. I'd really like to hear an objective analysis based on sound
> >reasoning.
>
> >Michael
>
> I hope you don't consider the above comments a flame. I think you're
> trying to get ahead of yourself and discover some revolutionary new
> principle before you have learned the fundamental laws of science. I
> admire your tenacity, but I think you'lll have better luck after you
> find out exactly what the rules are that you're trying to disprove.
>
> Bill Ward
There's nothing revolutionary about what I'm pursuing, and if Harry had
done his little experiment at the prep school I went to, he would have
been eaten alive by his not-quite-so-gullible students, or, at the very
least, me, as a student there.
No, I'm not looking at your civil, yet highly biased, responses, as
flames. Of course, I'm not exchanging personal emails with Harry, who
came "out of the gate" with me on the attack, either.
Best,
Michael
Michael Hannon
you're just a punk who knows science - otherwise, don't bother me -
you're not worth my time.
Michael
Robert Erck
In article <332F4C...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
>
> Funny, Harry - didn't you see the publications reffered to in either
> this NG, or oner of the other energy-related NG's from Lawrence
> Livermore directly contradicting what you have just said about H not
> being any energy source? You spend way too much time not seeing what is
> right in front of you. They have published whole studies, including
> charts, on the viability of H in cars, while you're sitting here playing with
> yourself. Tsk, Tsk, Tsk.
> It goes like this - you take nuclear sources electricity, generate hH in
> one of the new super-efficient generators, put it in a tank, and run
> your car's engine on it. Where are the fossil fuels in that?
> I saw a Taurus in the VA hospital parking lot the other day that had a
> small logo on the side of a leaf on a branch, with the letters FFV under
> it. What does that stand for?
> Michael
FFV stands for flexible fuel vehicle. It is likely that what you saw was
a hybrid between a Sable and a Taurus. Was it silver or dark blue?
The FFV is a Department of Energy project. The FFV car that runs around
our site burns, I think, M85, E85, and gasoline.
The blue van that usually sits in our parking lot is powered by compressed
natural gas, because you can see the tanks underneath.
Robert Erck
In article <332F47...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
-------------
Now, now, now. Let's have no overtly nasty words.
The pressure between *anyones* ears is at least that of the vapor pressure
of water, which would be about 47 torr.
Bill Ward
Michael Hannon <oha...@mailroom.worldnet.att.net> wrote:
Actually, it takes > 2 atm to move the pistons. You have one atm on the
R of A and one on the L of B, which must be balanced by at least 2 atm
between B & W pushing right on B. Once the pistons start to move, they
will continue till A hits the wall. The pressure between A and C will
remain 0, no matter what the distance J is, since there is no way any
gas can get between them (given your ideal assumptions). Pulling piston
C left also does not change the pressure for the same reason.
>you're nit-picking to stall the quiz.
And why would I do that?
>I'll, go back and critique what you did provide later
>- got to
>fix my car. It's becoming apparent, Bill, that you have little, if no,
>intention of
>acting in the spirit of this quiz, attempting to stall it at the getout,
>telling me
>to go study Azimov (it's been in my library for years, by the way),
>instead of con-
>tributing to its fruitful end.
>Michael
Mike, as far as I'm concerned, this _has_ come to its end. I only hope
it's been fruitful for you, but I fear not. I'm done.
Good luck
Bill (sucked in by the vacuum) Ward.
James C. Allison
Dear DGoncz,
Tried to see you website. It would not let me access your info about
sonoluminescense. I am a lumiast, vusician, and am interested in
anything that has to do with sound and light. E-Mail if you like.
--
Hang in there!
Regards and sincere best wishes
AllisonWonderland
---
CHECK OUT THE WEBPAGE AT
http://www.livingston.net/allison/home.htm
Comments welcome.
Michael Hannon
> : have no concept of what a vacuum is, or is capable of, and you should
> : learn, because maybe it will teach you a little wisdom about attacking
> : things you know little about, and that includes the whole Brown's gas
> : community. 400hp is CONSERVATIVE, Ian. A good vacuum engine can run on
> : .01 psi external pressure, never mind 15 psi. Suss it out.
>
> I humbly beseech you Michael, please tell me how it works. At the moment,
> you see, I teach thermodynamics to engineering undergraduates and I am
> obviously getting it all wrong - as well as every other engineer in the
> world. Your words of enlightenment would be enormously appreciated.
>
> Ian
DO THE QUIZ, Ian, especially QuizII, and stop wasting my time.
Michael
Michael Hannon
It was blue, Robert. The Lawrence-Livermore report I got(1 of 4 posted,
and then disappeared from the NG), by the way, is called Lean-Burn
Hydrogen Spark-Ignited Engines: the Mechanical Equivalent to the Fuel
Cell, by Salvador M. Aceves, and J. Ray Smith submitted to the American
Society of Mechanical Engineers, and has a code in the righthand corner:
UCRL-JC-124095 Rev1, in case you're interested.
