Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.
Dismiss

# Mathematics of reactionless drives?

381 views

### Nyrath the nearly wise

May 6, 2005, 12:00:18 PM5/6/05
to
This looks straightforwards to me, but so many
things with a similar appearance have blown up in my
face before.

Say you have an honest-to-John-Campbell reactionless drive,
that is, a black box that you feed electricity into
and the box convertes it into linear acceleration
without emitting any exhaust or otherwise using
Newton's laws.

How much electricity will it need?

It looks like all you have to do is mathematically
convert the energy of thrust into watts, and
assign some value for the efficiency of the black box.
Is that it, or are there further complications?

### chorned...@hushmail.com

May 6, 2005, 12:07:44 PM5/6/05
to

Yes, there is a complication.

How will the energy spent on changing the kinetic energy change when
you change the frame of observer?

### Mark Fergerson

May 6, 2005, 12:11:16 PM5/6/05
to
chorned...@hushmail.com wrote:
> Nyrath the nearly wise wrote:
>
>>This looks straightforwards to me, but so many
>>things with a similar appearance have blown up in my
>>face before.
>>
>>Say you have an honest-to-John-Campbell reactionless drive,
>>that is, a black box that you feed electricity into
>>and the box convertes it into linear acceleration
>>without emitting any exhaust or otherwise using
>>Newton's laws.
>>
>>How much electricity will it need?
>>
>>It looks like all you have to do is mathematically
>>convert the energy of thrust into watts, and
>>assign some value for the efficiency of the black box.
>>Is that it, or are there further complications?

Sounds about right to me, ignoring such mundanities (is that a
word?) as waste heat.

> Yes, there is a complication.
>
> How will the energy spent on changing the kinetic energy change when
> you change the frame of observer?

How is that different from an ordinary rocket?

Mark L. Fergerson

### IsaacKuo

May 6, 2005, 1:01:59 PM5/6/05
to

Mark Fergerson wrote:
>chorned...@hushmail.com wrote:
>>Nyrath the nearly wise wrote:

>>>This looks straightforwards to me, but so many
>>>things with a similar appearance have blown up in my
>>>face before.

>>>Say you have an honest-to-John-Campbell reactionless drive,
>>>that is, a black box that you feed electricity into
>>>and the box convertes it into linear acceleration
>>>without emitting any exhaust or otherwise using
>>>Newton's laws.

>>>How much electricity will it need?

>>>It looks like all you have to do is mathematically
>>>convert the energy of thrust into watts, and
>>>assign some value for the efficiency of the black box.
>>>Is that it, or are there further complications?

Yes. There are basically two approaches you can use:

1. There is a special frame of reference. In this
case, the "reactionless" drive is really pushing
against an infinitely massive special frame of
reference.

or

2. There is no special frame of reference. In this
case, the only way to sort of preserve
conservation of energy is to limit drive
efficiency to that of a photon drive. This is
not a very useful drive, though, since it has
the same (low) performance as a photon drive.

>>Yes, there is a complication.

>>How will the energy spent on changing the kinetic
>>energy change when you change the frame of observer?

>How is that different from an ordinary rocket?

Because an ordinary rocket has a "reaction". The
amount of kinetic energy added to the rocket by
a rocket thrust depends upon what frame of reference
you look at it. Indeed, there are plenty of frames
of reference where the rocket thrust subtracts
kinetic energy from the rocket! So you can't
meaningfully talk about THE amount of kinetic
energy added to the rocket. However, you CAN
meaningfully talk about how much kinetic energy the
rocket adds to the system because kinetic energy
is also added to (or subtracted from) the rocket
exhaust. No matter what frame of reference you use,
the total amount of kinetic energy in the rocket
plus the exhaust is increased by the same amount.

Isaac Kuo

### Hop David

May 6, 2005, 1:43:25 PM5/6/05
to

Nyrath the nearly wise wrote:

> This looks straightforwards to me, but so many
> things with a similar appearance have blown up in my
> face before.
>
> Say you have an honest-to-John-Campbell reactionless drive,
> that is, a black box that you feed electricity into
> and the box convertes it into linear acceleration
> without emitting any exhaust or otherwise using
> Newton's laws.

Tide locked tethers have been proposed. That is, the lower end of a
tether always points toward earth and the upper end points to the sky (a
tether enthusiast described our moon as such a tether, albeit a large
fat one). If the tether conducts electricity, running a current can push
against earth's magnetic field. I've also seen such tethers proposed for
Jupiter orbit as Jupiter has a healthy magnetic field.

