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# Putting numbers on Handwavium

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### Nyrath

Jan 24, 2006, 7:52:44â€¯PM1/24/06
to
Say that you are writing an SF story, and you really
need a bit of handwavium, that is, a gizmo that
does something necessary for the story but also
unfortunately violates one of the laws of physics.
For instance, a Dean Drive (which violates Newton's
laws of motion).
http://en.wikipedia.org/wiki/Dean_drive

Now the bad writer will say "since I've broken
one law, why not break them all?" and go on to
write some wretched science-less space fantasy.

A better writer will try to limit the damage, and
attempt to avoid breaking any more laws.

So how would one characterize the "efficiency"
of a Dean drive? In my naive view, I'd use the
K=1/2 MV^2 equation for kinetic energy as a measure of
how many joules would be required to make the
spacecraft move at velocity V.

This would be multiplied by the reciprocal of the
the drive efficiency in order to get a ballpark
estimate of the power output required of the power plant
( e.g., an efficiency of 0.1 would multiply the power
requirements by ten)

But in a feeble attempt at avoiding strains on the
reader's willing suspension of disbelief, the writer
should use the lowest possible efficiency. One that
was slightly better than the efficiency of whatever
conventional propulsion system is state of the art.

So the question arises: does anybody have any
ballpark figures of the efficiencies of some
conventional propulsion systems? By this I
mean the ratio between the power generated
and the amount of power that manages to wind
up in the kinetic energy of the moving spacecraft.

Efficiencies of a Saturn V F-1 engine, a Space Shuttle
main engine, the Smart-1 ion drive, a theoretical
Helicon double-layer ion drive, a DS4G, a KIWI Nerva,
that sort of thing.

Anybody have any data?

### Erik Max Francis

Jan 24, 2006, 8:19:01â€¯PM1/24/06
to
Nyrath wrote:

> Say that you are writing an SF story, and you really
> need a bit of handwavium, that is, a gizmo that
> does something necessary for the story but also
> unfortunately violates one of the laws of physics.
> For instance, a Dean Drive (which violates Newton's
> laws of motion).
> http://en.wikipedia.org/wiki/Dean_drive
>
> Now the bad writer will say "since I've broken
> one law, why not break them all?" and go on to
> write some wretched science-less space fantasy.
>
> A better writer will try to limit the damage, and
> attempt to avoid breaking any more laws.
>
> So how would one characterize the "efficiency"
> of a Dean drive? In my naive view, I'd use the
> K=1/2 MV^2 equation for kinetic energy as a measure of
> how many joules would be required to make the
> spacecraft move at velocity V.

The problem with trying to rationalize reactionless drives is that you
immediately start running into problems. Actually, some of those were
discussed in the thread "Mathematics of reactionless drives?" you
started back in May.

One immediate problem is that the power you have to use to accelerate is
frame dependent. So you basically need to introduce a universal frame,
in violation of special relativity (hey, you already broke Newtonian
mechanics, no big deal there), so that everyone agrees on the power
required to accelerate at a given speed. (And note that the power
required increases based on your speed, even if you're trying to
maintain the same acceleration ...) This is the kind of problem that
happens when you break a fundamental law of physics -- there is a chain
reaction of other laws that fall to the wayside as well. Pretty soon
you've got yourself an incoherent tangle.

So let's say you choose the absolute frame route. Well, it'd be
absolute luck if the frame were at rest with respect to the Solar
System, right? If you pick a perennial favorite of choices of absolute
frames and select the local cosmic background rest frame (of course
there are even problems with that as we discussed previously), you have
a big problem in terms of the drive being practical -- we're moving at
600 km/s with respect to it!

Okay, so how much energy would it take to use the drive? Kinetic energy is

K = (1/2) m v^2

The power needed is the derivative of kinetic energy:

P = dK/dt = m v dv/dt = m v a.

