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Date: 19 Sep 2007 00:18:00
From: Peter Webb
Subject: Proposed gravity drive - calculation of acceleration achievable
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This post works out the acceleration that could be derived from the
"emissionless drive" I described in a previous post.
Essentially, the drive works by having two (or more) large masses separated
by a long distance. If the gravitational field varies over time (as it does
in an elliptical orbit) and the gravitational field is non linear (which it
always is) then the force experienced by each mass will be different at
different times. By reducing the distance when the gravitational gradient is
high (through using a motor to bring the masses together) and increasing it
when it the gravitational gradient is low - at the top of the orbit - the
energy expended by the motor over a complete cycle is manifest as either
higher kinetic energy of the masses or a higher orbit, depending upon the
geometry.
So, how practical is it?
The first thing to note is that as long as the mass of the motor and rod is
insignificant compared to the two large masses, the net acceleration is
independent of the size of the masses, and depends only upon their
separation. To see this is true, you can imagine that two identical
spaceships orbiting side by side would experience the same acceleration as a
single spaceship with the double the mass at each end.
This very convenient, because as the mass falls out of the equation we need
only consider the differences in acceleration due to gravity.
Lets make the distance between the masses vary from zero to 100 kms.
Lets make the lowest point (perigee) of the orbit 7,000 kms from the centre
of the earth (ie 600 kms above the surface). Lets make the most distant
point (apogee) 20,000 kms. Lets make g = 10 m/s^2 at 7000 kms.
Using the inverse square law, the difference in acceleration between the two
masses at 7,000 kms is:
10 m/s^2 * (7,100,000/7,000,000 - 1)^2
= 0.28 m/s^2
At 20,000 kms, the difference in acceleration at the two ends is:
10 m/s^2 * (20,100,000/20,000,000 - 1)^2
= 0.1 m/s^2
This means we have an acceleration differential of 0.28 m/s^2 at perigee but
only 0.1 m/s^2 at apogee. This is manifest as the force the motor must use
to bring the masses together or separate them, its M x 0.28 m/s^2 at perigee
but only M x 0.1 ms/^2 at apogee.
So you pull the masses together at perigee and separate them at apogee. This
give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
0.18m/s^2. Because you have two masses which must share this acceleration,
the net acceleration is 0.9 m/s^2.
This a pretty reasonable rate. You could obviously build up to escape
velocity eventually.
There are a few problems. The first is keeping the masses pointing towards
the earth, which may require gyroscopes. Alternatively, there is no reason
why the spaceship needs to have only two masses on one axis; there could be
six masses on three axis. This would mean you wouldn't care if it tumbled,
as at least one axis would always be available. More to the point, having
multiple axis may allow the spacecraft to be steered, and the problem
avoided.
This almost seems to me to be a practical "gravity drive". Even if not used
to achieve escape velocity, it would seem to suffice for orbital adjustments
(unfortunately not for circular orbits though). A spaceship designed in this
manner could operate forever with no refueling, as long as it had access to
energy (eg solar energy).
Should I patent the "gravity drive" ? Or has somebody already done this!
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Date: 20 Sep 2007 18:34:31
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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On Sep 20, 4:34 pm, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> > There is nothing mysterious about conservation of angular momentum
> > implying that a change in the rotation rate is coupled with a change
> > in the orbital path. There's also nothing mysterious about the fact
> > that the change in the orbital path is utterly miniscule, so even if
> > you changed the rotation rate so much that the satellite was
> > literally pulling itself apart, the orbital change wouldn't be
> > significant.
> So you think that my drive is theoretically possible, but that with present
> day materials/technology would not give sufficient acceleration to be of
> practical use?
The acceleration is pretty irrelevant--the total amount of change you
can make to the orbit isn't significant. Whether you creep toward
"the wall" slowly and gradually or run straight for it quickly, you
can't get past "the wall". The wall you run up against is the point
at which the rotation of the masses pulling on the rod reach the
material strength limits.
It's like trying to stand on your tiptoes in order to fly. Your total
altitude is limited, no matter how slowly you lift yourself upward.
> I really was just trying to present a theoretical model of how a gravity
> drive could work. I was not attempting to produce a design spec on how
> strong and light the structural members (eg the tether rope) would have to
> be. Its more of an "in principle" model of a gravity drive rather than a
> serious contemplation of the materials that would be needed.
To the exact same extent, you can "in principle" fly upward by
standing on your tiptoes.
> Thanks for your input.
Isaac Kuo
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Date: 21 Sep 2007 12:24:49
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190338471.722467.52530@n39g2000hsh.googlegroups.com...
> On Sep 20, 4:34 pm, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> > There is nothing mysterious about conservation of angular momentum
>> > implying that a change in the rotation rate is coupled with a change
>> > in the orbital path. There's also nothing mysterious about the fact
>> > that the change in the orbital path is utterly miniscule, so even if
>> > you changed the rotation rate so much that the satellite was
>> > literally pulling itself apart, the orbital change wouldn't be
>> > significant.
>
>> So you think that my drive is theoretically possible, but that with
>> present
>> day materials/technology would not give sufficient acceleration to be of
>> practical use?
>
> The acceleration is pretty irrelevant--the total amount of change you
> can make to the orbit isn't significant. Whether you creep toward
> "the wall" slowly and gradually or run straight for it quickly, you
> can't get past "the wall". The wall you run up against is the point
> at which the rotation of the masses pulling on the rod reach the
> material strength limits.
>
> It's like trying to stand on your tiptoes in order to fly. Your total
> altitude is limited, no matter how slowly you lift yourself upward.
>
>> I really was just trying to present a theoretical model of how a gravity
>> drive could work. I was not attempting to produce a design spec on how
>> strong and light the structural members (eg the tether rope) would have
>> to
>> be. Its more of an "in principle" model of a gravity drive rather than a
>> serious contemplation of the materials that would be needed.
>
> To the exact same extent, you can "in principle" fly upward by
> standing on your tiptoes.
>
Its hardly a fair analogy, and its not to the "same extent". Its a
completely different problem. That you can't fly from standing on your
tiptoes does not derive from any physical limitation of the strength of your
toes. As I understand it, you agree that my drive would provide orbital
changes, but the limitations of the strengths of current materials means
that it isn't vaguely practical.
You cannot fly "in theory" by standing on your tiptoes; you cannot build a
gravity drive "in practice" because you can't build a rope strong enough to
get a decent effect. I would note that in the early days of rocketry many
people believed that you couldn't use a rocket to get into orbit, because
(amongst other reasons) there were no propellants strong enough. Solving
this materials problem took new materials (lox and hydrogen), more
sophisticated designs (multi-stage rockets), etc.
I think you agree that a propulsion effect occurs, but that the practical
limitations of the strength of the ropes means you can't get enough
propulsion to do anything useful. A bit like trying to build a space rocket
using the same materials and design that are used for a fireworks cracker,
just scaled up. If you only have access to materials and technology
available in the 19th Century, you couldn't build a rocket to get into
orbit. If you only have access to materials and technology available in the
early 21st Century, you can't build a practical gravity drive. Your
objections to my drive are similar to the objections raised against rockets
as a means of getting into orbit - physicists believed such a thing was
possible in theory (even though many people believed it couldn't be, because
there was nothing in space to "push against), but the limitations of 19th
Century technology and materials meant that it couldn't be done "in
practice".
I am very disappointed in the response to this idea. I had always believed
that to accelerate a spaceship, you needed some form of reaction mass. As
far as I am aware, nobody has ever described a spaceship drive that needed
only a gravity well, and did not require reaction mass. I have produced a
model for a gravity drive - a completely new concept for a spaceship drive,
as far as I am aware - and you are objecting that its not currently feasible
because we can't build ropes strong enough to get a large enough effect to
be useful. I believe a pure gravity drive that works in theory (subject to
having materials strong enough) is an interesting concept, even if materials
strength does not allow a useful spaceship to be built in practice.
Have you ever seen a design for an anti-gravity drive that worked even in
theory before? And if you haven't, is the best you can come up with is that
it wouldn't work in practice because we haven't got ropes strong enough?
Doesn't the fact that an anti-gravity drive is possible in theory if not in
practice have some interest to this newsgroup, given that as far as I am
aware nobody has previously been able to describe a mechanism by which an
antigravity drive could be built even in theory?
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Date: 20 Sep 2007 10:10:38
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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On Sep 20, 11:36 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> > You are having to do work moving the large weights around against the
> > force of gravity.
> I know. That is in fact how I calculated the acceleration that would be
> achieved, by calculating the force and the distance between the two weights
> and then using conservation of energy.
Too bad your calculation was wrong, because you failed to consider
most of the energy actually goes into rotating the spacecraft.
This should actually be really obvious, when you consider the rotation
of a tidally locked satellite. The reason why being tidally locked is
a stable configuration is precisely because it minimizes energy.
To change from tidally locked rotation to any other rotation rate
takes
energy. This is true whether you rotate one direction (which
slightly raises the orbit) or your rotate in the other direction
(which
slightly lowers the orbit). In other words, the kinetic energy of
the rotation overwhelms the change in the energy of the orbit
change.
There is nothing mysterious about conservation of angular momentum
implying that a change in the rotation rate is coupled with a change
in the orbital path. There's also nothing mysterious about the fact
that the change in the orbital path is utterly miniscule, so even if
you changed the rotation rate so much that the satellite was
literally pulling itself apart, the orbital change wouldn't be
significant.
Isaac Kuo
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Date: 21 Sep 2007 07:34:38
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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>
> There is nothing mysterious about conservation of angular momentum
> implying that a change in the rotation rate is coupled with a change
> in the orbital path. There's also nothing mysterious about the fact
> that the change in the orbital path is utterly miniscule, so even if
> you changed the rotation rate so much that the satellite was
> literally pulling itself apart, the orbital change wouldn't be
> significant.
So you think that my drive is theoretically possible, but that with present
day materials/technology would not give sufficient acceleration to be of
practical use?
I really was just trying to present a theoretical model of how a gravity
drive could work. I was not attempting to produce a design spec on how
strong and light the structural members (eg the tether rope) would have to
be. Its more of an "in principle" model of a gravity drive rather than a
serious contemplation of the materials that would be needed.
Thanks for your input.
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Date: 20 Sep 2007 09:59:37
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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On Sep 20, 11:19 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> "IsaacKuo" <mech...@yahoo.com> wrote in message
> > Now, instead of bouncing the masses off each other, you can
> > just let them pass by each other--barely missing. You can
> > pump in extra spring energy the same way as "perpetual
> > motion" toys (a well timed extra tug with electromagnets).
> > This propulsion scheme doesn't require any nonphysical
> > magic ropes or anything.
> There was nothing "non-physical" or "magic" about my ropes, they just can't
> be built light and strong enough yet. Any more than your system can.
The can't be build light and strong enough ever. The system
I describe is possible but impractical.
> Nor did it require electromagnetic springs, and two physically separate
> satellites, as yours does.
The fact that you don't separate the two masses limits the
usefulness of your drive to zero. The problem is the
consequences of conservation of angular momentum.
> Its also hard to see how your arrangement could be used for braking a
> spaceship when it reaches a distant target.
You simply do things in reverse. You start off with the two
masses entering in parabolic orbit. Then instead of giving
a "push" each time, you give a "pull" each time. The first
"pull" takes them down to an elliptical orbit. Subsequent
"pulls" lower the elliptical orbit each time.
> However, I think I do see a begrudging acceptance in your reply that my
> arrangement will produce a spacecraft capable of modifying its own orbit
> without expending any reaction mass, notwithstanding your observation that
> current ropes will not suffice.
Sigh. How many times do I have to repeat--your drive can
make a change in its orbit, but this change is negligible.
You can't just repeat giving small nudges each time and
have them accumulate. The problem is conservation of
angular momentum. For every nudge you make, angular
momentum must be conserved--so whatever small angular
momentum you add to your orbit must come with an
opposite angular momentum change to your spacecraft's
spin.
The result is that before you can accumulate a significant
change in the orbit, your run up against the strength limits
of the rope. If you consider the massive amount of
angular momentum in the orbit, the maximum angular
momentum your spacecraft's spin can have is dwarfed
in comparison.
In order for your drive to work, you'd need a rope so
strong that it's literally possible to fling things into
escape velocity with it.
You really need to come to grips with the fundamental
problem, here. It's conservation of angular momentum.
Isaac Kuo
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Date: 20 Sep 2007 08:00:52
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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On Sep 20, 2:24 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> I didn't ignore angular momentum. You just snipped the bit where I did:
> If we recast this in terms of angular momentum, this is what happens. The
> spherical symmetry of the black hole means there is no obvious mechanism to
> exchange angular momentum between the spaceship (or its two halves) and the
> black hole, as you identified earlier. In fact, there is no exchange of
> angular momentum with the black hole. The angular momentum of the spaceships
> trajectory past the black hole is converted into angular momentum of the two
> halves rotating around there common centre of gravity, as you previously
> identified. Angular momentum is conserved. However linear momentum is not;
> the two halves are on different trajectories and exchange a different amount
> of linear momentum with the black hole. The sum of the momentums after the
> close approach is different to the total momentum on the way down, because
> the orbits of the two halves are different. This allows you to reduce the
> spaceships momentum to close to zero in the inertial frame of reference of
> the black hole.
You're still ignoring the important consequence of conservation of
angular momentum--that the spinning of the masses around each
other puts a harsh limit on how much of an effect this can have
with a real life rope. You can't use this drive to do what you want,
which is to have cause a signficant change in the orbit.
> I am still curious to know whether in the world of perfect ropes etc this
> mechanism could be used to "brake" (and hence accelerate) a spaceship
> without expending reaction mass. If it can, this is a pure gravity drive.
It's not really a "pure gravity" drive, but rather a tether drive.
You're
confusing yourself by working with an overly complex example.
Consider a much simpler starting point--there are two masses
starting at zero velocity above Earth. If nothing is done, these
masses will simply fall straight down to Earth. However, instead
there's a spring between them which shoves them apart. Rather
than hitting Earth, the two masses go into elliptical orbits that
miss the Earth. After traveling 180 degrees, the two masses
spring against each other. By pumping extra energy into the
spring force, it's possible to bounce the masses off each other
at a faster speed. Instead of retracing the same elliptical orbit,
these masses bounce off each other onto a higher orbit.