Some proponents call this a reactionless drive. But it's probably not

>
> How much electricity will it need?
>
> It looks like all you have to do is mathematically
> convert the energy of thrust into watts, and
> assign some value for the efficiency of the black box.
> Is that it, or are there further complications?

--
Hop David
http://clowder.net/hop/index.html

### Damien R. Sullivan

May 6, 2005, 1:52:55 PM5/6/05
to

I'd probably either just looked at the kinetic energy of the ship, or use
E = pc, p being momentum, this being the formula for light and Robert
Forward's energy-momentum-conversion speculations.

-xx- Damien X-)

### Erik Max Francis

May 6, 2005, 6:42:17 PM5/6/05
to
Nyrath the nearly wise wrote:

If you're talking about a "true" reactionless drive, where energy is
converted directly into momentum (or angular momentum), then there are
lots of complications. Consider for instance that kinetic energy goes
as the square of the speed:

K = (1/2) m v^2.

So the power P you need to accelerate is dK/dt:

dK/dt = (1/2) m [2 v dv/dt] = m v a.

As you can see, the power is a function of not only the acceleration
that you want (which seems obvious), but also the _speed_ at which
you're currently travelling. The snag there is that your current speed
accelerating along nicely. At that point you pass someone who is
already coasting at nearly the same speed you are. He sees you using
much less power! Who's right?

The solution is that you either need to play by the rules of the game
and use reaction drives (even if it's just reaction momentum, like a
photon drive), or posit a special frame in violation of special
relativity. With the special frame, now there's a "correct" frame where
all the kinetic energy calculations are "official" and everyone agrees
on them.

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Little things / Cut like knives / Hurt and sting
-- Anggun

### John Schilling

May 6, 2005, 6:55:23 PM5/6/05
to
In article <2005050610...@eris.io.com>, Nyrath the nearly wise says...

There is the complication that "energy of the thrust"
is as meaningless a phrase as, e.g., "mass of the time".

Thrust is a force, not an energy. Force, *multiplied
by distance*, gives an energy. A force of one pound,
applied as an object moves over a distance of one foot,
equates to (unsurprisingly) one foot-pound of energy.
The same force, over a greater or lesser distance, comes
to proportionately more or less energy.

In MKS, by the way, that would be one Newton of force
over one meter of distance equals one Joule of energy.
If we assume constant force and motion, we can extend
that to one Newton of force applied constantly at a
velocity of one meter per second, equals a power of
one watt.

The question is, velocity *relative to what?*

If it is a rocket, the relevant velocity is that of the
rocket's own exhaust relative to the rocket itself. For
an "intertialess thruster", the answer isn't clear and
the power or energy associated with a given thrust will
change widely depending on what reference frame you use

Which is one facet of the reason "inertialess thrusters"
seem to be physically nonsensical. However, if you really
need one for some SFnal purpose, you could try either

A: the one universally invariant velocity in real physics.
That being the velocity of light, giving you a figure of
three hundred megawatts of power per Newton of thrust. A
tad high for most purposes, I thing, and functionally
equivalent to saying your thruster is a photon drive or
a (nearly-)massless-neutrino drive or a Dark Energy Rocket
or whatever.

B: the velocity of the spacecraft relative to some absolute
reference frame. Either a cosmic absolute, or a local
absolute tied e.g. to the nearest massive body or bodies
in whatever manner is most convenient to the story. This
is functionally equivalent to the old aetheric theories,
and you can mine those for ideas.

--
*John Schilling * "Anything worth doing, *
*Member:AIAA,NRA,ACLU,SAS,LP * is worth doing for money" *
*Chief Scientist & General Partner * -13th Rule of Acquisition *
*White Elephant Research, LLC * "There is no substitute *
*schi...@spock.usc.edu * for success" *
*661-951-9107 or 661-275-6795 * -58th Rule of Acquisition *

### Aaron Denney

May 6, 2005, 7:29:25 PM5/6/05
to
On 2005-05-06, Erik Max Francis <m...@alcyone.com> wrote:
> K = (1/2) m v^2.
>
> So the power P you need to accelerate is dK/dt:
>
> dK/dt = (1/2) m [2 v dv/dt] = m v a.
>
> As you can see, the power is a function of not only the acceleration
> that you want (which seems obvious), but also the _speed_ at which
> you're currently travelling. The snag there is that your current speed
> is frame dependent. Consider that you're already doing your job and
> accelerating along nicely. At that point you pass someone who is
> already coasting at nearly the same speed you are. He sees you using
> much less power! Who's right?
>
> The solution is that you either need to play by the rules of the game
> and use reaction drives (even if it's just reaction momentum, like a
> photon drive), or posit a special frame in violation of special
> relativity. With the special frame, now there's a "correct" frame where
> all the kinetic energy calculations are "official" and everyone agrees
> on them.

