Technical Notes

The project was started 1997-10-20. The first draft was finished approximately 1998-08-21. Final draft finished 1999-08-12.

Abbreviations used throughout: ly = light year; eyr = Earth year; e Eri = epsilon Eridani. ``Terra'' is generally used for Earth, the homeword of the humans, and ``Sol'' is its primary star. By contrast, ``the Sun'' is the primary star of whatever world you are on: epsilon Eridani for most of the story.

``E'' format numbers: you have a number (with or without a decimal point) followed by ``e'' and a (signed) integer. The number is multiplied by ten to the integer power. For example, 1e3 is one thousand while 1e-3 is one thousandth. ``E'' format is much easier to use than alternatives: for example, the luminosity of e Eri is 1.09e26 W, not ``one point zero nine times ten to the twenty-sixth power, watts''. Or even worse: ``one hundred and nine heptillion watts''; who can interpret that? In heptal radix, ``h'' is used instead of ``e''.

Epsilon Eridani System

e EriSole Eri/Sol
Distance10.8 ly
Spectral classK2G2
Absolute magnitude6.14.721.38
Apparent magnitude3.72.32 (from other star)
Luminosity1.09e26 W3.90e26 W0.281
Color temperature5040 K6420 K
Wavelength peak574 nm (yel-grn)451 nm (blue)

Likely the color of e Eri would appear similar to Sol but the sky would be much darker. For Sol the experimental peak is 450 to 480 nm due to absorbtion lines. For Sol spectral density 50% points are 330 to 800 nm.

Orbit radius, same power8.6e10 m1.50e11 m0.575 AU
Length of year1.810e7 sec3.156e7 sec0.573 eyr
Apparent radius of star0.75 deg0.53 deg1.41
(from equal power orbit)
Absolute radius of star1.13e9 m1.39e9 m0.81
Mass1.51e30 Kg1.991e30 Kg0.76
Age (but see below)2.69e9 eyr4.63e9 eyr0.581
Geocentric coordinates3h30m-9d38m15h30m+9d38m
In constellationEridanusSerpens (caput)

The apparent magnitude of Sol (as seen from epsilon Eridani) is very similar to alpha Serpentis, which is 2.5 to 3. In the story the Serpent is said to have two eyes; the stars are close, 2.5 degrees apart (neglecting parallax effects on constellation shapes, which really can't be neglected). For comparison, Luna is 0.5 degree across and Lyra is 7 x 4 degrees.

Note, in earlier calculations the absolute magnitude of Sol was taken as 2.5. Since the orbital period of Wotan is woven into the story in many places I have not adjusted the radius of the ``equal power'' orbit, which in fact gets 0.85 times the insolation of Terra.

The age of e Eri was picked somewhat arbitrarily; a younger but plausible age improves the ability of the heptapi to use nuclear power. After chapter 6 was written I found a hard statement of the age of e Eri: 0.1 x Sol rather than 0.58 x Sol (2.69e9 eyr) as the story says. 5e8 years is far too young to produce any life, much less to produce heptapi. Reference: Govert Schilling, News of the Week: ``Hints of a Nearby Solar System?'', Science, v281 p 152 (1998-07-10), quoting Jane Greaves (Joint Astronomy Center, Hawaii) speaking at the Protostars and Planets Conference, Santa Barbara, approx same date.

Reference for star data: Norton, Arthur P, Norton's Star Atlas 17th ed, Longman (Wiley in USA), 1986, ISBN 0-582-98898-5 in UK, 0-470-20678-0 in USA. Chapter 4, p. 94.

Planets of Epsilon Eridani

The Kuiper cloud of e Eri has been observed, but resolution is not good enough to detect planets if they were present. All this information, except for Terra, is fictional.
OrbitalLength ofIllumin-MassSurfaceName
(meters)(sec)(x Terra)(Kg)(meters)

Kuiper cloud of comets: radius 2.5e12 m to 4.5e12 m. Oort cloud: not sure; farther out. Estimated mass of Kuiper cloud material for Sol: 6e25 Kg (10 x Terra). The story says that the comet abundance at e Eri is about 1/10 Sol, which would be 1 earth mass. An update: William R. Ward and Joseph M. Hahn, ``Neptune's Eccentricity and the Nature of the Kuiper Belt'', Science, v280 (1998-06-26) p2104. Direct observations of Kuiper cloud objects are rare. In the article the mass in Sol's Kuiper cloud is estimated to be 1 x Terra in the radius range of 48 AU (Neptune's radius) to 75 AU = 1.13e13 m. The story remains with 1 Terran mass in e Eri's Kuiper cloud.

More on Kuiper cloud: The measured radius of e Eri's Kuiper cloud is about the same as for Sol (story says it's less), and the ``number of comets'' is around 1000 x Sol. Also the age of e Eri is given as 0.1 x Sol rather than 0.58 x Sol (2.69e9 eyr) as the story says. Reference: Govert Schilling, News of the Week: ``Hints of a Nearby Solar System?'', Science, v281 p 152 (1998-07-10), quoting Jane Greaves (Joint Astronomy Center, Hawaii) speaking at the Protostars and Planets Conference, Santa Barbara, approx same date.

Njord: Having a similar size and insolation as Venus, it has an Earthlike oxygen atmosphere. A large ocean was recently present but has mostly evaporated due to warming of the stratospheric cold trap, due to chlorofluorocarbon production. Sea-dwelling inhabitants, the heptapi (heptapus-es), will soon be extinct. In reality the effect of greenhouse gases on the stratospheric cold trap is very complex, depending on details of the absorbtion and emission spectra. The effect of CO2 in Terra's stratosphere is to cool it, not warm it as hypothesized for Njord. Frozen particles of nitric and sulfuric acid provide a site where chlorine oxides from the decomposition of chlorofluorocarbons can catalytically react with ozone.

Wotan: It's a gas giant, a little larger than Jupiter, but at a radius so the insolation is similar to Terra.

Thor: Moon of Wotan. It's rocky and generally Earthlike, though a little larger, but it lacks most volatiles. It has a very limited atmosphere and no surface water. Spin-orbit locked, like Luna.

Freyja: Gas giant, analogous to Saturn. No significant moons.

Loki: Cold, dimly illuminated rockball, like Mars in size.

Data about Thor

Rotation period1.38634e5 sec8.6400e4 sec(38.48 hours)
Mean density5.0 g/cc5.5 g/cc
Surface gravity10.3 m/s29.8 m/s2
Mass8.395e24 Kg5.983e24 Kg
Radius7.373e6 m6.37e6 m
Surface area6.83e14 m25.10e14 m2
Orbiter's velocity8627 m/s7823 m/s (at 150 Km)
Primary power from star1.395e3 W/m2(same)
Blackbody surface temp280 K(same)P/4 = s T4
Neighboring objectWotanTerraLuna
Radius of neighbor1.55e8 m6.37e6 m1.74e6 m
Distance from neighbor4.60e8 m3.84e8 m3.84e8 m
Neighbor subtends angle39 degrees1.90 deg0.52 deg
Thor's velocity vs. Wotan2.082e4 m/s
Orbiter's velocity over Wotan3.585e4 m/s(at 150 Km)
Gravity over Wotan8.30 m/s2(at 150 Km)

Star Travel

See below under ``CQMT Pusher Design'' for the equations from which much of this material was obtained.
Distance10.8 ly = 1.022e17 m
Average speed desired0.666 c
Maximum velocity0.85 c
Average time dilation0.713
Travel time16.2 eyr = 5.11e8 sec
Onboard duration11.6 eyr = 3.66e8 sec
Spacecraft mass (total)6680 Kg
Peak power2.12e12 W
Proton energy at this speed842 MeV

But see below; the shield mass was underestimated by about a factor of 3.5. Also, the power per kilogram was underestimated. However, 2.12e12 W is a very convenient number for the plot, so it wasn't corrected.

Lion Person Development

Time to manufacture a set of eggs:2 eyr
Time to sexual maturity:16 eyr
Generation time (minimum):18 eyr
Human lifetime:80 eyr
Lion lifetime:Unknown, longer

Story Outline

(Note: as actually written the story deviates from this outline. See the timeline for times of various events as actually appearing.)

  1. Tiger discovers the quantum connection effect or Tiger Hole (CQMT). Attempts to build a starship end with the Lion Foundation rejecting international participation and sending their own ship. Times: Effect discovered at age 41 yrs, ship launched at 53 yrs. On arrival the crew have not aged due to being frozen. Terra hears of arrival and existence of Heptapi at 80 yrs; on arrival the crew receive Terran data as of 58 yrs. Note: As written the story has just Surya at home when the effect is discovered. This would be at 44.7 yrs, not 41 yrs. Consider adding 4 yrs to all times below. Can Willie and Wilma survive to 88? Their interval of hard digging would end at 80-ish. Realistic? Or rewrite with 4 kittens still at home?
  2. The crew meet and negotiate with the Heptapi, who blow themselves to Kingdom Come. This takes 5 years. (Age 58 yrs; Terran input 63 yrs, Terra knows at 85 yrs.)
  3. The crew resolve to terraform Thor. 3 years are spent to build infrastructure in space and to manufacture a set of 8 lions. 8 otters follow 2 years later. The latter reach sexual maturity 16 years later, at which time: Age 79 years, Terran input 84 yrs, Terra knows at 106 yrs.
  4. First comet impact at 74 yrs, 16 yrs after start, lions aged 11 yrs. Comet bombardment continues for 20 years.
  5. Having launched at 82 yrs (29 yrs after the Lions, 2 yrs after existence of Heptapi was learned), the Chinese expedition arrives and gets itself integrated with the Lions. They arrive when the lions are 82. 8 humans are built finishing at 84 yrs; sexual maturity at 100 yrs. Lion kittens at 79 yrs, otters at 81 yrs, humans at 102 yrs.
  6. Willie and Wilma die at 84 yrs (surviving just long enough to hold the human babies, sniff, sniff).

