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(TFT) Re: Demon slingshot
Rumor has it that David Michael Grouchy II parted the dust of ages to
reveal this nugget of the archives, from the legendary pen of Ed
Thorn:
Anyone up for launching a demon into orbit?
Who needs to? A demon cleverly using teleport should be able to
launch *himself* into orbit (see Note B, far below). I'll assume the
question actually applies to a wizard wearing a "Fresh Air" item, or
a Proxy, or some other non-self-teleporting object, but will continue
to refer to it as a "demon".
Questions for the class: How many transits, (T), of a 1000-foot drop
shaft would an object need to make to achieve Earth-normal escape
velocity, assuming Earth-normal gravitational acceleration?
At the moment, I can't reference the archives to see what the
accumulated wisdom of ages since has said - apologies for any
duplication. I do, however, do this sort of thing for a living, from
time to time. So here I go:
Clarifying assumptions:
Earth-normal Acceleration: 9.8 m/s/s is a normal value.
However, in light of the world-building discussions, another critical
parameter is the gradient of that acceleration. That won't affect how
many transits the demon has to make to reach a given velocity, but it
*will* affect the "escape velocity" and/or the "orbital velocity" the
demon needs to reach and/or the "orbital radius" it would have to go
out to to be at "geosynchronous" altitude. A smaller, denser world
could have 9.8 m/s/s surface gravity which would fall off more
rapidly, requiring a lower escape velocity. Alternately, some of the
big hollow worlds we have talked about would have a lower gradient,
requiring a higher escape velocity.
Finally, rotation rate of the world directly drives
"geosynchronous" altitude. Rotating faster means a lower altitude.
As the question states, I'll assume a Cidri Earth-like in all
parameters. I'll list formulae as I go along, in case other worlds
are desired.
One other assumption: the exit velocity (including direction
of motion) from a gate pair depends on the orientation of the second
gate, not on the original direction of motion. In other words, you
can go into one gate of a pair travelling south, and come out the
other gate travelling east at the same speed. (If this is not true,
the drop tower described will produce a demon at escape velocity
travelling straight down, which is not going to do anybody much good.)
Escape Velocity:
For any central attracting body in a two-body problem, the
useful parameter in describing its gravitation is Mu, its mass
multiplied by the Gravitational constant. For Earth,
Mu = muEarth = 398600.44150*10^9 m^3/sec^2
Attraction due to gravity is given by this parameter divided by the
square of the radius from the center of the Earth. At the surface
(rEarth = 6378137 meters),
a = Mu / rEarth^2 = 398600.4 * 10^9 / 6378137.^2 = 9.8 m/s^2
which agrees with my claim above and with normal experience.
Gravitational force can also be treated (and it's really convenient
to do so) as the gradient of a potential function. The potential
function is
V = (- Mu / r)
In spherical coordinates, the gradient just becomes the derivative
with respect to r, or
a = Mu / r^2
as above (playing fast and loose with signs, and with apologies for
oversimplifying some vector math).
So going back to the potential, the potential goes to zero at r ->
infinity. Bodies always try to go to a lower (more negative)
potential by going to a smaller r.
This is really nice because it lets us calculate escape velocity
easily. It's just the velocity at which (specific) kinetic energy
plus (specific) potential energy equals zero.
(1/2) vEscape^2 + ( - Mu / r) = 0
As the body rises, it'll trade off kinetic energy for potential
energy. If it's exactly at vEscape, the sum will remain zero, and
velocity will reach zero at r -> infinity. (This will take a while
:-) .)
(Specific means "per unit mass". Bigger demon works out the same,
mass cancels from both terms.)
So for escape from Earth's surface,
(1/2) vEscape ^2 + ( - Mu / rEarth ) = 0
1/2 vEscape ^2 - 398600.4 * 10^9 / 6378137. = 0
vEscape = 11179.9 m/s
So we know how fast the demon has to go. At 9.8 m/s, collecting that
much velocity will require
( 11179.9 m/s ) / ( 9.8 m/s^2 ) = 1140.8 seconds.
At essentially uniform acceleration, during that time it will travel
(1/2) a t^2 = (1/2) 9.8 1140.8^2 = 6377014.07 meters = 20725295.726667 feet
which is a bit over
T = 20,725 transits
-------------------
of the 1000-foot drop. (Hey, why isn't this question posed in metric,
anyway? All the *rest* of Cidri is metric... must have come from the
United States of Cidri...)