Michael
Ian Johnston
Michael Hannon
Thanks, Bill, because you're obviously way outfield inrelation to not
only the spirit of this thing, but actual calculations as well. As I
said in reference to your post on quiz I, you're trying to tell us that
15 gauge psi (barely enough to half inflate a tire on my car) pushing on
piston B is basically enough to create a 0.42 Torr condition (that's
from an 1830/1 pressure reduction, or 770/1830, which is the
approximately the same condition which would be created by moving piston
C to the left 1830 times the distance J, which makes about as much
sense). If you had a piston with 1 atm in its chamber, which is 1831
in. long before it hit the head, and you pushed it up 1830 in. to within
1" of the head, what would the compression ratio be, and in that 1" of
space left, what would the pressure be- 760 + .42 Torr? I don't think
so. Try 1830 X 760 Torr., 1830 X 15 psi, or 27,000 odd psi. Thanks for
getting out while the getting is good, Bill.
Michael
Michael Hannon
> Thanks for a clear, well-described question. I'm going to take it
> seriously and answer it as clearly as I can.
>
> minus the .42 torr "vacuum" pressure, or about 1519.58 torr.
>
> mercury, or 1/760 atmosphere. The pressure inside a "vacuum" chamber is
> often measured in Torr, but it's still a pressure, just like
> atmospheric pressure. When you start pumping down a vacuum rig, you can
> watch the absolute pressure drop from 760 (or whatever the barometric
> pressure is that day) down to however low your equipment will go.
>
> What may be throwing you off a bit is the difference between "gauge" and
> absolute pressures. Absolute pressure refers to the actual pressure, not
> relative to any other pressure, while gauge pressure is the difference
> between atmospheric and the inside of the gauge. Gauge pressure will be
> negative (a "vacuum") if the absolute pressure inside the gauge is less
> than the outside atmospheric pressure. Absolute pressure can never be
> negative.
>
> In other words, vacuums don't suck, pressures push. When you suck on a
> straw and reduce the pressure in your mouth and the straw, atmospheric
> pressure pushes down on the soda, which pushes the liquid up into the
> lower pressure area in your mouth.
>
> So on your drawing, the difference in pressure across piston A is
> that atmospheric pressure E of 760 torr pushes. To remain in
> equilibrium, the force from P must equal the sum of those, or 1519.58
> torr times the area.
>
> I initially misread your diagram to be balancing the two forces, and
> caught the mistake right after posting. (of course :-) I apologize in
> advance for the confusion if I didn't cancel the bad post quick enough.
>
>
> Bill Ward
Thanks for the info, Bill, and we are, in fact, dealing with gauge
pressures and vacuum (sorry, it's always been this way as I was taught).
In this respect, there is definitely a Zero psi, which is at 1
amosphere, which, by the way, is different for me at a mile up, but
then, we can continue this discussion endlessly depending on the weather
(barometric pressure), so let's, just for sanity sake, say that all
externally variable parameters are stable for all quiz data.
Given this, when 15psi is placed in the B-W volume, piston A moves to
the right, creating a gauge vacuum (how's that?). I also understand that
an absolute vacuum is the REAL zero psi standard, but everyone I know
uses psi and inches of vacuum - you know - car stuff. So, now we're
talking Torr, and 15 psi = 760 Torr - OK. The rest, I'm afraid, escapes
me.
If the guage pressure in the volume left of A is .42 Torr, then the
total gauge pressure of P is barely 2 atm, or 30 psi? If that is so,
then what is the Torr left of after A has been moved to the right from
being flush with the closed left end of the container if the gauge
pressure P is 15 psi, as described in the quiz after moving piston B -
0.42 Torr? You mean
I only need to apply about 15 gauge psi to B to move it and A to the
right, so that the volume left of A is .42 Torr, or to put it another
way, in order to get .42 Torr (which is the result of an 1830/1 pressure
change) all I need to apply to B is barely enough pressure to
semi-inflate a tire on my car (15 gauge psi)?
Michael
Ian Johnston
Michael Hannon (oha...@mailroom.worldnet.att.net) wrote:
: Ian Johnston wrote:
: > What do you think the difference is, Michael? Is the force on a piston with
: > 15psi on one side and 0psi on the other more or less than the force on
: > a piston with 15psi on one side and 0.42 Torr on the other?
: Do the quizes, Ian - you may consider yourself a scientist, but to me,
: you're just a punk who knows science - otherwise, don't bother me -
: you're not worth my time.
Go on, answer the questions. Do you know what a Torr is?
Ian
Michael Hannon
else's work as fraudulent WITHOUT backing it up with FACTS based on
personal experiences and those of others doesn't deserve any more
substantiation than he gives. You call yourself a teacher? Do you know
how many real JERKS there are who teach? I was married to a teacher for
5 years, so don't try to pull that "I'm a professor" legitimizing caca
on me. Some of the biggest lowlifes around are teachers with tenure. I
was molested by one when I was 12.
A long time ago, when you fist started attacking Brown and other people,
I asked you for substantiation - REAL SUBSTANTIATION - not your HEARSAY
and BASELESS PERSONAL OPINION kind, which you like to spew out, and you
have YET to provide ANY. So, do I need to answer stupid, pompous,
deriding questions from a bagpipe of hot air like yourself, who refuses
to play on an even field because he says he's a "professor?"
Get lost. You don't deserve what you yourself are not willing, and
unable, to give. Go play "you-have-to-answer-to-me" bigshot authority
figure with your students or someone else who you can sucker into
believing your hustle.
