Long cables to power "ioncraft" to orbit?

Groups trimmed.

Robert Clark wrote:
George Dishman wrote:
"Robert Clark" wrote in message
oups.com...
snipped due to indentation fault in OE

The point of this thread is the savings in power you would
get by using a lifter thruster method.
Look at the table near the bottom on this page:

Lifter Theory.
http://jnaudin.free.fr/html/lf=ADtheory.htm

The last line in this table labled Thrust(g)/Power(W) ratio gives
the
weight that could be lifted for given power with the air density
available at ground level. It is given as 0.509, or about 2 to 1
for
power in watts required to lift a weight in grams.
This is at ground level.

This is also at zero speed. Energy is force times
distance and power is force times speed so 0.5g
rising vertically at 202 m/s would double the power
needed. You seem to be neglecting that, but at
orbital speeds it is going to be far greater than
the figures you are quoting.

This is the same problem you had with the thrust
equations, you are forgetting the basic conservation
laws for momentum and energy.

On a practical front, it is trivial to reach high
altitude, just launch from a balloon, they need no
power at all. The hard part is reaching orbital
velocity once you get up there, and lifters aren't
going to work well in a near vacuum.

the only benefit they give you is the reaction mass
of the surrounding air at low altitude but this
advantage will soon be outweighed by the inefficiency
of burning fuel to run a generator which in turn
powers the lifter as the air gets thinner.

If you look at the pages describing the drive, the electrical power
is
a means to create airflow. The thrust is due to this airflow.

Or as I said the lift comes from "the reaction
mass of the surrounding air".

I'm
calculating the thrust as you do with the rocket equation.

I haven't seen you calculate the thrust, you
just quoted the page. You are making the same
mistake you did last time with the pipe and
nozzle, you have read the static force that
has been claimed. You are again ignoring the
simple fact that once it starts moving you
have to supply more power for the same force.

The value in the table you quote is like
tethering a helicopter to the runway with a
spring balance in between. It will tell you
how hard the helicopter can pull if it isn't
rising. Release the tether and that force
divided by the mass will give the initial
upwards acceleration of the helicopter, but
you cannot assume the acceleration will stay
constant until the helicopter is moving
vertically at Mach 15 !

For constant force, power is proportional to
speed, and speed doesn't even appear in the
table you cited. They have only tested the
static case.

If you want to calculate thrust, work out how
long the ions will be between the plates at
different air speeds and hence how much
reaction there will be. If it doesn't fall
with speed, you have missed something.

What would be dependent on the speed is the drag. That would be a
rather complicated dependence on the shape of the vehicle. As a first
guess you could give the craft the aerodynmic shape of a supersonic
or
hypersonic vehicle and use the same type of intakes on those
vehicles.

Drag is irrelevant, if the device works by
accelerating ionised air, what happens when
it exceeds the speed of sound? Think about=20
the basics Bob.

George

Dishman wrote:
I'm
calculating the thrust as you do with the rocket equation.

I haven't seen you calculate the thrust, you
just quoted the page. You are making the same
mistake you did last time with the pipe and
nozzle, you have read the static force that
has been claimed. You are again ignoring the
simple fact that once it starts moving you
have to supply more power for the same force.

The value in the table you quote is like
tethering a helicopter to the runway with a
spring balance in between. It will tell you
how hard the helicopter can pull if it isn't
rising. Release the tether and that force
divided by the mass will give the initial
upwards acceleration of the helicopter, but
you cannot assume the acceleration will stay
constant until the helicopter is moving
vertically at Mach 15 !

For constant force, power is proportional to
speed, and speed doesn't even appear in the
table you cited. They have only tested the
static case.


The thrust and power calculations are analogous to those for a jet or
a rocket. The power and thrust produced by the engines is going to be
the same no matter the mass of the craft or its speed. If you get a
craft with a very high payload its acceleration could be close to zero,
so the speed the craft attains could be small. But the thrust produced
by the engines will be the same and the power of the engines as
measured by the mass of the reaction mass moved (air or propellant) and
the speed of that reaction mass will be the same. Likewise the thrust
and power for the lifter is calculated from how much the mass of air is
accelerated with respect to the craft.
However, there is a dependence on density of air for the size of the
thrust for the lifter. For a gas, density is dependent on pressure. And
by Bernoulli's principle pressure decreases at high velocity, so it is
possible that the thrust produced will be less on that basis. I'll
check on that.


Bob Clark

"Robert Clark" wrote in message
oups.com...
Dishman wrote:
I'm
calculating the thrust as you do with the rocket equation.

I haven't seen you calculate the thrust, you
just quoted the page. You are making the same
mistake you did last time with the pipe and
nozzle, you have read the static force that
has been claimed. You are again ignoring the
simple fact that once it starts moving you
have to supply more power for the same force.

The value in the table you quote is like
tethering a helicopter to the runway with a
spring balance in between. It will tell you
how hard the helicopter can pull if it isn't
rising. Release the tether and that force
divided by the mass will give the initial
upwards acceleration of the helicopter, but
you cannot assume the acceleration will stay
constant until the helicopter is moving
vertically at Mach 15 !