Names of Characters

Each Novanima is identified by a letter representing the species, a unique serial number in decimal, hyphen, and an octal encoding of the recognition scent. The latter has fourteen odor components which can be present or absent. Two indicate the species (and are both absent in lions). Females smell of allspice while males smell of cinnamon, and the 2000 bit encodes both scents, being one for females and zero for males.
NameCodeSpeciesSexMateJobName means
TigerL6-3512LionFSimbaMission Commander
SimbaL7-1340LionMTigerBiotech, Medic
Willie--HumanMWilmaMechanical Engr
Wilma--HumanFWillieGeology, Astronomy

Kittens (born in chapter 11 and 12):
IrisL1001-1212LionMLucentStudent LeaderRainbow (Greek)

Iris is Tiger's favorite because she feels guilty about aborting Iris-3. Iris has a problem with short attention span. This retards his lesson progress, and he feels unworthy because of it. He thinks before acting, but is decisive and determined. He's willing to break rules and suffer the consequences if he sees a benefit to it. He loves to explore.
JacinthL1002-3027 LionFKenGem zircon

Jacinth is impulsive and aggressive. In chapter 18 she takes the lead in piloting spacecraft. She ends up taking the lead in cooking.
KenL1003-0522LionMJacinthEconomicsJapanese sword

Ken is the smartest of the kittens. He is reserved, timid and tends toward introspection. Nonetheless he can be courageous when really needed. He has a problem with passive-resistive behavior and depression. One of his major contributions is the money system (economics).

Lucent is intelligent and prudent. She thinks before acting. She can be forceful if needed but prefers to avoid confrontation. (A major Terran corporation has this name.)
MicaO1005-0275OtterMNightPapermakerSparkly rock

Tiger and Simba remember the sparkly black mica crystals in a rock they sat on, on their wedding day.
NightO1006-3431OtterFMicaBall bearingsBlack fur

Like other otters, Night is prudent and thoughtful and social, but is courageous when it's needed. She decides on a career manufacturing rotating machinery.
OsoU1009-1530'UomiMPetraAgricultureBear (Spanish)

Named in honor of El Oso, padre of Mariposa and Coyote, who was killed in Simba Leones. Stonecutter and farmer.
PetraU1010-3037'UomiFOsoStone (Latin)

Several incidents of brat behavior and overreaction.

Interested in genetic engineering. Plastic producer for the colony. His name refers to the quinones in the Chang bushes.
RoseJ1013-3400JaguarFOrionMedicHas thorns

Rose is smart. She is Iris' close friend because of a match in lesson progress. Medical training.

Selen is always happy and is highly social. Unbalanced about sex. His work is craft manufacturing: pottery and textiles. He becomes the colony's psychologist. (At least one Swede has the family name of Selen. Original origin unknown.)
TitaniaO1008-3026OtterFSelenArtist, MedicThe metal

Titania is fierce defending her position. She is an excellent artist and consumes Mica's paper. Medical training. Works in genetic engineering.
OrionJ1014-1210JaguarMRoseMighty hunter

Name should start with U but none suitable. Orion is strong and aggressive and boisterous. He's a little behind the others in lesson progress.
ValeriaU1012-2423'UomiFQuinMetalworkerRoman legion

Valeria has the nickname Vulcan because of her manufacturing and metalworking specialty. In Chapter 13 she is greatly disturbed because she's a fake human, not real, but gets this worked out. The last Roman legion to stand against chaos in Britain was Valeria Victrix.
WolfJ1015-1725JaguarMXenaWillie's son

Wolf is less aggressive and more introspective than Orion, and not as muscular.

Xena takes her Warrior Princess role seriously, being the colony's military expert. (Name is normally spelled Xenia, from the Greek. In a currently popular TV series Xena is a warrior princess.)

The crew's kittens on Terra:

Lion Button Panels

1SweatExtend genitalia
2Wake up (reset 3 and 4)Urinate
3Blood pressure downSex

CQMT Pusher Design

Through the Tiger hole (name changed to Alembic d'Alimentation Trans-Spatiale or AATS), null-mass (photonic) signals take a very small amount of time longer than they would have taken on a normal space geodesic path between the endpoints. Non-null effects are best analyzed in a non-accelerating reference frame such as Terra's.

To operate a CQMT pusher (beside overhead and losses) requires power as if a mechanical connection existed, i.e. Force*Velocity. Power is in fact transferred through the hole to the spacecraft and is subject to the Doppler shift, but it's easier to analyses as if the active end were on Terra and acted on mass inflated by the Lorentz-Fitzgerald contraction. Excess power cannot be dissipated safely on the spacecraft and is returned through the hole to Terra.
mSpacecraft rest mass
VVelocity of spacecraft (relative to Terra)
aAcceleration of spacecraft
FForce exerted by pusher
pPower supplied to spacecraft (excluding overhead and losses).
cSpeed of light

Evaluation for power fixed on Terra by sundipper output. We're referring to longitudinal acceleration. At zero velocity the acceleration at finite power is technically infinite; the initial transient is ignored during which the force on the pusher is the limiting factor.
F dt = m/(1-(v/c)2)0.5 dvIf transverse
F dt = m/(1-(v/c)2)1.5 dvIf longitudinal

(Reference: Eddington, sir Arthur, The Mathematical Theory of Relativity, Cambridge University Press, 1922, p. 31)
a = v' = (1-(v/c)2)1.5 (p/mv)F = ma, solve for a
Let b = v/c
db/dt = b' = (p/mc2) (1-b2)1.5/bThe same, dimensionless
Let T = mc2/p
dt = T b/(1-b2)1.5 dbSeparate variables
t = T / (1-b2)0.5Integrate both sides
(T/t)2 = 1-b2Solve for b
b = (1 - (T/t)2)0.5Finish solving for b
x = c intg (b dt)Integrate to get position
w = T/t, dw = -T dt/t2, dt = -T dw/w2Change of variables
x = -cT intg (1-w2)0.5/w2 dwFinish the integral
  = -(cT) ((1-(T/t)2)0.5t/T + arcsin(T/t)) from t = T to infinity.
  = -(cT) (bt/T + arcsin(T/t))
a = c db/dt = c/T (1-b2)1.5/b

The following table contains a numerical integration of the starship's motion. All quantities except arc length are evaluated in the reference frame of the start point (Terra). Quantities are "dimensionless", that is, velocities are normalized by the speed of light, and times (equivalently distances) are normalized by how long it would take for the power fed to the ship (assumed constant in the start point reference frame) to equal its rest mass. The trajectory shown is accelerating only. For the actual journey the ship would accelerate for half the time, then do a symmetrical deceleration.

These are the columns in the table:

Numerical Integration of Starship Motion
TimeArclenDistanceSpeedAccelAvg Speed
T s X v a v(avg)
1.00 0.000 0.0000.0000 0.000 0.000
1.10 0.095 0.0290.4166 1.803 0.286
1.20 0.182 0.0780.5528 1.047 0.388
1.30 0.262 0.1380.6390 0.712 0.458
1.40 0.336 0.2050.6999 0.521 0.512
1.50 0.405 0.2770.7454 0.398 0.554
1.60 0.470 0.3530.7806 0.313 0.589
1.70 0.531 0.4330.8087 0.252 0.618
1.80 0.588 0.5150.8315 0.206 0.644
1.90 0.642 0.5990.8503 0.171 0.666 <-
2.00 0.693 0.6850.8660 0.144 0.685
2.10 0.742 0.7720.8793 0.123 0.702
2.20 0.788 0.8610.8907 0.105 0.717
2.30 0.833 0.9500.9005 0.091 0.731
2.40 0.875 1.0410.9091 0.080 0.743
2.50 0.916 1.1320.9165 0.070 0.755
2.60 0.956 1.2240.9231 0.062 0.765
2.70 0.993 1.3170.9289 0.055 0.774
2.80 1.030 1.4100.9340 0.049 0.783
2.90 1.065 1.5030.9387 0.044 0.791
3.00 1.099 1.5970.9428 0.039 0.799

When the spacecraft slows down, the decrease in kinetic energy relative to the power nexus is sent back through the hole (minus losses and overhead). It is assumed that the equipment is sized to accommodate the same power coming out as coming in, but no more. Thus the relevant journey is halfway to the target, at which point the spacecraft reverses thrust for an equal time. It is desired to make the trip of 10.8 ly at an average speed of 2/3 c, or a travel time of 16.2 yr. From the table above, this means t/T has to run from 1.0 to 1.90, v(max) = 0.85 c, x = 0.599 cT = 5.4 ly (half way then turn around), so T = 9.0 yr. At maximum velocity, acceleration is 0.154 M/s2 (i.e. very small).