In massic terms, P/m = v a. So to accelerate at 10 m/s^2, you need to
supply 6 MW/kg of power. Install the drive in a 2 t car and your total
energy requirements for accelerating at a bit over one gee are 12 GW.
Getting up to 65 mph, or about 30 m/s, requires 37 TJ.

Even if physical-law-violating means are available, that still doesn't
necessarily make them practical ...

--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Forgiveness is another word for letting go.
-- Matthew Fox

### Nyrath

Jan 24, 2006, 9:53:15â€¯PM1/24/06
to
Erik Max Francis wrote:
> Actually, some of those were
> discussed in the thread "Mathematics of reactionless drives?" you
> started back in May.

I'm starting to become concerned with these "senior
moments" that I'm having. I had forgotten

### Robert Shaw

Jan 25, 2006, 6:26:15â€¯AM1/25/06
to
Erik Max Francis wrote:

> So let's say you choose the absolute frame route. Well, it'd be
> absolute luck if the frame were at rest with respect to the Solar
> System, right? If you pick a perennial favorite of choices of absolute
> frames and select the local cosmic background rest frame (of course
> there are even problems with that as we discussed previously), you have
> a big problem in terms of the drive being practical -- we're moving at
> 600 km/s with respect to it!
>
> Okay, so how much energy would it take to use the drive? Kinetic energy is
>
> K = (1/2) m v^2
>
> The power needed is the derivative of kinetic energy:
>
> P = dK/dt = m v dv/dt = m v a.
>

In that last line, you've got a scalar product.

Accelerate at right angles to v, and the power needed will be zero.
This makes the drive anisotropic, which could be interesting.

--
Matter is fundamentally lazy:- It always takes the path of least effort
Matter is fundamentally stupid:- It tries every other path first.
That is the heart of physics - The rest is details.- Robert Shaw

### chorned...@hushmail.com

Jan 25, 2006, 12:06:35â€¯PM1/25/06
to

Robert Shaw wrote:
> Erik Max Francis wrote:
>
> > So let's say you choose the absolute frame route. Well, it'd be
> > absolute luck if the frame were at rest with respect to the Solar
> > System, right? If you pick a perennial favorite of choices of absolute
> > frames and select the local cosmic background rest frame (of course
> > there are even problems with that as we discussed previously), you have
> > a big problem in terms of the drive being practical -- we're moving at
> > 600 km/s with respect to it!
> >
> > Okay, so how much energy would it take to use the drive? Kinetic energy is
> >
> > K = (1/2) m v^2
> >
> > The power needed is the derivative of kinetic energy:
> >
> > P = dK/dt = m v dv/dt = m v a.
> >
>
> In that last line, you've got a scalar product.
>
>
> Accelerate at right angles to v, and the power needed will be zero.
> This makes the drive anisotropic, which could be interesting.
>
Indeed. What an even more important effect of the anisotropy, is that
you can actually get the power needed to be negative, by accelerating
against v.

Effectively, you will be sailing on the ether wind, and making use of
the fact that the Earth and Solar System is not stationary. You can use
deploying and undeploying the drive to use gravity from Sun and planets
- in effect, slowing down the movement of the Solar System with respect
to ether.

Actually, it would be a bad luck if there were no ether wind...

### rap...@netscape.net

Jan 25, 2006, 1:35:30â€¯PM1/25/06
to

Erik Max Francis wrote:

> Nyrath wrote:
> > So how would one characterize the "efficiency"
> > of a Dean drive? In my naive view, I'd use the
> > K=1/2 MV^2 equation for kinetic energy as a measure of
> > how many joules would be required to make the
> > spacecraft move at velocity V.
>
> The problem with trying to rationalize reactionless drives is that you
> immediately start running into problems. Actually, some of those were
> discussed in the thread "Mathematics of reactionless drives?" you
> started back in May.
>
> One immediate problem is that the power you have to use to accelerate is
> frame dependent. So you basically need to introduce a universal frame,
> in violation of special relativity (hey, you already broke Newtonian
> mechanics, no big deal there), so that everyone agrees on the power
> required to accelerate at a given speed. (And note that the power
> required increases based on your speed, even if you're trying to
> maintain the same acceleration ...) This is the kind of problem that
> happens when you break a fundamental law of physics -- there is a chain
> reaction of other laws that fall to the wayside as well. Pretty soon
> you've got yourself an incoherent tangle.