Michael
Dan Doner
Michael Hannon wrote:
> It was blue, Robert. The Lawrence-Livermore report I got(1 of 4 posted,
> and then disappeared from the NG), by the way, is called Lean-Burn
> Hydrogen Spark-Ignited Engines: the Mechanical Equivalent to the Fuel
> Cell, by Salvador M. Aceves, and J. Ray Smith submitted to the American
> Society of Mechanical Engineers, and has a code in the righthand corner:
> UCRL-JC-124095 Rev1, in case you're interested.
> Michael
There is no mechanical equivelant of a fuel cell. None. Not even
close. I even dare say there never will be. Read a bit about Carnot
efficiency, then read a bit about electrochemistry.
Dan
Robert Erck
In article <333184...@mailroom.worldnet.att.net>,
Now Michael, you must understand that science is a learning process, and
there is a give and take. But first one needs to establish that he knows
what he is talking about, so there is a common basis for understanding.
When someone asks "Do you know what a Torr is?" you should tell him the
answer. It is not an unreasonable question. It is inappropriate to reply
by stating that the questioner is a "slut."
Now Ian, I think you will get nowhere by acting like a professor. You'll
have much more luck "connecting" to Michael by acting like figures he
understands. Like Mr. Rogers, or Barney.
Robert Erck
Ah, ok. I got confused. (Brain is getting frazzled). The silver
Sable-Taurus-SHO is a lightweight almost all-aluminum vehicle. I don't
know how many exist.
FFV's come in many styles and burn ethanol/methanol/gasoline. Then
there's *alternative* fuelled vehicles, which burn fuels like propane and
CNG.
I suspect that "Lean-Burn Hydrogen Spark-Ignited Engines: the Mechanical
Equivalent to the Fuel Cell" is a bit of hyperbole from an engineer who
wishes to emphasize that his engine has much high efficiency. Lean-burn
gasoline engines have been around for ages (they have their own problems),
so it seems likely that a lean-burn hydrogen engine concept would be
explored.
One of the features of a fuel cell is absence of noise or moving parts.
I'm not sure how one could have "mechanical equivalent" of a fuel cell.
Some people refer to flywheels as "mechanical batteries."
------------
In article <3330C2...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
Michael Hannon
Robert Erck wrote:
>
> Ah, ok. I got confused. (Brain is getting frazzled). The silver
> Sable-Taurus-SHO is a lightweight almost all-aluminum vehicle. I don't
> know how many exist.
>
> FFV's come in many styles and burn ethanol/methanol/gasoline. Then
> there's *alternative* fuelled vehicles, which burn fuels like propane and
> CNG.
>
> I suspect that "Lean-Burn Hydrogen Spark-Ignited Engines: the Mechanical
> Equivalent to the Fuel Cell" is a bit of hyperbole from an engineer who
> wishes to emphasize that his engine has much high efficiency. Lean-burn
> gasoline engines have been around for ages (they have their own problems),
> so it seems likely that a lean-burn hydrogen engine concept would be
> explored.
>
> One of the features of a fuel cell is absence of noise or moving parts.
> I'm not sure how one could have "mechanical equivalent" of a fuel cell.
> Some people refer to flywheels as "mechanical batteries."
> ------------
> In article <3330C2...@mailroom.worldnet.att.net>,
> oha...@mailroom.worldnet.att.net wrote:
Uh huh.
Michael
Michael Hannon
workable formulas. So far, I've gotten someone saying that it would take
as much pressure (a little over 15 psi to, or enough to half-inflate a
tire on my car, to counteract .42 Torr on the other side of piston A).
Now, let's see your genius in this. Let's see YOUR expertise in
calculating rational figures for pressure vs. vacuum, in quiz II, as
well as I - YOU'RE the BIG VACUUM EXPERT - LET'S SEE IT. What's it going
to be from you? 20 psi? NONE of you EXPERTS, SCIENTISTS, can even figure
out what is going on here, and YOU want to know what a Torr is! YOU ARE
ALL a FRIGGIN JOKE! It's really fun watching you "SCIENTISTS" make fools
of yourselves on these quizes. Tell me, Bob - how much pressure does it
take to oppose an 1860/1 pressure reduction, as in quiz II? 25 psi?
760.4259871 Torr. DO ANY OF YOU IDIOTS KNOW WHAT A TORR REALLY IS?
What a joke!!! NO wonder you all want to hide behind your anal-retentive
terminology - NONE OF YOU KNOWS WHAT YOU"RE DOING!!!!! What a laugh!!
Michael,
Do I know what a Torr is???? Ha! Do any of you????
"laughing his "uninformed" butt off!!!!"
See you all in space, with your exploding spacesuits and craft!!!
Now, THIS is hilarious. Thanks for the entertainment, folks - you're a
laugh a minute. Wait til you get a load of quiz Ia. You think you look
stupid now?
Michael Hannon
Dan Doner wrote:
>
> Michael Hannon wrote:
> > It was blue, Robert. The Lawrence-Livermore report I got(1 of 4 posted,
> > Hydrogen Spark-Ignited Engines: the Mechanical Equivalent to the Fuel
> > Society of Mechanical Engineers, and has a code in the righthand corner:
> > UCRL-JC-124095 Rev1, in case you're interested.