However, it's not quite so simple. Air drag in the gate pair will add
up, even if it's what would normally be considered a pretty hard
vacuum in the drop shaft. One might imagine sealing up the whole
shaft in glass, then putting a "pump gate" in series, whose rule is
to pass air only to outside of the shaft. Eventually, air entrained
or compressed by the demon's falling body would push its way through
that gate and out of the shaft. But this gate would have to be
totally leak-proof in the reverse direction, and I'm not at all sure
it doesn't amount to a "Maxwell's demon" piece of equipment out of
which one could get unlimited energy. One might use repeated "Open
Tunnel" on the air inside the drop shaft, but this approach has a
diminishing returns problem. One might create a special purpose
potion similar to universal solvent which only works on gases,
causing all gasses in contact with it to precipitate into solution. A
couple of bucketfuls of that in the bottom of the drop shaft would
eventually suck down all of the air (and create some pretty highly
"carbonated" solvent)!. (I guess one would not want to leave a bucket
of this stuff out in the open. It'd eat all the air on Cidri, or
explode trying...) I don't know exactly how to get a good vacuum, but
I'll treat this as outside the scope of the question.
And then, there's the direction problem. Since the world is turning
(one rev/24 hours), the 1000-foot shaft is turning. At the equator,
after 1140 seconds, the bottom end will be 83 feet away from where it
started relative to the top end. Something pretty frictionless would
have to keep the demon centered in the shaft. This problem could be
overcome by placing the shaft at either north or south pole, and
there are other possible solutions. For now we will table discussion
of this as a workable detail.
Given T
from above, assuming Gate instability beginning on the 100th transit of
any particular gate, and taking into account the one-minute period of
gate function despite instability, derive number of Gate pairs, (G),
needed to attain T.
Well, the cheap answer is *one* gate pair, with a wizard who knows
"control gate" (or "create gate") to restabilize it during the one
minute of flickering. That would require enough apprentices or ST
batteries to cast a 50-ST spell once a minute for just about 19
minutes. However, it's impossible to guarantee that the wizard won't
have a string of bad luck on the "Control Gate" spells. Personally,
once the demon had built up 10 km/sec of velocity or so, I would not
stand within several kilometers of "ground zero" without a pretty
ironclad guarantee that the thing would be somewhere else when it
hit. However, I assume that once the demon has accelerated to escape
velocity, a controlling wizard utters a word which causes the demon
to pass through the lower gate(s) and fall onto a "launch gate", the
other end of which redirects him upward and outward. If this is the
setup, then a failure of the accelerator gate pair will just mean a
premature launch, with minimal consequences for the wizard(s) around
the launch facility.
If you assume 100 transits causes failure of a gate pair and need to
have all of the gate pairs in place prior to starting, it's more
complicated. Using the formula for distance under uniform
acceleration, where v1 is starting velocity (useful for calculating
time for 100 transits on second and later gate pairs):
s = v1 t + 1/2 a t^2
or
a / 2 t^2 + v1 t - s = 0
which can be solved using the quadratic formula as:
t = (-v1 +/- sqrt (v1^2 + 4 (a/2) s) )/ a
The negative root will not be useful, so in each case we just take
the positive root. (ie replace +/- with + in the above.)
First 100 transits = 100,000 feet = 30,770 m -> 78.8 seconds.
Flickering period for first gate pair: 60 seconds.
Second gate pair starts working at 138.8 seconds. Time for a spreadsheet:
time start pair start velocity seconds/100 trans. time
at gate failure
0 0 78.829355435 138.82935543
138.82935543 1361.9159768 20.819187336 219.64854277
219.64854277 2154.7522046 13.71715655 293.36569932
293.36569932 2877.9175103 10.406420559 363.77211988
363.77211988 3568.604496 8.4431706177 432.2152905
432.2152905 4240.0319998 7.129818293 499.34510879
499.34510879 4898.5755172 6.1839259358 565.52903473
565.52903473 5547.8398307 5.4675995867 630.99663431
630.99663431 6190.0769826 4.9049459592 695.90158027
695.90158027 6826.7945025 4.4505288155 760.35210909
760.35210909 7459.0541902 4.0753871946 824.42749628
824.42749628 8087.6337385 3.7601418195 888.1876381
888.1876381 8713.1207298 3.4913107106 951.67894881
951.67894881 9335.9704878 3.2592109084 1014.9381597
1014.9381597 9956.5433469 3.0567004508 1077.9948602
1077.9948602 10575.129578 2.8783913407 1140.8732515
So,
G = 16 "accelerator" pairs plus one "launch pair".