What about a "magic" material that can have any velocity without it
actually changing position relative to whatever it is attached to. One
option for such a material would be a fluid and some micro wormholes
connecting all the edges. If the fluid flows left it will flow into
the left hand set of wormhole mouths and out the right hand mouths.
The effect is that is can continuously flow to the left without
actually moving. (In practice, this wouldn't actually work as the left
hand wormhole mouths would increase in mass, but it is likely to be a
more invisable handwave :) ). It also means that the centre of mass of
a closed system can be changed.

Using the system to boost to orbit depends on assumptions. If the
fluid is 10 times less massive than the ship, then it needs to be
accelerated at 10g downwards just to maintain a hover. Assuming a
1000kg ship, v = 1km/s and a = 2g, the power is

P = m.v.a = 1000kg*1000m/s*20m/s/2 = 20MW

That is just to raise the ship, the fluid must also be accelerated

P = m.v.a = 100kg*10000m/s*200m/s/s = 200MW

OK, so the moral of the story is that the fluid should be the same mass

This drops it to total 40MW required. This would probably be
sufficient to get to orbit by running the reactor at constant power and
reducing acceleration.

One small problem is that if you switch on the magic effect while the
ship is waiting on the pad, the magic matter will gain downward
velocity without limit. Also, if you started the fluid without an
upward velocity, all that would be required would be to break against
the fluid to get to orbit.

### Peter D. Tillman

Jan 25, 2006, 2:41:37â€¯PM1/25/06
to
In article <gYmdnTDL74C...@io.com>,
Nyrath <nyr...@projectrho.com.INVALID> wrote:

Oh, that's OK. You'll forget this, too... <evil grin>

--
THOG'S MASTERCLASS, Detached Viewpoint Dept: `Isaac threw up his face
and swung it around him, desperately searching for light.' (China
Mieville, _Perdido Street Station_, 2000)

### JimboCat

Jan 25, 2006, 4:17:37â€¯PM1/25/06
to
Erik Max Francis wrote:

>If you pick a perennial favorite of choices of absolute
>frames and select the local cosmic background rest frame (of course
>there are even problems with that as we discussed previously), you have
>a big problem in terms of the drive being practical -- we're moving at
>600 km/s with respect to it!
>
>Okay, so how much energy would it take to use the drive? Kinetic energy is
>
> K = (1/2) m v^2
>
>The power needed is the derivative of kinetic energy:
>
> P = dK/dt = m v dv/dt = m v a.
>
>In massic terms, P/m = v a. So to accelerate at 10 m/s^2, you need to
>supply 6 MW/kg of power. Install the drive in a 2 t car and your total
>energy requirements for accelerating at a bit over one gee are 12 GW.
>Getting up to 65 mph, or about 30 m/s, requires 37 TJ.

That 600 km/s really kills you here, doesn't it? I would guess you,
the OP, were looking for something more like a magic anti-gravity
drive: "Dean Drive" is not specific as to mechanism, since the thing
never did actually *work*.

An automobile gets up to a speed of 65 mph on the highway with lots
less than 37 TJ. That's because it builds up its kinetic energy by
pushing against an effectively infinite mass (the whole earth). If your
magic drive can push against, say, the entire mass of the universe
(don't ask me how!), then your energy requirements are a lot lower. All
the energy you use goes directly into adding to your own kinetic
energy.

One Newton of thrust gives 1kg an acceleration of 1 m/s^2. So,
integrating force over distance for that one second, you've moved half
a meter, and used half a Watt of power: you get a Newton of thrust from
half a Watt.