Ah, this explains why my reactionless drive works so poorly:
The preferred frame is moving very quickly relative to Earth!

--
Aaron Denney
-><-

### John Schilling

May 6, 2005, 7:09:48 PM5/6/05
to
In article <CGMee.3576\$Fa1.120@fed1read02>, Mark Fergerson says...

In an ordinary rocket, both the kinetic energy of the rocket and the
kinetic energy of the exhaust will change. Different observers will
disagree about the absolute change of each, but will agree about the
*net* change in kinetic energy, and so energy conservation can be
enforced.

Example: A hundred-kilogram satellite ejects one gram of nitrogen
through a cold-gas thruster at a velocity, relative to the spacecraft,
of one hundred meters per second.

An observer at rest relative to the initial position of the spacecraft
will see it accelerate to 0.001 meters per second, increasing its kinetic
energy by 0.05 millijoules. The exhaust will be observed to accelerate
to 99.9995 meters per second, with resulting kinetic energy of 4.99995
Joules. The total kinetic energy increase, ,provided by the expanding gas,
comes to 5 Joules.

An observer zipping along in the opposite direction at 1,000,000 meters
per second, will see both the spacecraft and the propellant as having had
an initial velocity of 1,000,000 meters per second, and an initial kinetic
energy of 50,000,000,000,000 Joules and 500,000,000 Joules, respectively.
The spacecraft accelerates to 1,000,000.001 meters per second, giving it
a new kinetic energy of 50,000,000,100,000 Joules - a gain of 100,000
Joules. Far cry from the .05 millijoules the stationary observer had
thought the spacecraft acquired.

But the moving observer will have seen the slug of exhaust gas *decelerate*
from 1,000,000 m/s to 999,900.0005 meters per second, with a new kinetic
energy of 499,900,005 Joules. That's a loss of 99,995 Joules. So the
net change in energy is, spacecraft +100,000.0, exhaust -99,995.0, or
plus 5.0 Joules. Both observers agree on conservation of energy. And,
for that matter, momentum.

If there were only the spacecraft involved, they'd be arguing about the
missing hundred kilojoules.

### Paul F. Dietz

May 6, 2005, 7:45:56 PM5/6/05
to
Erik Max Francis wrote:

> With the special frame, now there's a "correct" frame where
> all the kinetic energy calculations are "official" and everyone agrees
> on them.

You can call that frame the 'road', because you're effectively
reacting against it (as if it had very large mass). So it really
isn't reactionless after all.

Paul

### Paul F. Dietz

May 6, 2005, 7:47:07 PM5/6/05
to
Aaron Denney wrote:

> Ah, this explains why my reactionless drive works so poorly:
> The preferred frame is moving very quickly relative to Earth!

Ah, but that means you should be able to run it backwards,
and have a reactionless generator!

Paul

### Raghar

May 7, 2005, 7:55:52 AM5/7/05
to
Nyrath the nearly wise <nyr...@projectrho.com> wrote in
news:2005050610...@eris.io.com:

The real problem is, could reactionless drives change direction
without changing orientation?

### IsaacKuo

May 7, 2005, 10:51:51 AM5/7/05
to

In any case, if you have this sort of "reactionless" drive, I
think it's far more interesting to think of the possibilities
of a "reactionless" BRAKE than a "reactionless" drive. For
instance, you can boost off of Earth 12 out of 24 hours by
applying the brake. At that point, you're in a highly eliptical
orbit. The Earth's and/or the Moon's gravity can the "bend"
your trajectory towards where you want to go. You probably
want to make timed applications of the brake to alter the shape
of your orbit depending on where you want to go.

Figuring out the sort of navigational maneuvers you'd use with
an "aether brake" is rather interesting.

Isaac Kuo

### Chuck Stewart

May 7, 2005, 1:06:54 PM5/7/05
to
On Sat, 07 May 2005 07:51:51 -0700, IsaacKuo wrote:

> Figuring out the sort of navigational maneuvers you'd use with an "aether
> brake" is rather interesting.