Radiation Issues

Protons at 0.85 c have an energy of 842 MeV vs. rest. Shielding cross section is 2e-2 barn/electron (asymptotic over 100 MeV) = 2e-30 m2/el. Water has a density of 3.3e29 el/m3 so the e-folding depth is 1.5 m. In space the dose from (non-solar) cosmic rays is 5e-4 gray/day. For the planned journey the nondirectional dose is 2.96 gray. Reference for cosmic rays and shielding: Gregory, A. and R.W. Clay, ``Cosmic Radiation'', in: Weast, R.C. (ed.), Handbook of Chemistry and Physics, 63rd ed. (1982), CRC, page F-175.

Density of interstellar matter: 1e-24 g/cc = 1e-21 Kg/m3 (average over patches) = 6e5 protons/m3. Radiation power: 1.9e4 W/m2 = 12.6 gray/sec. (over 1.5 m e-folding distance). Times journey duration this gives 4.8e12 gray. To reduce this dose to equal the nondirectional dose requires e-folding of 28, or 42 m of water. Suppose we can squeeze everything into 0.5 m2 shadow area.
Hydrogen intercepted8.02e-5 Kg
Energy at 0.85c7.22e12 J
Radiation dose (1.5 m e-folding)4.81e12 gray
Assumed metallicity2%
Dust intercepted1.61e-6 Kg
Dust grain mass3.16e-16 Kg
Kinetic energy at 0.85c25.5 J (mc2(1/(1-(v/c)2)0.5-1))
Density of grains (approx)4.8e-8 #/m3
Quantity of grains hit6.12 #/sec (0.5 m2 area)
2.45e9 grains
7.75e-7 Kg

Reference: M. Baguhl et al, Space Sci Rev #72 p471 (1995), quoted in: Colwell, J.E., Horanyi, M, Grun, E., ``Capture of Interplanetary and Interstellar Dust by the Jovian Magnetosphere'', Science, v280 #5360 (1998-04-03). Measurement of dust impacts by Galileo and Ulysses show that beyond 3 AU the dust flux is dominated by interstellar grains with a flux of 1e-8 #/cm2 sec parallel to the local interstellar wind. Their mean mass is 10(-12.5±1.5) gram. The velocity of the grains is approximately equal to Jupiter's orbital velocity, 2.079e3 m/sec.

If the spacecraft loses power when traveling at maximum speed, the radiation power (if independent of velocity, which is not accurate but isn't too bad of an approximation) would bring it to rest in about 6e9 eyr. The radius of the universe is approximately 1.6e10 eyr, so in reality the spacecraft would continue moving at considerable speed, though not 0.85c, ``to the utter end of the universe''.

Spacecraft Mass Budget

4 people280 Kg
Supplies and stuff1000 Kg
Pusher, power conversion500 Kg
Structure500 Kg
Radiation shield21000 Kg
Total23080 Kg

For T = 9.0 yr, m = 6680 Kg, then p = 7.32e12 w (formerly 2.12e12 watts). Conversion: p = mc2/T = 3.17e8 m (mass in Kg, power in watts). Note that the plot was established for a prior underestimate of the shield mass (total 6680 Kg vs. 23080 Kg). The 3.5x increase in needed power would be inconvenient, though, and so the story was not corrected.

Rescuing the Telescope (Chapter 2)

Mass of telescope10 tons
Cross sectional area10 m2
Nominal orbital altitude400 Km (250 U.S. miles)
Scale height of exosphere6500 m

The disaster is assumed to have been an impulsive discharge of maneuvering fuel which lowered the perigee. For the plot, the telescope's orbit will decay in about 8 days = 7e5 sec. Preliminary calculation show a perigee around 100 Km. The major effect of atmospheric drag would be to gradually lower the apogee, circularizing the orbit, upon which the telescope spirals in rapidly. It should be noted that after the story was written, AXAF (renamed the Chandra X-ray Observatory) was not placed in low Earth orbit but instead well outside the van Allen belts. This orbital placement, however, would necessitate quite a bit of changes to the plot.
Energy in nominal orbit-5.8937e11 J
Energy in 100-400 orbit-6.0273e11 J. delta = 1.33e10 J
Energy in 100 Km orbit-6.1670e11 J, delta = 2.73e10 J
Power to take down satellite2.00e4 W

Let's assume that atmospheric drag is concentrated over 1/4 of the orbit (actually an overestimate given the scale height). Drag = RAv2 (R = density). The drag is inferred from the loss power, and the perigee altitude is inferred from the air density necessary to give this drag.
Orbital velocity7850 m/s
Drag force (average)2.5 Nt
Drag force (peak)10.2 Nt
Air density1.65e-8 Kg/m3
Perigee altitude125 Km

To rescue the satellite we'll need to apply twice the drag power. The pusher chip efficiency is assumed to be 85% but half of that is dissipated on the ground.
Heat sink power3e3 W
Heat sink temperature70 C = 343 K
Radiance at that temp.785 W/m2
Radiator area3.8 m2
Square flat sheet, 2 sides1.38 m edge length
Power per chip30 W
Size of case2 cm approximately square
Number of chips100 (plus 100 partners)
Radiator area per chip.038 m2
Radius of circular patch11 cm
Pitch of square grid19 cm

To calculate the thickness of the sheet: the temperature drop is

intg (r1,r2) P dr/(2pi rtC) = P/(2pi tC) ln(r2/r1)

where P = power, C = thermal conductivity, t = thickness, r1-r2 = patch radii.
Target conductive drop15 K
Conductivity of copper394 W/mK
Thickness (for 30 W)1.9 mm
Mass of copper57 Kg

Reference for atmospheric data: U.S. Standard Atmosphere (1976), NOAA-NASA-USAF, in: Weast, R.C. (ed.), Handbook of Chemistry and Physics, 63rd edition (1982), CRC, page F-164.

Burmese History (Chapter 3)

Alaungpaya ``The Coming Buddha'' reigned 1752-1760, reuniting (or re-conquering) all of Burma plus at least some of Thailand. He was killed in battle during the Thailand campaign. In 1886 the King was deposed and Burma was annexed by Britain. Independence was gained in 1948, after which there was a sequence of various parliamentary, socialist and outright brutal regimes.

People's Skills (Chapter 3)

SkillMain personBackup
Plasma PhysicsWillie*Simba*
Mechanical EngineeringWillieWilma*
Quantum Computer DesignTigerSimba*
ProgrammingSimba, TigerEveryone

* = needs to be learned.

Chang Bush Varieties (Chapter 3)

The varieties of grain grown on Terra, in order of importance (quantity?), are wheat, maize, rice, oats, barley, rye, millet, sorghum. Chang bushes are bioengineered to produce simulations of all these grains. In addition the Chang bushes are said to produce soybean and azuki bean flavors. Additional starch crops (not Chang varieties) are yams, sweet potato, Inca potato, and cassava (taro).

The Chang bushes economize on water by using sunlight energy to isomerize quinones, and then at night opening the stomata, taking in CO2 and releasing O2. The bushes also actively transport water from the air. In Simba Leones a net positive water balance was alleged during summer in the Panamint valley. Cacti also use the strategy of storing energy during the day and opening the stomata at night, although they don't use quinones and don't actively take up water.

Chang bushes are host to symbiotic acacia ants. These defend their plants fiercely against herbivores. If a pheromone is sprayed in a box set near a plant, the ants will bring seeds (ripe ones only) to the box. The plants have hollow spaces in their stems for the ants to live in, and secrete food for the ants. Some acacia species actually have this arrangement with specialized ant species (not including seed collection).

Metals in Comets (Chapter 4)

Data is used for Terra's crust. It's kind of fudged up with added volatiles; quantities are guessed. It's well known that there are substantial segregation effects, so the abundance in comet dust will be different. Carbon is assumed about equal to oxygen. Fractions are by mass. Fractions add up to about 1.043, sorry.
O (Total)0.35 (Sum of O content of above items)

More authoritative cosmic abundances (by count of atoms):
H1 - Y - Z (0.95?)

Chang Bush Growth (Chapter 4)

Mass of seed (dry)0.025 g
Mass of seedling10 g
Doubling time30 days
Time to harvest 1st seeds90 days / 80 g
Full size plant10 Kg
Production (immature plant)1e-3 of plant mass per day =
10% of plant mass every 100 days
Production (mature plant)2e-3 per day = 20 gram/day
Seed cycle30 days (seeds develop asynchronously).