And after traveling another 180 degrees, the masses bounce
off each other again--providing another opportunity to pump in
more energy. In this fashion, it's possible to repeatedly
boost the orbit upward, until ultimately the masses escape
the planet (into parabolic or hyperbolic paths).
Now, instead of bouncing the masses off each other, you can
just let them pass by each other--barely missing. You can
pump in extra spring energy the same way as "perpetual
motion" toys (a well timed extra tug with electromagnets).
This propulsion scheme doesn't require any nonphysical
magic ropes or anything. It doesn't suffer ugly conservation
of angular momentum limitations because the angular
momentum is zero.
Does it make sense to call this a "gravity" drive? Not really;
it was the electromagnets which did the work. If you replace
the electromagnets with a tether, then I'd argue the drive is
really a sort of tether drive.
Isaac Kuo
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Date: 21 Sep 2007 02:19:05
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190300452.882621.269700@50g2000hsm.googlegroups.com...
> On Sep 20, 2:24 am, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>
>> I didn't ignore angular momentum. You just snipped the bit where I did:
>
>> If we recast this in terms of angular momentum, this is what happens. The
>> spherical symmetry of the black hole means there is no obvious mechanism
>> to
>> exchange angular momentum between the spaceship (or its two halves) and
>> the
>> black hole, as you identified earlier. In fact, there is no exchange of
>> angular momentum with the black hole. The angular momentum of the
>> spaceships
>> trajectory past the black hole is converted into angular momentum of the
>> two
>> halves rotating around there common centre of gravity, as you previously
>> identified. Angular momentum is conserved. However linear momentum is
>> not;
>> the two halves are on different trajectories and exchange a different
>> amount
>> of linear momentum with the black hole. The sum of the momentums after
>> the
>> close approach is different to the total momentum on the way down,
>> because
>> the orbits of the two halves are different. This allows you to reduce the
>> spaceships momentum to close to zero in the inertial frame of reference
>> of
>> the black hole.
>
> You're still ignoring the important consequence of conservation of
> angular momentum--that the spinning of the masses around each
> other puts a harsh limit on how much of an effect this can have
> with a real life rope. You can't use this drive to do what you want,
> which is to have cause a signficant change in the orbit.
>
>> I am still curious to know whether in the world of perfect ropes etc this
>> mechanism could be used to "brake" (and hence accelerate) a spaceship
>> without expending reaction mass. If it can, this is a pure gravity drive.
>
> It's not really a "pure gravity" drive, but rather a tether drive.
> You're
> confusing yourself by working with an overly complex example.
>
> Consider a much simpler starting point--there are two masses
> starting at zero velocity above Earth. If nothing is done, these
> masses will simply fall straight down to Earth. However, instead
> there's a spring between them which shoves them apart. Rather
> than hitting Earth, the two masses go into elliptical orbits that
> miss the Earth. After traveling 180 degrees, the two masses
> spring against each other. By pumping extra energy into the
> spring force, it's possible to bounce the masses off each other
> at a faster speed. Instead of retracing the same elliptical orbit,
> these masses bounce off each other onto a higher orbit.
>
> And after traveling another 180 degrees, the masses bounce
> off each other again--providing another opportunity to pump in
> more energy. In this fashion, it's possible to repeatedly
> boost the orbit upward, until ultimately the masses escape
> the planet (into parabolic or hyperbolic paths).
>
> Now, instead of bouncing the masses off each other, you can
> just let them pass by each other--barely missing. You can
> pump in extra spring energy the same way as "perpetual
> motion" toys (a well timed extra tug with electromagnets).
>
> This propulsion scheme doesn't require any nonphysical
> magic ropes or anything.
There was nothing "non-physical" or "magic" about my ropes, they just can't
be built light and strong enough yet. Any more than your system can.
Nor did it require electromagnetic springs, and two physically separate
satellites, as yours does.
Its also hard to see how your arrangement could be used for braking a
spaceship when it reaches a distant target.
However, I think I do see a begrudging acceptance in your reply that my
arrangement will produce a spacecraft capable of modifying its own orbit
without expending any reaction mass, notwithstanding your observation that
current ropes will not suffice.
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Date: 21 Sep 2007 00:31:31
From: ike
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f29d47$0$14405$afc38c87@news.optusnet.com.au...
Any ideas about tapping into the free energy of a vacuum....zero point?
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Date:
From: Martin Brown
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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Date: 21 Sep 2007 02:36:33
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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>> >> Why do you think my drive won't work? It doesn't seem to contravene
>> >> the
>> >> laws
>> >> of conservation of momentum, angular momentum and energy. Where is the
>> >> flaw
>> >> in my (theoretical) design?
>
> You are having to do work moving the large weights around against the
> force of gravity.
>
I know. That is in fact how I calculated the acceleration that would be
achieved, by calculating the force and the distance between the two weights
and then using conservation of energy.
> Your proposal is no more interesting than a little girl pumping energy
> into a playground swing by moving her body weight up and down at the
> right time.
Funnily enough I used exactly that analogy elsewhere inthis thread,
explaining the second mass was analogous to the child's legs. I also had
this in mind when I modelled the spaceship as having the weights extended on
the downswing and retracted on the backswing.
>Parametric excitation of a periodic oscillator isn't even
> remotely new.
>
> http://en.wikipedia.org/wiki/Parametric_oscillator
>
Have you ever heard of this being used to create a spaceship drive? I
wondered in my original post whether the concept of using this effect for
this purpose was new. I couldn't find any relevant Google hits. Whilst I
agree that there are obvious practical difficulties, this would seem to
qualify as a theoretically possible means of accelerating a spaceship
without expending reaction mass - a "gravity drive". However practical it
is, if this is the first time that a spaceship drive which doesn't require a
reaction mass has been described, I think it definitely is something new. I
had always assumed they were impossible. Now I do not.
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Date: 21 Sep 2007 00:29:49
From: ike
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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Right On !! You go.
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Date: 19 Sep 2007 17:27:27
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
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On Sep 19, 6:17 pm, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> "IsaacKuo" <mech...@yahoo.com> wrote in message
> news:1190218209.411323.214890@d55g2000hsg.googlegroups.com...
> >> > Bottom line--you need to explain how your proposed drive
> >> > applies a torque on the Earth.
> >> My spaceship doesn't need to apply any torque to the earth, because it
> >> doesn't necessarily have to rotate. There is no need for transfer of
> >> angular
> >> momentum.
> > Yes, you do. There is a massive amount of angular momentum in the
> > orbit itself. If you want to raise or lower the orbit, you need to
> > add or
> > remove angular momentum to the orbit. There's just no way around it.
I notice you didn't respond to this.
> >> Here is a slightly different form of my gravity drive, used for the
> >> purposes
> >> of braking and not achieving orbital velocity, which may make the central
> >> physical mechanism clearer.
> >> Imagine a hyperbolic orbit, such that your spaceship comes close enough
> >> to a
> >> black hole for your outgoing path to be deflected say 10 degrees from
> >> your
> >> incoming path as it passes by the black hole. From a distance, you are
> >> coming almost directly at the black hole. I put it to you that a slight
> >> variation on my drive can slow down the spaceship so that it is captured
> >> into an elliptical orbit around the black hole.
> >> This is how I do it. When I am well away from the black hole, I split my
> >> spaceship in two halves, connected by a very, very long rope. One half
> >> continues on its original orbit, which has its trajectory changed by only
> >> 10
> >> degrees as a result of passing the black hole. The other half gets a
> >> slight
> >> sidewise nudge, so that it falls slightly deeper into the gravity well
> >> and
> >> is whipped around almost 180 degrees.
> >> Both halves pass the black hole and move away from the black hole, but
> >> one
> >> half ends up with its course deflected by 10 degrees and the other by
> >> almost
> >> 180 degrees. They have identical speeds, but the two halves are moving in
> >> almost opposite directions. Considering the two halves as a single
> >> system,
> >> the total momentum with respect to the black hole is almost zero, as the
> >> two
> >> halves have the same speed but in almost opposite directions.
> >> They are connected by a rope. You use a motor to pull the rope in. By the
> >> conservation of momentum, the two halves will end up almost stationary
> >> with
> >> respect to the black hole.
> > Oh god, you do NOT want to know how much energy this is going to
> > take. Conservation of angular momentum is nasty here. Note that
> > not only will the inner mass get deflected by 180 degrees but it will
> > also be shifted sideways so there's a large moment arm.
> I think this is wrong on a couple of counts.
> Firstly, I don't actually care about the energy that is required. My
> spaceship has effectively unlimitted energy available. I am eliminating the
> need for reaction mass, not energy.
This doesn't make what I said the slightest bit wrong.
> Secondly, its energy neutral anyway.
No, it isn't.
> If the spaceship is moving at speed v
> at distance d from the black hole, then after the two masses separate and
> are each again at distance d from the black hole they will each still be
> moving at speed v, albeit moving in different directions. When the rope
> becomes taut, you use regenerative breaking to recover the kinetic energy
> and store it in a battery. Or you just let the rope "snap" taut, and
> disipate the kinetic energy as heat. Either way, the momentum equation is
> the same - both halves are moving at speed v but in almost opposite
> directions. Whether you recover that energy or not, the momentum of the two
> halves of the spaceship is almost equal in size and opposite in direction,
> so the net momentum is small.
You seem to think that you can just look at conservation of momentum
and this somehow makes conservation of angular momentum
irrelevant.
But it isn't. Both conservation of momentum and conservation of
angular momentum must always be satisfied.
If you define moving slowly as the center of gravity is moving slowly,
then congratulations--your spaceship is moving slowly. But at the
same time, both of the individual masses are moving very quickly.
Either that, or the individual masses are extremely far apart from
each other (assuming an unlimited length rope).
It's all about conservation of angular momentum.
> You will note that my gravity drive must dissipate energy somehow - either
> by charging a battery through regenerative breaking, or by throwing it away
> as friction, or some other mechanism. The whole point of this drive is to
> reduce speed; the energy has to go somewhere.
The problem isn't dissipating energy. The problem was always and
remains conservation of angular momentum.
> >> Do you agree that such a mechanism could be used to slow down or stop a
> >> spaceship, with a long enough rope and a deep enough gravity well? Even
> >> though the orbit of the spaceship (and indeed each half after it
> >> separates)
> >> is hyperbolic?
> > This depends on your definition of "slow down". What will
> > happen is that as you pull on the rope, the two masses will
> > continue "orbiting" around the black hole but will do so with
> > increasing speed. The action of the rope is essentially
> > massively increasing the inward pull. Assuming the rope
> > isn't perfectly centered and the masses aren't perfectly
> > moving in perfect sync, then the motion will actually be
> > rather chaotic. But assuming the masses neither collide
> > with each other or the black hole in the chaos, there's
> > no avoiding the severe increase in velocity to conserve
> > angular momentum.
> Indeed. But my argument is that the spaceship can potentially be captured by
> the black hole, despite it approaching in a hyperbolic orbit. It does this
> without reaction mass.
I'm not sure what amazing thing you think you've discovered
here. It's the equivalent of two bodies approaching a planet
in parallel and purposefully colliding on the far side. You can,
in principle, use some sort of magical rope or spring to
cushion this controlled collision.
Note that in this simpler example, the angular momentum
starts and ends as zero. The two bodies have the equal
and opposite angular momentums because the moment
arms are opposite vectors. At the end of the controlled
collision, the two bodies are sitting at zero velocity.
> Lets look at this in the inertial frame of the black hole. On approach, at
> distance d the two halves are together and each half has kinetic energy
> 0.5*m*v^2 and momentum mv. After the close approach, one half continues in
> more or less the same direction and has kinetic energy at distance d of
> 0.5mv^2, and momentum mv. The other half is whipped around the black hole
> and is deflected by almost 180 degrees. At distance d it also has energy
> 0.5*mv^2, but momentum -mv. When the rope snaps taut (or the relative speed
> difference is reduced to zero by regenerative breaking), the net momentum of
> the spaceship as a whole relative to the black hole is zero. With a perfect,
> inelastic rope the relative speed of the two halves will become the same,
> and be almost at rest with respect to the black hole.
You're ignoring conservation of angular momentum again.
In your example, the bodies start off with angular momentum.
That can't change. If you let a stupendously long rope snap taut,
then the relative velocities of the masses will NOT be zero.
Instead, the masses will be revolving around each other.
> A remark you made earlier provides a key insight as to why this works.
> Essentially, if this is a two body problem (the spaceship and the black
> hole), there is nothing you can do to reduce the speed or the magnitude of
> the momentum vector. By splitting the spaceship into two, connected by a
> rope, this is changed into a three body problem, with the two halves coupled
> by a rope rather than by mutual gravitational attraction. This allows a huge
> range of orbital solutions other than continuing in a hyperbolic orbit.
It only allows a huge range of solutions because of the magic
rope. With a real rope, the range becomes very extremely small.
> > No physical rope is that strong. And the physics
> > limitations of rope strength are what limits your drive
> > concept.
> So, with a strong enough and long enough rope it will work?
This depends on your definition of "work". If you mean it's
possible to produce a possibly measurable change in your
orbit (limited by the strength of the rope), then it works.
But if you mean you can keep on changing the orbit gradually
indefinitely, then it doesn't work. It will only work up until
the point where the masses revolve around each other
so quickly that the rope snaps.
> > You can trade angular momentum between
> > the rotation of the ship and the orbit's angular momentum,
> > but this only works until the rotation of the ship breaks
> > the strength limits of the rope.
> I know why you want to recast this as a problem in angular momentum. You
> want to use conservation of angular momentum to "prove" its impossible.
I don't "want" to. I already did it.
> You
> don't have to use angular momentum to demonstrate this seemingly paradoxical
> result - the same argument applies to "normal" momentum.
Sorry, but you can't ignore conservation of angular momentum.