Anti-Acceleration Drive
"Black Destroyer" A. E. Von Vogt (1939)

> Isaac Kuo

--
Chuck Stewart
"Anime-style catgirls: Threat? Menace? Or just studying algebra?"

May 7, 2005, 1:24:14 PM5/7/05
to
On Fri, 06 May 2005 10:43:25 -0700, Hop David
<hopspageHA...@tabletoptelephone.com> wrote:

>
>
>Nyrath the nearly wise wrote:
>> This looks straightforwards to me, but so many
>> things with a similar appearance have blown up in my
>> face before.
>>
>> Say you have an honest-to-John-Campbell reactionless drive,
>> that is, a black box that you feed electricity into
>> and the box convertes it into linear acceleration
>> without emitting any exhaust or otherwise using
>> Newton's laws.
>
>Tide locked tethers have been proposed. That is, the lower end of a
>tether always points toward earth and the upper end points to the sky (a
>tether enthusiast described our moon as such a tether, albeit a large
>fat one). If the tether conducts electricity, running a current can push
>against earth's magnetic field.

That's not reactionless. Pushing on the Earth's magnetic field
pushes on the Earth.

Demonstration: Hold magnet A in your hand with the north pole
pointing out. Hold magnet B with its north pole near magnet A's north
field. You'll feel a force on magnet B pushing away from magnet A.

>I've also seen such tethers proposed for
>Jupiter orbit as Jupiter has a healthy magnetic field.
>
>Some proponents call this a reactionless drive.

I certainly don't see how.

>But it's probably not
>
>>
>> How much electricity will it need?
>>
>> It looks like all you have to do is mathematically
>> convert the energy of thrust into watts, and
>> assign some value for the efficiency of the black box.
>> Is that it, or are there further complications?

May 7, 2005, 2:06:17 PM5/7/05
to
On Fri, 6 May 2005 11:00:18 -0500, Nyrath the nearly wise
<nyr...@projectrho.com> wrote:

>This looks straightforwards to me, but so many
>things with a similar appearance have blown up in my
>face before.
>
>Say you have an honest-to-John-Campbell reactionless drive,
>that is, a black box that you feed electricity into
>and the box convertes it into linear acceleration
>without emitting any exhaust or otherwise using
>Newton's laws.
>
>How much electricity will it need?

An infinite amount. :)

If I had to answer it, I'd calculate it as if it were an electic
car on Earth, discounting tire friction and air resistance (presuming
you're in the vacuum of outer space), which seems to be the same thing
you decsribe below:

>It looks like all you have to do is mathematically
>convert the energy of thrust into watts, and
>assign some value for the efficiency of the black box.
>Is that it, or are there further complications?

Mainly that this is a fictional device.

I have a strong urge to make something "based in reality," no
matter how tenuous that reality. Have it generate and spit out
neutrinos (and not equally in all directions) or something. I also
feel better with the 'ether brake' or something similar that you can
claim it IS a reaction drive, and it's pushing against every particle
in the Universe. But this is still a problem, you've replaced one kind
of magic (reactionless drive) with another kind (a star-trek-like
tractor beam that shines on everything outside the ship), both things
that have no explanation when you keep asking how it works.

Speaking of the Dork Side, and for those who don't already read
slashdot:
http://hardware.slashdot.org/article.pl?sid=05/05/05/1535230

### Dr John Stockton

May 7, 2005, 6:30:10 PM5/7/05
to
JRS: In article <2005050610...@eris.io.com>, dated Fri, 6 May
2005 11:00:18, seen in news:rec.arts.sf.science, Nyrath the nearly wise
<nyr...@projectrho.com> posted :

>
>Say you have an honest-to-John-Campbell reactionless drive,
>that is, a black box that you feed electricity into
>and the box convertes it into linear acceleration
>without emitting any exhaust or otherwise using
>Newton's laws.
>
>How much electricity will it need?
>
>It looks like all you have to do is mathematically
>convert the energy of thrust into watts, and
>assign some value for the efficiency of the black box.
>Is that it, or are there further complications?

Since a reactionless drive conflicts fundamentally with our present
understanding of the universe, it is in conflict with all that we know

One might imagine an apparently-reactionless drive, one that pushed
against the ultimate space-time foundations of the entire universe; the
energy used would then just be the kinetic energy gained (with respect
to those foundations) divided by the drive efficiency (assuming the mass
of the vehicle to be negligible in comparison with that of the
universe). Unfortunately, there do not yet seem to be any such
foundations.