Initially the story was written with 1 gram seeds, but that's the size of a small Lima bean, not realistic for rice, wheat and oat simulations. The time between planting and the 10 gram stage would actually be at least a month, but too much already depends on the schedule below, so it's assumed to be instantaneous.
Planned Schedule
DaysKg perTotalSeeds
Most of planting done300.015000 (planted 50 Kg)
Start thinning weak plants900.0420000 (immature seeds start)
First harvest1200.0820002.0
Mature plants33010.020004.0

Actual Schedule
DaysKg perTotalSeeds
Initial test sprouts50.010.080
Most of planting done300.01400 (planted 4 Kg)
First harvest1200.083200.32
Full production2000.520002.0
Mature plants33010.020004.0

Human Nutrition (Chapter 4)

Lions are assumed to have the same need for calories as a human. However, they are autotrophic: they can make all needed vitamins (not minerals) and can manufacture protein in their livers.
Various Human Requirements
Adult maleProtein56 g/day = 0.8 g/ (70 Kg male)
Adult femaleProtein46 g/day
LumberjackEnergy3700 kcal/day = 1.55e7 J/day
Household WorkerEnergy2100 kcal/day = 8.78e6 J/day
Geneva ConventionEnergy1500 kcal/day = 6.27e6 J/day
Florence on DietEnergy1000 kcal/day = 4.18e6 J/day
Killer WhaleEnergy55 kcal/ = 2.3e5 J/

Composition of Foods
Kg/KgKg/KgKg/KgMcal/KgJ/Kg x1e6
Millet (dry).103.029.6553.2813.7
Rice (dry).075.020.7803.6015.0
Wheat (seeds, dry).137.028.6293.3013.8
Soybean flour, no fat.470.010.0323.2613.6
Soybean flour, full fat.371.200.2294.6619.5
Chicken, meat only.233----1.35 5.7
Sea otter, homogenized------1.81 7.6
Pork lard--1.0--8.8537.0
Fat (on the hoof)--.925--8.1934.0

The last item refers to fat on a living body, for diet calculations. Reference for killer whale (Orcinus orca) and sea otter (Enhydra lutris): J.A. Estes et al, "Killer Whale Predation on Sea Otters Linking Oceanic and Nearshore Ecosystems", Science, vol. 282 nbr 5388 (1998-10-16), page 473.

Rice in CIMMYT experimental plots under ideal conditions of fertilization, pesticide application, water and weather can produce 2 tons (edible portion) per hectare 3 times a year. This corresponds to 1.67 grams per m2 per day or 6 Kcal/ or 2.5e4 J/ This data is from around 1970. In 1999 yields up to 22 tons/ha.yr are occasionally obtained, but 12 to 15 tons/ha.yr is more typical of what real farmers get with the best new varieties of wheat.

In Simba Leones, Simba tells the Stanford board plan people that he properly eats 7 MJ/day, and can handle 5 MJ/day uncomfortably without losing mass. Assuming Chang bush production is equal to the CIMMYT performance, this corresponds to:
Percent full power100%71%
Kcal/day17001200 (rounded)
Kg/day Chang seeds0.507 (500g)0.362 (360g)
Nbr of mature Chang plants2518
Floor area (mature plants)279 m2199 m2
Floor area (immature plants)558 m2398 m2

The proportions of C,N,P average 106 C / 16 N / 1 P with little variation, in worldwide samples of marine particulate matter (microscopic plants). The proportions in the soft tissue (not? counting bones) of humans is nearly the same. Reference: {C. and G.} Copin-Montegut, Deep-Sea Research v30 p31 (1983). Quoted in: Paul G. Falkowski, Richard T. Barber, Victor Smetacek, ``Biogeochemical Controls and Feedbacks on Ocean Primary Productivity'', Science, v281 p200 (1998-07-10).

Interplanetary Travel Times (Chapter 5)

aacceleration = 3 m/s2,
ttravel time.

Basic formula: d = .5at2, t = (2d/a)0.5

Plan: constant acceleration half way, then turn over and constant deceleration. Thus d -> d/2, t -> t/2.

Formula: t = 2 (2(d/2)/a)0.5 = 2 (d/a)0.5
FromToDistance (m)Time (eday)
Kuiper CloudNjord4.5e1228.3
Kuiper CloudWotan4.5e1215.5

All assuming acceleration at 3 m/s2 except the last, in which the miner accelerates at 10 m/s2.

Planet Exploration (Chapter 4 and 5)

B = Tiger's age in eyr; N = newsfeed date. See the main timeline for details on these units.
Arrival050.450.5Start chapter 4
Get oriented, arrive comet1050.450.5
Finish comet activities3050.550.8
Arrive at Njord6050.651.2Start chapter 5
Finish Njord9050.651.6
Arrive at Wotan9450.751.7
Finish Wotan and Thor12550.752.1harvest starts,160 g/day
Arrive at Freyja13050.852.2
Finish Freyja14050.852.3
Arrive at Loki14550.852.4harvest 320 g/day at 150 days
Finish Loki16550.952.7
Arrive at Njord again17050.952.7end of chapter 5
180(harvest 640 g/day)
200(1000 g/day, full production)

Ground Imaging Telescope (Chapter 5)

lwavelength (lambda) = 500 nm
ddiameter of mirror = 30 cm
rdistance to target = 150 Km
ediffraction limit
l/d1.6e-6 rad
erl/d = 0.25 m

However, Spot Image sells pictures with 25 m resolution. A spy satellite (with better mirrors than Tiger has) can show a U-2 on the ground, probably with 1 to 2 meter feature size. Let's assume 25 m resolution. An image of 2k x 2k pixels would cover (50 Km)2. Complete tiled coverage would take 273200 images or 1.09e12 pixels. Let's assume 8 multispectral channels of 8 bits each; gives 8 terabytes. Note, figures are for Thor. Njord is ``a bit smaller than Earth'' whereas Thor is a bit larger.

Orbital velocity: 7823 m/s, which means 6 seconds per frame. How long to cover the whole planet? Over-coverage due to orbital convergence (on a polar orbit) increases the time by pi /2. 2 orbiters but half the planet is in night. Thus: 28.5 days.

How does the coverage time vary with planetary parameters? Assume a fixed patch size of p2. Other than through the patch size we'll neglect the height above the surface.
v = (ra)0.5 = (Gm/r)0.5(orbital velocity)
v = ((4pi /3)Gqr2)0.5(q = density)
A = 4pi r2(surface area)
t = (pi /4) (A/p2) (p/v)(time to cover the whole surface)
t = (pi /4) (12pi /Gq)0.5 (r/p)

Hunting for Comets (Chapter 5)

A comet that's rich in dust, for easy mining, will be very hard to see in visible light because the dust is black. However, it will radiate in thermal infrared.
Radius of Kuiper cloud (avg.)3.5e12 m
Power from e Eri1.09e26 W
Power density at Kuiper cloud0.708 W/m2
Stefan-Boltzmann const.5.67e-8 W/m2K4
Blackbody temperature42 K (P/4 = s T4)
Peak of normalized Planck distr.hf/kT = 2.821
Wavelength of maximum power1.21e-4 m (hc/(2.821)kT)

The power density is divided by 4 because the comet is spherical, not flat (integral of sin2) and it faces away from the star half the time. Thanks to Dr. Peter Wang, president of Fermionics, Inc., for suggesting the detector design: photoconductive silicon doped with low-lying impurity states. It would need to be cooled to 4K.

Altitudes on Njord (Chapter 5 and 6)

Original sea level6260830 geopotential meters.
Sea level (truncated 2 digits)60830
Lower limit of erosion60000
Plate surface; atmosphere base57000, varies a lot.

Charlie's Government Service

Chapter 5 (40% through): He's nominated to the SEC but the committee won't approve it, and a supporter has them adjourn sine die rather than have an embarrassing rejection. (Which committee handles the SEC, and is that proper Senate procedure?)

Chapter 9 (90%): He's nominated to the SEC and confirmed.

Chapter 12 (73%): He's nominated as Chairman the FRB and confirmed. (Handled by the Committee on Banking, Housing, and Urban Affairs, per Senate rules.)

On 1998-08-17 I sent a message to asking for help on the above questions. 1998-11-20 I got a form letter reply saying the message had been received.

A prof in the political science department suggested that I read committee transcripts, which are on a web site. I decided that for a relatively minor subplot that degree of research was anally retentive. The sections in chapters 5 and 9 have been rewritten so the parliamentary maneuvers are handled informally by the chairman, which is plausible. If the Senator comes through, the sections can be revised again.

References indicate that the Chairman is appointed by the President, saying nothing about advice and consent of the Senate. However, recently a FRB member was promoted to Vice-Chairman and Senate confirmation was obtained. The story was rewritten accordingly.

Timeline of Chapter 6 to 9

61Collect air and water samples, attacked by heptapus, find truck
62Water radioactivity. Imaging program.
63Observe heptapi by sonar. Decode their sonar output.
74Meet bravo heptapus. Meet commander.
75More water samples. Look for mines. Language progress.
76Visit by the Nagus. Survey Eastern ocean.
87Armageddon. (Day 0 on chapter 9 timeline.)
88Debate replanting.
89Continue debate. Gaia plan. Tentative decision to replant.
910Final decision to replant. Design colonists.
911Begin execution phase, no longer day by day.
931End of chapter: arrive at Thor.
10?Smelting ore
1040Heptapus archaeology

Uranium Decay (Chapter 6)

Symbolize U-238 by ``U'' and U-235 by ``X''. Let [U] and [X] = concentration of the species in the ore. Let u and x represent the e-folding times for decay (1.44 times the half life = half life over ln(2)). t = elapsed time; [U0] = initial concentration. It is assumed that the nuclear processes in a supernova (R process) that synthesize the uranium isotopes produce [X0]/[U0] which is similar everywhere, i.e. Terra and Njord.