> > Michael
>
> close. I even dare say there never will be. Read a bit about Carnot
> efficiency, then read a bit about electrochemistry.
>
> Dan
Uh huh.
Michael
Michael Hannon
Tell me, as well, Robert - how is that you get a fuel cell to propel you
down the street - oh, yes - I guess that you WOULD have to use some
silly, noisy type of engine, motor, and transmission, or something,
wouldn't you, and the control circuitry and mechanisms involved - I
wonder how much all of that "stuff," when put it together with that fuel
cell, would weigh, and consume on the way to getting you "doon th'
rood" compared to that "wishful thinking" pipedream these Livermore
engineers came up with. That'd be an interesting study, but, y'know,
I'll bet you didn't look that report up to see what was going on, and,
lo and behold, without even seeing it, let alone studying it, you've
already got their whole method and results aaaalll figured out, don't
you?
Been a'workin on those quizes, Bob, or have you got them aaaalll figured
out as well? What a joke....
Michael
Michael
Robert Erck
(snip)
If I understand the diagram correctly:
Atmospheric pressure is 760 torr
The pressure to the left of piston A is 0.42 torr
The pressure to the right of piston A is atmospheric
The pressure to the right of piston B is atmospheric
The seals are perfect and frictionless, and the rod is very thin.
You want to know what the pressure is on the left of piston B.
My answer: 1519.58 torr
This is the same answer as Bill Ward's.
Robert Erck
In article <333228...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
>(snip)
> That's very nice, Robert, but let's see you answer the quizes with
> workable formulas. So far, I've gotten someone saying that it would take
> as much pressure (a little over 15 psi to, or enough to half-inflate a
> tire on my car, to counteract .42 Torr on the other side of piston A).
> Now, let's see your genius in this. Let's see YOUR expertise in
> calculating rational figures for pressure vs. vacuum, in quiz II, as
> well as I
(snip)
Whew, quiz 2 is more difficult. Here is my analysis. AM I CORRECT??
> What is the pressure
> difference between the partial vacuum in the volume between pistons A
> and C, when the pressure between piston B and wall W is 15psi (equal to
> the pressure of the open air outside)?
----> A and C are in contact. The push on each other with a force of 15psi.
----> Note to Bill Ward: *of course* you can have pressure with no gap -
there is no gap between my shoe and the floor, but there is pressure!
> During this operation, piston C has been held firmly stationary,
> while piston A has moved a distance(J).
----> No, it hasn't moved.
If we now lock pistons A and B
> in their positions, maintaining the partial vacuum between A and C, and
> the pressure between B and W, turn on the winch, and pull piston C
> backwards a distance equal to 1830 X the distance J (which piston A has
> traveled from piston C previously), what is the pressure between A and C
> now?
-----> Because there is no trapped air between A and C, as soon as they
are pulled apart, a perfect vacuum will be formed. The pressure is zero.
> How many psi pumped between piston B and wall W
> will it take to keep piston B in its same position after pistons A and B
> are unlocked, and are free to respond to the vacuum created between
> pistons A and C after C has been moved that distance?
----> When the movable piston is unlocked, it will remain stationary.
Nowhere is there 0.42 torr, as far as I can see.
AM I CORRECT??
Robert Erck
*Whew* this is much more complicated. Here is my analysis.
> What is the pressure
> difference between the partial vacuum in the volume between pistons A
> and C, when the pressure between piston B and wall W is 15psi (equal to
> the pressure of the open air outside)?
----> A and C are in contact, and they push on each other with a force of
15psi.
----> Note to Bill Ward: you can have pressure with no gap - there is no
gap between my shoe and the floor, but there is pressure!
> During this operation, piston C has been held firmly stationary,
> while piston A has moved a distance(J).
----> No, it hasn't moved.
If we now lock pistons A and B
> in their positions, maintaining the partial vacuum between A and C, and
> the pressure between B and W, turn on the winch, and pull piston C
> backwards a distance equal to 1830 X the distance J (which piston A has
> traveled from piston C previously), what is the pressure between A and C
> now?
-----> Because there is no trapped air between A and C, as soon as they
are pulled apart, a perfect vacuum will be formed. The pressure is zero.
> How many psi pumped between piston B and wall W
> will it take to keep piston B in its same position after pistons A and B
> are unlocked, and are free to respond to the vacuum created between
> pistons A and C after C has been moved that distance?
----> When the movable piston is unlocked, it will remain stationary.
Nowhere is there 0.42 torr, as far as I can see.
>
Kevin Jones
Michael Hannon wrote:
> > > If those pistons (whose sizes are identical) are to remain
> > > in exactly the positions they are, how much Pressure (P)
> > > is required to counteract a Vacuum (V) of (777/1830) 0.42 Torr?
> >
> > > (S) Open air (W)
> > > |_____________________\|||______________________
> > > | |-------| ||| >>>P>>> |-------|
> > > | Vacuum | |Vent||| Pressure | |
> > > | <<<<<< | | ||| | |
> > > | V | Piston|----|||---Rod----| Piston| Open (E)
> > > | V | A |----|||----------| B | air
> > > | <<<<<< | | ||| | |
> > > | Vacuum | | ||| Pressure | |
> > > | |_______|Vent||| >>>P>>> |_______|
> > > |--------------------- |||----------------------
> > > (S) Open air/ (W)
> >
<<<Excellent reply by Bill Ward snipped>>>
If the guage pressure in the volume left of A is .42 Torr, then the
> total gauge pressure of P is barely 2 atm, or 30 psi?