--
Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
Do not Mail News to me. Before a reply, quote with ">" or "> " (SoRFC1036)

### Mark Fergerson

May 7, 2005, 11:05:08 PM5/7/05
to
IsaacKuo wrote:
> Mark Fergerson wrote:
>
>>chorned...@hushmail.com wrote:
>>
>>>Nyrath the nearly wise wrote:

>>>>This looks straightforwards to me, but so many
>>>>things with a similar appearance have blown up in my
>>>>face before.

>>>>Say you have an honest-to-John-Campbell reactionless drive,
>>>>that is, a black box that you feed electricity into
>>>>and the box convertes it into linear acceleration
>>>>without emitting any exhaust or otherwise using
>>>>Newton's laws.

>>>>How much electricity will it need?

>>>>It looks like all you have to do is mathematically
>>>>convert the energy of thrust into watts, and
>>>>assign some value for the efficiency of the black box.
>>>>Is that it, or are there further complications?

> Yes. There are basically two approaches you can use:
>
> 1. There is a special frame of reference. In this
> case, the "reactionless" drive is really pushing
> against an infinitely massive special frame of
> reference.

Sorry, I forgot to mention that this was my assumption; one of
Campbell's favorite phrases for such drives was "digs its fingers
and toes into the very warp and weft of spacetime and _pulls_!".
This sounds Machian to me, hence when your ship moves _this_ way,
the entire rest of the Universe moves _that_ way a teensy bit. The
ship/Universe center-of-mass doesn't, though; where's it got to go?

> or
>
> 2. There is no special frame of reference. In this
> case, the only way to sort of preserve
> conservation of energy is to limit drive
> efficiency to that of a photon drive. This is
> not a very useful drive, though, since it has
> the same (low) performance as a photon drive.

Then it's emitting _something_ that carries momentum away, hence
isn't "really" reactionless.

>>>Yes, there is a complication.

>>>How will the energy spent on changing the kinetic
>>>energy change when you change the frame of observer?

>>How is that different from an ordinary rocket?

> Because an ordinary rocket has a "reaction". The
> amount of kinetic energy added to the rocket by
> a rocket thrust depends upon what frame of reference
> you look at it.

The ship/Universe COM.

> No matter what frame of reference you use,
> the total amount of kinetic energy in the rocket
> plus the exhaust is increased by the same amount.

Not if the Machian assumption is valid; observers in all frames
of reference will have to figure in the "Machian dragging" induced
on them. No, we don't seem to live in that kind of Universe, but
that wasn't the original question.

Mark L. Fergerson

### Aaron Denney

May 8, 2005, 5:16:56 AM5/8/05
to

Are you crazy? The amount of energy delivered before it burnt out would
be like a small nuke. No way am I trying that.

May 13, 2005, 5:33:07 PM5/13/05
to
>Say you have an honest-to-John-Campbell reactionless drive,

Then you can build a perpetual motion machine that produces energy.

Clearly, you either have to abandon conservation of energy, in which
case the efficiency of the drive becomes irrelevant, or the energy consumed
by the drive must increase to make conservation of energy still work,
which means that "efficiency" for the drive becomes dependent upon totally
other things and, effectively, a meaningless term (because it will be whatever
is needed to conserve energy globally).

If you want a universe with reactionless drives, you are best off abandoning
conservation of energy for the laws of physics in that creation.

Now for story purposes, to rate various drive implementations, consider "effectiveness"
rather than efficiency. Where effectiveness may have something to do with controllability,
steerability, transient demands of electricity (or whatever) or perhaps the use of
some rare element or compound or crystal to make the device useable. Like in "Seetee"
where the ships used "tuning diamonds" and higher quality diamonds allowed faster travel,
lifting greater masses &c.

May 13, 2005, 5:37:18 PM5/13/05
to
>Yes. There are basically two approaches you can use:
>
>1. There is a special frame of reference. In this
> case, the "reactionless" drive is really pushing
> against an infinitely massive special frame of
> reference.
>
>or
>
>2. There is no special frame of reference. In this
> case, the only way to sort of preserve
> conservation of energy is to

or

3. No special frame of reference AND abandon conservation
of energy.
Of course, at that point, it may be more aptly classified as
fantasy rather than SF.