Isaac Kuo
|
| | |
Date: 20 Sep 2007 17:24:17
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190248047.456924.22710@k79g2000hse.googlegroups.com...
> On Sep 19, 6:17 pm, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> "IsaacKuo" <mech...@yahoo.com> wrote in message
>> news:1190218209.411323.214890@d55g2000hsg.googlegroups.com...
>
>> >> > Bottom line--you need to explain how your proposed drive
>> >> > applies a torque on the Earth.
>
>> >> My spaceship doesn't need to apply any torque to the earth, because it
>> >> doesn't necessarily have to rotate. There is no need for transfer of
>> >> angular
>> >> momentum.
>
>> > Yes, you do. There is a massive amount of angular momentum in the
>> > orbit itself. If you want to raise or lower the orbit, you need to
>> > add or
>> > remove angular momentum to the orbit. There's just no way around it.
>
> I notice you didn't respond to this.
>
>> >> Here is a slightly different form of my gravity drive, used for the
>> >> purposes
>> >> of braking and not achieving orbital velocity, which may make the
>> >> central
>> >> physical mechanism clearer.
>
>> >> Imagine a hyperbolic orbit, such that your spaceship comes close
>> >> enough
>> >> to a
>> >> black hole for your outgoing path to be deflected say 10 degrees from
>> >> your
>> >> incoming path as it passes by the black hole. From a distance, you are
>> >> coming almost directly at the black hole. I put it to you that a
>> >> slight
>> >> variation on my drive can slow down the spaceship so that it is
>> >> captured
>> >> into an elliptical orbit around the black hole.
>
>> >> This is how I do it. When I am well away from the black hole, I split
>> >> my
>> >> spaceship in two halves, connected by a very, very long rope. One half
>> >> continues on its original orbit, which has its trajectory changed by
>> >> only
>> >> 10
>> >> degrees as a result of passing the black hole. The other half gets a
>> >> slight
>> >> sidewise nudge, so that it falls slightly deeper into the gravity well
>> >> and
>> >> is whipped around almost 180 degrees.
>
>> >> Both halves pass the black hole and move away from the black hole, but
>> >> one
>> >> half ends up with its course deflected by 10 degrees and the other by
>> >> almost
>> >> 180 degrees. They have identical speeds, but the two halves are moving
>> >> in
>> >> almost opposite directions. Considering the two halves as a single
>> >> system,
>> >> the total momentum with respect to the black hole is almost zero, as
>> >> the
>> >> two
>> >> halves have the same speed but in almost opposite directions.
>
>> >> They are connected by a rope. You use a motor to pull the rope in. By
>> >> the
>> >> conservation of momentum, the two halves will end up almost stationary
>> >> with
>> >> respect to the black hole.
>
>> > Oh god, you do NOT want to know how much energy this is going to
>> > take. Conservation of angular momentum is nasty here. Note that
>> > not only will the inner mass get deflected by 180 degrees but it will
>> > also be shifted sideways so there's a large moment arm.
>
>> I think this is wrong on a couple of counts.
>
>> Firstly, I don't actually care about the energy that is required. My
>> spaceship has effectively unlimitted energy available. I am eliminating
>> the
>> need for reaction mass, not energy.
>
> This doesn't make what I said the slightest bit wrong.
>
>> Secondly, its energy neutral anyway.
>
> No, it isn't.
>
>> If the spaceship is moving at speed v
>> at distance d from the black hole, then after the two masses separate and
>> are each again at distance d from the black hole they will each still be
>> moving at speed v, albeit moving in different directions. When the rope
>> becomes taut, you use regenerative breaking to recover the kinetic energy
>> and store it in a battery. Or you just let the rope "snap" taut, and
>> disipate the kinetic energy as heat. Either way, the momentum equation is
>> the same - both halves are moving at speed v but in almost opposite
>> directions. Whether you recover that energy or not, the momentum of the
>> two
>> halves of the spaceship is almost equal in size and opposite in
>> direction,
>> so the net momentum is small.
>
> You seem to think that you can just look at conservation of momentum
> and this somehow makes conservation of angular momentum
> irrelevant.
>
> But it isn't. Both conservation of momentum and conservation of
> angular momentum must always be satisfied.
>
> If you define moving slowly as the center of gravity is moving slowly,
> then congratulations--your spaceship is moving slowly. But at the
> same time, both of the individual masses are moving very quickly.
> Either that, or the individual masses are extremely far apart from
> each other (assuming an unlimited length rope).
>
> It's all about conservation of angular momentum.
>
>> You will note that my gravity drive must dissipate energy somehow -
>> either
>> by charging a battery through regenerative breaking, or by throwing it
>> away
>> as friction, or some other mechanism. The whole point of this drive is to
>> reduce speed; the energy has to go somewhere.
>
> The problem isn't dissipating energy. The problem was always and
> remains conservation of angular momentum.
>
>> >> Do you agree that such a mechanism could be used to slow down or stop
>> >> a
>> >> spaceship, with a long enough rope and a deep enough gravity well?
>> >> Even
>> >> though the orbit of the spaceship (and indeed each half after it
>> >> separates)
>> >> is hyperbolic?
>
>> > This depends on your definition of "slow down". What will
>> > happen is that as you pull on the rope, the two masses will
>> > continue "orbiting" around the black hole but will do so with
>> > increasing speed. The action of the rope is essentially
>> > massively increasing the inward pull. Assuming the rope
>> > isn't perfectly centered and the masses aren't perfectly
>> > moving in perfect sync, then the motion will actually be
>> > rather chaotic. But assuming the masses neither collide
>> > with each other or the black hole in the chaos, there's
>> > no avoiding the severe increase in velocity to conserve
>> > angular momentum.
>
>> Indeed. But my argument is that the spaceship can potentially be captured
>> by
>> the black hole, despite it approaching in a hyperbolic orbit. It does
>> this
>> without reaction mass.
>
> I'm not sure what amazing thing you think you've discovered
> here. It's the equivalent of two bodies approaching a planet
> in parallel and purposefully colliding on the far side. You can,
> in principle, use some sort of magical rope or spring to
> cushion this controlled collision.
>
> Note that in this simpler example, the angular momentum
> starts and ends as zero. The two bodies have the equal
> and opposite angular momentums because the moment
> arms are opposite vectors. At the end of the controlled
> collision, the two bodies are sitting at zero velocity.
>
>> Lets look at this in the inertial frame of the black hole. On approach,
>> at
>> distance d the two halves are together and each half has kinetic energy
>> 0.5*m*v^2 and momentum mv. After the close approach, one half continues
>> in
>> more or less the same direction and has kinetic energy at distance d of
>> 0.5mv^2, and momentum mv. The other half is whipped around the black hole
>> and is deflected by almost 180 degrees. At distance d it also has energy
>> 0.5*mv^2, but momentum -mv. When the rope snaps taut (or the relative
>> speed
>> difference is reduced to zero by regenerative breaking), the net momentum
>> of
>> the spaceship as a whole relative to the black hole is zero. With a
>> perfect,
>> inelastic rope the relative speed of the two halves will become the same,
>> and be almost at rest with respect to the black hole.
>
> You're ignoring conservation of angular momentum again.
> In your example, the bodies start off with angular momentum.
> That can't change. If you let a stupendously long rope snap taut,
> then the relative velocities of the masses will NOT be zero.
> Instead, the masses will be revolving around each other.
>
>> A remark you made earlier provides a key insight as to why this works.
>> Essentially, if this is a two body problem (the spaceship and the black
>> hole), there is nothing you can do to reduce the speed or the magnitude
>> of
>> the momentum vector. By splitting the spaceship into two, connected by a
>> rope, this is changed into a three body problem, with the two halves
>> coupled
>> by a rope rather than by mutual gravitational attraction. This allows a
>> huge
>> range of orbital solutions other than continuing in a hyperbolic orbit.
>
> It only allows a huge range of solutions because of the magic
> rope. With a real rope, the range becomes very extremely small.
>
>> > No physical rope is that strong. And the physics
>> > limitations of rope strength are what limits your drive
>> > concept.
>
>> So, with a strong enough and long enough rope it will work?
>
> This depends on your definition of "work". If you mean it's
> possible to produce a possibly measurable change in your
> orbit (limited by the strength of the rope), then it works.
> But if you mean you can keep on changing the orbit gradually
> indefinitely, then it doesn't work. It will only work up until
> the point where the masses revolve around each other
> so quickly that the rope snaps.
>
>> > You can trade angular momentum between
>> > the rotation of the ship and the orbit's angular momentum,
>> > but this only works until the rotation of the ship breaks
>> > the strength limits of the rope.
>
>> I know why you want to recast this as a problem in angular momentum. You
>> want to use conservation of angular momentum to "prove" its impossible.
>
> I don't "want" to. I already did it.
>
>> You
>> don't have to use angular momentum to demonstrate this seemingly
>> paradoxical
>> result - the same argument applies to "normal" momentum.
>
> Sorry, but you can't ignore conservation of angular momentum.
>
> Isaac Kuo
>
I didn't ignore angular momentum. You just snipped the bit where I did:
If we recast this in terms of angular momentum, this is what happens. The
spherical symmetry of the black hole means there is no obvious mechanism to
exchange angular momentum between the spaceship (or its two halves) and the
black hole, as you identified earlier. In fact, there is no exchange of
angular momentum with the black hole. The angular momentum of the spaceships
trajectory past the black hole is converted into angular momentum of the two
halves rotating around there common centre of gravity, as you previously
identified. Angular momentum is conserved. However linear momentum is not;
the two halves are on different trajectories and exchange a different amount
of linear momentum with the black hole. The sum of the momentums after the
close approach is different to the total momentum on the way down, because
the orbits of the two halves are different. This allows you to reduce the
spaceships momentum to close to zero in the inertial frame of reference of
the black hole.
I am still curious to know whether in the world of perfect ropes etc this
mechanism could be used to "brake" (and hence accelerate) a spaceship
without expending reaction mass. If it can, this is a pure gravity drive.
|
| |
Date: 19 Sep 2007 09:10:09
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Sep 19, 10:46 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> > You keep on making these claims, but I have yet to see you
> > demonstrate any method by which the proposed drive can
> > apply any torque on the Earth. Without this, conservation of
> > angular momentum implies your drive won't work.
> > Gyroscopes can have interesting effects, but they can't
> > create or destroy angular momentum.
> > Bottom line--you need to explain how your proposed drive
> > applies a torque on the Earth.
> My spaceship doesn't need to apply any torque to the earth, because it
> doesn't necessarily have to rotate. There is no need for transfer of angular
> momentum.
Yes, you do. There is a massive amount of angular momentum in the
orbit itself. If you want to raise or lower the orbit, you need to
add or
remove angular momentum to the orbit. There's just no way around it.
> Here is a slightly different form of my gravity drive, used for the purposes
> of braking and not achieving orbital velocity, which may make the central
> physical mechanism clearer.
> Imagine a hyperbolic orbit, such that your spaceship comes close enough to a
> black hole for your outgoing path to be deflected say 10 degrees from your
> incoming path as it passes by the black hole. From a distance, you are
> coming almost directly at the black hole. I put it to you that a slight
> variation on my drive can slow down the spaceship so that it is captured
> into an elliptical orbit around the black hole.
> This is how I do it. When I am well away from the black hole, I split my
> spaceship in two halves, connected by a very, very long rope. One half
> continues on its original orbit, which has its trajectory changed by only 10
> degrees as a result of passing the black hole. The other half gets a slight
> sidewise nudge, so that it falls slightly deeper into the gravity well and
> is whipped around almost 180 degrees.
> Both halves pass the black hole and move away from the black hole, but one
> half ends up with its course deflected by 10 degrees and the other by almost
> 180 degrees. They have identical speeds, but the two halves are moving in
> almost opposite directions. Considering the two halves as a single system,
> the total momentum with respect to the black hole is almost zero, as the two
> halves have the same speed but in almost opposite directions.
> They are connected by a rope. You use a motor to pull the rope in. By the
> conservation of momentum, the two halves will end up almost stationary with
> respect to the black hole.
Oh god, you do NOT want to know how much energy this is going to
take. Conservation of angular momentum is nasty here. Note that
not only will the inner mass get deflected by 180 degrees but it will
also be shifted sideways so there's a large moment arm.
When pulling on the rope, angular momentum must be conserved.
This means that however fast your approach velocity was, multiply
that velocity by the ratio with which you're reducing the rope
length. If you reduce the rope length to 1/100 of the original
length, then you increase the velocity of the weights by 100.
This requires energy input equal to 9999 times the original
kinetic energy.
> Do you agree that such a mechanism could be used to slow down or stop a
> spaceship, with a long enough rope and a deep enough gravity well? Even
> though the orbit of the spaceship (and indeed each half after it separates)
> is hyperbolic?
This depends on your definition of "slow down". What will
happen is that as you pull on the rope, the two masses will
continue "orbiting" around the black hole but will do so with
increasing speed. The action of the rope is essentially
massively increasing the inward pull. Assuming the rope
isn't perfectly centered and the masses aren't perfectly
moving in perfect sync, then the motion will actually be
rather chaotic. But assuming the masses neither collide
with each other or the black hole in the chaos, there's
no avoiding the severe increase in velocity to conserve
angular momentum.
Of course, I'm assuming physics thought experiment
equipment--the infinitely strong rope is required.
No physical rope is that strong. And the physics
limitations of rope strength are what limits your drive
concept. You can trade angular momentum between
the rotation of the ship and the orbit's angular momentum,
but this only works until the rotation of the ship breaks
the strength limits of the rope.
Isaac Kuo
|
| | |
Date: 20 Sep 2007 09:17:57
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190218209.411323.214890@d55g2000hsg.googlegroups.com...
> On Sep 19, 10:46 am, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> > You keep on making these claims, but I have yet to see you
>> > demonstrate any method by which the proposed drive can
>> > apply any torque on the Earth. Without this, conservation of
>> > angular momentum implies your drive won't work.
>
>> > Gyroscopes can have interesting effects, but they can't
>> > create or destroy angular momentum.
>
>> > Bottom line--you need to explain how your proposed drive
>> > applies a torque on the Earth.