By George, I think he's finally got it!!
If you do all your calculations in "absolute", you'll see how simple it
is Mike. Then you'll get the idea there is no "sucking" force at all,
just atmospheric pressure working against some pressure *less* than
atmospheric.
If that is so,
> then what is the Torr left of after A has been moved to the right from
> being flush with the closed left end of the container if the gauge
> pressure P is 15 psi, as described in the quiz after moving piston B -
> 0.42 Torr? You mean
> I only need to apply about 15 gauge psi to B to move it and A to the
> right, so that the volume left of A is .42 Torr, or to put it another
> way, in order to get .42 Torr (which is the result of an 1830/1 pressure
> change) all I need to apply to B is barely enough pressure to
> semi-inflate a tire on my car (15 gauge psi)?
Now you understand why 15psig simulates the exact same thing as your
"implosion" engine, and why the "power" developed from one is limited to
small numbers.
K. Jones
The opinions expressed are my own, and not those of my employer, or
anyone else
Michael Hannon
Robert Erck wrote:
>
> In article <333228...@mailroom.worldnet.att.net>,
> oha...@mailroom.worldnet.att.net wrote:
>
> >(snip)
> > That's very nice, Robert, but let's see you answer the quizes with
> > workable formulas. So far, I've gotten someone saying that it would take
> > as much pressure (a little over 15 psi to, or enough to half-inflate a
> > tire on my car, to counteract .42 Torr on the other side of piston A).
> > Now, let's see your genius in this. Let's see YOUR expertise in
> > calculating rational figures for pressure vs. vacuum, in quiz II, as
> > well as I
> (snip)
>
>
> Atmospheric pressure is 760 torr
> The pressure to the left of piston A is 0.42 torr
> The pressure to the right of piston A is atmospheric
> The pressure to the right of piston B is atmospheric
> The seals are perfect and frictionless, and the rod is very thin.
>
> You want to know what the pressure is on the left of piston B.
>
> My answer: 1519.58 torr
>
> This is the same answer as Bill Ward's.
Robert Erck
Michael Hannon
Robert Erck wrote:
>
> (snip)
> > Here goes:
> > In a cylindrical container as illustrated below,
> > whose left end (S) is completely closed and sealed,
> > while its right end (E) is open to the atmosphere and
> > which has vents to the atmosphere, as shown, between
> > piston (A) and a middle immovable wall which (W) is
> > well-sealed at the point where the moveable rod of the
> > solid rod-piston assembly (pistons A and B) passes through it,
> > and there is a vacuum to the left of piston A (pulling it to
> > the left) and pressure left of piston B, (pushing it to the
> > right) acting in opposition to that vacuum. (The pistons,
> > rigidly fixed to the rod, can slide freely to the left or
> > right, in unison, and are well-sealed at the points where
> > they contact the chamber)
> >
> > in exactly the positions they are, how much Pressure (P)
> > is required to counteract a Vacuum (V) of (777/1830) 0.42 Torr?
> >
> > (S) Open air (W)
> > |_____________________\|||______________________
> > | |-------| ||| >>>P>>> |-------|
> > | Vacuum | |Vent||| Pressure | |
> > | <<<<<< | | ||| | |
> > | V | Piston|----|||---Rod----| Piston| Open (E)
> > | V | A |----|||----------| B | air
> > | <<<<<< | | ||| | |
> > | Vacuum | | ||| Pressure | |
> > | |_______|Vent||| >>>P>>> |_______|
> > |--------------------- |||----------------------
> > (S) Open air/ (W)
> >
>
> If I understand the diagram correctly:
>
> Atmospheric pressure is 760 torr
> The pressure to the left of piston A is 0.42 torr
> The pressure to the right of piston A is atmospheric
> The pressure to the right of piston B is atmospheric
> The seals are perfect and frictionless, and the rod is very thin.
>
> You want to know what the pressure is on the left of piston B.
>
> My answer: 1519.58 torr
>
> This is the same answer as Bill Ward's.
Michael Hannon
Kevin Jones wrote:
>
> Michael Hannon wrote:
>
> > > > If those pistons (whose sizes are identical) are to remain
> > > > in exactly the positions they are, how much Pressure (P)
> > > > is required to counteract a Vacuum (V) of (777/1830) 0.42 Torr?
> > >
> > > > (S) Open air (W)
> > > > |_____________________\|||______________________
> > > > | |-------| ||| >>>P>>> |-------|
> > > > | Vacuum | |Vent||| Pressure | |
> > > > | <<<<<< | | ||| | |
> > > > | V | Piston|----|||---Rod----| Piston| Open (E)
> > > > | V | A |----|||----------| B | air
> > > > | <<<<<< | | ||| | |
> > > > | Vacuum | | ||| Pressure | |
> > > > | |_______|Vent||| >>>P>>> |_______|
> > > > |--------------------- |||----------------------
> > > > (S) Open air/ (W)
> > >
>
>
> > total gauge pressure of P is barely 2 atm, or 30 psi?