That's a *lot* better than the photon drive, which needs 300 MW of
power for a Newton of thrust (which is even worse than Erik's example
because c = 300 Megameters/s is lots bigger than his 600 km/s). And it
is even applicable to an actually conceivable-in-principle space drive:
all you'd need is a cable strung between your origin and your
destination :-)

Your 2 t car at 30 m/s has a kinetic energy of about 1000/2*30^2 = 450
kJ. If you've got sufficiently magic handwavium, that's all the energy
you need to provide. Conservation of momentum is satisfied by the rest
of the universe moving (very, very slowly) in the opposite direction
(whatever that means...).

Jim Deutch (JimboCat)
--
"For escape velocity at one gravity of horizontal acceleration,
"one radian is needed." - John Stockton

### Paul Colquhoun

Jan 25, 2006, 7:39:08â€¯PM1/25/06
to

^^^^^^^
Surely you meant "with"?

| upward velocity, all that would be required would be to break against
| the fluid to get to orbit.

OK, so now we *do* turn the unit on while the ship is waiting on the pad.
Leave it run until the fluid has picked up enough energy, then we quickly
rotate it 180 degrees. *Now* we can (as you suggested) brake against the
moving fluid and send the ship into orbit.

We cauld also wait a bit longer before launch, and store up enough energy
to enable us to do any required manouvers, and land again afterwards.

(Yes, I am aware that the "rotate 180 degrees" step would be like
turning a high-speed flywheel. We *are* just trying to hide the
handwaving here, arn't we?)

--
Reverend Paul Colquhoun, ULC. http://andor.dropbear.id.au/~paulcol
http://catb.org/~esr/faqs/smart-questions.html#intro

### Erik Max Francis

Jan 25, 2006, 8:56:30â€¯PM1/25/06
to
Robert Shaw wrote:

> Accelerate at right angles to v, and the power needed will be zero.
> This makes the drive anisotropic, which could be interesting.

Right. In vector form (v, a vectors),

K = (1/2) m v dot v

P = dK/dt = m v dot dv/dt = m v dot a.

If v and a point in the same general direction, then P is positive. If
v and a point in the opposite general direction, then P is negative. If
they're orthogonal, then P is zero. The negative case was discussed the
last time this thread came up; theoretically you could use such a drive
as a power generator (and it could conceivably generate _lots_ of power)
by using it as a brake.

Unfortunately that still makes the drive pretty impractical, unless you
just happen to want to travel in a direction that is luckily
perpendicular to the direction of the apparent cosmic background
anistropy ...

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

Laws are silent in time of war.
-- Cicero

Jan 25, 2006, 9:31:14â€¯PM1/25/06
to
"JimboCat" <10313...@compuserve.com> wrote:

> One Newton of thrust gives 1kg an acceleration of 1 m/s^2. So,
> integrating force over distance for that one second, you've moved half
> a meter, and used half a Watt of power: you get a Newton of thrust
> from half a Watt.

So in order to levitate in a 1 gee field, it'd only cost 5 W/kg ?

--
>;k

### Erik Max Francis

Jan 25, 2006, 11:13:18â€¯PM1/25/06
to

> So in order to levitate in a 1 gee field, it'd only cost 5 W/kg ?

No, it need not cost anything at all. Work is force times distance, so
no work is done when you hold something up. Right now the chair I'm
sitting in is doing a pretty good illustration of the effect.

--
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

### Wayne Throop

Jan 25, 2006, 11:31:35â€¯PM1/25/06
to
: Erik Max Francis <m...@alcyone.com>
: No, it need not cost anything at all. Work is force times distance, so
: no work is done when you hold something up. Right now the chair I'm
: sitting in is doing a pretty good illustration of the effect.