>
>> My spaceship doesn't need to apply any torque to the earth, because it
>> doesn't necessarily have to rotate. There is no need for transfer of
>> angular
>> momentum.
>
> Yes, you do. There is a massive amount of angular momentum in the
> orbit itself. If you want to raise or lower the orbit, you need to
> add or
> remove angular momentum to the orbit. There's just no way around it.
>
>> Here is a slightly different form of my gravity drive, used for the
>> purposes
>> of braking and not achieving orbital velocity, which may make the central
>> physical mechanism clearer.
>
>> Imagine a hyperbolic orbit, such that your spaceship comes close enough
>> to a
>> black hole for your outgoing path to be deflected say 10 degrees from
>> your
>> incoming path as it passes by the black hole. From a distance, you are
>> coming almost directly at the black hole. I put it to you that a slight
>> variation on my drive can slow down the spaceship so that it is captured
>> into an elliptical orbit around the black hole.
>
>> This is how I do it. When I am well away from the black hole, I split my
>> spaceship in two halves, connected by a very, very long rope. One half
>> continues on its original orbit, which has its trajectory changed by only
>> 10
>> degrees as a result of passing the black hole. The other half gets a
>> slight
>> sidewise nudge, so that it falls slightly deeper into the gravity well
>> and
>> is whipped around almost 180 degrees.
>
>> Both halves pass the black hole and move away from the black hole, but
>> one
>> half ends up with its course deflected by 10 degrees and the other by
>> almost
>> 180 degrees. They have identical speeds, but the two halves are moving in
>> almost opposite directions. Considering the two halves as a single
>> system,
>> the total momentum with respect to the black hole is almost zero, as the
>> two
>> halves have the same speed but in almost opposite directions.
>
>> They are connected by a rope. You use a motor to pull the rope in. By the
>> conservation of momentum, the two halves will end up almost stationary
>> with
>> respect to the black hole.
>
> Oh god, you do NOT want to know how much energy this is going to
> take. Conservation of angular momentum is nasty here. Note that
> not only will the inner mass get deflected by 180 degrees but it will
> also be shifted sideways so there's a large moment arm.
>
I think this is wrong on a couple of counts.
Firstly, I don't actually care about the energy that is required. My
spaceship has effectively unlimitted energy available. I am eliminating the
need for reaction mass, not energy.
Secondly, its energy neutral anyway. If the spaceship is moving at speed v
at distance d from the black hole, then after the two masses separate and
are each again at distance d from the black hole they will each still be
moving at speed v, albeit moving in different directions. When the rope
becomes taut, you use regenerative breaking to recover the kinetic energy
and store it in a battery. Or you just let the rope "snap" taut, and
disipate the kinetic energy as heat. Either way, the momentum equation is
the same - both halves are moving at speed v but in almost opposite
directions. Whether you recover that energy or not, the momentum of the two
halves of the spaceship is almost equal in size and opposite in direction,
so the net momentum is small.
You will note that my gravity drive must dissipate energy somehow - either
by charging a battery through regenerative breaking, or by throwing it away
as friction, or some other mechanism. The whole point of this drive is to
reduce speed; the energy has to go somewhere.
> When pulling on the rope, angular momentum must be conserved.
> This means that however fast your approach velocity was, multiply
> that velocity by the ratio with which you're reducing the rope
> length. If you reduce the rope length to 1/100 of the original
> length, then you increase the velocity of the weights by 100.
> This requires energy input equal to 9999 times the original
> kinetic energy.
>
Again, I don't care about the energy equation, I care about the momentum
equation. In any event, if the two halves are moving in opposite directions
but connected by a rope , then this energy is recoverable anyway by using
the extending rope to drive a generator.
>> Do you agree that such a mechanism could be used to slow down or stop a
>> spaceship, with a long enough rope and a deep enough gravity well? Even
>> though the orbit of the spaceship (and indeed each half after it
>> separates)
>> is hyperbolic?
>
> This depends on your definition of "slow down". What will
> happen is that as you pull on the rope, the two masses will
> continue "orbiting" around the black hole but will do so with
> increasing speed. The action of the rope is essentially
> massively increasing the inward pull. Assuming the rope
> isn't perfectly centered and the masses aren't perfectly
> moving in perfect sync, then the motion will actually be
> rather chaotic. But assuming the masses neither collide
> with each other or the black hole in the chaos, there's
> no avoiding the severe increase in velocity to conserve
> angular momentum.
>
Indeed. But my argument is that the spaceship can potentially be captured by
the black hole, despite it approaching in a hyperbolic orbit. It does this
without reaction mass.
Lets look at this in the inertial frame of the black hole. On approach, at
distance d the two halves are together and each half has kinetic energy
0.5*m*v^2 and momentum mv. After the close approach, one half continues in
more or less the same direction and has kinetic energy at distance d of
0.5mv^2, and momentum mv. The other half is whipped around the black hole
and is deflected by almost 180 degrees. At distance d it also has energy
0.5*mv^2, but momentum -mv. When the rope snaps taut (or the relative speed
difference is reduced to zero by regenerative breaking), the net momentum of
the spaceship as a whole relative to the black hole is zero. With a perfect,
inelastic rope the relative speed of the two halves will become the same,
and be almost at rest with respect to the black hole.
A remark you made earlier provides a key insight as to why this works.
Essentially, if this is a two body problem (the spaceship and the black
hole), there is nothing you can do to reduce the speed or the magnitude of
the momentum vector. By splitting the spaceship into two, connected by a
rope, this is changed into a three body problem, with the two halves coupled
by a rope rather than by mutual gravitational attraction. This allows a huge
range of orbital solutions other than continuing in a hyperbolic orbit.
More generally, this is how my gravity drives works. Because the two masses
can move independently (up to the limit of the length of the rope or the
length of the connecting rod), its really a three body problem disguised as
a two body problem. This allows additional solutions to the equations of
motion. Specifically each half exchanges a different amount of momentum with
the black hole depending upon its exact orbit. If you are recovering energy
by using the extending rope to power a motor when they are separating, then
you will recover all the kinetic energy, but the total momentum of the
system is now almost zero (0.5*m*v plus 0.5*m*(-v)).
> Of course, I'm assuming physics thought experiment
> equipment--the infinitely strong rope is required.
>
Of course.
> No physical rope is that strong. And the physics
> limitations of rope strength are what limits your drive
> concept.
So, with a strong enough and long enough rope it will work?
> You can trade angular momentum between
> the rotation of the ship and the orbit's angular momentum,
> but this only works until the rotation of the ship breaks
> the strength limits of the rope.
>
I know why you want to recast this as a problem in angular momentum. You
want to use conservation of angular momentum to "prove" its impossible. You
don't have to use angular momentum to demonstrate this seemingly paradoxical
result - the same argument applies to "normal" momentum. On approach, when
the masses are together, the momentum of the spaceship is 0.5mv + 0.5mv.
After the two halves have passed the black hole, one mass continuues in an
almost unchanged trajectory (momentum 0.5mv) and the other has the same
speed but almost opposite direction of travel (momentum -0.5mv). The rope
snaps taut, or the kinetic energy is recovered using regenerative braking,
and the total momentum relative to the black hole is zero. The spaceship
comes to rest (or close to it) relative to the black hole. A hyperbolic
orbit has been changed to an elliptic orbit, albeit possibly with the
spaceship also rotating around its centre of gravity.
If we recast this in terms of angular momentum, this is what happens. The
spherical symmetry of the black hole means there is no obvious mechanism to
exchange angular momentum between the spaceship (or its two halves) and the
black hole, as you identified earlier. In fact, there is no exchange of
angular momentum with the black hole. The angular momentum of the spaceships
trajectory past the black hole is converted into angular momentum of the two
halves rotating around there common centre of gravity, as you previously
identified. Angular momentum is conserved. However linear momentum is not;
the two halves are on different trajectories and exchange a different amount
of linear momentum with the black hole. The sum of the momentums after the
close approach is different to the total momentum on the way down, because
the orbits of the two halves are different. This allows you to reduce the
spaceships momentum to close to zero in the inertial frame of reference of
the black hole.
Again, I stress that allowing the two masses to move independently changes a
two body problem into a three body problem, and hence derive many more
solutions to the equations of motion. The solution you get depends upon the
boundary conditions - the length of the rope, whether you use the rope to to
drive a generator to bleed off relative speed (which determines the
"attraction" between the two halves), etc.
In my original drive design, I limitted the distance between the masses to a
maximum of 100 kms for a satellite orbiting the earth in an elliptic orbit.
Because 100 kms is a short length compared to the diameter of the earth, the
trajectories of the two masses cannot be wildly different and you can't
recover generate much net acceleration - I calculated about 0.09 m/s^2 in a
previous post. This is a "2.001 body problem" - its almost but not quite a
two body problem. With a long enough rope - say many times larger than the
diameter of the black hole - you can generate enough acceleration to almost
completely stop the spaceship when the masses are again reconnected (through
the rope snapping taut, or by using the extending rope to recover the energy
until the two halves are at relative rest). This is a three body solution,
at least until you run out of rope. At that point, it becomes a two body
problem again, but the time that the two masses were separated allowed them
to independently exchange a different amount of momentum with the black hole
(depending on the orbit they take) which means the momentum of the spaceship
as a whole is changed.
Energy is conserved; momentum is conserved; angular momentum is conserved.
But a spaceship on a hyperbolic orbit on the way down ends up in orbit
around the black hole; its net momentum is reduced (although the momentum of
the black hole is increased by the same amount). My gravity drive has
operated as a "brake".
Do you agree that in the perfect world of thought experiments with perfectly
flat billiard tables and arbitrarily long and strong ropes of zero weight,
this arrangement can be used to provide a net acceleration to the spaceship
(in this case, reducing its net momentum relative to the black hole to
almost zero)? If so, its only a tiny step to using a similar argument/design
to generate acceleration in the direction of motion (instead of the opposite
direction), as does the original formulation of my gravity drive.
|
| |
Date: 19 Sep 2007 07:57:23
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
One aside--I'm generally sympathetic to novel space drive concepts
based on known physics, and I greatly empathize with your frustration
at having to repeatedly explain the concept to people who simply
"don't get it". I've had that experience many times, both with my own
concepts and with explaining other concepts. My current pet peeve
is constantly having to re-explain Dr. Young Bae's "Photonic Laser
Thruster" concept.
Your drive concept is something which is based on normal physics
principles and isn't "kooky"...and I remain convinced that something
involving shifting masses should work in conjunction with three
body orbital mechanics (i.e. Earth+Moon+spacecraft). I just think
the drive concept you're presenting in a two body situation doesn't
work.
Isaac Kuo
|
| |
Date: 19 Sep 2007 07:46:01
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Sep 19, 9:16 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> You seem to be arguing two completely opposite points of view.
> On one hand, you say that all that will happen is that the masses will start
> rotating or tumbling. This obviously involves a transfer of angular momentum
> from the earth to the spaceship.
No, it doesn't. It involves trading angular momentum of the orbit
with
the angular momentum of the ship's rotation. However, the change
in the orbit is insignificant because the moment arm is so much
bigger.
> However, then you say that the spaceship can't rob the earth of angular
> momentum.
> Obviously, only one of these can be true.
> I believe the whole subject of the spaceship tumbling/rotating to be a
> complete red herring. Firstly, this is potentially controllable through
> gyroscopes, and as I am only discussing "in principle" I can easily put them
> into my spaceship. Secondly, the arrangement will still generate lift (in
> theory) even if the masses are also revolving around their centre of
> gravity. Thirdly, if you tweak the geometry to deliberately cause the
> spaceship to start revolving, you have a similar paradoxical result except
> it applies to angular momentum instead (linear) momentum - the gravity drive
> has transferred momentum from a spherically symmetric earth to the
> spaceship.
You keep on making these claims, but I have yet to see you
demonstrate any method by which the proposed drive can
apply any torque on the Earth. Without this, conservation of
angular momentum implies your drive won't work.
Gyroscopes can have interesting effects, but they can't
create or destroy angular momentum.
Bottom line--you need to explain how your proposed drive
applies a torque on the Earth.
I repeat--explain how your proposed drive applies a torque
on the Earth.
Isaac Kuo
|
| | |
Date: 20 Sep 2007 01:46:51
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
> You keep on making these claims, but I have yet to see you
> demonstrate any method by which the proposed drive can
> apply any torque on the Earth. Without this, conservation of
> angular momentum implies your drive won't work.
>
> Gyroscopes can have interesting effects, but they can't
> create or destroy angular momentum.
>
> Bottom line--you need to explain how your proposed drive
> applies a torque on the Earth.
>
> I repeat--explain how your proposed drive applies a torque
> on the Earth.
>
> Isaac Kuo
>
My spaceship doesn't need to apply any torque to the earth, because it
doesn't necessarily have to rotate. There is no need for transfer of angular
momentum.
Here is a slightly different form of my gravity drive, used for the purposes
of braking and not achieving orbital velocity, which may make the central
physical mechanism clearer.
Imagine a hyperbolic orbit, such that your spaceship comes close enough to a
black hole for your outgoing path to be deflected say 10 degrees from your
incoming path as it passes by the black hole. From a distance, you are
coming almost directly at the black hole. I put it to you that a slight
variation on my drive can slow down the spaceship so that it is captured
into an elliptical orbit around the black hole.
This is how I do it. When I am well away from the black hole, I split my
spaceship in two halves, connected by a very, very long rope. One half
continues on its original orbit, which has its trajectory changed by only 10
degrees as a result of passing the black hole. The other half gets a slight
sidewise nudge, so that it falls slightly deeper into the gravity well and
is whipped around almost 180 degrees.
Both halves pass the black hole and move away from the black hole, but one
half ends up with its course deflected by 10 degrees and the other by almost
180 degrees. They have identical speeds, but the two halves are moving in
almost opposite directions. Considering the two halves as a single system,
the total momentum with respect to the black hole is almost zero, as the two
halves have the same speed but in almost opposite directions.
They are connected by a rope. You use a motor to pull the rope in. By the
conservation of momentum, the two halves will end up almost stationary with
respect to the black hole.
