I just shoot them up....Who cares where they come down?

ANIM8Rfsk wrote:

 > Or are you saying that if I throw a rock to port, it will go to port,
 > and if I
 > throw a rock to starboard, it will go to starboard?

I think he's suggesting that if you throw a rock, it will land spinward,
regardless of which direction you threw it in (port, starboard, or even
antispinward). However, I don't think his analysis is complete. His
analysis neglects the fact that angular momentum (which is indeed
conserved) is a vector quantity, not a scalar one.

--
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 >> Stay informed about: I just shoot them up....Who cares where they come down? 

[Followups to rec.arts.sf.science added.]

David Lesher wrote:

 > So I got into a Ringworld discussion over pizza with some other
 > Linux types the other day. They'd not heard of it but we used "units"
 > to make some area guesses, etc.
 >
 > One aspect was launches of sub-orbital objects. The professor-type
 > promptly brought something we've not really explored: conservation
 > of angular momentum. His followup was:
  ...
 > 6) The rock's angular momentum will be L=mVr/Rr, where m is
 > the
 > mass of the rock. This quantity is conserved while in solar orbit.
 > Therefore, as the rock moves closer to the sun, its angular velocity
 > must
 > correspondingly increase. And since its initial angular velocity was
 > equal to that of Ringworld, its angular velocity will be greater than
 > that
 > of Ringworld at all points along its solar orbit. And therefore, the
 > rock
 > will land on Ringworld at a point forward that of the rock thrower.
 >
 > Reactions?

An immediate problem is that in his angular momentum analysis he's
totally ignoring the fact that after launch, the projectile will have a
different angular momentum than the Ringworld -- it's in a different
orbit now, after all. It _is_ true that after launch, its angular
momentum will remain constant. But the impulse is totally selectable.
Consider a trivial example: Apply a deltavee to the projectile that
completely cancels the Ringworld's tangential velocity, and adds a small
sunward component. In the heliocentric coordinate system, this
projectile moves radially sunward (while the Ringworld rotates
underneath it), reaches its peak, then falls back down and hits the
Ringworld. But since the Ringworld rotated underneath it, its impact
point is well antispinward of the launching point. Right there we have
a counterexample to his claim.

Now, granted, that would take an utterly ridiculous amount of deltavee.
The point here, though, is his analysis is incorrect, since he didn't
take into account how the deltavee of the projectile can change the
mechanics of the situation -- in fact, he totally ignored the velocity
and angular momentum of the projectile as opposed to the Ringworld.

Another important point is missing in his analysis. The Ringworld
rotates very, very, very (is that enough verys?) fast. I mean, really
fast. So fast, in fact, that for any reasonable path of a normal
projectile, solar gravity is totally negligible -- it's moving way too
fast for gravity to take effect. The effect is so minor that ignoring
gravity only makes a difference in the fourth significant digit of the
calculations (in terms of range and maximum height). The fact that the
effect of gravity is so small should indicate that his analysis doesn't
make sense -- even if his arguments were sound, other effects, such as
the initial relative velocity of the projectile -- can change the
outcome.

I mean, think about it. Solar gravity at Earth orbit is about 6 x 10^-3
m/s^2. If your projectile is in solar orbit for, say, an hour, it can't
get very far, and the total amount of deltavee the Sun could possibly
apply to it is about 2 m/s. With deltavees measured in km/s, it should
be pretty clear that only minor changes in the way the projectile is
launched will completely swamp out any effect due to solar gravity. So
the analysis in terms of solar gravity and angular momentum can't be
relevant.

Instead of going through the drudgery of plotting orbits, or better yet
plotting secant lines after proving that gravitational effects are
negligible, I just wrote a little orbital mechanics simulator (which
ignores the gravity of the Ringworld itself, the effect of which is
going to be a whole lot less than that of the Sun). The results
certainly convince me that it's easy to fire a projectile at an angle so
that it lands antispinward, even with reasonable deltavees. (I'll send
the Python code for the simulator to anyone who wants to see it.)

The simulator nominally takes the deltavee of the projectile (relative
to the launching point) in km/s, and the azimuthal angle in rad at which
it should be launched. The angle is measured relative to the spinward
direction, so pi/2 is directly upward (toward the sun), pi/4 is 45
degress to spinward, 3 pi/4 is 45 degrees to antispinward, etc. You can
also select the time slice, which defaults to 1 s, for the Euler
integrator, and optionally turn off gravity. It returns a 3-tuple of
numbers, which are 1. the range (in metres) the projectile manages, as
measured along the arc of the Ringworld -- positive means it lands
spinward, and negative means it lands antispinward), 2. the maximum
height achieved, and 3. the time it is in flight.