Yes, but that's only because the chair you are sitting on is
accelerating you upwards at 1g. If you weren't resting on anything,
ie, "hovering", and had only this inertialess wossname to provide
this acceleration, you'd need to spend the energy.

Now oridinarily, this wouldn't make sense, to spend energy
when you aren't gaining either potential or kinetic energy.
But then, you are throwing out energy conservation and
sensible physics right off the bat.

Or so it seems to me.

Wayne Throop thr...@sheol.org http://sheol.org/throopw

### Erik Max Francis

Jan 26, 2006, 12:06:59â€¯AM1/26/06
to
Wayne Throop wrote:

> Yes, but that's only because the chair you are sitting on is
> accelerating you upwards at 1g. If you weren't resting on anything,
> ie, "hovering", and had only this inertialess wossname to provide
> this acceleration, you'd need to spend the energy.
>
> Now oridinarily, this wouldn't make sense, to spend energy
> when you aren't gaining either potential or kinetic energy.
> But then, you are throwing out energy conservation and
> sensible physics right off the bat.
>
> Or so it seems to me.

My point was, the chair is doing it without expending energy. It can do
that because it is doing no work on me. It is certainly true that it is
resisting the force of gravity, but on the contrary, the force of
gravity is what would be doing work on me if there were no chair below
me. That is, it does not necessarily require a constant expenditure of
energy to stop something from falling in a gravitational field.
Stopping work from being done does not necessarily require work to be done.

### Erik Max Francis

Jan 26, 2006, 12:07:42â€¯AM1/26/06
to
Erik Max Francis wrote:

> My point was, the chair is doing it without expending energy. It can do
> that because it is doing no work on me. It is certainly true that it is
> resisting the force of gravity, but on the contrary, the force of
> gravity is what would be doing work on me if there were no chair below
> me. That is, it does not necessarily require a constant expenditure of
> energy to stop something from falling in a gravitational field.
> Stopping work from being done does not necessarily require work to be done.

I should point out that I am talking more generally than the specific
hypothetical drive we are talking about. I'm just pointing out that
hovering does not necessarily require power at all.

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

Never make a promise or plan / Take a little love where you can
-- Florence, _Chess_

### rap...@netscape.net

Jan 26, 2006, 5:15:31â€¯AM1/26/06
to

Paul Colquhoun wrote:

> | One small problem is that if you switch on the magic effect while the
> | ship is waiting on the pad, the magic matter will gain downward
> | velocity without limit. Also, if you started the fluid without an
> ^^^^^^^
> Surely you meant "with"?

Yeah.

>
> | upward velocity, all that would be required would be to break against
> | the fluid to get to orbit.
>
>
> OK, so now we *do* turn the unit on while the ship is waiting on the pad.
> Leave it run until the fluid has picked up enough energy, then we quickly
> rotate it 180 degrees. *Now* we can (as you suggested) brake against the
> moving fluid and send the ship into orbit.
>
> We cauld also wait a bit longer before launch, and store up enough energy
> to enable us to do any required manouvers, and land again afterwards.
>

I think there would have to be something like a pressure is created in
the wormhole mouths that cancels gravity. Basically, the work required
to move a particle of fluid up the gravity field from the bottom mouth
to the top mouth would have to be equal to the work required going
through the mouths. Otherwise, the effect kills conservation of
energy.

### Raghar

Jan 26, 2006, 2:05:32â€¯PM1/26/06
to
Erik Max Francis <m...@alcyone.com> wrote in
news:jKWdnQSHLJ_...@speakeasy.net:

> Robert Shaw wrote:
>
>> Accelerate at right angles to v, and the power needed will be
>> zero. This makes the drive anisotropic, which could be
>> interesting.
>
> Right. In vector form (v, a vectors),
>
> K = (1/2) m v dot v
>
> P = dK/dt = m v dot dv/dt = m v dot a.
>
> If v and a point in the same general direction, then P is
> positive. If v and a point in the opposite general direction,
> then P is negative. If they're orthogonal, then P is zero. The
> negative case was discussed the last time this thread came up;
> theoretically you could use such a drive as a power generator
> (and it could conceivably generate _lots_ of power) by using it
> as a brake.