### Bob Lyle

May 19, 2005, 3:54:05 AM5/19/05
to
On Sat, 7 May 2005 23:30:10 +0100, Dr John Stockton
<sp...@merlyn.demon.co.uk> wrote:

>
>One might imagine an apparently-reactionless drive, one that pushed
>against the ultimate space-time foundations of the entire universe; the
>energy used would then just be the kinetic energy gained (with respect
>to those foundations) divided by the drive efficiency (assuming the mass
>of the vehicle to be negligible in comparison with that of the
>universe). Unfortunately, there do not yet seem to be any such
>foundations.
>

http://www.grc.nasa.gov/WWW/bpp/pdf/Cramer-JPC.pdf

### sigi...@yahoo.com

May 19, 2005, 9:38:57 AM5/19/05
to

Erik Max Francis wrote:

> As you can see, the power is a function of not only the acceleration
> that you want (which seems obvious), but also the _speed_ at which
> you're currently travelling. The snag there is that your current
speed
> is frame dependent.

[...]

> The solution is that you either need to play by the rules of the game

> and use reaction drives (even if it's just reaction momentum, like a
> photon drive), or posit a special frame in violation of special
> relativity.

What if you choose a special frame that's not in violation of special
relativity?

For instance, suppose each drive must have an "anchor" unit. We can
visualize this as a small sphere of unobtainium, sitting quietly on a
shelf somewhere. Speed is always measured relative to the anchor.

Would that work?

Doug M.

### Erik Max Francis

May 19, 2005, 11:32:02 AM5/19/05
to
Bob Lyle wrote:

We're not even sure our Universe is Machian. Einstein, for instance,
was inspired by Mach's principle, but general relativity itself is not a
Machian theory.

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

Love is, above all, the gift of oneself.
-- Jean Anouilh

### Erik Max Francis

May 19, 2005, 11:33:34 AM5/19/05
to
sigi...@yahoo.com wrote:

> What if you choose a special frame that's not in violation of special
> relativity?
>
> For instance, suppose each drive must have an "anchor" unit. We can
> visualize this as a small sphere of unobtainium, sitting quietly on a
> shelf somewhere. Speed is always measured relative to the anchor.
>
> Would that work?

No. What we're talking about is what determines what _other_ frames see
as power expenditures in order to move the ship. If they see kinetic
energy changes which indicate that the ship is accelerating relative to
some fixed frame, then that frame must be special, regardless of how it
got that way.

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

### Michael Ash

May 19, 2005, 1:56:39 PM5/19/05
to
Erik Max Francis <m...@alcyone.com> wrote:
> sigi...@yahoo.com wrote:
>
>> What if you choose a special frame that's not in violation of special
>> relativity?
>>
>> For instance, suppose each drive must have an "anchor" unit. We can
>> visualize this as a small sphere of unobtainium, sitting quietly on a
>> shelf somewhere. Speed is always measured relative to the anchor.
>>
>> Would that work?
>
> No. What we're talking about is what determines what _other_ frames see
> as power expenditures in order to move the ship. If they see kinetic
> energy changes which indicate that the ship is accelerating relative to
> some fixed frame, then that frame must be special, regardless of how it
> got that way.

It seems to me that it ought to be possible to use the OP's scenario with
some modifications. Instead of the anchor simply defining a frame, it acts
as reaction mass. The ship gets a force in one direction, and the anchor
gets the same force in the opposite direction. This is pretty much what
happens with the EM force in a rocket, just with a much greater distance.
Instead of leaving the anchor on a shelf, you bury it in a whole bunch of
concrete. I have a feeling that this is violating something too, but I
can't see it; where would this scheme break known physics?

May 19, 2005, 7:14:33 PM5/19/05
to
Michael Ash <mi...@mikeash.com> wrote:

> It seems to me that it ought to be possible to use the OP's scenario
> with some modifications. Instead of the anchor simply defining a
> frame, it acts as reaction mass. The ship gets a force in one
> direction, and the anchor gets the same force in the opposite
> direction. This is pretty much what happens with the EM force in a
> rocket, just with a much greater distance. Instead of leaving the
> anchor on a shelf, you bury it in a whole bunch of concrete. I have a
> feeling that this is violating something too, but I can't see it;
> where would this scheme break known physics?

Or 'tractor beams' the ship can use to lock onto distant objects
and pull itself towards them (and objects toward it), slinging
along like some sort of interplanetary Spiderman...

--
>;k

### Anthony Buckland

May 20, 2005, 9:38:04 AM5/20/05
to
Michael Ash wrote:

> ...