For constant force, power is proportional to
speed, and speed doesn't even appear in the
table you cited. They have only tested the
static case.

The thrust and power calculations are analogous to those for a jet or
a rocket. The power and thrust produced by the engines is going to be
the same no matter the mass of the craft or its speed.

Learn the basic physics Bob, the power needed
is speed times mass. After you have accelerated
an object, it has kinetic and/or potential
energy that can be recovered. Conservation of
energy means you must supply that and in your
suggestion, that has to come from the electrical
feed, there is no other source.

George

Robert Clark wrote:
The thrust and power calculations are analogous to those for a jet
or
a rocket. The power and thrust produced by the engines is going to be
the same no matter the mass of the craft or its speed. If you get a
craft with a very high payload its acceleration could be close to
zero,
so the speed the craft attains could be small.

I dont think that is correct.

My understanding of how it works is that a (relatively) stationary
nitrogen atom is ionised near one of the terminals. This is then
accelerated towards the other terminal. When it hits the other
terminal it is neutralised.

While the ion is in motion, the terminals are applying a force on the
ion. The ion is slowed down by collisions with the air. The slower
the ion moves, the longer the force is applied. This means that
ideally you want the ion to move as slowly as possible.

However, if the air is moving, then the interaction with the air will
never reduce the speed of the ion to below that amount.

The effective speed of the ion would probably be something like

Veff = V + Vstationary

Vstationary is the speed the ion moves at when the system is hovering
and V is the speed of the lifter.

The thrust is inversely proportional to the ion speed giving something
like

T = Tstationary * ( Vstationary) / ( V + Vstationary )

(feel free to take the above with as many grains of salt as you want
)

Anyway, assuming that the ion moves very quickly during hover, then
Vstationary will be alot bigger than V. This means that the thrust
will only reduce slowly as the lifter increases in speed. OTOH, if
Vstationary is low, then increasing speed will greatly reduce thrust.

wrote in message
oups.com...

Robert Clark wrote:
The thrust and power calculations are analogous to those for a jet
or
a rocket. The power and thrust produced by the engines is going to be
the same no matter the mass of the craft or its speed. If you get a
craft with a very high payload its acceleration could be close to
zero,
so the speed the craft attains could be small.

I dont think that is correct.

My understanding of how it works is that a (relatively) stationary
nitrogen atom is ionised near one of the terminals. This is then
accelerated towards the other terminal. When it hits the other
terminal it is neutralised.

While the ion is in motion, the terminals are applying a force on the
ion. The ion is slowed down by collisions with the air. The slower
the ion moves, the longer the force is applied. This means that
ideally you want the ion to move as slowly as possible.

Right, and it also means that the energy given to the
ion is shared with the other molecules increasing the
effective mass. See the discussion on the page Robert
quoted where they find the effective mass is much
greater than individual molecules could account for.

The benefit is that momentum is proportional to speed
while kinetic energy is propotional to the square so
a lower speed means more thrust per watt.

However, if the air is moving, then the interaction with the air will
never reduce the speed of the ion to below that amount.

Also, if the air is moving quickly, there is less time
to accelerate the ions. There are many reasons why the
thrust will drop off rapidly with increasing speed.

George

George Dishman wrote:
wrote in message
oups.com...

Robert Clark wrote:
The thrust and power calculations are analogous to those for a
jet
or
a rocket. The power and thrust produced by the engines is going to
be
the same no matter the mass of the craft or its speed. If you get
a
craft with a very high payload its acceleration could be close to
zero,
so the speed the craft attains could be small.

I dont think that is correct.

My understanding of how it works is that a (relatively) stationary
nitrogen atom is ionised near one of the terminals. This is then
accelerated towards the other terminal. When it hits the other
terminal it is neutralised.

While the ion is in motion, the terminals are applying a force on
the
ion. The ion is slowed down by collisions with the air. The
slower
the ion moves, the longer the force is applied. This means that
ideally you want the ion to move as slowly as possible.

Right, and it also means that the energy given to the
ion is shared with the other molecules increasing the
effective mass. See the discussion on the page Robert
quoted where they find the effective mass is much
greater than individual molecules could account for.

The benefit is that momentum is proportional to speed
while kinetic energy is propotional to the square so
a lower speed means more thrust per watt.

However, if the air is moving, then the interaction with the air
will
never reduce the speed of the ion to below that amount.

Also, if the air is moving quickly, there is less time
to accelerate the ions. There are many reasons why the
thrust will drop off rapidly with increasing speed.

George

From that page the drift velocity of the ion is around 2.8 km/s
(assumes 40kV and 3 cm spacing). If that is true, then the drop in
thrust at orbital velocity of around 8km/s would be ( 2.8 ) / ( 2.8+8)
= 26% of normal thrust, so maybe not fatal.

George Dishman wrote:
"Robert Clark" wrote in message
oups.com...
Dishman wrote:
I'm
calculating the thrust as you do with the rocket equation.

I haven't seen you calculate the thrust, you
just quoted the page. You are making the same
mistake you did last time with the pipe and
nozzle, you have read the static force that
has been claimed. You are again ignoring the
simple fact that once it starts moving you
have to supply more power for the same force.