[U] = [U0] exp(-t/u) and similarly for X.

[X]/[U] = ([X0]/[U0]) exp(-t(1/x - 1/u))

Terra was assembled 4.63e9 years ago. It is not known how long elapsed between Terra's supernova and assembly of Terra, but almost certainly it was well under 1e9 years, probably no more than 1e8 years. Thus I neglect that time.

Presently, [X]/[U] = 7.11e-3

u.5 = 4.51e9 yr; x.5 = 7.1e8 yr (half lives)

1/(1/x - 1/u) = 1.216e9 yr (= 1 / 8.23e-10) (e-folding)

exp(-t(1/x - 1/u)) = exp(-3.81) = 0.0222 with t = age of Terra.

[X0]/[U0] = 0.321

[X]/[U] = 0.035 would occur at a time of 2.69e9 years

Refrigerants (Chapter 6)

Common NameFreon-11
Boiling point23.82 C
Average molecular mass137.37 amu
Assuming most common isotopes:
Total molecular mass135.90495

Heptapus Mathematics (Chapter 7)

Numbers are in heptal (base 7). In Heptapus language the numeric format is unit, exponent, separator (H), integer part, point, fractional part. In English we use the opposite order (1.23H4 Unit) using H as the marker for a heptal exponent.

Length unit: tentacle lengths (about 2 meters).

Time unit: swim strokes (about 1.5 seconds).

Conversion procedure on calculator:

Nuclear Decay (Chapter 8 and 18)

(All energies in KeV)
Halflife5.26 eyr
Decay energy2819
Beta rays 315 (99.87%), a few at 663 and 1488
Gamma rays1173 (99.88%)
1332 (100%)
2158 (0.001%)
Halflife28.1 eyr
Decay energy546
Beta rays546 (100%)
Gamma raysNone
Halflife30.23 eyr
Decay energy1176
Beta rays511 (94%)
1176 (6%)
Gamma rays662 (85%)

Reference: Handbook of Chemistry and Physics, Table of the Isotopes, B-264; ibid, Gamma energies and intensities of radionuclides, B-340

Oceans of Njord and Thor (Chapter 8)

This data is in fact for Terra. Thor will be terraformed similarly.
Area of ocean3.61e14 m2 (71% of surface) (For Terra)
Mean depth of ocean3794 m
Maximum depth10430 m
Volume of ocean1.37e18 m3 (area * mean depth)
1.66e18 m3 (mass of hydrosphere)
Water buried in lithosphere2e18 ton approx.
Water in mantle as OH2e20 ton approx.
Mass of atmosphere5.14e18 Kg
Mass of Njord oceans (puddles)2.5e14 ton (fiction)

Some geologists also believe that there is a substantial amount of hydrogen dissolved in the Earth's core, like several percent by mass, and if the redox state of the mantle were to change, that could become available as surface water.

Reference for Terran data: ``The Earth: its Mass, Dimensions and Other Related Quantities'' from NASA TT-F-533, translation of Russian sources or data table with references dated mostly 1950-1959. In: Weast, R.C. (ed.), Handbook of Chemistry and Physics, 63rd ed (1982), CRC, page F-154.

Potential Energies (Chapter 8)

This is Willie's assignment. Value is the potential energy to go from place A to place B, in units of (m/s)2.

Potential energy to refill Njord puddle from Wotan: 1.79e23 J. This is equal to ship's power for 2683 eyr.

Volatiles in Comets (Chapter 8)

This is Wilma's assignment. Comet infall time (Kepler's law):

R = semimajor axis of comet's orbit (radius of circular orbit)

G = gravitational constant

M = mass of primary (e Eri)

t = period of orbit

For circular orbit (orbital period for comet will be the same):

w = 2pi /t (angular frequency)

a = R w2 = GM/R2 (Acceleration)

w = (GM/R3)0.5

t = 2pi (GM)-1/2 R3/2 = 6.261e-10 R3/2(for e Eri)

If the comet of mass m needs to be moved laterally a distance dx within

(arbitrary) time T, the force F needed to do it is:

dx = 1/2 F/m T2

m = 1/2 F/dx T2

F = 2m dx/T2

We're approximating that all comets have the same semimajor axis and all have a very high eccentricity (so periapsis is well within Wotan's orbit). Each comet has a particular location in phase space, which is the direct product of position space (3D) and momentum space (3D). In the story there's a maximum force F which can be used to move comets laterally so they hit Thor, and a (fixed) time T during which the force can act. It's evident that the product (m dX) is fixed; thus it's irrelevant whether a few large comets or many small ones are chosen. The set of points (hypothetical comets) that will hit Thor by themselves within a time T is topologically equivalent to a 4D ball. Given the upper bound on impulse, a 5D tubular region of phase space can be pushed to hit Thor. The problem is thus equivalent to arraying the comets on a 2D sphere of radius 2R and saying that one point on that sphere will drop on Thor, and comets within dX of that point can be pushed into Thor. Of course comets in the outbound portion of the orbit can be pushed for a longer time, or shorter inbound, so the available impulse will average out.

The available impulse can be used to move many little comets or fewer bigger ones, or all the comets in a small tube vs. some of the comets in a big tube. But for optimum performance one should clean out all the comets in the smaller tube. For a radius dX the average displacement dx = 2/3 dX.

Also a comet could hit Thor on the inbound or outbound portion of its orbit, and could also be steered via Freyja and Wotan (in the latter case hitting Thor after a turn around e Eri), so there are six independent regions of the cloud that can be made to collide. Thus the density of the cloud should be inflated by a factor of six. So if the total mass of the Kuiper cloud is K (and doubling R since it's a radius):
m= 6K pi dX2 / (16pi R2) = K/16 (dX/R)2 (mass in radius dX)
dX= R (8/3 m/K)0.5
F= 2m (2/3 dX) / T2 = 8/3 (2/3)0.5 R m1.5/(K0.5 T2)
2RAverage rad. of Kuiper cloud3.5e12 m (must divide by 2)
1/2 Period (infall time)7.25e8 sec = 22.96 eyr
Velocity at Wotan's orbit4.84e4 m/s (from potential energy)
TPush time demanded by plot3.15e8 sec = 10 eyr
mRequired comet mass9.25e17 Kg (neglecting composition)
KMass of Kuiper cloud5.983e24 Kg (1 x Terra)
FForce required1.40e10 Nt
Number of large chips (3000 Nt)4.66e6 chips
Radius of equivalent sphere9.03e4 m (density 0.3 g/cc)
Number of comets, 1 Km radius3.08e6 comets

The above is to put an atmosphere on Thor so people could breathe pure oxygen with helmets, not suits, on the plate surface, not just in the subduction trench, which means a density at the plate surface of 18% Terran normal. To refill the Njord puddle would require 2.5e17 Kg H2O, 1e18 Kg total, which takes about the same resources to do. Using the assumptions on production rate in, the B factories would put out 4.247e+06 chips in 12 years. Of course the chips are also being used to bring in water. Details, details. The story assumes a favorable orbit for the big comet that provides most of the atmosphere, so the indicated number of chips is an overestimate, plotwise.

Building an Atmosphere (Chapter 8)

Atomic composition of incoming comet material, from cosmic abundances (count of atoms):
ElementRel. to HRelative
H30 (combined only)

Estimated composition and proportion of molecules. ``Rel'' gives the proportion by molecules (unnormalized) and the mass percent. It's assumed (without objective evidence) that half the carbon is in CH4 and half is in CO2.
CompoundRel.By MassKcal/MoleJ/Mole
CH4315%-12.1-5.06e4(half the carbon)
CH2O---38.6-1.62e5(biomass, eth glycol)
CH2-4.93-2.04e4(per additional unit)

Desired composition of atmosphere plus ocean, imitating the Terran atmosphere and ocean but not subterranean water. An approximately equal amount of water is buried in the lithosphere and 100 times more is dissolved in the mantle as OH.
CompoundQty KgQty Moles
CO2A lot

H2 is not gravitationally bound and if produced in large quantities will exit the planet before reacting with atmospheric oxygen. In the story water is deposited in orbit around the planet and is photolysed by solar ultraviolet.

Reaction: CH4 + 2 H2O -> CO2 + 4 H2 (1.30e5 J/M endothermic) gives:

Reaction: CH4 + CO2 -> 2 CH2O (biomass) -> 2 H2O + 2 C (coal): Perfect!

Reaction: n CH4 + m CO2 -> C(n+m)H(4n-4m) + 2m H2O

Reaction: 2 NH3 + 3 CH4 + 3 CO2 -> N2 + 6 H2O + C6H6

Reaction: C + 2 H2O -> CO2 + 2 H2

If we assume CH4 + CO2 -> 2C + 2 H2O, creating an Earth-type ocean (1.66e24 Kg H2O) requires the following inputs with 2:3:3:6 mix.

An equal amount of water is produced in the reaction, plus an equal number of moles of carbon.