---------------------------------------------------
Given G from above, assuming Guild-standard 25-ST
working days, and two apprentices per master, derive number of
master/days to produce G. Show all work.
17 pairs at 100 ST /pair = 1700 fST.
1700 fST / 75 fST (per master and 2 apprentices) = 22.667 master-days.
For extra credit: Volunteer to ride that puppy, so we can open a Gate at
apogee.
Not yet.
Problem A: I'm not convinced the vacuum problem has been solved
inside the drop tube. I am convinced it has not been solved at the
exit of the launch gate pair. Trying to put a solid body through air
- any air a wizard and apprentice could breathe to emplace a gate
exit - at 11 km/sec. will create a frighteningly devastating shock
wave and spectacular (and somewhat dangerous in itself) light show.
It'll also shred the body in very short order. Avoiding this means
placing the exit of the launch gate in vacuum. But to do that, a
wizard with a ST battery has to be able to climb to vacuum on his own.
A Fresh Air item (self-powered) and a really really well-tailored
silver suit of fine plate armor *might* keep him alive, but I'm not
so sure. And what carries him up there? Maybe a flying carpet (with
another pressure-suited wizard driving it)? Maybe a Self-Powered
Flight item? (But once we have this technology, can't we already go
to geosynchronous without the drop tower?) Possibly the Wizard could
Teleport up, create the gate, teleport back down *near* the ground,
and use a parachute or Flight item for final descent. He could not
teleport back *to* the ground - the 10 seconds it took him to create
the Gate would have accelerated him up to around 90 m/s, a fatal
impact velocity. However, there's good evidence he could withstand
vacuum for 10 seconds, so he might dispense with the silver suit, and
air drag could safely slow him from 90 m/s of velocity. That would
then only require 90 (plus a couple for spares) points of fST battery
or Aid, plus a reliable parachute or plus a Flight item and a few
more points of fST for the landing.
Problem B: What goes up... well, if it's truly at or above escape
velocity (or if it's close and uses the Moon for a gravitational
slingshot), then it isn't going to come down at all. (But that's
another class of problem, which would also dissuade me from
volunteering.) But if it's below escape velocity, it'll be back, and
at that point it'll have a reentry and landing problem which we have
not discussed yet. One possible solution would be to teleport the
demon from the atmosphere reentry point to the *bottom* end of a
similar 1000-foot drop tower on the far side of the planet, with
gates there set to return to the bottom of the tower from the top.
Let the demon rise repeatedly through that tower for the next 1140
seconds until gravity drags it to a halt, then teleport it out to
terra firma. Same discussions as for the launching drop tube for air,
direction, number of gates, etc.
Volunteers should be able to answer the following question:
Will the Gate created in orbit be geosynchronous? Explain if not, and
describe.
Gates and Long-Distance Teleport both work over "unlimited" distance.
This strongly implies they can work from one side of a spinning
planet to the other. But (for an Earth-like Cidri), the opposite
sides of the planet are moving at something like 3000 km/hr compared
to each other (both counter-clockwise around the pole, as seen from
the north pole). If velocity of the teleporting body is preserved in
any inertial coordinate system, it will arrive moving 3000 km/hr
relative to its surroundings. That's bad. The only way I see to
prevent this problem from occurring is to assume that velocity is
preserved during teleportation in the *rotating* (ie non-inertial)
coordinate system attached to the planet. This specifically means
that energy of the teleporting body is *not* conserved (see also Note
A).
For a pair of gates, there is some "sense" to this. Velocity of the
body relative to the exit gate is similar to velocity of the body
relative to the entry gate. The gates themselves are moving with
respect to one another due to rotation of the planet they are
emplaced upon, and the delta-V of the body during transit is equal to
the relative velocity of the gates.