Do you agree that such a mechanism could be used to slow down or stop a
spaceship, with a long enough rope and a deep enough gravity well? Even
though the orbit of the spaceship (and indeed each half after it separates)
is hyperbolic?
|
| | | |
Date: 19 Sep 2007 16:08:26
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f14451$0$1030$afc38c87@news.optusnet.com.au...
: > You keep on making these claims, but I have yet to see you
: > demonstrate any method by which the proposed drive can
: > apply any torque on the Earth. Without this, conservation of
: > angular momentum implies your drive won't work.
: >
: > Gyroscopes can have interesting effects, but they can't
: > create or destroy angular momentum.
: >
: > Bottom line--you need to explain how your proposed drive
: > applies a torque on the Earth.
: >
: > I repeat--explain how your proposed drive applies a torque
: > on the Earth.
: >
: > Isaac Kuo
: >
:
: My spaceship doesn't need to apply any torque to the earth,
Not only do you want perpetual motion, you want to gain
free energy as well, and if you don't, where are you getting it from?
|
| |
Date: 19 Sep 2007 06:05:24
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Sep 19, 2:37 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> "IsaacKuo" <mech...@yahoo.com> wrote in message
> news:1190183492.129890.306150@g4g2000hsf.googlegroups.com...
> > Your drive doesn't work because of conservation of angular
> > momentum.
> My drive is not intended to rotate at all. If the spaceship does rotate, its
> an unintended consequence.
As far as I can tell, the only thing your drive achieves is this
unintended consequence. You're simply wishing that the
energy pumped in goes into orbital energy, when it actually
goes into rotation of the spacecraft.
> If you mean that the angular momentum of the spaceship as a whole has
> increased because its moving faster, this is not neccesarily true. It would
> also depend upon the eccentricity of the orbit. If it does increase, this
> doesn't really worry me; its robbing the earth of angular momentum, just as
> the moon is.
The problem is that you don't demonstrate any method by
which the Earth is robbed of angular momentum. If
the ship doesn't put any torque on the Earth, the ship
can't rob the Earth of angular momentum. It's as simple
as that.
> Let me ask you a question.
> If you have two masses separated by a 100kms at apogee, pointing towards the
> earth, and when you reach perigee (the bottom of the orbit) you pull them
> together, then significant work is done. When the two masses are at apogee
> again, you allow them to separate so they are 100 kms apart again apart.
> This provides less energy than was used to pull them together at apogee, as
> the gravitational gradient is less. Where has the excess energy that you
> used to pull the masses together gone?
It mostly goes into revolving the masses around each other.
> Why do you think my drive won't work? It doesn't seem to contravene the laws
> of conservation of momentum, angular momentum and energy. Where is the flaw
> in my (theoretical) design?
Conservation of angular momentum. Unless you can figure out
how the drive puts a torque on the Earth (which it might), then it
can't
be robbing the Earth of angular momentum.
Isaac Kuo
|
| | |
Date: 20 Sep 2007 00:16:27
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190207124.057797.32060@g4g2000hsf.googlegroups.com...
> On Sep 19, 2:37 am, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> "IsaacKuo" <mech...@yahoo.com> wrote in message
>> news:1190183492.129890.306150@g4g2000hsf.googlegroups.com...
>
>> > Your drive doesn't work because of conservation of angular
>> > momentum.
>
>> My drive is not intended to rotate at all. If the spaceship does rotate,
>> its
>> an unintended consequence.
>
> As far as I can tell, the only thing your drive achieves is this
> unintended consequence. You're simply wishing that the
> energy pumped in goes into orbital energy, when it actually
> goes into rotation of the spacecraft.
>
>> If you mean that the angular momentum of the spaceship as a whole has
>> increased because its moving faster, this is not neccesarily true. It
>> would
>> also depend upon the eccentricity of the orbit. If it does increase, this
>> doesn't really worry me; its robbing the earth of angular momentum, just
>> as
>> the moon is.
>
> The problem is that you don't demonstrate any method by
> which the Earth is robbed of angular momentum. If
> the ship doesn't put any torque on the Earth, the ship
> can't rob the Earth of angular momentum. It's as simple
> as that.
>
>> Let me ask you a question.
>
>> If you have two masses separated by a 100kms at apogee, pointing towards
>> the
>> earth, and when you reach perigee (the bottom of the orbit) you pull them
>> together, then significant work is done. When the two masses are at
>> apogee
>> again, you allow them to separate so they are 100 kms apart again apart.
>> This provides less energy than was used to pull them together at apogee,
>> as
>> the gravitational gradient is less. Where has the excess energy that you
>> used to pull the masses together gone?
>
> It mostly goes into revolving the masses around each other.
>
>> Why do you think my drive won't work? It doesn't seem to contravene the
>> laws
>> of conservation of momentum, angular momentum and energy. Where is the
>> flaw
>> in my (theoretical) design?
>
> Conservation of angular momentum. Unless you can figure out
> how the drive puts a torque on the Earth (which it might), then it
> can't
> be robbing the Earth of angular momentum.
>
> Isaac Kuo
>
You seem to be arguing two completely opposite points of view.
On one hand, you say that all that will happen is that the masses will start
rotating or tumbling. This obviously involves a transfer of angular momentum
from the earth to the spaceship.
However, then you say that the spaceship can't rob the earth of angular
momentum.
Obviously, only one of these can be true.
I believe the whole subject of the spaceship tumbling/rotating to be a
complete red herring. Firstly, this is potentially controllable through
gyroscopes, and as I am only discussing "in principle" I can easily put them
into my spaceship. Secondly, the arrangement will still generate lift (in
theory) even if the masses are also revolving around their centre of
gravity. Thirdly, if you tweak the geometry to deliberately cause the
spaceship to start revolving, you have a similar paradoxical result except
it applies to angular momentum instead (linear) momentum - the gravity drive
has transferred momentum from a spherically symmetric earth to the
spaceship.
|
| |
Date: 18 Sep 2007 23:31:32
From: IsaacKuo
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Sep 18, 11:34 pm, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> "Sam Wormley" <sworml...@mchsi.com> wrote in message
> news:2q1Ii.109840$Fc.38232@attbi_s21...
> > Peter Webb wrote:
> >> There is already a "spaceship" which uses a similar mechanism - earth's
> >> moon. Its altitude is increasing due to differential gravity between the
> >> near side and the far side of the earth.
> > No... Friction of the rotating earth under the tidal bulges gives
> > the bulges a positive phase with respect to the earth-moon line
> > "leading" the moon into a higher orbit.
> > Conservation of Angular Momentum.
> I said it was a "similar mechanism", not the same mechanism. The moons orbit
> increased without it expending reaction mass; its merely a demonstration
> that the concept is not impossible.
The moon gains angular momentum at the expense of the Earth
losing angular momentum. This is due to the torque applied to
the Earth by the tides.
> Do you understand how my drive works? If you don't, what is unclear? If you
> do, can you see any theoretical reason it won't work?
Your drive doesn't work because of conservation of angular
momentum. Your drive doesn't apply any torque on the Earth.
As such, you can only increase or decrease the orbit by the
insignificant amount you can counterbalance with the angular
momentum of the rotating spacecraft.
Isaac Kuo
|
| | |
Date: 19 Sep 2007 17:37:30
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"IsaacKuo" <mechdan@yahoo.com > wrote in message
news:1190183492.129890.306150@g4g2000hsf.googlegroups.com...
> On Sep 18, 11:34 pm, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> "Sam Wormley" <sworml...@mchsi.com> wrote in message
>> news:2q1Ii.109840$Fc.38232@attbi_s21...
>
>> > Peter Webb wrote:
>
>> >> There is already a "spaceship" which uses a similar mechanism -
>> >> earth's
>> >> moon. Its altitude is increasing due to differential gravity between
>> >> the
>> >> near side and the far side of the earth.
>
>> > No... Friction of the rotating earth under the tidal bulges gives
>> > the bulges a positive phase with respect to the earth-moon line
>> > "leading" the moon into a higher orbit.
>
>> > Conservation of Angular Momentum.
>
>> I said it was a "similar mechanism", not the same mechanism. The moons
>> orbit
>> increased without it expending reaction mass; its merely a demonstration
>> that the concept is not impossible.
>
> The moon gains angular momentum at the expense of the Earth
> losing angular momentum. This is due to the torque applied to
> the Earth by the tides.
>
>> Do you understand how my drive works? If you don't, what is unclear? If
>> you
>> do, can you see any theoretical reason it won't work?
>
> Your drive doesn't work because of conservation of angular
> momentum.
My drive is not intended to rotate at all. If the spaceship does rotate, its
an unintended consequence.
If you mean that the angular momentum of the spaceship as a whole has
increased because its moving faster, this is not neccesarily true. It would
also depend upon the eccentricity of the orbit. If it does increase, this
doesn't really worry me; its robbing the earth of angular momentum, just as
the moon is.
> Your drive doesn't apply any torque on the Earth.
> As such, you can only increase or decrease the orbit by the
> insignificant amount you can counterbalance with the angular
> momentum of the rotating spacecraft.
>
> Isaac Kuo
>
Its not intended to transfer angular momentum as such.
Let me ask you a question.
If you have two masses separated by a 100kms at apogee, pointing towards the
earth, and when you reach perigee (the bottom of the orbit) you pull them
together, then significant work is done. When the two masses are at apogee
again, you allow them to separate so they are 100 kms apart again apart.
This provides less energy than was used to pull them together at apogee, as
the gravitational gradient is less. Where has the excess energy that you
used to pull the masses together gone?
Why do you think my drive won't work? It doesn't seem to contravene the laws
of conservation of momentum, angular momentum and energy. Where is the flaw
in my (theoretical) design?
|
| | | |
Date: 19 Sep 2007 21:54:20
From: Andrew Smallshaw
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On 2007-09-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au > wrote:
>
> If you have two masses separated by a 100kms at apogee, pointing towards the
> earth, and when you reach perigee (the bottom of the orbit) you pull them
> together, then significant work is done. When the two masses are at apogee
> again, you allow them to separate so they are 100 kms apart again apart.
> This provides less energy than was used to pull them together at apogee, as
> the gravitational gradient is less. Where has the excess energy that you
> used to pull the masses together gone?
>
> Why do you think my drive won't work? It doesn't seem to contravene the laws
> of conservation of momentum, angular momentum and energy. Where is the flaw
> in my (theoretical) design?
I have to admit that I don't completely understand your proposal,
but the problem seems to lie right here: when you pull the two
masses together, yes, you are doing work. What happens when you
stop them again, either by them colliding with each other or via
some other mechanism? You are doing work again. The momentum of
the two operations cancels each other out. Bear in mind that
momentum is proportional to mass, not weight, and so any changes
to the strength of gravity are irrelevant to the conservation of
momemtum. It's here that I believe your logic is lacking.
Things get slightly more interesting if we allow one mass to to
pass through the other (e.g. one of the masses is hoop-shaped), so
that we can seperate them again without needing to do any more
work. However, to repeat the process we them have to apply two
lots of momemtum to each body. One lot to bring them back to rest
with respect to each other, and a second lot to send them back in
the opposite direction. After applying the first lot we are
basically back to square one, albeit with the masses at opposite
sides of the system. Repeat the process again and everything is as
it was in the first instance.
--
Andrew Smallshaw
andrews@sdf.lonestar.org
|
| | | | |
Date: 20 Sep 2007 23:08:48
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Andrew Smallshaw" <andrews@sdf.lonestar.org > wrote in message
news:slrnff36jl.qoa.andrews@sdf.lonestar.org...
> On 2007-09-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
>>
>> If you have two masses separated by a 100kms at apogee, pointing towards
>> the
>> earth, and when you reach perigee (the bottom of the orbit) you pull them
>> together, then significant work is done. When the two masses are at
>> apogee
>> again, you allow them to separate so they are 100 kms apart again apart.
>> This provides less energy than was used to pull them together at apogee,
>> as
>> the gravitational gradient is less. Where has the excess energy that you
>> used to pull the masses together gone?
>>
>> Why do you think my drive won't work? It doesn't seem to contravene the
>> laws
>> of conservation of momentum, angular momentum and energy. Where is the
>> flaw
>> in my (theoretical) design?
>
> I have to admit that I don't completely understand your proposal,
> but the problem seems to lie right here: when you pull the two
> masses together, yes, you are doing work. What happens when you
> stop them again, either by them colliding with each other or via
> some other mechanism? You are doing work again. The momentum of
> the two operations cancels each other out.
No, or at least this is irrelevant.
I gave in another post a different description of a very similar mechanism
used for braking (which is of course also a form of acceleration).
My drive works by transfering momentum from the earth to the spaceship.
Here's how it does it.
If you have a single mass in an elliptical orbit around the earth, it
constantly exchanges momentum with the earth. For example, it moves quickly
at perigee and slowly at apogee. The laws of physics (and in particular the
conservation of energy) says that if the satellite is at distance d from the
earth, it will always have the same speed v, irrespective of the path
(orbit) that it takes to get to that distance d. Because the speed is a
function only of the distance d from the earth, so is the magnitude of the
momentum vector. While the size of the momentum vector at distance d is
independent of the path, the direction the vector points certainly does
depend upon the path taken.
If you have two masses 100 kms apart, they will try and take two slightly
different orbits around the earth. This will be limitted to 100 kms
different, but it will still be different. By the time the two masses are
again at apogee, the momentum vectors for each of the two masses will have
the same magnitude, but slightly different directions. When you add these -
by considering the two masses as a single connected object - the magnitude
of the sum of the momentum vectors is slightly less than the sum of the
magnitudes, as the vectors point in slightly different directions. This is
exactly what happens when you use your motor to slowly bring the masses
together. Your spaceship has lost momentum, and hence speed. This is a
second order effect. If the distance between the masses is very small
compared to the distance from the centre of the earth, then its a very small
effect indeed.
For a 100 kms separation ("kms" = "kilometres" where I come from), and an
eccentric orbit with a perigee of 7,000 kms from the centre of the earth, I
calculate you get a sustained acceleration of about 0.09 m/s. This is about
1% of the acceleration due to gravity at this altitude.