So, for instance, a projectile with an initial relative speed of 10 km/s
and an azimuth angle of 90 degrees (straight upward), you have:

   >>> simulate(10e3, math.pi/2)
(112202.51286125183, 5134472.9930725098, 2054.0)

That is, it lands 112 km antispinward, reaches a maximum height of about
5130 km, and takes 34.2 minutes. To see how the numbers change if you
turn off gravity:

   >>> simulate(10e3, math.pi/2, doGravity=False)
(112075.88988924026, 5131577.6446838379, 2053.0)

To make sure this isn't the 1 s timeslice, pick one with 10 ms and see
the numbers don't change substantially:

   >>> simulate(10e3, math.pi/2, doGravity=True, dt=0.01)
(112202.51943397522, 5134475.861114502, 2053.7599999984081)
   >>> simulate(10e3, math.pi/2, doGravity=False, dt=0.01)
(112075.91344070435, 5131578.0784912109, 2052.5999999983828)

Now then, if we fire it an acute angle it should get more range at the
expense of height:

   >>> simulate(10e3, math.pi/4)
(10193042.700881243, 2537803.9789733887, 1436.0)

Now the range is about 1020 km spinward, with a height of 2540 km, and a
duration of 23.9 minutes. So what if it's fired at an obtuse angle?

   >>> simulate(10e3, 3*math.pi/4)
(-10354100.695576191, 2597313.6339111328, 1470.0)

If it's fired at an azimuth of 45 degrees "backward," i.e. along the
antispinward line rather than spinward, then after 24.5 minutes, it
lands 1040 km _antispinward_, after reaching a maximum height of 2600
km.

--
Erik Max Francis && max DeleteThis @alcyone.com && <a style='text-decoration: underline;' href="http://www.alcyone.com/max/" target="_blank">http://www.alcyone.com/max/</a>
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-- Carl Sagan<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

Erik Max Francis wrote:
 >
 > [Followups to rec.arts.sf.science added.]
 >
 > David Lesher wrote:
 >
  > > So I got into a Ringworld discussion over pizza with some other
  > > Linux types the other day. They'd not heard of it but we used "units"
  > > to make some area guesses, etc.
  > >
  > > One aspect was launches of sub-orbital objects. The professor-type
  > > promptly brought something we've not really explored: conservation
  > > of angular momentum. His followup was:
 > ...
  > > 6) The rock's angular momentum will be L=mVr/Rr, where m is
  > > the
  > > mass of the rock. This quantity is conserved while in solar orbit.
  > > Therefore, as the rock moves closer to the sun, its angular velocity
  > > must
  > > correspondingly increase. And since its initial angular velocity was
  > > equal to that of Ringworld, its angular velocity will be greater than
  > > that
  > > of Ringworld at all points along its solar orbit. And therefore, the
  > > rock
  > > will land on Ringworld at a point forward that of the rock thrower.
  > >
  > > Reactions?
 >
 > An immediate problem is that in his angular momentum analysis he's
 > totally ignoring the fact that after launch, the projectile will have a
 > different angular momentum than the Ringworld -- it's in a different
 > orbit now, after all. It _is_ true that after launch, its angular
 > momentum will remain constant. But the impulse is totally selectable.
 > Consider a trivial example: Apply a deltavee to the projectile that
 > completely cancels the Ringworld's tangential velocity, and adds a small
 > sunward component. In the heliocentric coordinate system, this
 > projectile moves radially sunward (while the Ringworld rotates
 > underneath it), reaches its peak, then falls back down and hits the
 > Ringworld. But since the Ringworld rotated underneath it, its impact
 > point is well antispinward of the launching point. Right there we have
 > a counterexample to his claim.

Indeed. But substitute the more modest conclusion "it will land to
spinward of where a person on the surface would expect it to land" and
Lesher is correct. He's just re-derived the Coriolis effect, though in a
cylinder as big as the Ringworld it will be too small to notice unless the
thrower can throw his projectile "up" more than a negligible fraction of
the way to the sun.<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

John David Galt wrote:

 > Indeed. But substitute the more modest conclusion "it will land to
 > spinward of where a person on the surface would expect it to land" and
 > Lesher is correct.

Right. And that was a great deal more modest than the original claim
(which, admittedly, was his friend's claim, not actually his).

 > Indeed. But substitute the more modest conclusion "it will land to
 > spinward of where a person on the surface would expect it to land" and
 > Lesher is correct. He's just re-derived the Coriolis effect, though
 > in a cylinder as big as the Ringworld it will be too small to notice
 > unless the thrower can throw his projectile "up" more than a
 > negligible fraction of the way to the sun.

Right. Even for orbital mechanics type speeds, the difference is pretty
small -- only about a second different in total trip time, and only a
few kilometres difference in total range (out of thousands).