>
> Instead of the anchor simply defining a frame, it acts
>as reaction mass. The ship gets a force in one direction, and the anchor
>gets the same force in the opposite direction. This is pretty much what
>happens with the EM force in a rocket, just with a much greater distance.
>Instead of leaving the anchor on a shelf, you bury it in a whole bunch of
>concrete. I have a feeling that this is violating something too, but I
>can't see it; where would this scheme break known physics?
>

You're describing a non-reactionless drive, it seems, and therefore it
wouldn't violate anything. You just want to extend the distance over
which the reaction takes place, as the "big solar sail, with a normungous
laser back on the home planet" does.

Sorry if that's all been said already: I'm a late arrival in this thread.

### IsaacKuo

May 20, 2005, 11:40:42 AM5/20/05
to

>Or 'tractor beams' the ship can use to lock onto distant objects
>and pull itself towards them (and objects toward it), slinging
>along like some sort of interplanetary Spiderman...

I prefer "pressor beams", using special photon-like
particles which move at C but interact with atoms
in an unusual way--they reflect off as if they were
hitting a perfect "corner mirror" (the particle
bounces off doing a perfect 180).

In deep space, a pressor beam is no more effective
than a photon drive--in other words, not very!
However, near a planet you shoot a pressor beam
at the planet and the particles bounce back and
forth between the planet's atmosphere (if any)
and the ship.

Actually, in deep space you can use a pressor beam
as an efficient rocket by shooting it at some
"propellant". This "propellant" is a hunk of mass
that you sacrifice. Ignoring diffraction and
thermal vibration losses, the pressor beam will
keep on bouncing back and forth between the ship
and the propellant until it gets red shifted
out of existence. Theoretically, it perfectly
translates the beam's energy into kinetic energy
of the ship and propellant.

Isaac Kuo

### Raghar

May 20, 2005, 1:47:47 PM5/20/05
to
Nyrath the nearly wise <nyr...@projectrho.com> wrote in
news:2005050610...@eris.io.com:

>
> It looks like all you have to do is mathematically
> convert the energy of thrust into watts, and
> assign some value for the efficiency of the black box.
> Is that it, or are there further complications?

First we should get rid of that pesky possiblility to gain energy
by simple applying a acceleration before kicking up a reactionless
drive.

1. We might say that just to use a reactionless drive you'd need
expend (2 *c)^2 energy * unit of mass. This would partly kick out
any of such attempts.

2. Also you could say that to use an reaction less drive you need
to expend energy to anchor with a underlying structure. Of course
that amount would be |a| so it would look like underlying structure
would always move away from you. So the neccessary energy expended
would be E = 1/2 m * (desired speed + anchor speed) ^2

Of course there is a little question, would there be problems if
underlying structure would be reachable easily? If it would provide
no resitance to pushing (no momentum), and it would need multiple
ships to alow a movement, there would be no net gain of energy,
aside energy from that poor planet that would be spiraling to the
sun. It might be probably nice to add a nice black hole in a center
of a galaxy to the anchor station and by done with it. 1/2 * m *
(200 * c)^2 is still much less than inertia of that small black
hole.

Actually you could see a reaction less drive as a drive that would
connect two ships moving in opposite direction at the same speed,
and alow them to increase speed even more.

We might call it a centered drive, or an anchored drive.

Even if it would need to expend some energy to add to the effect of
a underlying structure. It needn't be so bad. As Silvertooth
experiment showed us, it might be more telling about a global frame
of reference than about space structure, the global reference frame
is moving around 200 km/s.

### Raghar

May 20, 2005, 2:04:35 PM5/20/05
to
Anthony Buckland <buck...@direct.ca> wrote in
news:SZKdnQwbyMO...@look.ca:

This remminds me of a PM. An old saying was about moving, but
because all planets would be someting like PM and everyone decided
it was that device that would need to be hard obtain, they added
ARTIFICALLY a requirement about energy gain.
Reactionless drive is in SF a drive that doesn't need a REACTION
MASS (to carry within itself, and doesn't suck particles from
surrounding normal space). Translated it doesn't need to carry any
fuel.
The other somehow less common deffinition was a drive that doesn't
use an pushing towards an object in normal space. This means if it
push against subspace, or throught something not in normal space
pushes against some object it qualifies as reactionless drive.
For example something is teleported through quantum tunneling into
ship, used for acceleration, and teleported back.