For Long-Distance Teleport, the connection is a bit more tenuous. The
change in velocity of the body from origin to destination is perhaps
governed by displacement vector of the teleportation, combined with
rotation vector of the local terrain at the origin of the
teleportation (I think the actual equation would be a vector cross
product, Delta V = Rotation x Displacement). If so, a teleported body
always ends up moving in a circle about the rotation axis of the
planet, at a velocity which matches the local velocity of the planet
matter at the destination.
For getting into geosynchronous orbit, the consequence is that
Long-Distance Teleport may be the tool of choice. To place a
destination gate at geosynch, we would have to first climb to the
correct altitude, *then* accelerate to geosynchronous orbit velocity,
*then* create the Gate. The Drop Tower could give us velocity, which
we could then trade for altitude (if we solve the air drag problem).
But it could not *then* give us more velocity to go from a transfer
orbit to a circular geosynchronous orbit. Of course, if the gate
automatically refers itself to the rotating coordinate frame of the
nearest planet at creation, it might solve the second part of the
problem for us. In this case, we'd shoot the wizard up to
geosynchronous altitude. He would then create the gate, which would
immediately go zipping away from him at some 3.071 km/sec (see below
for velocity in geosynchronous orbit). The wizard would then fall
back to the surface, at which point see problem B above. (Note also
that all of this would have to take place at the equator, which rules
out our polar solution to the "side-of-the-drop-tube" problem.)
Long-Distance Teleport, if it's governed as I suggest by rotation at
the origin and vector distance of transmission, can do both. Simply
teleport a body up to the correct radius, and it'll have the same
rotational velocity vector as the rest of the planet - which *at that
radius from the planet center, and on the equator only*, will place
it in geosynchronous orbit.
The correct velocity can be derived by finding the radius at which
angular rate in a circular orbit matches the rotation rate of the
planet. Circular orbit velocity is found by equating "centrifugal
force" (really centripetal acceleration) to gravitational attraction.
Using Mu as above for the gravitational attraction of the planet, and
w (meaning small omega) as the angular rate in the orbit, and working
with specific force (ie per unit mass),
Mu / r^2 = v^2 / r = w^2 r
For Earth, where wEarth = 2 Pi radians / day = 2 Pi / 86400 seconds =
wEarth = 7.2722 * 10^-5 radians/sec,
398600.4 * 10^9 / r^2 = (7.2722 * 10^-5)^2 / r
r = 42241094 m (42,241 km)
and the velocity to maintain this orbit is just
v = r x w = 42241094. m x 7.2722 * 10^-5 radians/sec = 3071.8568569314 m/s.
Long Distance Teleport rules state, "...to anyplace else, provided
the wizard has been to the destination or can see it (or a clear
memory of it) by some means, magical or otherwise. ...". Obviously, a
wizard could see geosynchronous orbit on any clear night; however,
it's not too clear whether that is adequate, as one patch of empty
space looks a lot like another. A series of Teleportations of Proxies
could do a pretty good job of bracketing the distance. Too far, and
the Proxy will be in too high an orbit at too high a velocity, and
will go into too large an orbit. It will initially recede from the
planet, and by the time it has gone all the way around (as referenced
to distant stars) one orbit, the planet will have turned more than
once. Too close, and the Proxy will be in too small an orbit and will
initially fall closer to the planet. By the time it has gone around
one orbit, the planet will not yet have turned a full revolution. By
observing what the Proxy sees, the owning wizard could tell the
teleporter how to correct for the next attempt.
One other consequence of that principle, if it holds beyond
geosynchronous altitude, is that it is actually very easy to achieve
escape velocity. Simply teleport to a location *higher* than
geosynchronous. Circular orbital velocity and escape velocity from
that altitude will be lower, but the speed due to the rotation of the
coordinate system (r x wEarth) will be higher. The exact radius
(rEsc) is where (positive) kinetic energy due to the velocity
imparted by the rotation of the coordinate frame equals (negative)
potential energy due to the gravitational field:
-Mu / rEsc + (1 / 2) (wEarth rEsc)^2 = 0
or rEsc = 53,220,420 m (which is not that far past geosynch).
Back to the original problem, which I will now divide into two
separate problems.
A) getting into low circular orbit. I think a working configuration
(which I **might** be willing to ride) would consist of
1) Demon must be vacuum-proof. Self-powered Fresh Air item and Fine
Plate Armor, with lots of layers of rubberized canvas to help hold
pressure, is my best shot at that. Could also be a large "diving
bell" type pressure vessel, with appropriate items, but that will
have to fit through the Drop Tower gates.