Wilma's Second Plan (Chapter 8)

We'll build an Earth-density atmosphere from comet impacts and later bring in the rest of the H2O and CO2 after fractionation on Freyja, leaving behind the NH3 and CH4. Mass of Terra's atmosphere: 5.14e18 Kg, 79% N2.
N2 + O25.14e181.78e20Total atmosphere when finished
NH34.79e182.82e20Composition of comet material

Oxygen will be derived from CO2 + H2O -> CH2O + O2, endothermic 4.70e5 J/M or 113.31 Kcal/M. This plan is changed later to ultraviolet dissociation of H2O and NH3 whereupon the H2 escapes to Wotan.
Total energy to produce oxygen1.76e25 J
Solar power6.37e18 W (incl 50% atmos loss only)
Time to produce oxygen (50% eff)2.76e6 sec = 31.9 days :-)
Mass of oxygen per area2.36e3 Kg/m2
Biomass per area2.21e3 Kg/m2
Tree production rate10 ton/hectare-eyr = 1 Kg/m2.eyr
Time to produce oxygen (biomass)2.21e3 eyr
Disposing of reduced species:
CH4 + H2O -> CH2O + 2 H230.2 Kcal/M, 1.26e5 J/M endothermic
Total energy to dispose of CH45.33e25 J
Mass of CH4 per area1.33e4 Kg/m2
Biomass per area2.50e4 Kg/m2
Time to dispose (biomass)2.50e4 eyr
2 NH3 -> N2 + 3 H2, per N: 3.9 Kcal/M, 1.63e4 J/M endothermic
Total energy to produce N2:4.60e24 J

Size of Genome (Chapter 9)

The coding regions of all human genes extend over about 120 Mb. Reference: David G. Wang et al (27 authors), ``Large-Scale Identification, Mapping and Genotyping of Single-Nucleotide Polymorphisms in the Human Genome'', Science, v280 (1998-05-15) p. 1077; in the article a reference is given which may be the original source of this number.

The total size of the human genome is about 3 Gb. Essential areas in addition to coding regions include upstream regulatory domains, introns, transfer and ribosomal RNA genes, centromeres and telomeres. Introns tend to be larger than really necessary for their regulatory and evolutionary functions. In addition there are numerous classes of repeated sequences, pseudogenes, retrotransposons, integrated viral genome fragments and so on, collectively termed ``junk DNA''.

The cystic fibrosis gene is about 30 Kb long but is strung out over 1.5 Mb with many large introns. Some genes have 40 to 50 introns. Most genes are present in a single copy, but gene duplication events are common too, and the pairs tend to diverge in sequence and function. (In some cases one copy becomes inoperative; this is a "pseudogene".) Some plants such as maize and rice have clearly duplicated their entire genomes in nature, being quadruploid or even octuploid (doubled twice), though the copies aren't exact.

The figure of ``about 300 Mb'' for the lion genome is probably reasonable. If the issue comes up, we'll use 312 Mb as the exact size.

Adiabatic Lapse Rate (Chapter 9)

The initial site for the colony is in a subduction trench 1e4 meters deep. The atmosphere is a mixture of argon (mass 40, monatomic) and nitrogen (mass 28, diatomic). What is the pressure at the bottom of the trench as additional material is added at the top, uniformly diluting the argon, assuming the trench holds a negligible fraction of the total atmosphere?

Pv = RTEquation of state
Cv = (n[j]+1/2)RIdeal gas specific heat, const volume
Cp = (n[j]+3/2)RSame at constant pressure
y = Cp/Cv = (n[j]+3)/(n[j]+1)Normal symbol: gamma.

The dependence of Cv and Cp on the number of atoms per molecule is almost exact for n=1 and 2, less exact for n=3, and poor for more atoms.

In an atmosphere heated from below, the temperature and pressure as a function of altitude z will be such that when a unit of gas moves from point A to point B, the change of gravitational potential energy will exactly balance the change of thermal energy (and a corresponding statement about derivatives). Less temperature gradient and there's no energy to drive convection; more gradient and the convection will accelerate until the gradient is smoothed out. (Reference: Reif, Fundamentals of Statistical and Thermal Physics, McGraw-Hill (NY), 1965, p. 159.)
Pvy = KConstant in adiabatic convection.
v = K(1/y) P(-1/y)
dP/dz = -gr = -gm/v = -mgK(-1/y)P(1/y)
P(-1/y)dP = -mgK(-1/y)dz
P(1-1/y) = -mgK(-1/y)(1-1/y)zValid for negative z; z = 0 when T = 0
K = P(RT/P)y = Ry Ty P(1-y)evaluated at one point, P0 and T0.
K(-1/y) = P0((y-1)/y) / (R T0)
P = P0 (((y-1)/y)mg/(R T0) (-z))(y/(y-1))
z0 = -(y/(y-1)) R T0 / mg
P = P0 (z/z0)(y/(y-1))with both z, z0 negative.
T0 = 280 K (blackbody temperature).
v = RT/P
P(1-y)Ty = const.
T = T0 (P/P0)((y-1)/y) = T0 z/z0(!)

In the following table, the pressure and temperature are given at the bottom of the trench assuming adiabatic conditions, and the ``pressure ratio'' has the plate surface pressure in the denominator. In reality, radiative cooling will produce a superadiabatic lapse rate, decreasing the temperature and increasing the pressure, but this effect is not referred to explicitly in the story.
CvgammaMol wtScalePressPressTempComposition
J/MKKg/Mht, MratioNt/m2Kpercent Earth normal
16.621.5034.00-1.99e+43.381.03e+4420Ar 1.5 N2 1.5
15.251.5536.00-1.78e+43.541.61e+4437Ar 3.0 N2 1.5
16.621.5034.00-1.99e+43.382.06e+4420Ar 3.0 N2 3.0
18.001.4632.00-2.24e+43.222.94e+4405Ar 3.0 N2 6.0
25.351.3328.99-3.16e+43.055.53e+4369Ar 3.0 NH3 3.7 CH4,CO2 5.6
20.741.4028.95-2.73e+42.983.01e+5383O2 21 N2 78 Ar 0.96

The first row is the initially chosen atmosphere at 3% Earth normal. The resulting pressure at Gondolin is too low to be practical. So let's fiddle with the composition by adding more argon: the second row. This is what's used in the story. Rows 3 and 4 show the effect of gradual addition of nitrogen. Row 5 is the atmosphere after some major comet bombardment; P0 (on plate surface) is 18% Earth normal = 1.81e4 Nt/m2; better than the original trench bottom. The last row is Earth normal.

Terra's surface oxygen partial pressure is 2.12e4 Pa. Andean people handle 1.0e4 Pa after major adaptation. Pure O2 at 1.64e4 Pa is equivalent to an altitude on Terra of about 2100 M (6900 ft); 1600 M is handled by ``ordinary'' people after modest adaptation.

The atmosphere mass after the comet (row 5) is about 9.44e17 Kg.

Timeline for Events in Chapter 9

Day Event
0Heptapi blow themselves up (in ch. 8; day 7 on ch. 6 timeline)
1Planning about replanting (chapter 8)
2Continue debate. Gaia plan. Tentative decision (chapter 8).
3Final decision to replant (chapter 9)
3Production schedule and Novanima design (chapter 9)
Begin search for heptapus cities. Start comet miner returning.
9Find heptapus village. Inspect heptapus city.
19Comet miner arrives.
20Depart for Thor.
24Arrive at Thor.

Time to Make Equipment (Chapter 9)

These items are made in the Japanese chip prototyper (proto) and the B factory, which is about 4 times faster. Times in Earth days.
2.5Small pusher chip, 100 Nt (includes partner)
102.5Large pusher chip, 3000 Nt (includes partner)
2.5Video camera
2.5W32 CPU chip
2.5Memory 64 Mb total
15APX detector
15X-ray detector
15Interplanetary bee prototype (4 sections) (not including partner)
150.45Production bee (chiller, or pusher 10000 Nt)
150.45Power nexus for set of four full-size bees

Concrete (Chapter 10)

From Encyclopedia Brittanica, ``Concrete'', the proportions to make it are:
Aggregate (rocks and sand)9 units (by mass)
Cement2 units
Water1 unit
Total12 units
Plus carbon dioxide2.5 units (releasing the 1 unit of water)

The carbon dioxide is absorbed during about thirty days after pouring. The amount was computed by assuming that all the water hydrates oxides, and the hydroxide is entirely replaced by carbonate (probably an overestimate):

Day Phase (Chapter 10)

Thor's rotation period (also orbital period) is 1.385e5 sec = 38.48 Earth hours. The crew decide on a 3:2 day cycle with the activity cycle 9.235e4 secs long (25.7 Earth hours). Pro forma, a Thor Hour is day/36 = 3848 secs.

Eclipses: Wotan subtends 39 degrees, so the longest eclipse would last 3.90 (Thor) hours = 1.50e4 seconds. Most or all days have eclipses; the inclination of Thor's orbit hasn't been specified but it's nonzero. The timings given below are for longitude 0 (sub-Wotan point) where the eclipse is centered in the day.