However note the |v dot a/t| could be always positive, so this
drive would need to expend energy for going in any direction.
Quantum locking and unlocking for example.

Also I'm unsure if you can do v dot a/t instead of |v dot v| / t

As far I know no enegy was moved to energy grid by recent return of
interplanet probe. So that lots of energy generated at efficiency
1/(c^2 - 1) could be somehow VERY unintereting.

### Erik Max Francis

Jan 26, 2006, 3:52:49â€¯PM1/26/06
to
Raghar wrote:

> However note the |v dot a/t| could be always positive, so this
> drive would need to expend energy for going in any direction.
> Quantum locking and unlocking for example.
>
> Also I'm unsure if you can do v dot a/t instead of |v dot v| / t

I just did the derivation above. K = (1/2) m |v|^2 = (1/2) m v dot v.
Its derivative is m v dot a. Not m |v dot a|.

>
> As far I know no enegy was moved to energy grid by recent return of
> interplanet probe. So that lots of energy generated at efficiency
> 1/(c^2 - 1) could be somehow VERY unintereting.

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

### sam kayley

Jan 26, 2006, 6:23:50â€¯PM1/26/06
to

"Nyrath" <nyr...@projectrho.com.INVALID> wrote in message
news:r8-dnY7Aze1AVkve...@io.com...

> Say that you are writing an SF story, and you really
> need a bit of handwavium, that is, a gizmo that
> does something necessary for the story but also
> unfortunately violates one of the laws of physics.
> For instance, a Dean Drive (which violates Newton's
> laws of motion).
> http://en.wikipedia.org/wiki/Dean_drive
>
> Now the bad writer will say "since I've broken
> one law, why not break them all?" and go on to
> write some wretched science-less space fantasy.
>
> A better writer will try to limit the damage, and
> attempt to avoid breaking any more laws.
>...

Why not go for a tractor beam, whatever you point it at provides the frame
of reference?

### Logan Kearsley

Jan 26, 2006, 6:54:23â€¯PM1/26/06
to
"Erik Max Francis" <m...@alcyone.com> wrote in message
news:jKWdnQSHLJ_...@speakeasy.net...

> Robert Shaw wrote:
>
> > Accelerate at right angles to v, and the power needed will be zero.
> > This makes the drive anisotropic, which could be interesting.
>
> Right. In vector form (v, a vectors),
>
> K = (1/2) m v dot v
>
> P = dK/dt = m v dot dv/dt = m v dot a.
>
> If v and a point in the same general direction, then P is positive. If
> v and a point in the opposite general direction, then P is negative. If
> they're orthogonal, then P is zero. The negative case was discussed the
> last time this thread came up; theoretically you could use such a drive
> as a power generator (and it could conceivably generate _lots_ of power)
> by using it as a brake.

Like Niven's gravity brake (if I recall the name correctly), where the local
gravitational field provides the reference frame. You just deploy a very big
radiator, turn it on, and wait to come to rest wrt the largest closest mass,
typically a planet.

> Unfortunately that still makes the drive pretty impractical, unless you
> just happen to want to travel in a direction that is luckily
> perpendicular to the direction of the apparent cosmic background
> anistropy ...

-l.
------------------------------------
My inbox is a sacred shrine, none shall enter that are not worthy.

### Erik Max Francis

Jan 26, 2006, 7:30:23â€¯PM1/26/06
to
Logan Kearsley wrote:

> Like Niven's gravity brake (if I recall the name correctly), where the local
> gravitational field provides the reference frame. You just deploy a very big
> radiator, turn it on, and wait to come to rest wrt the largest closest mass,
> typically a planet.