--
Erik Max Francis && max.RemoveThis@alcyone.com && <a style='text-decoration: underline;' href="http://www.alcyone.com/max/" target="_blank">http://www.alcyone.com/max/</a>
__ San Jose, CA, USA && 37 20 N 121 53 W && &tSftDotIotE
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\__/ Why do we give ourselves away / 'Till only emptiness remains
-- Chante Moore<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

On Wed, 26 Nov 2003 14:22:19 -0800, Erik Max Francis <max DeleteThis @alcyone.com>
wrote:

 >ANIM8Rfsk wrote:
 >
  >> Or are you saying that if I throw a rock to port, it will go to port,
  >> and if I
  >> throw a rock to starboard, it will go to starboard?
 >
 >I think he's suggesting that if you throw a rock, it will land spinward,
 >regardless of which direction you threw it in (port, starboard, or even
 >antispinward). However, I don't think his analysis is complete. His
 >analysis neglects the fact that angular momentum (which is indeed
 >conserved) is a vector quantity, not a scalar one.

Plus he's leaving out the all-important "wind resistance".

Note that all these theoretical discussions always have the assumption
of "in a vacuum", which of course does not apply to anyone standing on
the Rignworld inner surface.<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

Mark Landin wrote:

 > Plus he's leaving out the all-important "wind resistance".
 >
 > Note that all these theoretical discussions always have the assumption
 > of "in a vacuum", which of course does not apply to anyone standing on
 > the Rignworld inner surface.

Sure, but it doesn't apply to people on the surface of the Earth,
either.

--
Erik Max Francis && max.RemoveThis@alcyone.com && <a style='text-decoration: underline;' href="http://www.alcyone.com/max/" target="_blank">http://www.alcyone.com/max/</a>
__ San Jose, CA, USA && 37 20 N 121 53 W && &tSftDotIotE
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-- Lamya<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

"David Lesher" <wb8foz DeleteThis @panix.com> wrote in message
news:bq2k1d$t80$1@reader2.panix.com...
 > ...that's not my department, said Werner Von Braun...
 >
 > Tom Lehrer
 >
 >
 > So I got into a Ringworld discussion over pizza with some other
 > Linux types the other day. They'd not heard of it but we used "units"
 > to make some area guesses, etc.
 >
 > One aspect was launches of sub-orbital objects. The professor-type
 > promptly brought something we've not really explored: conservation
 > of angular momentum. His followup was:
 >
 > Orbital problems always require meticulous definition of at least
 > two coordinate systems (and usually a good many more).
 >
 > 1) Ringworld describes a circle around the sun of radius Rr, and
 > Ringworld orbits the sun with velocity Vr. Therefore Ringworld orbits the
 > sun with angular velocity Vr/Rr.
 >
 > 2) Let there be a topocentric Cartesian coordinate system (x',y')
 > for the rock thrower, which origin is where he stands, and which y'-axis
 > always points toward the sun, and which x'-axis always points in the
 > direction of rotation. This coordinate system rotates with angular
 > velocity Vr/Rr.
 >
 > 3) Let there be a heliocentric Cartesian coordinate system (x,y)
 > which origin is at the sun and which axes are fixed, and which y-axis
 > points at the rock thrower at the moment he throws, and which x-axis lies
 > in the ecliptic (the plane defined by the orbit of Ringworld).
 >
 > 4) Let someone in Ringworld throw a rock straight up with initial
 > vertical velocity Vs. Then the initial velocity vector will be Vo'=(0,Vs)
 > in the topocentric frame, and Vo=(Vr,-Vs) in the heliocentric frame.
 >
 > 5) If someone in Ringworld throws a rock (in any direction) then
 > said rock will be (at least for a short time) in solar orbit, and its path
 > in the heliocentric frame will be that of a conic section (e.g. ellipse).
 > If the direction of throw lies within the ecliptic, then the orbit will
 > lie entirely within the circle defined by Ringworld, and intersect it at
 > exactly two points.
 >
 > 6) The rock's angular momentum will be L=mVr/Rr, where m is the
 > mass of the rock. This quantity is conserved while in solar orbit.
 > Therefore, as the rock moves closer to the sun, its angular velocity must
 > correspondingly increase. And since its initial angular velocity was
 > equal to that of Ringworld, its angular velocity will be greater than that
 > of Ringworld at all points along its solar orbit. And therefore, the rock
 > will land on Ringworld at a point forward that of the rock thrower.
 >
 >
 >
 >
 > Reactions?
 >
 >

What you need to do as a person on the ring who wants to launch into space
is build your launcher within 500 miles of a rim wall and launch your
vehical so that it makes it high enough and far enough to clear the wall and
fall away in free space.

No way a chemical rocket will have enough oomph to acheive solar orbit, the
rotational speed of the ringworld is far too high, but once you are in free
space beyond the ring you can use ion drives or light sails to make a very
long lorbit away from the star and then back to its vicinity 20 or 30 years
later.

Allen W.<!-- ~MESSAGE_AFTER~ -->
 >> Stay informed about: I just shoot them up....Who cares where they come down? 

<< From: "Allen W. McDonnell" tanada.RemoveThis@provide.net >>


<< What you need to do as a person on the ring who wants to launch into space
is build your launcher within 500 miles of a rim wall and launch your
vehical >>

Of course, if you're half a million miles from a rim wall, there's probably
plenty to do without worrying about space travel.
 >> Stay informed about: I just shoot them up....Who cares where they come down?