The value in the table you quote is like
tethering a helicopter to the runway with a
spring balance in between. It will tell you
how hard the helicopter can pull if it isn't
rising. Release the tether and that force
divided by the mass will give the initial
upwards acceleration of the helicopter, but
you cannot assume the acceleration will stay
constant until the helicopter is moving
vertically at Mach 15 !

For constant force, power is proportional to
speed, and speed doesn't even appear in the
table you cited. They have only tested the
static case.

The thrust and power calculations are analogous to those for a jet
or
a rocket. The power and thrust produced by the engines is going to
be
the same no matter the mass of the craft or its speed.

Learn the basic physics Bob, the power needed
is speed times mass. After you have accelerated
an object, it has kinetic and/or potential
energy that can be recovered. Conservation of
energy means you must supply that and in your
suggestion, that has to come from the electrical
feed, there is no other source.

George

How are you measuring the power of the engines? According to the mass
of the craft and speed of the craft?
Clearly after the engines start on the space shuttle but it is still
latched to the launch pad, the engines are generating a tremendous
amount of thrust and power but the speed of the vehicle is zero.


Bob Clark

wrote:
George Dishman wrote:

Also, if the air is moving quickly, there is less time
to accelerate the ions. There are many reasons why the
thrust will drop off rapidly with increasing speed.

For ref, the page in queston is:

http://jnaudin.free.fr/html/lftheory.htm

From that page the drift velocity of the ion is around 2.8 km/s
(assumes 40kV and 3 cm spacing).

I make it 280m/s.

If that is true, then the drop in
thrust at orbital velocity of around 8km/s would be ( 2.8 ) / (
2.8+8)
= 26% of normal thrust, so maybe not fatal.

My value would result in a 50% reduction at 626mph
other factors being equal. However I don't think
that is the whole story.

The equation assumes the charge moves linearly from
source to collector but with an airflow the motion
will be the vector sum of the air velocity and the
drift velocity.

The method relies on the asymmetry of a small
source creating a high field which produces the
ionisation together with a large collector to
avoid a reverse flow. Thinking of the toy example
of a tip and ring, ions from the tip will be swept
through the ring when the air is moving resulting
in a build-up of charge _behind_ the device which
will reduce its efficiency more rapidly than the
static equation. This would obviously depend on
the design but at vehicle speeds above the drift
velocity the charge once past the collector can
never catch up, so there is a good chance that the
drift speed would be a top limit.

George

Robert Clark wrote:
George Dishman wrote:
Learn the basic physics Bob, the power needed
is speed times mass. After you have accelerated
an object, it has kinetic and/or potential
energy that can be recovered. Conservation of
energy means you must supply that and in your
suggestion, that has to come from the electrical
feed, there is no other source.

How are you measuring the power of the engines? According to the
mass
of the craft and speed of the craft?

The mass and speed of the craft give you the power
needed to supply the kinetic energy that the craft
is gaining. Dividing the power supplied to the
engines by that figure is a measure of efficiency.

Clearly after the engines start on the space shuttle but it is still
latched to the launch pad, the engines are generating a tremendous
amount of thrust and power but the speed of the vehicle is zero.

Correct, you are supplying lots of power which all
goes into heating an accelerating the exhaust gas
while none goes into craft kinetic energy. You have
an efficiency of zero.

Using the mass and speed gives you a lower limit and
the engine power will always be higher because no
design has 100% efficiency. For example, after the
lifter has passed some point, the air that was there
has been accelerated downwards and probably heated a
bit through the ionisation process, both of which are
losses in the system. Drag losses are on top of that.

George

Dishman wrote:
wrote:
George Dishman wrote:

Also, if the air is moving quickly, there is less time
to accelerate the ions. There are many reasons why the
thrust will drop off rapidly with increasing speed.

For ref, the page in queston is:

http://jnaudin.free.fr/html/lftheory.htm

From that page the drift velocity of the ion is around 2.8 km/s
(assumes 40kV and 3 cm spacing).

I make it 280m/s.

I may have messed up the units:

w = k*E/d

E = 40kV
k = 2.1(cm^2)/Vs
d = 30mm = 3cm

w = 40kV*2.1(cm^2)/Vs/3cm = 40*2.1/3*(kV*cm^2)/(Vs)/(cm) =

40*2.1/3 k(cm)/s = 28000 cm/s = 2800 m/s


My value would result in a 50% reduction at 626mph
other factors being equal. However I don't think
that is the whole story.

The equation assumes the charge moves linearly from
source to collector but with an airflow the motion
will be the vector sum of the air velocity and the
drift velocity.

Agreed, but that is what the equation takes into account. The
effective speed is Vair + Vstationary. The point was that drift speed
is probably related to the maximum speed. I was just trying to get a
BOE value.

Also, drag against surfaces increases by v^2 so you would have to go to
less dense air for high speed. This would mean that the drift speed
would increase due to less interactions with other gas molecules.

Since drag is proportional to gas density, it is possible that that
wouldn't help anyway. You would get a decrease in drag and also a
decrease in thust in proportion.