Under a variety of scoring variations, where waking in daylight and sleeping in night are preferred, it's optimal to wake at sunrise on one of the days and pessimal to wake six hours later, or at sunset two days later. The scores are periodic over 12 hours. Phase 6 has about 3/4 the ``correctly lighted'' hours of phase 0.
1SunriseEclipse: wake+8Sunrise+16Sunset: sleep+2
2Sunset+6Sunrise: wake+12Sunrise+4Eclipse: sleep+4
3Sunrise+12Sunset: wake+6Sunset+10(none)

At Gondolin Wotan hangs in the west (ch. 10) so it's at maybe +45 to 50 degrees (East). In chapter 21-23 the day phase is critical and the plot is designed for 0 degrees (sub-Wotan point). Gondor doesn't have timezones; hour zero is sunrise at the sub-Wotan point.
Longitude0 deg+50 deg (East)
Sunrise (type 1)0 hrs19 hrs (= -5 hrs)
Eclipse (center)9 hrs9 hrs
(Range)7 - 11 hrs7 - 11 hrs
Eclipse (after sunrise)9 hrs14 hrs
(Range)7 - 11 hrs12 - 16 hrs

Calendar (Chapter 10 and 11)

Unless specified, time units refer to Thor, except Earth seconds are always used (there is no special second adapted to Thor's rotation).
Seconds per day92423 approx.
Hours (Thor) per day24
Hours (Earth) per day25.673
Thor rotations per day2/3
Seconds per (Wotan) year1.810e7
Days per year195.83
Days per month28 (1.016 x average Earth month)
Months per year7 (minus 1 day on unleap years).
Seconds per Earth year3.156e7
Earth years per Wotan year0.574 = 1 / 1.743

The year starts on the day containing the vernal equinox. Most years are 196 days long; about 1/6 have an anti-leap day and are 195 days long.

Timeline for Chapter 10

50Smelting ore at Thor
45Travel back to Njord
429Heptapus archaeology at Njord
451Travel to Thor
155Test dig at the Gondolin site

Gondolin Site Geometry (Chapter 10)

Gondolin is in a subduction trench about 1e4 meters below the surrounding plate surface (ocean floor, on Terra). On Terra there would be 3000 to 4000 meters of ocean (mean 3794 m) above the plate surface but the trench would also be partially filled with copious sediment. The trench axis runs southeast to northwest, or, convergence is northeast to southwest. The northeast plate is going down. It has about a twenty degree slope, though the land is cracked and irregular, meaning that there's a horizontal distance of about 2.7e4 meters to the plate surface, or 3e4 meters path length not counting going around or through the numerous deep canyons. The southwest side is a continental block. Its edge has about a 40 degree slope and is dominated by landslides into the trench. Gondolin is established somewhat to the southeast of the trench center on the descending (northeast) plate, and Sirion flows northwest.

Snowball Gathering Cycle (Chapter 11)

One way Thor to Freyja1.81e4 sec(5 hours, at 10 G)
Thor to Wotan7.52e2 sec(at 10 G)
Freeze a snowball5e4 sec
One way Freyja to Thor9.07e4 sec(with snowball, at 2 G)
Wotan to Thor3.75e3 sec(at 2 G)
Total (Thor <-> Freyja)1.59e5 sec
Total (Thor <-> Wotan)5.45e4 sec
Potential energy, Freyja/Wotan0.874(vs. Thor, ratio)
Ratio (PE/trip time)0.300
Division of chips0.231 Wotan, 0.718 Freyja
Mass of snowball0.5 Kg
Number of butterflies7100(at start of chapter 11)
Mass flow (Freyja + Wotan)3.22e-2 Kg/sec= 1.72e-2 + 1.50e-2
Power (Freyja -> Thor only)1.17e7 watts(Wotan -> Thor is equal)
Time per snowball arrival15.5 sec
Time per chip1.10e5 sec(avg round trip time)

Timeline in Chapter 11

Partway through chapter 12 the integrated implanter gun lets Willie make implanter factories in 2.5e5 secs each. This timeline is replicated in an analysis for chapter 16.

Kitten Development (Chapter 12)

Ages are in Earth months, for Novanima species and Homo sapiens.
93Ready to implant next set of kittens
16--First handsigns
1915-20Toilet training (goes as needed, and presses own buttons)
2013First words (voice)
3024Useful conversation
40?Can pay attention to stories
5042Reading on their own
Age ofEvent
0Lions implanted (start of chapter 11)
7Lions born (end of ch 11)
8Finish dome 2
90Otters implanted
145Iris finds a jacinth crystal (ch 12)
156Lions ejected from pockets
167Otters born
167Selen plays pattycake with Lucent (ch 12)
19101Finish dome 3
25167'Uomi born
25167Swimming lessons (ch 12)
271890Jaguars implanted; work on other life
2920112Finish dome 4
3425167Jaguars born
39302113Simba finishes air plant; finish dome 5
42332415Jaguars ejected from pockets
43342516Reading lesson (ch 12)
49403122Finish dome 6

Domes, Food and Oxygen (Chapter 12)

Domes are 24 m diameter, floor area 452 m2, except 1/3 is occupied by mats, tables, tunnel ports, emergency cans, foundations, etc. Effectively 300 m2. With practice it takes 0.8 eyr to build one residential dome, with tunnels. An agricultural dome takes 0.3 eyr (it's less elaborate, less digging.) It takes 1000 m2 of floor area to feed four adult lions and humans (260 Kg biomass). Kittens count double due to higher metabolism. This table assumes all 16 kittens were born at the same time, ignoring the actual 27 month skew. (See ``Human Nutrition'', chapter 4.)

Crops in Domes (Chapter 12)

Each dome has at least a small amount of each plant species, for disaster recovery, but the major occupants go like this (in chapter 12):
1AChang bushes
2AChang bushes
4ANonfoods: trees, sagebrush and feather grass

Kittens in Domes (Chapter 12)

Kittens are segregated by sex but, as much as possible, not by species. This table indicates what combinations of species and sexes don't have a dome in which all can legally meet. In the failed list, sets with fewer species are listed first.

Theorem: If there are M species there are 2M combinations of sexes. To accommodate every one requires 2M domes. In N domes the number of M-species combinations that can't be realized is 2M - N. Empirically it's possible to get sex assignments, for 4 species in 5 domes, such that this relation holds for every subset of species, i.e. 4x(8-5) 3-species combinations plus (16-5) 4-species combinations are illegal. Similarly for 4 species in 6 domes. All combinations of two species are legal in 4 or more domes. In 4 to 6 domes a relatively few combinations of three species are illegal.

(For reference: 6 domes 4 species took 1.5 days to compute, in PERL, hiss, boo.)

This is the assignment finally adopted. (H) indicates the home dome.

Given two Novanima of specified species and sex, which dome(s) can they meet in?

Dome assignments during emergency (chapter 14):
6No Cover

Dome assignments after mate selection (chapter 17 and 18): * indicates moved.
2Open space
3Quin and Valeria*
Wolf* and Xena
4Oso* and Petra
Orion and Rose*
5Iris and Lucent*
Selen* and Titania
6Ken* and Jacinth
Mica and Night*

Timeline for Chapter 13 and 14

0Chapter 13 starts, quoted age 19.7 tyr (11.3 eyr). Tail docked.
5Valeria is angry with Tiger.
40Chapter 14. Iris insists on being ready to do damage control.
47Comet impact. Iris is hit by a meteor.
48Rain and flooding.

Lesson Progress (Chapter 13)

Numbers are age when the skill is learned, in eyr from assembly or begetting (subtract 0.75 eyr for human age from birth). Grade and Homo sapiens progress rates refers to public schools with ``ordinary'' kids.
1761 or 2 digit add
287Simple multiply
396Writing (unstructured)
5117-8Formal grammar
61210Long division
71311Syntactic analysis
71311Writing paragraphs
81412Writing essays

European Union Currency (Chapter 15)

The colony's coinage is patterned after the European Union. The 1 and 2 Euro coins are bimetallic, with brass on the outside and inside, respectively, and German silver for the other half. This brass has the composition Cu(75%) Zn(20%) Ni(5%) and the German silver is Cu(75%) Ni(25%).

The 0.1, 0.2 and 0.5 Euro coins are in Nordic Gold, which is a brass consisting of Cu(89%) Al(5%) Zn(5%) Sn(1%). It is said to be very resistant to tarnish, probably due to the aluminum.

The 0.01, 0.02 and 0.05 Euro coins are steel jacketed with (pure?) copper.

Population Growth (Chapter 16)

Let g = e-folding time for population growth, and p(t) = population as a function of time. Suppose the population is known at two times:

Let r = age of lions at the begetting of the first kitten and k = number of kittens per couple, assumed one per eyr. In the story they rush it: r = 18 eyr, k = 8. Two parents are needed. Taking a linear average,

Total population as a function of time (Tiger's ``B'' age), assuming all the kittens were assembled at once and reproduction in each generation is simultaneous.
54.716 (kittens)

Later (chapter 21) it's assumed that kittens are produced once every 2 Thor years (14 months), whereas on Terra it's once every eyr (12 months). The above model goes by 1 per eyr.

Factory Production (Chapter 16)

AdultKittenTimeMakingTechnological Advance
52.25e5B facWillie makes B factories 50% of 2000 hr/eyr
3.89e4chipTime for B factory to make a butterfly chip
56.012.5e5B facIntegrated implanter gun
58.741e5B facAutomated assembly of most of B factory
60.768e3B facAutomated assembly of most of smelter
65.0102.5e5A facManual assembly of A factory
5.0e5B facTime for A factory to make B factory
70.0152.0e5A facSimplification of A factory
75.0208.0e3A facE factory prototype, A factory finished by hand
100455.0e5E facTime for E factory to make itself
2.5e5A facTime for E factory to make A factory

Ages are in eyr.