2) A Wizard, similarly vacuum-proof, to accompany the Demon. This
Wizard must know "create gate" and must have pre-placed the exit end
of the "landing" gate described below. (This wizard could make the
trip alone, halving the needed number of Drop gates and Rise gates.)
3) Demon and Wizard Teleport into a Vacuum-filled Drop Tower at the
north pole, with 24 (12 * 2) paired gate sets to accelerate both
Wizard and Demon (holding hands) to just over 7784 m/s (which is the
speed for low circular orbit - see below, near the end of note B, and
above for the table of Drop-Tower speeds) plus the origin of a
"launch gate" to move them out to low orbit.
4) Exit end of the "launch gate" placed 100 km above the Drop Tower,
facing horizontal. This would have to have been previously created by
a Wizard who can get to an altitude of 100 km. See Problem A above
for discussions of how that happens.
5) When ready to de-orbit, Wizard waits until crossing above the Drop
Tower, then creates the origin of the "landing" gate directly in
front of the orbiting demon and of the wizard himself. The gate
freezes into the slowly rotating planetary reference frame, and the
wizard and demon slam through it at 7 km/sec.
Note: this means that the opportunity to de-orbit occurs only once
per orbit, so the Demon and Wizard *have* to be able to survive
around 100 minutes of vacuum. This has to be done at the north or
south pole, so that the motion of the planet doesn't cause the Wizard
and Demon to have "sideward" velocity on exiting the landing gate in
the base of the Rise Tube.
6) Destination end of the "landing" gate has been pre-set at the base
of a 1000-foot "Rise Tower" also at the north pole, pointing up. This
tower has the same requirements for vacuum and also has 24 gate
pairs, to slow the returning demon and wizard back to reasonable
speeds.
7) Demon and wizard fall upward, slowing, until they come to a halt,
and begin to fall back downward. At this point, they can teleport to
land outside the rise tower, or:
8) Bottom end of "Rise Tower" has one end of a "touchdown" gate, the
other end of which spits them out 500 feet or so above an open plain.
9) Both demon and wizard use parachutes, magic carpet, or Flight item
for final descent and touchdown on the landing plain.
Note: there's a more complicated version of this that'll also work
placing the Rise Tower at the South Pole and the launch gate exit at
the equator. However it requires pretty good knowledge of where the
equator is; this version is simpler.
B) To place a Demon in geosynchronous orbit:
1) Wizard and Demon must be Vacuum-proof, as above, and must have
Flight Items or parachutes for final touchdown, as above.
2) Wizard makes a series of Proxies (probably with Far Vision).
3) Wizard uses Long-Distance Teleport to send the Proxies to what he
thinks is geosynch.
4) Wizard uses the observed behavior of the Proxies, plus what they
can see, to target a Long-Distance Teleport to place the Demon in
true geosynch.
5) Wizard Long-Distance Teleports himself to the Demon's side.
6) Wizard Long-Distance Teleports Demon back to a location above a
landing plain
7) Wizard Long-Distance Teleports himself to same location
8) Both use Flight Items to land.
As before, teleporting back to Earth should be done from the *same*
point in the orbit as the arrival point (referenced to inertial
space) to minimize the velocity difference between the returning
travellers and local air. Even a small eccentricity in the orbit will
result in a fatal airspeed if they teleport out to perigee and back
from apogee, but if they go out and back at the same part of the
orbit (as determined by sighting past Earth's limb at distant stars),
they will be relatively safe. This means de-orbit opportunities occur
only once every 24 hours.
----------------------
Note A: Gates already present something of a problem for energy
analysis, because one end of a gate can be at a substantially
different potential energy (in any coordinate system I've been able
to imagine) from the other. There is no obvious way that the energy
added to a body by transiting the gate in the "uphill" direction gets
supplied by any visible agency. A perpetual motion machine can easily
be constructed this way. (Big rock rolls into the "downhill" end of a
gate, rolls out the "uphill" end, down a trough, and back into the
"downhill" end.) One might argue that the limited lifetime of the
gates, combined with the fST cost to create them, constitutes the
energy "input", but even here there are problems. Consider the first
and last gates of the "launch tower" table above. The first gate gets
about even numbers of transits before going unstable and during its
flickering period. The last gate gets around 30 times *more* transits
during its flickering period. Since energy imparted is proportional
to number of transits, the last gate provided about 15 times more
energy to the demon than the first gate - but cost no more to create.