>
>
> If you do all your calculations in "absolute", you'll see how simple it
> is Mike. Then you'll get the idea there is no "sucking" force at all,
> just atmospheric pressure working against some pressure *less* than
> atmospheric.
>
> > then what is the Torr left of after A has been moved to the right from
> > being flush with the closed left end of the container if the gauge
> > pressure P is 15 psi, as described in the quiz after moving piston B -
> > 0.42 Torr? You mean
> > I only need to apply about 15 gauge psi to B to move it and A to the
> > right, so that the volume left of A is .42 Torr, or to put it another
> > way, in order to get .42 Torr (which is the result of an 1830/1 pressure
> > change) all I need to apply to B is barely enough pressure to
> > semi-inflate a tire on my car (15 gauge psi)?
>
> "implosion" engine, and why the "power" developed from one is limited to
> small numbers.
>
> K. Jones
> The opinions expressed are my own, and not those of my employer, or
> anyone else
A very foolish answer, Kevin, and in time you're also going to eat every
word you have said and every calculation you have done on that V6 as
well. You have absolutely no idea what you're dealing with - not a clue.
The best thing I can say is to get a bicycle tire pump and do try
Roamer's example.
Michael
Michael Hannon
Robert Erck wrote:
>
> *Whew* this is much more complicated. Here is my analysis.
>
> > What is the pressure
> > difference between the partial vacuum in the volume between pistons A
> > and C, when the pressure between piston B and wall W is 15psi (equal to
> > the pressure of the open air outside)?
>
> ----> A and C are in contact, and they push on each other with a force of
> 15psi.
>
> ----> Note to Bill Ward: you can have pressure with no gap - there is no
> gap between my shoe and the floor, but there is pressure!
You can have volume with things touching as well, and unless what is
called "flat" surfaces are infinitely smooth, of which I know none,
except for the "seamless portrayal of reality" you, Bill, and some of
the others here seem to have worked out for yourselves, and everyone
else, there is always a gap.
>
> > During this operation, piston C has been held firmly stationary,
> > while piston A has moved a distance(J).
>
> ----> No, it hasn't moved.
>
> If we now lock pistons A and B
> > in their positions, maintaining the partial vacuum between A and C, and
> > the pressure between B and W, turn on the winch, and pull piston C
> > backwards a distance equal to 1830 X the distance J (which piston A has
> > traveled from piston C previously), what is the pressure between A and C
> > now?
>
> -----> Because there is no trapped air between A and C, as soon as they
> are pulled apart, a perfect vacuum will be formed. The pressure is zero.
My, My, My, what a seamless portrayal of reality - I guess REAL GENIUS
such as this has no bounds - the very of it brings tears to these worn
eyes.....ecxuse me, I'll be back in a moment....
Kevin Jones
Michael Hannon wrote:
>
> Kevin Jones wrote:
> Blah, Blah, Blah.......or shut up, Mike.
> >
> > K. Jones
>
> Sorry, Kevin....no time to waste on you this century.
Can't put up, eh Mike??
K. Jones
The opinions expressed are my own, not those of my employer, or anyone
else.
Kevin Jones
DGoncz wrote:
>
> The figure of 16,000 HP means that using a regular engine would outperform
> a vacuum engine, even using math that wildly inflates estimates of power
> output. I was trying to show that the math being used was too simple.
> Yes, a two minute calculation can show that the proposed vacuum engine
> should not be built, or at least that it won't provide enough power to
> make a profit in the marketplace. What bothered me was that the short
> calculation was compared to a real working device.
They were two totally different examples. Knowing Mr. Hannon would have
difficulty with the "math" concept (as he still doesn't understand the
concept of a vacuum, and I
unfortunately have a wealth of previous experience with him), I was
trying to demonstrate that "real life" IC reciprocating engines use
hundreds of pounds of pressure to produce less HP than his wild numbers,
how could he expect atmospheric pressure to do more?? The short
calculation was shown to demonstrate the validity of my claim that it
could only produce little amounts of power, in anticipation of his
"prove it" claims, (ie, I'm full of hot air, don't know what I'm saying,
etc)
For what reason would I go further, when the point was made easily??
> Beliefs cannot be effectively challenged outside the belief system
> of the believer. That is where all the name-calling comes from. I try to
> respect the belief systems of people I meet on the various NGs. I think
> that instead of trying to directly attack the belief that Brown's Gas
> implodes, it is more reasonable and fair to work within the belief system
> of the believers and let them make up their own minds.
I've never stated an opinion about "Browns' Gas", one way or another.
Do the name search, on Deja News, as someone else suggested, and find
out for yourself where the "name calling" and "unreasonable" bits stem
from.
> I am arguring for more
> tolerance and depth, not more words. Extensive quotes and a two line reply
> are not the way to spread tolerance and respect. In depth replies help
> more. I have a lot of respect for belief systems, scientific or not. It's
> like religious tolerance, a fundamental American personal right. Just keep
> trying to work from the known to the unknown.
Tolerance only goes so far. Stick around, post a few replies showing
contradictions in his out-and-out lies, and watch the flame-thrower come
out, and his inablity to remain civil when any errors are politely
pointed out. Time will tell.
K. Jones
The opinions expressed are my own, and not those of my employer, or
anyone else.