If you're referring to his gravity polarizer, I don't recall him ever
going into significant detail about its mode of operation in his
canonical stories, except of course to make it clear that it indeed was
a reactionless drive.

His later Known Space stories also have thrusters, the human
reactionless drive, whose operational basis was also unspecified.
(Probably quite wisely.)

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

Ipsa scientia potestas est. "Knowledge itself is power."
-- a Latin proverb

### Nyrath

Jan 26, 2006, 8:19:47â€¯PM1/26/06
to
Logan Kearsley wrote:
> Like Niven's gravity brake (if I recall the name correctly), where the local
> gravitational field provides the reference frame. You just deploy a very big
> radiator, turn it on, and wait to come to rest wrt the largest closest mass,
> typically a planet.

Technically, he called it a "gravity drag".
From "Flatlander" by Larry Niven.

By then we were close enough to use the gravity drag to slow us. The
beautiful thing about a gravity drag is that it uses very little power.
It converts a ship's momentum relative to the nearest powerful mass into
heat, and all you have to do is get rid of the heat. Since the ST8's
hull would pass only various ranges of radiation corresponding to what
the puppeteers' varied customers considered visible light, the
shipbuilders had run a great big radiator fin out from the gravity drag.
It glowed dull red behind us.

### Jake

Jan 26, 2006, 11:19:00â€¯PM1/26/06
to

I was thinking the same thing. Tractor/repulsor beams don't require
breaking any conservation laws IIRC.

### Erik Max Francis

Jan 27, 2006, 1:18:48â€¯AM1/27/06
to
Nyrath wrote:

> By then we were close enough to use the gravity drag to slow us. The
> beautiful thing about a gravity drag is that it uses very little power.
> It converts a ship's momentum relative to the nearest powerful mass into
> heat, and all you have to do is get rid of the heat. Since the ST8's
> hull would pass only various ranges of radiation corresponding to what
> the puppeteers' varied customers considered visible light, the
> shipbuilders had run a great big radiator fin out from the gravity drag.
> It glowed dull red behind us.

Ah, cool, I didn't remember that variation.

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

In principle I am against principles.
-- Tristan Tzara

### Raghar

Jan 27, 2006, 3:47:23â€¯PM1/27/06
to
Erik Max Francis <m...@alcyone.com> wrote in
news:mMednaP6r-I...@speakeasy.net:

> Raghar wrote:
>
>> However note the |v dot a/t| could be always positive, so this
>> drive would need to expend energy for going in any direction.
>> Quantum locking and unlocking for example.
>>
>> Also I'm unsure if you can do v dot a/t instead of |v dot v|
>> / t
>
> I just did the derivation above. K = (1/2) m |v|^2 = (1/2) m v
> dot v. Its derivative is m v dot a. Not m |v dot a|.
>

However my point was the principle of reactionless drive could
force the equation into m * |v^2| / 2 state. This would cause P =
change K / per T to be P = m * |v^2| / 2*T and roughly
aproximated into m * |v| * |a| / 2 if T >= 0

This equation would also satisfy condition there is very unlikely
something like a free lunch.

### Erik Max Francis

Jan 27, 2006, 6:08:32â€¯PM1/27/06
to
Raghar wrote:

> However my point was the principle of reactionless drive could
> force the equation into m * |v^2| / 2 state. This would cause P =
> change K / per T to be P = m * |v^2| / 2*T and roughly
> aproximated into m * |v| * |a| / 2 if T >= 0

(1/2) m |v^2| is the same as (1/2) m |v|^2 which is the same as (1/2) m
v dot v.

If v is a vector, v^2 is a shorthand for |v|^2 = v dot v.

> This equation would also satisfy condition there is very unlikely
> something like a free lunch.

To basically get the equations not to work as written above and in
previous posts, you'd basically have to modify Newtonian mechanics. And
you'd have some troubles explaining why it works so well as it is.

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

He who knows how to be poor knows everything.
-- Jules Michelet

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