This is small, but is achieved without using any reaction mass, which makes
it a "gravity drive".
You can use this mechanism to lose momentum, and hence provide braking,
again without expending any reaction mass. You can also run this in reverse
to increase momentum/speed/energy, by using a motor to force the masses
apart when the natural orbit is trying to pull them together.
> Bear in mind that
> momentum is proportional to mass, not weight, and so any changes
> to the strength of gravity are irrelevant to the conservation of
> momemtum. It's here that I believe your logic is lacking.
>
No. When you are in orbit around the earth, you are constantly exchanging
momentum with the earth, because there is a force between the satellite and
the earth (gravity). If you have two connected masses orbiting the earth,
then the force of gravity (and hence the amount of momentum exchanged) will
typically be different for the two bodies. By changing the distance between
the masses, you can change the magnitude of the effect at different points
in the orbit and use it to adjust the momentum of the spaceship as a whole.
All of the stuff you/we learned at school about two bodies interacting under
gravity - for example the only possible orbits being conic sections
(circular, elliptical, parabolic or hyperbolic) really only apply to point
or spherical masses. This has never before been a practical problem. All of
the large, rigid bodies in the universe are spheres. However, we can now
contemplate building a space station consisting of two large masses,
separated by a 100 kms tether and revolving around their common centre of
gravity - a dumbell shape. That would not have an elliptical orbit; not even
the mid point of the rope would have an elliptical orbit. It does not orbit
as a single body concentrated at its "centre of gravity"; because the force
of gravity falls off as the inverse square (and not linearly) its a far more
chaotic motion.
This is another way of looking at my "gravity drive". You allow the masses
to be together and hence have an elliptical orbit when it suits you, and you
seperate them and hence use the equations of motion of a large dumbell when
this suits you better.
If you are trying to achieve an increase in altitude/momentum, the strategy
I used for my calculation (and I believe the optimal strategy) is to
separate the masses by the maximim possible distance on the way "down", and
pull them together on the way up. You will achieve higher and higher apogees
on each orbit, until you reach escape velocity. You haven't violated
conservation of energy, because you are expending more energy pulling the
masses together than you could have gained even in theory from letting them
separate. You are continually pumping energy into the system.
You are, however, continually increasing your speed without apparently
"pushing against" something - a reaction mass. In reality, the push comes
from the fact that the two masses have different gravitational attraction
with the earth, and the equal and opposite reaction is against the earth
itself. Hence my term "gravity drive", because it works in a sense by using
gravity to "push" against the earth.
The unbelievably great thing is of course that it requires no reaction mass,
only energy which you can get from solar panels. Its slow, but it will run
for ever. Its like a sail boat compared to a power boat. You could use this
effect to boost your altitude enough to get into a gravtitational slingshot
around Venus or Mars, and from their you can coast to Jupiter. Then you have
a much deeper gravity well to play with, and you can potentially scoot
around the outer solar system quite quickly.
I not only believe this is possible in theory, I believe it could even be
very practical. We cannot deploy a sattelite consisting two large masses
joined by a 100kms retractable tether at the moment, but I can certainly
believe that one day we will be able to. At that point, we have built a
gravity drive.
|
| | | | |
Date: 20 Sep 2007 11:38:24
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Andrew Smallshaw" <andrews@sdf.lonestar.org > wrote in message
news:slrnff36jl.qoa.andrews@sdf.lonestar.org...
: On 2007-09-19, Peter Webb <webbfamily@DIESPAMDIEoptusnet.com.au > wrote:
: >
: > If you have two masses separated by a 100kms at apogee, pointing towards
the
: > earth, and when you reach perigee (the bottom of the orbit) you pull
them
: > together, then significant work is done. When the two masses are at
apogee
: > again, you allow them to separate so they are 100 kms apart again apart.
: > This provides less energy than was used to pull them together at apogee,
as
: > the gravitational gradient is less. Where has the excess energy that you
: > used to pull the masses together gone?
: >
: > Why do you think my drive won't work? It doesn't seem to contravene the
laws
: > of conservation of momentum, angular momentum and energy. Where is the
flaw
: > in my (theoretical) design?
:
: I have to admit that I don't completely understand your proposal,
: but the problem seems to lie right here: when you pull the two
: masses together, yes, you are doing work. What happens when you
: stop them again, either by them colliding with each other or via
: some other mechanism? You are doing work again. The momentum of
: the two operations cancels each other out. Bear in mind that
: momentum is proportional to mass, not weight, and so any changes
: to the strength of gravity are irrelevant to the conservation of
: momemtum. It's here that I believe your logic is lacking.
:
: Things get slightly more interesting if we allow one mass to to
: pass through the other (e.g. one of the masses is hoop-shaped), so
: that we can seperate them again without needing to do any more
: work. However, to repeat the process we them have to apply two
: lots of momemtum to each body. One lot to bring them back to rest
: with respect to each other, and a second lot to send them back in
: the opposite direction. After applying the first lot we are
: basically back to square one, albeit with the masses at opposite
: sides of the system. Repeat the process again and everything is as
: it was in the first instance.
:
: --
: Andrew Smallshaw
: andrews@sdf.lonestar.org
He has two masses, both at apogee, separated by 100 km seconds
(although I suspect he means kilometres) and therefore not both at
the same apogee. Then he anticipates both will arrive at perigee and he'll
"pull them together", but since they have different orbital periods
they are not going to arrive at perigee AT THE SAME TIME and still
be "pointing towards the earth", by which we presume he means
the Earth and the two masses all lie on a straight line. It would be a
simple matter to program Webb's craziness to show him it won't
work but that would involve the programmer in doing his stupidity
for him. It is up to Webb to prove his "gravity drive", not others
to do the work for him. It very much seems he's trying to pick
himself up by his own bootstraps, imagining that pulling on them
uses energy. He doesn't seem to realise that jumping up in the air
results in falling back down again unless he can jump fast enough
to achieve escape velocity in one leap.
|
| |
Date: 19 Sep 2007 03:22:03
From: Sam Wormley
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
Peter Webb wrote:
> This post works out the acceleration that could be derived from the
> "emissionless drive" I described in a previous post.
>
> Essentially, the drive works by having two (or more) large masses
> separated by a long distance. If the gravitational field varies over
> time (as it does in an elliptical orbit) and the gravitational field is
> non linear (which it always is) then the force experienced by each mass
> will be different at different times. By reducing the distance when the
> gravitational gradient is high (through using a motor to bring the
> masses together) and increasing it when it the gravitational gradient is
> low - at the top of the orbit - the energy expended by the motor over a
> complete cycle is manifest as either higher kinetic energy of the masses
> or a higher orbit, depending upon the geometry.
>
And have you analyzed your "drive" in terms of conservation of energy?
The second Law of thermodynamics? What about Newton's third law?
|
| | |
Date: 19 Sep 2007 14:04:59
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Sam Wormley" <swormley1@mchsi.com > wrote in message
news:vB0Ii.109787$Fc.10675@attbi_s21...
> Peter Webb wrote:
>> This post works out the acceleration that could be derived from the
>> "emissionless drive" I described in a previous post.
>>
>> Essentially, the drive works by having two (or more) large masses
>> separated by a long distance. If the gravitational field varies over time
>> (as it does in an elliptical orbit) and the gravitational field is non
>> linear (which it always is) then the force experienced by each mass will
>> be different at different times. By reducing the distance when the
>> gravitational gradient is high (through using a motor to bring the masses
>> together) and increasing it when it the gravitational gradient is low -
>> at the top of the orbit - the energy expended by the motor over a
>> complete cycle is manifest as either higher kinetic energy of the masses
>> or a higher orbit, depending upon the geometry.
>>
>
> And have you analyzed your "drive" in terms of conservation of energy?
> The second Law of thermodynamics? What about Newton's third law?
>
>
>
Sure. Indeed, it relies heavily on the law of conservation of energy. The
energy expended in bringing the masses together at the bottom of the orbit
is larger than the energy recovered by separating them at the top of the
orbit. This extra energy expended by the motor is the source of the
increased altitude.
I don't believe the second law of thermodynamics enters into the idealised
model (no friction, etc).
Newton's third law is also conserved. The equal and opposite force is
against the earth, just as it is with the acceleration of any ballistic
object (eg a satellite). The key insight is that the relative gravitational
coupling between the earth and the two masses changes during the orbit,
giving us something to pull and push against.
There is already a "spaceship" which uses a similar mechanism - earth's
moon. Its altitude is increasing due to differential gravity between the
near side and the far side of the earth. The moon, being spherically
symmetric and in a nearly circular orbit, has to rely on tidal forces. My
spaceship is a little more cleverly designed and relies on a similar second
order effect.
That the moon can increase its orbital radius without ejecting reaction mass
makes this mechanism practical in theory, given a large enough spaceship (eg
the size of the moon). My spaceship is stripped down so it is only two
highly separated masses, maximising the effect.
The calculations above suggest that the effect is strong enough to be
useful. Given that the spaceship can operate in perpetuity without reaction
mass (just as the moon has been receding for billions of years without
ejecting reaction mass), you don't need much sustained acceleration to build
up to a respectable velocity if you are prepared to wait long enough.
The analogy I have previously used (and its only an analogy) is how a child
builds up speed on a swing by extending their legs on the down cycle and
tucking their legs in at the top of the cycle. Their legs are one of the
masses and their body is the other mass; the ropes are analogous to gravity.
I was gobsmacked when I thought of this drive mechanism. It gives every
indication of being impossible, as its hard to see how a spaceship can
achieve acceleration (thrust) without a reaction mass to push against. My
drive actually uses the earth as the reaction mass, via gravitational
coupling.
Do you have a mental picture of how my drive works, or are you skeptical
just because it sounds impossible? I agree that it sounds impossible, but I
don't believe it is impossible (in theory at least).
|
| | | |
Date: 19 Sep 2007 04:18:06
From: Sam Wormley
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
Peter Webb wrote:
>
> There is already a "spaceship" which uses a similar mechanism - earth's
> moon. Its altitude is increasing due to differential gravity between the
> near side and the far side of the earth.
No... Friction of the rotating earth under the tidal bulges gives
the bulges a positive phase with respect to the earth-moon line
"leading" the moon into a higher orbit.
Conservation of Angular Momentum.
|
| | | | |
Date: 19 Sep 2007 14:34:54
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Sam Wormley" <swormley1@mchsi.com > wrote in message
news:2q1Ii.109840$Fc.38232@attbi_s21...
> Peter Webb wrote:
>
>>
>> There is already a "spaceship" which uses a similar mechanism - earth's
>> moon. Its altitude is increasing due to differential gravity between the
>> near side and the far side of the earth.
>
> No... Friction of the rotating earth under the tidal bulges gives
> the bulges a positive phase with respect to the earth-moon line
> "leading" the moon into a higher orbit.
>
> Conservation of Angular Momentum.
I said it was a "similar mechanism", not the same mechanism. The moons orbit
increased without it expending reaction mass; its merely a demonstration
that the concept is not impossible.
Do you understand how my drive works? If you don't, what is unclear? If you
do, can you see any theoretical reason it won't work?
|
| | | |
Date: 18 Sep 2007 21:14:52
From: N:dlzc D:aol T:com \(dlzc\)
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
Dear Peter Webb:
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in
message news:46f09fd1$0$31115$afc38c87@news.optusnet.com.au...
...
> Do you have a mental picture of how my drive works,
> or are you skeptical just because it sounds
> impossible?
It is possible with only a single mass and no drive motor. We
use it frequently to boost the speed of satellites / probes. But
we do not start from a circular orbit, and depending on which
side of the planetary body we approach on, controls whether we
boost or brake.
> I agree that it sounds impossible, but I don't believe
> it is impossible (in theory at least).
If you think you can keep from imparting spin, or that it is even
desirable to do so, yes it is impossible. As someone pointed
out, the Moon is getting a boost simply because the center of
mass of the tides and the Earth-proper are not co-aligned.
The "inner" weight will be orbitting slightly faster. Reducing
the weight-to-weight spacing will merely speed up the spin
rotation (conservation of angular momentum).
You mention "theory". I have not seen your math yet. If you are
going to apply for a patent (or whatever), you have (now) one
year and you will need math...
David A. Smith
|
| | | | |
Date: 19 Sep 2007 14:47:48
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"N:dlzc D:aol T:com (dlzc)" <dlzc1@cox.net > wrote in message
news:Im1Ii.130562$xx1.40838@newsfe09.phx...
> Dear Peter Webb:
>
> "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
> news:46f09fd1$0$31115$afc38c87@news.optusnet.com.au...
> ...
>> Do you have a mental picture of how my drive works,
>> or are you skeptical just because it sounds
>> impossible?
>
> It is possible with only a single mass and no drive motor. We use it
> frequently to boost the speed of satellites / probes. But we do not start
> from a circular orbit, and depending on which side of the planetary body
> we approach on, controls whether we boost or brake.
>
>> I agree that it sounds impossible, but I don't believe
>> it is impossible (in theory at least).
>
> If you think you can keep from imparting spin, or that it is even
> desirable to do so, yes it is impossible. As someone pointed out, the
> Moon is getting a boost simply because the center of mass of the tides and
> the Earth-proper are not co-aligned.
>
> The "inner" weight will be orbitting slightly faster. Reducing the
> weight-to-weight spacing will merely speed up the spin rotation
> (conservation of angular momentum).
>
> You mention "theory". I have not seen your math yet. If you are going to
> apply for a patent (or whatever), you have (now) one year and you will
> need math...
>
I posted some maths indicating the size of the effect above. I made some
simplifications which make it an "order of magnitude" calculation only, but
do not affect whether it is theoretically possible.
You correctly point out that the increased energy may be evident in the body
spinning. This is an equally paradoxical result. If you consider two
successive orbits, at the apogee (furthest point) the masses are the same
distance apart, but the spaceship has increased angular momentum. You can
use this effect to spin up a satellite to any speed without ejecting
reaction mass - a seemingly paradoxical result. If you don't want spin, this
can be controlled with gyroscopes.