Results (program

Chinese Vocabulary in Chapter 18

The digit after each word is the tone (in the Mandarin or guo2 yue3 "national language" dialect):
1High even

Available Power (Chapter 19)

Ship power (SS David Franck)2.12e12 W
Ship power (Chinese ship)2.5e12 (with same chip area)
Solar power (at Terra)1e4 W/m2
Solar power at sundipper orbit100 x Terra
Efficiency of sundippers39%
Total area of sundippers5.43e6 m2
Number of B factories:7.409e+04
Time per chip:38880 sec
Area per chip:10 x 10 cm
Time to make chips:2.852e8 sec = 9.06 eyr = 15.8 tyr
Water rate:1.913e+10 Kg/eyr = 1.096e10 Kg/tyr
Potential energy, Freyja -> Thor6.789e8 J/Kg
Power4.123e11 W
Fraction of starship power0.194

(This analysis is oversimplified since some water comes from Wotan and the chips would be redeployed to the longer Freyja -> Thor run, reducing the water flow. This is mentioned in the story.)

Land Use Ratios (Chapter 21)

Suppose a fraction W of the land is designated for large contiguous wilderness patches, P is a similar fraction for people patches, and the remainder (1-W-P) is to be split up in the same proportions recursively at smaller scales. In the remainder, all the land will eventually be in one or the other category, just in larger or smaller patches. Thus the ratio of wilderness to people area is just W/P.

For the purpose of the story, W = 0.25 and P = 0.25.

Epoch and Population for Chapter 21

The story originally described ``about 50 villages, a few isolated homesteads, and our permanent capital''. Assume 20 families (an average of 6 people each counting kittens) per village, 100 families in isolated homesteads, and 300 families in the capital. Population:

Using the population growth model developed for chapter 16, it would take 100 years, until 154.1, to achieve this population. This won't do. Lan Ying would be 112 eyr old. We need her to be about 70 eyr old which makes the date 111.7. At that point the population would be 580 people. This corresponds to about 2.6 generations at 8 kittens per 2 parents.

Character Names in Chapter 21

NameSernoAge eyrSexRole
TalismanJ1282-044024MaleParent, crazy
JoyJ1297-200324FemaleParent, driven mad
ArdentJ1548-14416.7MaleSenior kitten
CinderJ1580-04224.3MaleTurned back, fell in hole
DiabloJ1581-32343.2FemaleRenamed Dragon
FlameJ1583-115310moMalePocket kitten

Timeline for Chapter 21

See ``Day Phase'' for chapter 10, above. The listed eclipses are for latitude 0 (sub-Wotan point). Ardent mentions that the eclipse occurs after they started running away, helping them evade Talisman. ``Hr'' = hour of day. ``Lt'' = lighting phase: day vs. night.
1108181dKittens run away from home
0?dTiger starts hiking
221220--6nTiger meets kittens
7nRescue Cinder from hole
8nBack to Mithrim
10nPick up Simba from trail
11nConfront Talisman; he runs
19dTalisman collects supplies
33--None60dTalisman at Mithrim, hides
1dTalisman tries to rape Tiger
4dDeath ceremony for Talisman

Talisman's Farm (Chapter 21)

Each mature Chang bush produces (10 Kg bush mass)*(2e-3 per day) = 20 grams of seeds per day. A 70 Kg person consumes 507 grams of seeds per day, the output from 25.4 plants. 360 grams per day is the lower bound for starvation. (See Chang plant growth schedule, chapter 4.) Kittens count double by mass, due to higher metabolism. The Fire Kittens are undersized due to insufficient food. (See Food and Oxygen Production, chapter 12.)
NameAge eyrMass Kg
Ardent6.720(Kittens count double mass)
Blaze5.518(Deceased, not counted)
Flame10mo3(Age after begetting, not birth)
Total mass248 Kg
Food needed1.80 Kg/day
Starvation1.28 Kg/day
Plants90mature healthy plants (minimum)
Floor area1000 m2for 90 plants

We'll say they have 100 plants, but not healthy ones, due to ant flu and inadequate fertilization. Talisman believes fertilizer is a government plot to pollute his precious bodily fluids.

Timeline for Tiger in the River

Two times are given for each event: B = Tiger's biological age (where the journey counts as zero elapsed, since Tiger is frozen); N = what would have been Tiger's age had she stayed on Terra, minus speed of light delay. The latter is the age of Tiger's friends when they sent reports in the news feed then being received by Tiger. The term "friend" means someone, like the other original lions, assembled at the same time Tiger was. Units are eyr (Earth years). Decimal numbers are eyrs and decimals.

Lightspeed delay:
10.8 eyr
Journey time:
16.2 eyr (in Terra / e Eri reference frame)
5.4 eyr (when she reads messages from her friends, they are 5.4 eyr older than she is)
27.0 eyr (Friends reading her reports are 27 eyr older than Tiger when she sent them)
16.2 eyr (In a simultaneous reference frame, friends differ in age by the journey time)
21.6 eyr (An immediate response to news from Tiger reaches Tiger after this time, twice lightspeed)

Months: Tiger was assembled in March so that's age 0.0.

Upon arrival, newsfeed is read at 5x speedup for one year until they catch up. N dates are corrected for this and are marked *. If A = arrival date, N = (B-A)/5 + A.
00Tiger assembled (in March). (Birth 195 Earth days later.)
16.016.0Tiger and Simba marry. (Simba is 1 month younger.)
17.517.5Tiger starts college
21.321.3Attila begotten (other kittens 1 year apart). Finish college.
28.328.3Surya begotten (zero age for him).
42.342.3Anansi begotten (grandkitten: Alex x Attila)
44.644.6Begin ch 1: discover CQMT. Surya and Holly get into MIT.
45.045.0Ch 2: A Rescue in Space. Stock market collapse.
45.545.5Surya starts college (MIT) (his age 17.2 eyr)
47.847.8Ch 3: UNESCO Conference, starship announced.
49.349.3Ch 3: Test flight; test freezing process.
49.549.5Ch 3: Wolf starts college (Wooly 1 year later).
50.450.4End of Ch 3: Launch. 16.2 years pass frozen.
50.450.4*Ch 4: Arrive at epsilon Eridani.
50.4650.7*Election results (not a presidential election) (ref. in ch 8)
50.651.2*End ch 4: grow plants, find comet, mine it, go to Njord, ch 5.
50.651.3*Wolf finishes college
50.751.4*Chinese launch starship.
50.752.0*Charlie nominated for SEC, not confirmed.
50.852.3*Wooly finishes college
50.852.4*Chinese lose contact with ship.
50.8652.7*Election results (presidential). Faraldo out, Martin in.
50.952.9*Ch 6, 7, 8, 9. Heptapi discovered, destroyed.
51.053.3*Surya finishes grad school (4+4 eyr), thesis published
51.053.4*Charlie appointed to SEC, confirmed.
51.053.4*Revolution in China, Maoists in power.
51.053.4*End of ch 9.
51.254.4*End of ch 10: start work on Gondolin.
54.760.1Ch 11: implant lion kittens.
55.160.5Anansi starts college. (Not mentioned in story.)
55.260.6Lion kittens born (end of ch 11, start ch 12). Dome 2 done.
55.460.8Implant otters.
56.261.6Implant uomi.
56.361.7Dome 3 done (takes 1.1 yr due to kitten care).
57.062.4Implant jaguars.
57.162.5Dome 4 done. (Not mentioned in story.)
57.663.0Jaguars born.
57.963.3Dome 5 done. (Not mentioned in story.)
58.463.8Jaguars out of pockets. Kittens learn to read.
58.463.8Last date for return to Terra.
58.664.0Charlie appointed chairman of the FRB, confirmed.
58.764.1Dome 6 done. Food crisis, begin building greenhouse domes.
59.765.1Four ag domes built. Iris asks for reading lesson. End ch. 12.
62.067.4First comet impact.
66.071.4Big comet impact. Chapter 13, 14, 15. (Lions 11.3 eyr)
66.772.1Night suggests putting snowballs in orbit. Chapter 16.
67.372.7Chinese launch second starship.
71.276.6Iris sexually mature. Ken delayed until 71.9.
71.977.3Lions and otters get married. Chapter 17.
71.977.3Revolution in China, Maoists out of power.
72.077.4Tiger sees Terran reaction to arrival at e Eri. (Not in story.)
72.577.9Terran reaction to Heptapus debacle. Obliquely mentioned in 17.
72.778.1Chinese starship arrives. Chapter 18. Lan Ying's age 31 eyr.
72.778.1Uomi and jaguars get married. Chapter 19.
76.381.7Terran reaction to colony formation orders. (Not in story.)
77.582.9Wilma dies. Described (flashback) in chapter 21.
77.883.2Willie dies.
100.0105.4E factory capable of making A factory and smelter.
106.0111.4Full atmosphere mass of 5.14e18 Kg. Chip production restrained.
111.7117.1Chapter 21, 22, 23.
126.0131.4Full ocean mass of 1.66e21 Kg.