Kevin Jones
More of Mikeys postive, contributive, non-attacking, non-obfuscating
posts he's so fond of (preaching to others to do) lately. Once again,
Mikey was asked a simple question (requiring him to be precise, not make
up stuff out of thin air), and since he doesn't know the answer, he
launches into verbal diarrhea, attacking ones character. You're
"obfuscating" Mike, I thought you didn't like that.
The "attack of the disruptive child" will surely launch again.
K. Jones
The opinions expressed are my own, not those of my employer, or anyone
else.
Harry H Conover
Kevin Jones (mah...@rohmhaasNOSPAM.com) wrote:
:
: Knowing Mr. Hannon would have
: difficulty with the "math" concept (as he still doesn't understand the
Sad but true! Evidently Michael managed to sleep through his 7th Grade
General Science classes -- at least, most of them.
Like a small child, he appears to think of a vacuum as having
some sort of 'suck power,' evidently unable to comprehend the fact that
a vacuum is nothing more than a reduction in or lack of pressure. This
is possibly why he doesn't understand why vacuum is measured in Torr
(a unit of pressure).
Also, like a spoiled small child, he appears to have a tantrum when
anyone attempt to explain his errors to him, a character trait that is
hardly conducive to learning.
I had always wondered about the kind of person that would fall victim
to the misguided purveyors of pseudo-scientific products like BG
Generators, Fuel Line Magnets, Free Energy Devices, and perpetual
motion schemes. Michael provides us all a service by supplying
one form of answer to this question.
Harry C.
Robert Erck
In article <33335F...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
> Robert Erck wrote:
(snip)
>
> You can have volume with things touching as well, and unless what is
> called "flat" surfaces are infinitely smooth, of which I know none,
> except for the "seamless portrayal of reality" you, Bill, and some of
> the others here seem to have worked out for yourselves, and everyone
> else, there is always a gap.
Actually, if you take two atomically smooth and clean epitaxial films and
push them together, there will be no volume between them. Of course,
you'll never get the surfaces apart again!
>
> >
(snip)
> >
> > -----> Because there is no trapped air between A and C, as soon as they
> > are pulled apart, a perfect vacuum will be formed. The pressure is zero.
>
> My, My, My, what a seamless portrayal of reality - I guess REAL GENIUS
> such as this has no bounds - the very of it brings tears to these worn
> eyes.....ecxuse me, I'll be back in a moment....
>
(snip)
All these calculations assume ideal, perfect mechanisms, and ideal gases.
If you want calculations using imperfect mechanisms, and real gases, then
that will be much more difficult. I don't want to make all that effort.
The difference may be only a percent or two, depending how well you can
build your piston machine. I think the seals will be a problem. The best
vacuum on the the left, I would suspect would be a few torr, due to gases
inside crevices.
Robert Erck
In article <333355...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
> Robert Erck wrote:
> >
> (snip)
> >
> > My answer: 1519.58 torr
> >
> > This is the same answer as Bill Ward's.
> Wrong.
Wrong? Please provide me with your estimate of the answer, and I will try
to understand it.
Robert Erck
> Robert Erck wrote:
> >
> > (snip)
> > >
> > My answer: 1519.58 torr
> >
> > This is the same answer as Bill Ward's.
> Wrong.
Please tell us all what the *correct* answer is.
Robert Erck
In article <333356...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
(snip)
> A very foolish answer, Kevin, and in time you're also going to eat every
> word you have said and every calculation you have done on that V6 as
> well. You have absolutely no idea what you're dealing with - not a clue.
> The best thing I can say is to get a bicycle tire pump and do try
> Roamer's example.
> Michael
Michael,
I've got about ninety-thousand dollars worth of vacuum equipment in my
lab. Perhaps you can suggest an experiment that I can perform which will
demonstrate some of these marvelous ideas. If it looks revolutionary, I
will publish the results in a technical journal.
Jean Luc Deaux
In article <333268...@mailroom.worldnet.att.net>,
oha...@mailroom.worldnet.att.net wrote:
> Tell me, as well, Robert - how is that you get a fuel cell to propel you
> down the street - oh, yes - I guess that you WOULD have to use some
> silly, noisy type of engine, motor, and transmission, or something,
That 'something' being an electric motor, by chance?
> wouldn't you, and the control circuitry and mechanisms involved - I
> wonder how much all of that "stuff," when put it together with that fuel
> cell, would weigh, and consume on the way to getting you "doon th'
> rood" compared to that "wishful thinking" pipedream these Livermore
> engineers came up with.
It weighs a hell of a lot less than the enormous bank of batteries
necessary to power an electric car further than to the end of your
driveway.
> That'd be an interesting study, but, y'know,
> I'll bet you didn't look that report up to see what was going on, and,
> lo and behold, without even seeing it, let alone studying it, you've
> already got their whole method and results aaaalll figured out, don't
> you?
Generally, if something makes sense, obeys the laws of physics, and
obeys the laws of thermodynamics, a lot of us engineers are willing
to take them on their word -- to a point. When they start asking for
money to fund their research or to sell us some magical device, we
often like to see data before we pull out our checkbooks. If they
claim to violate the laws of thermodynamics (but, strangely, can't
prove it -- but still claim to violate them) we just have to laugh
in their faces and try to ensure that our own children don't receive
such a lousy science education.