I believe that where this effect is most practical would be in using it in
reverse, as a method of dropping a sattelite into a lower orbit. This
actually generates usable energy in the process. It would effectively allow
atmospheric breaking for bodies with no atmosphere.
|
| | | | | |
Date: 19 Sep 2007 07:37:31
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f0a9da$0$22253$afc38c87@news.optusnet.com.au...
:
: "N:dlzc D:aol T:com (dlzc)" <dlzc1@cox.net > wrote in message
: news:Im1Ii.130562$xx1.40838@newsfe09.phx...
: > Dear Peter Webb:
: >
: > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
: > news:46f09fd1$0$31115$afc38c87@news.optusnet.com.au...
: > ...
: >> Do you have a mental picture of how my drive works,
: >> or are you skeptical just because it sounds
: >> impossible?
: >
: > It is possible with only a single mass and no drive motor. We use it
: > frequently to boost the speed of satellites / probes. But we do not
start
: > from a circular orbit, and depending on which side of the planetary body
: > we approach on, controls whether we boost or brake.
: >
: >> I agree that it sounds impossible, but I don't believe
: >> it is impossible (in theory at least).
: >
: > If you think you can keep from imparting spin, or that it is even
: > desirable to do so, yes it is impossible. As someone pointed out, the
: > Moon is getting a boost simply because the center of mass of the tides
and
: > the Earth-proper are not co-aligned.
: >
: > The "inner" weight will be orbitting slightly faster. Reducing the
: > weight-to-weight spacing will merely speed up the spin rotation
: > (conservation of angular momentum).
: >
: > You mention "theory". I have not seen your math yet. If you are going
to
: > apply for a patent (or whatever), you have (now) one year and you will
: > need math...
: >
:
: I posted some maths indicating the size of the effect above. I made some
: simplifications which make it an "order of magnitude" calculation only,
but
: do not affect whether it is theoretically possible.
:
: You correctly point out that the increased energy may be evident in the
body
: spinning. This is an equally paradoxical result. If you consider two
: successive orbits, at the apogee (furthest point) the masses are the same
: distance apart, but the spaceship has increased angular momentum. You can
: use this effect to spin up a satellite to any speed without ejecting
: reaction mass - a seemingly paradoxical result. If you don't want spin,
this
: can be controlled with gyroscopes.
:
: I believe that where this effect is most practical would be in using it in
: reverse, as a method of dropping a sattelite into a lower orbit. This
: actually generates usable energy in the process. It would effectively
allow
: atmospheric breaking for bodies with no atmosphere.
:
Practical?
Hahaha!
Webby, when we slingshot a body past a planet, the planet slows down.
Here's the math:
Mv = mV
There are no paradoxes.
|
| | | | | | |
Date: 19 Sep 2007 17:50:08
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
> : I believe that where this effect is most practical would be in using it
> in
> : reverse, as a method of dropping a sattelite into a lower orbit. This
> : actually generates usable energy in the process. It would effectively
> allow
> : atmospheric breaking for bodies with no atmosphere.
> :
> Practical?
> Hahaha!
> Webby, when we slingshot a body past a planet, the planet slows down.
> Here's the math:
>
> Mv = mV
> There are no paradoxes.
>
I know that. I also know that my gravity drive doesn't involve any real
paradox, however paradoxical it may seem at first.
You will of course note that my gravity drive does not use the slingshot
effect; it uses a different mechanism.
|
| |
Date: 18 Sep 2007 17:38:36
From: Chris L Peterson
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au > wrote:
>So you pull the masses together at perigee and separate them at apogee. This
>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
>0.18m/s^2. Because you have two masses which must share this acceleration,
>the net acceleration is 0.9 m/s^2.
You sure about that?
_________________________________________________
Chris L Peterson
Cloudbait Observatory
http://www.cloudbait.com
|
| | |
Date: 19 Sep 2007 11:28:17
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Chris L Peterson" <clp@alumni.caltech.edu > wrote in message
news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
> On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
> <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
>
>>So you pull the masses together at perigee and separate them at apogee.
>>This
>>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
>>0.18m/s^2. Because you have two masses which must share this acceleration,
>>the net acceleration is 0.9 m/s^2.
>
> You sure about that?
>
Well spotted.
I realised that 0.18 / 2 = 0.09 after I posted, but as it was at the end of
the post and didn't affect the argument I didn't correct it.
|
| | | |
Date: 19 Sep 2007 07:27:31
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f07b17$0$11059$afc38c87@news.optusnet.com.au...
:
: "Chris L Peterson" <clp@alumni.caltech.edu > wrote in message
: news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
: > On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
: > <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
: >
: >>So you pull the masses together at perigee and separate them at apogee.
: >>This
: >>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
: >>0.18m/s^2. Because you have two masses which must share this
acceleration,
: >>the net acceleration is 0.9 m/s^2.
: >
: > You sure about that?
:
: Well spotted.
:
: I realised that 0.18 / 2 = 0.09 after I posted, but as it was at the end
of
: the post and didn't affect the argument I didn't correct it.
:
I knew I was wrong before you pointed out my error but didn't
bother to correct it until you noticed I'd goofed, hoping to get
away with it.
Q. A normal golf ball is driven 200 yards on a normal driving range
with a normal club by a normal golfer. No quirks, gimmicks or special cases.
What acceleration is needed?
|
| | | | |
Date: 19 Sep 2007 17:47:33
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Androcles" <Engineer@hogwarts.physics > wrote in message
news:Db4Ii.239942$p7.16657@fe2.news.blueyonder.co.uk...
>
> "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
> news:46f07b17$0$11059$afc38c87@news.optusnet.com.au...
> :
> : "Chris L Peterson" <clp@alumni.caltech.edu> wrote in message
> : news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
> : > On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
> : > <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
> : >
> : >>So you pull the masses together at perigee and separate them at
> apogee.
> : >>This
> : >>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
> : >>0.18m/s^2. Because you have two masses which must share this
> acceleration,
> : >>the net acceleration is 0.9 m/s^2.
> : >
> : > You sure about that?
> :
> : Well spotted.
> :
> : I realised that 0.18 / 2 = 0.09 after I posted, but as it was at the end
> of
> : the post and didn't affect the argument I didn't correct it.
> :
>
> I knew I was wrong before you pointed out my error but didn't
> bother to correct it until you noticed I'd goofed, hoping to get
> away with it.
It doesn't (and didn't) affect the argument.
I may also point out another slight error in my maths; the acceleration of
the two ends at apogee has to be reduced by a factor of (7000/20000)^2. This
increases the relative difference in the force on the two masses, and
increases the thrust by another 30 or 40%. Again, it doesn't affect the
argument.
> Q. A normal golf ball is driven 200 yards on a normal driving range
> with a normal club by a normal golfer. No quirks, gimmicks or special
> cases.
> What acceleration is needed?
>
Do you mean "initial acceleration" of the golfball? This would depend upon
how long the head of the golfclub is in contact with the ball; the shorter
the period the higher the acceleration required.
Returning to my question, can you see any reason that my gravity drive
wouldn't work in theory at least?
>
>
|
| | | | | |
Date: 19 Sep 2007 08:26:44
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f0d49b$0$10708$afc38c87@news.optusnet.com.au...
:
: "Androcles" <Engineer@hogwarts.physics > wrote in message
: news:Db4Ii.239942$p7.16657@fe2.news.blueyonder.co.uk...
: >
: > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
: > news:46f07b17$0$11059$afc38c87@news.optusnet.com.au...
: > :
: > : "Chris L Peterson" <clp@alumni.caltech.edu> wrote in message
: > : news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
: > : > On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
: > : > <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
: > : >
: > : >>So you pull the masses together at perigee and separate them at
: > apogee.
: > : >>This
: > : >>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2,
or
: > : >>0.18m/s^2. Because you have two masses which must share this
: > acceleration,
: > : >>the net acceleration is 0.9 m/s^2.
: > : >
: > : > You sure about that?
: > :
: > : Well spotted.
: > :
: > : I realised that 0.18 / 2 = 0.09 after I posted, but as it was at the
end
: > of
: > : the post and didn't affect the argument I didn't correct it.
: > :
: >
: > I knew I was wrong before you pointed out my error but didn't
: > bother to correct it until you noticed I'd goofed, hoping to get
: > away with it.
:
: It doesn't (and didn't) affect the argument.
:
: I may also point out another slight error in my maths; the acceleration of
: the two ends at apogee has to be reduced by a factor of (7000/20000)^2.
This
: increases the relative difference in the force on the two masses, and
: increases the thrust by another 30 or 40%. Again, it doesn't affect the
: argument.
:
:
: > Q. A normal golf ball is driven 200 yards on a normal driving range
: > with a normal club by a normal golfer. No quirks, gimmicks or special
: > cases.
: > What acceleration is needed?
: >
:
: Do you mean "initial acceleration" of the golfball?
I said: "No quirks, gimmicks or special cases."
That means exactly what happens on a driving range.
That's what I mean. No more. No less.
It's a fuckin' simple enough question, what's the answer?
: This would depend upon ...
Never mind that crap, Webby. I asked a simple practical question
and you go off into cloud cuckoo land.
Look, I'll make it easier for you.
http://members.aye.net/~bspen/ballistics.html
Is your "pull the masses together at perigee and separate them at
apogee" "proposed gravity drive - calculation of acceleration achievable"
0.09 m/s^2 going to propel a golf ball 200 fuckin' yards?
|
| | | | | | |
Date: 19 Sep 2007 18:33:29
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Androcles" <Engineer@hogwarts.physics > wrote in message
news:835Ii.240139$p7.1700@fe2.news.blueyonder.co.uk...
>
> "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
> news:46f0d49b$0$10708$afc38c87@news.optusnet.com.au...
> :
> : "Androcles" <Engineer@hogwarts.physics> wrote in message
> : news:Db4Ii.239942$p7.16657@fe2.news.blueyonder.co.uk...
> : >
> : > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
> : > news:46f07b17$0$11059$afc38c87@news.optusnet.com.au...
> : > :
> : > : "Chris L Peterson" <clp@alumni.caltech.edu> wrote in message
> : > : news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
> : > : > On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
> : > : > <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
> : > : >
> : > : >>So you pull the masses together at perigee and separate them at
> : > apogee.
> : > : >>This
> : > : >>give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2,
> or
> : > : >>0.18m/s^2. Because you have two masses which must share this
> : > acceleration,
> : > : >>the net acceleration is 0.9 m/s^2.
> : > : >
> : > : > You sure about that?
> : > :
> : > : Well spotted.
> : > :
> : > : I realised that 0.18 / 2 = 0.09 after I posted, but as it was at the
> end
> : > of
> : > : the post and didn't affect the argument I didn't correct it.
> : > :
> : >
> : > I knew I was wrong before you pointed out my error but didn't
> : > bother to correct it until you noticed I'd goofed, hoping to get
> : > away with it.
> :
> : It doesn't (and didn't) affect the argument.
> :
> : I may also point out another slight error in my maths; the acceleration
> of
> : the two ends at apogee has to be reduced by a factor of (7000/20000)^2.
> This
> : increases the relative difference in the force on the two masses, and
> : increases the thrust by another 30 or 40%. Again, it doesn't affect the
> : argument.
> :
> :
> : > Q. A normal golf ball is driven 200 yards on a normal driving range
> : > with a normal club by a normal golfer. No quirks, gimmicks or special
> : > cases.
> : > What acceleration is needed?
> : >
> :
> : Do you mean "initial acceleration" of the golfball?
>
> I said: "No quirks, gimmicks or special cases."
> That means exactly what happens on a driving range.
> That's what I mean. No more. No less.
> It's a fuckin' simple enough question, what's the answer?
>
> : This would depend upon ...
>
> Never mind that crap, Webby. I asked a simple practical question
> and you go off into cloud cuckoo land.
> Look, I'll make it easier for you.
> http://members.aye.net/~bspen/ballistics.html
>
> Is your "pull the masses together at perigee and separate them at
> apogee" "proposed gravity drive - calculation of acceleration achievable"
> 0.09 m/s^2 going to propel a golf ball 200 fuckin' yards?
It only works if you are in orbit already.
If you are, you can use it to produce any speed up to escape velocity.
You don't understand the concept, do you?
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Date: 19 Sep 2007 11:04:29
From: Androcles
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au > wrote in message
news:46f0debf$0$10708$afc38c87@news.optusnet.com.au...
:
: "Androcles" <Engineer@hogwarts.physics > wrote in message
: news:835Ii.240139$p7.1700@fe2.news.blueyonder.co.uk...
: >
: > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
: > news:46f0d49b$0$10708$afc38c87@news.optusnet.com.au...
: > :
: > : "Androcles" <Engineer@hogwarts.physics> wrote in message
: > : news:Db4Ii.239942$p7.16657@fe2.news.blueyonder.co.uk...
: > : >
: > : > "Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> wrote in message
: > : > news:46f07b17$0$11059$afc38c87@news.optusnet.com.au...
: > : > :
: > : > : "Chris L Peterson" <clp@alumni.caltech.edu> wrote in message
: > : > : news:5ao0f354sh6s7p3ab3e4k59j4s4g03jjkk@4ax.com...
: > : > : > On Wed, 19 Sep 2007 00:18:00 +1000, "Peter Webb"
: > : > : > <webbfamily@DIESPAMDIEoptusnet.com.au> wrote:
: > : > : >
: > : > : >>So you pull the masses together at perigee and separate them at
: > : > apogee.
: > : > : >>This
: > : > : >>give you a net difference of acceleration of 0.28m/s^2 -
0.1m/s^2,
: > or
: > : > : >>0.18m/s^2. Because you have two masses which must share this
: > : > acceleration,
: > : > : >>the net acceleration is 0.9 m/s^2.
: > : > : >
: > : > : > You sure about that?
: > : > :
: > : > : Well spotted.
: > : > :
: > : > : I realised that 0.18 / 2 = 0.09 after I posted, but as it was at
the
: > end
: > : > of
: > : > : the post and didn't affect the argument I didn't correct it.
: > : > :
: > : >
: > : > I knew I was wrong before you pointed out my error but didn't
: > : > bother to correct it until you noticed I'd goofed, hoping to get
: > : > away with it.