### Michael Ash

May 20, 2005, 2:47:47 PM5/20/05
to

The Big Solar Sail is effected by things like red shift, delays in
reaction times, etc. I'm basically wondering how many of those effects are
necessary in order to retain conservation of energy. Some kind of "red
shift" would be necessary to provide the rising energy requirements for
constant acceleration. A lightspeed delay would be needed to prevent FTL
communication using the device. Anything else?

> Sorry if that's all been said already: I'm a late arrival in this thread.

It appears you're the first. :)

### Erik Max Francis

May 20, 2005, 4:45:58 PM5/20/05
to
Michael Ash wrote:

> It seems to me that it ought to be possible to use the OP's scenario with
> some modifications. Instead of the anchor simply defining a frame, it acts
> as reaction mass. The ship gets a force in one direction, and the anchor
> gets the same force in the opposite direction. This is pretty much what
> happens with the EM force in a rocket, just with a much greater distance.
> Instead of leaving the anchor on a shelf, you bury it in a whole bunch of
> concrete. I have a feeling that this is violating something too, but I
> can't see it; where would this scheme break known physics?

It wouldn't violate anything. The legitimacy of tractor and pressor
beams have been discussed from time to time in the newsgroup.

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

God grant me to contend with those that understand me.
-- Thomas Fuller

### Paul F. Dietz

May 20, 2005, 7:58:43 PM5/20/05
to
Michael Ash wrote:

> It seems to me that it ought to be possible to use the OP's scenario with
> some modifications. Instead of the anchor simply defining a frame, it acts
> as reaction mass. The ship gets a force in one direction, and the anchor
> gets the same force in the opposite direction. This is pretty much what
> happens with the EM force in a rocket, just with a much greater distance.
> Instead of leaving the anchor on a shelf, you bury it in a whole bunch of
> concrete. I have a feeling that this is violating something too, but I
> can't see it; where would this scheme break known physics?

Yes, it does break known physics.

The problem is that unless the anchor is at the same place
as the ship, the 'gets the same force in the opposite direction'
will, in most reference frames, happen at a different time
than the ship gets its application of force. So you violate
conservation of energy and momentum for a while.

It also violates conservation of angular momentum, even in
a newtonian universe.

Paul

### Erik Max Francis

May 20, 2005, 8:28:15 PM5/20/05
to
Paul F. Dietz wrote:

> The problem is that unless the anchor is at the same place
> as the ship, the 'gets the same force in the opposite direction'
> will, in most reference frames, happen at a different time
> than the ship gets its application of force. So you violate
> conservation of energy and momentum for a while.
>
> It also violates conservation of angular momentum, even in
> a newtonian universe.

Well, he never really implied it was instantaneous. If one makes the
force behave as, say, electromagnetism (though it's not
electromagnetism), then there shouldn't be a problem.

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis

Walk into a room and make the / Whole interior inferior
-- Ice Cube

### Paul F. Dietz

May 20, 2005, 9:19:25 PM5/20/05
to
Erik Max Francis wrote:

> Well, he never really implied it was instantaneous. If one makes the
> force behave as, say, electromagnetism (though it's not
> electromagnetism), then there shouldn't be a problem.

Another possibility is a coupling involving tachyons.

Paul

### Anthony Buckland

May 21, 2005, 12:59:23 PM5/21/05
to
Raghar wrote:

> ...

>
>Reactionless drive is in SF a drive that doesn't need a REACTION
>MASS (to carry within itself, and doesn't suck particles from
>surrounding normal space). Translated it doesn't need to carry any
>fuel.
>The other somehow less common deffinition was a drive that doesn't
>use an pushing towards an object in normal space. This means if it
>push against subspace, or throught something not in normal space
>pushes against some object it qualifies as reactionless drive.
>

I figure a reactionless drive is a drive that violates Newton's "For
every action, there is
an equal and opposite reaction" law.

I recall one story that used a gate device. Back on the home planet, a
big rocket engine
with a big fuel supply operated full blast. The reaction force from the
rocket was
directed against a surface in the gate, which in turn shoved the ship.
No free ride,
but no engine or fuel to carry in the ship. What kept the rocket engine
in place
escapes my memory. Another thing you can do with a gate is to use it to
send
fuel. You still have a reaction drive, but the reaction mass and the
energy to
push it keep arriving through the gate instead of having to be carried
along.
Constructing a gate is left as an exercise for the student.

0 new messages