> Been a'workin on those quizes, Bob, or have you got them aaaalll figured
> out as well? What a joke....
Yes, Michael, those 'quizzes' are a joke. And so are you. All they have
done is make it abundantly clear that you haven't a clue about such basic
concepts as vacuum, and absolute and gage pressures. You demonstrated
your profound ignorance before you even got out of the gate. But like
all devotees of Brown's Gas or Stan Meyer's Water Fuel Fraud, that
doesn't stop you from explaining to us how and why they work and why
countless scientists have been wrong for decades.
Jean Luc Deaux
> Robert Erck wrote:
> > My answer: 1519.58 torr
> > This is the same answer as Bill Ward's.
> Wrong.
Would you kindly provide us ignorant scientists with the 'correct'
answer? Please show your calculations and reasoning.
And here's a quiz for you, Michael. A piston in a cylinder separates
two volumes, A and B. The piston has a cross-sectional area of 1
square inch (the faces facing volumes A and B). Initially, the
piston is held stationary. Volume A initially a perfect vacuum, but is
then filled with air to an absolute pressure of 0.001 psia. Volume B
is initially filled with air at atmospheric pressure (assume 15 psia),
but then is pumped down with a vacuum pump to a gage pressure of
-14.999 psig. Once the pressure in A and the vacuum in B are set,
please calculate the force necessary to hold the piston in place.
If the restraining force is removed, which direction will the piston
move?
--------------------------------------------------------
| |xxxx| |
| |xxxx| |
| |xxxx| |
| A |xxxx| B |
| |xxxx| |
| |xxxx| |
| |xxxx| |
--------------------------------------------------------
Donald Whisenhunt
In article <bob_erck-210...@et212pc113.et.anl.gov>,
bob_...@qmgate.anl.gov (Robert Erck) wrote:
<snip>
>
> My answer: 1519.58 torr
>
> This is the same answer as Bill Ward's.
This is the same answer I got. Sorry, me detailed calculations
didn't make it to the newsgroup as my employer decided to terminate
USENET access. Oh, well.
Donald
Harry H Conover
Michael Hannon (oha...@mailroom.worldnet.att.net) wrote:
:
: You can have volume with things touching as well, and unless what is
: called "flat" surfaces are infinitely smooth, of which I know none,
: except for the "seamless portrayal of reality" you, Bill, and some of
: the others here seem to have worked out for yourselves, and everyone
: else, there is always a gap.
Ask a machinist friend to demonstate the use of "gauge blocks" for
you.
Harry C.
Kevin Jones
Michael Hannon wrote:
> > Actually, Mike, yes. Science is all about proving things wrong. The
> > catch is that you have to know all the basics and much more to do it.
> > Ignorance of basics is not a recipe for success.
> >
> No, they won't. Their ego's won't let them. I've been Harry's and
> Kevin's types most of my life - they are a clique, depending on
> self-congratulation and protection. Have you EVER seen one of them argue
> with another on this NG about ANYTHING? Is this natural?
I've posted a few retractions and apologies *publically* in Usenet when
I've made errors (have you??!?). That's one reason for trying to keep
your words civil Mike, it doesn't take as big a man to admit to ones
mistakes then, my little friend.
I've yet to see Harry (or many of the "others" you believe are
*conspiring* "againts" you)post something that doesn't make sense, or is
not verifiable. If *anyone* (that includes Harry) posts something that
doesn't "add up", I'll question it, and ask for more data, as anyone
else would.
--
K. Jones
Wearing flame-resistant clothing, and mirrored shades
The opinions expressed aren't necessarily my own, and are not those of
my employer, and anyone else can make up their own minds.
Kevin Jones
Michael Hannon wrote:
> > If that is so,
> > > then what is the Torr left of after A has been moved to the right from
> > > being flush with the closed left end of the container if the gauge
> > > pressure P is 15 psi, as described in the quiz after moving piston B -
> > > 0.42 Torr? You mean
> > > I only need to apply about 15 gauge psi to B to move it and A to the
> > > right, so that the volume left of A is .42 Torr, or to put it another
> > > way, in order to get .42 Torr (which is the result of an 1830/1 pressure
> > > change) all I need to apply to B is barely enough pressure to
> > > semi-inflate a tire on my car (15 gauge psi)?
> >
> > Now you understand why 15psig simulates the exact same thing as your
> > "implosion" engine, and why the "power" developed from one is limited to
> > small numbers.
> >
> > K. Jones
> > The opinions expressed are my own, and not those of my employer, or
> > anyone else
>
> A very foolish answer, Kevin, and in time you're also going to eat every
> word you have said and every calculation you have done on that V6 as
> well. You have absolutely no idea what you're dealing with - not a clue.
> The best thing I can say is to get a bicycle tire pump and do try
> Roamer's example.
> Michael
Facinating how the little vacuum pump I have on my test bench, is really
a several hundred horsepower machine, and it runs on 115V! Amazing.
You are my hero Mike. Now finish your homework and go to bed.
Robert Erck
In article <5h3ru8$7...@news-central.tiac.net>, con...@tiac.net (Harry H
Conover) wrote:
Gauge blocks would be a useful demonstration for Hannon. Wringed gage
blocks probably have a thin film of oil between the surfaces.