Note B: Real "Demons" can teleport themselves. A simple algorithm for
a vacuum-proof demon to get into a low orbit is:
1) teleport to someplace at least 55 km (for Earth-like Cidri) above
the atmosphere.
2) Fall back toward the atmosphere until it starts to feel warm.
3) teleport back up to the original place.
4) repeat 2-3 until the desired velocity is reached. At escape
velocity, the teleports will be coming (55 km) / (11 km/sec) = 5
seconds apart, which is still a legal rate for a demon. Any orbital
velocity will be less than this.
5) Teleport to a place 1/4 of the way around the planet, and still
above the atmosphere.
The demon's velocity in the new location will now be horizontal,
instead of vertical. It will be at the perigee (pericid?) of an orbit
or of a hyperbolic flyby trajectory, depending on what velocity it
stored up in step 4. Velocity for circular orbit (see Note C) depends
on the radius of the orbit, and can be calculated from the equation
above,
Mu / r^2 = v^2 / r
If my surmise above about rotating coordinate frames and delta
velocities is correct, the demon will need to fine-tune the above
algorithm to take account of his velocity due to rotation of the
planet (or else do this algorithm starting above the north or south
poles, and teleporting to perigee above the equator). The vector math
there is a bit complicated and depends on latitude of the starting
point, so I'll skip it unless requested to dig into it.
Of course, for geosynchronous orbit, the easy way is to simply
teleport directly there, still assuming I'm correct about rotating
coordinate frames for teleportation and assuming that the same rule
continues to apply all the way out to geosynchronous orbit altitude.
If this turns out not to be the case, the above algorithm will work,
except that it has to be started above the equator (so that the final
velocity vector will be in the plane of the equator).
If the demon knows *where* geosynchronous orbit is, it can time step
4 to accumulate the correct velocity, then teleport out to
geosynchronous altitude 1/4 of the way around the planet. Even if it
guesses wrong, it should be able to fine-tune its way into perfect
geosynch by successive approximation.
A suitably vacuum-proof wizard with a big enough Strength Battery
could do the same, of course, but he should refine the algorithm to
minimize the number of teleports. For a low orbit, he should teleport
out to a location somewhat more than twice the altitude of the
desired orbit, fall back almost to the atmosphere, then teleport 1/4
of the way around the planet. For geosynch, he should teleport to a
somewhat lower altitude, fall back, then teleport to geosynch
altitude 1/4 of an orbit around.
The correct altitude for a given desired orbital velocity is
calculated based on the desired velocity and the potential energies
at the teleport destination and the top of the atmosphere (which I'll
arbitrarily call 100 km altitude). Using k for kinetic energy,
subscript "Terminal" for the condition at the final teleport (just
reaching the top of the atmosphere), and subscript "Initial" for the
destination of the first teleport, and as usual working with specific
energy,
kTerminal = (- Mu / rInitial) - ( - Mu / rTerminal)
kTerminal = 1/2 vTerminal^2
1/2 vTerminal^2 = (- Mu / rInitial ) - (-Mu / rTerminal)
Substitute in rTerminal = rEarth + 100 km = 6478137 m, and choose
vTerminal for the appropriate orbit.
For geosynch, vTerminal = 3071 m/s --> rInitial = 7,016,140 m.
For low orbit, say vTerminal = 7784 m/s --> rInitial = 12,761,400 m.
Of course, escape velocity requires an infinitely high initial
altitude to accomplish on a single fall, so the wizard will need at
least three teleports total for that.
Note C:
It is curious that Escape velocity at a given altitude is
always just the square root of 2 times circular velocity for the same
altitude. From Note B for circular orbit,
Mu / r^2 = vCirc^2 / r
or
Mu / r = vCirc^2
and from the discussion of escape velocity,
(1/2) vEscape^2 + ( - Mu / r) = 0
or
2 Mu / r = vEscape^2
so combining,
vEscape = Sqrt(2) vCirc
--
- Mark
Cell Phone: 210-379-4635
office: 210-522-6025
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