: > :
: > : It doesn't (and didn't) affect the argument.
: > :
: > : I may also point out another slight error in my maths; the
acceleration
: > of
: > : the two ends at apogee has to be reduced by a factor of
(7000/20000)^2.
: > This
: > : increases the relative difference in the force on the two masses, and
: > : increases the thrust by another 30 or 40%. Again, it doesn't affect
the
: > : argument.
: > :
: > :
: > : > Q. A normal golf ball is driven 200 yards on a normal driving range
: > : > with a normal club by a normal golfer. No quirks, gimmicks or
special
: > : > cases.
: > : > What acceleration is needed?
: > : >
: > :
: > : Do you mean "initial acceleration" of the golfball?
: >
: > I said: "No quirks, gimmicks or special cases."
: > That means exactly what happens on a driving range.
: > That's what I mean. No more. No less.
: > It's a fuckin' simple enough question, what's the answer?
: >
: > : This would depend upon ...
: >
: > Never mind that crap, Webby. I asked a simple practical question
: > and you go off into cloud cuckoo land.
: > Look, I'll make it easier for you.
: > http://members.aye.net/~bspen/ballistics.html
: >
: > Is your "pull the masses together at perigee and separate them at
: > apogee" "proposed gravity drive - calculation of acceleration
achievable"
: > 0.09 m/s^2 going to propel a golf ball 200 fuckin' yards?
:
:
: It only works if you are in orbit already.
I am in orbit already, I orbit the sun. I also play golf.
: If you are, you can use it to produce any speed up to escape velocity.
All right then, put the fucking golf ball into orbit around Mars
with your widget, all I asked for was 200 yards.
: You don't understand the concept, do you?
Pulling the mass of a golf ball together with the mass of a 3 iron
sounds simple enough and gets a whole lot more acceleration
than 0.09 m/s^2. What is it about 9.8 m/s^2, 100 times your
value, that you don't understand?
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Date: 18 Sep 2007 21:44:10
From: Sitav
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
On Sep 18, 10:18 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au > wrote:
> This post works out the acceleration that could be derived from the
> "emissionless drive" I described in a previous post.
>
> Essentially, the drive works by having two (or more) large masses separated
> by a long distance. If the gravitational field varies over time (as it does
> in an elliptical orbit) and the gravitational field is non linear (which it
> always is) then the force experienced by each mass will be different at
> different times. By reducing the distance when the gravitational gradient is
> high (through using a motor to bring the masses together) and increasing it
> when it the gravitational gradient is low - at the top of the orbit - the
> energy expended by the motor over a complete cycle is manifest as either
> higher kinetic energy of the masses or a higher orbit, depending upon the
> geometry.
>
> So, how practical is it?
>
> The first thing to note is that as long as the mass of the motor and rod is
> insignificant compared to the two large masses, the net acceleration is
> independent of the size of the masses, and depends only upon their
> separation. To see this is true, you can imagine that two identical
> spaceships orbiting side by side would experience the same acceleration as a
> single spaceship with the double the mass at each end.
>
> This very convenient, because as the mass falls out of the equation we need
> only consider the differences in acceleration due to gravity.
>
> Lets make the distance between the masses vary from zero to 100 kms.
>
> Lets make the lowest point (perigee) of the orbit 7,000 kms from the centre
> of the earth (ie 600 kms above the surface). Lets make the most distant
> point (apogee) 20,000 kms. Lets make g = 10 m/s^2 at 7000 kms.
>
> Using the inverse square law, the difference in acceleration between the two
> masses at 7,000 kms is:
>
> 10 m/s^2 * (7,100,000/7,000,000 - 1)^2
> = 0.28 m/s^2
>
> At 20,000 kms, the difference in acceleration at the two ends is:
>
> 10 m/s^2 * (20,100,000/20,000,000 - 1)^2
> = 0.1 m/s^2
>
> This means we have an acceleration differential of 0.28 m/s^2 at perigee but
> only 0.1 m/s^2 at apogee. This is manifest as the force the motor must use
> to bring the masses together or separate them, its M x 0.28 m/s^2 at perigee
> but only M x 0.1 ms/^2 at apogee.
>
> So you pull the masses together at perigee and separate them at apogee. This
> give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
> 0.18m/s^2. Because you have two masses which must share this acceleration,
> the net acceleration is 0.9 m/s^2.
>
> This a pretty reasonable rate. You could obviously build up to escape
> velocity eventually.
>
> There are a few problems. The first is keeping the masses pointing towards
> the earth, which may require gyroscopes. Alternatively, there is no reason
> why the spaceship needs to have only two masses on one axis; there could be
> six masses on three axis. This would mean you wouldn't care if it tumbled,
> as at least one axis would always be available. More to the point, having
> multiple axis may allow the spacecraft to be steered, and the problem
> avoided.
>
> This almost seems to me to be a practical "gravity drive". Even if not used
> to achieve escape velocity, it would seem to suffice for orbital adjustments
> (unfortunately not for circular orbits though). A spaceship designed in this
> manner could operate forever with no refueling, as long as it had access to
> energy (eg solar energy).
>
> Should I patent the "gravity drive" ? Or has somebody already done this!
now i feel really stupid and normally i could process this
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| | |
Date: 18 Sep 2007 23:34:54
From: Sam Wormley
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
Sitav wrote:
> On Sep 18, 10:18 am, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>> This post works out the acceleration that could be derived from the
>> "emissionless drive" I described in a previous post.
>>
>> Essentially, the drive works by having two (or more) large masses separated
>> by a long distance. If the gravitational field varies over time (as it does
>> in an elliptical orbit) and the gravitational field is non linear (which it
>> always is) then the force experienced by each mass will be different at
>> different times. By reducing the distance when the gravitational gradient is
>> high (through using a motor to bring the masses together) and increasing it
>> when it the gravitational gradient is low - at the top of the orbit - the
>> energy expended by the motor over a complete cycle is manifest as either
>> higher kinetic energy of the masses or a higher orbit, depending upon the
>> geometry.
>>
>> So, how practical is it?
>>
>> The first thing to note is that as long as the mass of the motor and rod is
>> insignificant compared to the two large masses, the net acceleration is
>> independent of the size of the masses, and depends only upon their
>> separation. To see this is true, you can imagine that two identical
>> spaceships orbiting side by side would experience the same acceleration as a
>> single spaceship with the double the mass at each end.
>>
>> This very convenient, because as the mass falls out of the equation we need
>> only consider the differences in acceleration due to gravity.
>>
>> Lets make the distance between the masses vary from zero to 100 kms.
>>
>> Lets make the lowest point (perigee) of the orbit 7,000 kms from the centre
>> of the earth (ie 600 kms above the surface). Lets make the most distant
>> point (apogee) 20,000 kms. Lets make g = 10 m/s^2 at 7000 kms.
>>
>> Using the inverse square law, the difference in acceleration between the two
>> masses at 7,000 kms is:
>>
>> 10 m/s^2 * (7,100,000/7,000,000 - 1)^2
>> = 0.28 m/s^2
>>
>> At 20,000 kms, the difference in acceleration at the two ends is:
>>
>> 10 m/s^2 * (20,100,000/20,000,000 - 1)^2
>> = 0.1 m/s^2
>>
>> This means we have an acceleration differential of 0.28 m/s^2 at perigee but
>> only 0.1 m/s^2 at apogee. This is manifest as the force the motor must use
>> to bring the masses together or separate them, its M x 0.28 m/s^2 at perigee
>> but only M x 0.1 ms/^2 at apogee.
>>
>> So you pull the masses together at perigee and separate them at apogee. This
>> give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
>> 0.18m/s^2. Because you have two masses which must share this acceleration,
>> the net acceleration is 0.9 m/s^2.
>>
>> This a pretty reasonable rate. You could obviously build up to escape
>> velocity eventually.
>>
>> There are a few problems. The first is keeping the masses pointing towards
>> the earth, which may require gyroscopes. Alternatively, there is no reason
>> why the spaceship needs to have only two masses on one axis; there could be
>> six masses on three axis. This would mean you wouldn't care if it tumbled,
>> as at least one axis would always be available. More to the point, having
>> multiple axis may allow the spacecraft to be steered, and the problem
>> avoided.
>>
>> This almost seems to me to be a practical "gravity drive". Even if not used
>> to achieve escape velocity, it would seem to suffice for orbital adjustments
>> (unfortunately not for circular orbits though). A spaceship designed in this
>> manner could operate forever with no refueling, as long as it had access to
>> energy (eg solar energy).
>>
>> Should I patent the "gravity drive" ? Or has somebody already done this!
>
> now i feel really stupid and normally i could process this
>
Don't -- the posting you are responding to is bullshit!
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| | | |
Date: 19 Sep 2007 11:30:34
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Sam Wormley" <swormley1@mchsi.com > wrote in message
news:ygZHi.92498$Xa3.50121@attbi_s22...
> Sitav wrote:
>> On Sep 18, 10:18 am, "Peter Webb"
>> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
>>> This post works out the acceleration that could be derived from the
>>> "emissionless drive" I described in a previous post.
>>>
>>> Essentially, the drive works by having two (or more) large masses
>>> separated
>>> by a long distance. If the gravitational field varies over time (as it
>>> does
>>> in an elliptical orbit) and the gravitational field is non linear (which
>>> it
>>> always is) then the force experienced by each mass will be different at
>>> different times. By reducing the distance when the gravitational
>>> gradient is
>>> high (through using a motor to bring the masses together) and increasing
>>> it
>>> when it the gravitational gradient is low - at the top of the orbit -
>>> the
>>> energy expended by the motor over a complete cycle is manifest as either
>>> higher kinetic energy of the masses or a higher orbit, depending upon
>>> the
>>> geometry.
>>>
>>> So, how practical is it?
>>>
>>> The first thing to note is that as long as the mass of the motor and rod
>>> is
>>> insignificant compared to the two large masses, the net acceleration is
>>> independent of the size of the masses, and depends only upon their
>>> separation. To see this is true, you can imagine that two identical
>>> spaceships orbiting side by side would experience the same acceleration
>>> as a
>>> single spaceship with the double the mass at each end.
>>>
>>> This very convenient, because as the mass falls out of the equation we
>>> need
>>> only consider the differences in acceleration due to gravity.
>>>
>>> Lets make the distance between the masses vary from zero to 100 kms.
>>>
>>> Lets make the lowest point (perigee) of the orbit 7,000 kms from the
>>> centre
>>> of the earth (ie 600 kms above the surface). Lets make the most distant
>>> point (apogee) 20,000 kms. Lets make g = 10 m/s^2 at 7000 kms.
>>>
>>> Using the inverse square law, the difference in acceleration between the
>>> two
>>> masses at 7,000 kms is:
>>>
>>> 10 m/s^2 * (7,100,000/7,000,000 - 1)^2
>>> = 0.28 m/s^2
>>>
>>> At 20,000 kms, the difference in acceleration at the two ends is:
>>>
>>> 10 m/s^2 * (20,100,000/20,000,000 - 1)^2
>>> = 0.1 m/s^2
>>>
>>> This means we have an acceleration differential of 0.28 m/s^2 at perigee
>>> but
>>> only 0.1 m/s^2 at apogee. This is manifest as the force the motor must
>>> use
>>> to bring the masses together or separate them, its M x 0.28 m/s^2 at
>>> perigee
>>> but only M x 0.1 ms/^2 at apogee.
>>>
>>> So you pull the masses together at perigee and separate them at apogee.
>>> This
>>> give you a net difference of acceleration of 0.28m/s^2 - 0.1m/s^2, or
>>> 0.18m/s^2. Because you have two masses which must share this
>>> acceleration,
>>> the net acceleration is 0.9 m/s^2.
>>>
>>> This a pretty reasonable rate. You could obviously build up to escape
>>> velocity eventually.
>>>
>>> There are a few problems. The first is keeping the masses pointing
>>> towards
>>> the earth, which may require gyroscopes. Alternatively, there is no
>>> reason
>>> why the spaceship needs to have only two masses on one axis; there could
>>> be
>>> six masses on three axis. This would mean you wouldn't care if it
>>> tumbled,
>>> as at least one axis would always be available. More to the point,
>>> having
>>> multiple axis may allow the spacecraft to be steered, and the problem
>>> avoided.
>>>
>>> This almost seems to me to be a practical "gravity drive". Even if not
>>> used
>>> to achieve escape velocity, it would seem to suffice for orbital
>>> adjustments
>>> (unfortunately not for circular orbits though). A spaceship designed in
>>> this
>>> manner could operate forever with no refueling, as long as it had access
>>> to
>>> energy (eg solar energy).
>>>
>>> Should I patent the "gravity drive" ? Or has somebody already done this!
>>
>> now i feel really stupid and normally i could process this
>>
>
> Don't -- the posting you are responding to is bullshit!
You don't understand it either, then.
Why do you think its bullshit? Perhaps I can find out what you don't
understand in the post and explain it better?
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Date: 18 Sep 2007 14:44:09
From: Sam Wormley
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
Coupled with rocket propulsion, we've been using gravity drive
to get around the solar system for forth years--transfer orbits
gravity assist, etc.
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Date: 18 Sep 2007 18:48:29
From: Jim Jam Gee
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
>
> Coupled with rocket propulsion, we've been using gravity drive
> to get around the solar system for forth years--transfer orbits
> gravity assist, etc.
LOL..forth years? try forthy years starting with VGER...
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Date: 19 Sep 2007 01:39:49
From: Peter Webb
Subject: Re: Proposed gravity drive - calculation of acceleration achievable
|
"Sam Wormley" <swormley1@mchsi.com > wrote in message
news:ZuRHi.91979$Xa3.6589@attbi_s22...
>
> Coupled with rocket propulsion, we've been using gravity drive
> to get around the solar system for forth years--transfer orbits
> gravity assist, etc.
Yes, but mine operates with only a single body. It could be used by a
satellite orbiting the earth to achieve escape velocity. You can't do that
with a standard gravitational slingshot.
I might also add that mine works by a completely different principle.
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