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

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.

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.

	Orbital	Length of	Illumin-	Mass	Surface	Name
	Radius	Year	ation		Radius
	(meters)	(sec)	(x Terra)	(Kg)	(meters)
I	6.02e10	1.06e7	2.04	5.38e24	6.64e6	Njord
II	8.60e10	1.810e7	1.0	2.99e27	1.55e8	Wotan
IIa	4.60e8	1.38634e5	1.0	8.395e24	7.373e6	Thor
III	1.38e11	3,66e7	0.39	4.28e26	1.05e8	Freyja
IV	2.41e11	8.48e7	0.13	5.62e23	3.22e6	Loki
(	1.50e11	3.156e7	1.0	5.983e24	6.37e6	Terra)

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

Star Travel

See below under ``CQMT Pusher Design'' for the equations from which much of this material was obtained.

Distance	10.8 ly = 1.022e17 m
Average speed desired	0.666 c
Maximum velocity	0.85 c
Average time dilation	0.713
Travel time	16.2 eyr = 5.11e8 sec
Onboard duration	11.6 eyr = 3.66e8 sec
Spacecraft mass (total)	6680 Kg
Peak power	2.12e12 W
Proton energy at this speed	842 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

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.

Name	Code	Species	Sex	Mate	Job	Name means
Tiger	L6-3512	Lion	F	Simba	Mission Commander
Simba	L7-1340	Lion	M	Tiger	Biotech, Medic
Willie	--	Human	M	Wilma	Mechanical Engr
Wilma	--	Human	F	Willie	Geology, Astronomy

Kittens (born in chapter 11 and 12):

Iris

L1001-1212

Lion

M

Lucent

Student Leader

Rainbow (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.

Jacinth

L1002-3027

Lion

F

Ken

Gem zircon

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

Ken

L1003-0522

Lion

M

Jacinth

Economics

Japanese 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

L1004-3565

Lion

F

Iris

Shining

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.)

Mica

O1005-0275

Otter

M

Night

Papermaker

Sparkly rock

Tiger and Simba remember the sparkly black mica crystals in a rock they sat on, on their wedding day.

Night

O1006-3431

Otter

F

Mica

Ball bearings

Black 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.

Oso

U1009-1530

'Uomi

M

Petra

Agriculture

Bear (Spanish)

Named in honor of El Oso, padre of Mariposa and Coyote, who was killed in Simba Leones. Stonecutter and farmer.

Petra

U1010-3037

'Uomi

F

Oso

Stone (Latin)

Several incidents of brat behavior and overreaction.

Quin

U1011-1624

'Uomi

M

Valeria

Biotech

Quinone

Interested in genetic engineering. Plastic producer for the colony. His name refers to the quinones in the Chang bushes.

Rose

J1013-3400

Jaguar

F

Orion

Medic

Has thorns

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

Selen

O1007-0160

Otter

M

Titania

Psychologist

Moon

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.)

Titania

O1008-3026

Otter

F

Selen

Artist, Medic

The metal

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

Orion

J1014-1210

Jaguar

M

Rose

Mighty 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.

Valeria

U1012-2423

'Uomi

F

Quin

Metalworker

Roman 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.

Wolf

J1015-1725

Jaguar

M

Xena

Willie's son

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

Xena

J1016-3062

Jaguar

F

Wolf

Warrior

Stranger

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:

• Wolf: marries Wendy. (Ch. 5). (Willie and Wilma's son.)
• Wooly: Dating Wallace, who's Jamaican (Ch. 5) (Willie and Wilma's daughter.)
• Attila: Marries Alex (Charlie's #1); grandkitten Anansi. Yes, Attila is female; see Simba Leones, in which she earns her name several times over.
• Bear
• Claude: Marries Demeter; ends up as financial manager for Leones Family Trust.
• Diamond
• Emerald: Competitive with his supervisor, Tiger
• Fox
• G: The ``G'' kitten has never been named in either story.
• Helios: Uses nickname of Surya. Marries Holly (Adam and Elsa's #8).
• Becket: Charlie and Diana's #2 (Simba is the gene father), marries Aurora, Leo and Alice's #1. Very good on writing lessons; eventually replaces Simba as Lion Disc editor.

Lion Button Panels

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.

m	Spacecraft rest mass
V	Velocity of spacecraft (relative to Terra)
v	abs(V)
a	Acceleration of spacecraft
F	Force exerted by pusher
p	Power supplied to spacecraft (excluding overhead and losses).
c	Speed 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 dv	If transverse
F dt = m/(1-(v/c)2)1.5 dv	If 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/b	The same, dimensionless
Let T = mc2/p
dt = T b/(1-b2)1.5 db	Separate variables
t = T / (1-b2)0.5	Integrate both sides
(T/t)2 = 1-b2	Solve for b
b = (1 - (T/t)2)0.5	Finish solving for b
x = c intg (b dt)	Integrate to get position
w = T/t, dw = -T dt/t2, dt = -T dw/w2	Change of variables
x = -cT intg (1-w2)0.5/w2 dw	Finish 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:

• T: Time (rectangular coordinate), for convenience starting at 1.
• s: Arc length, the elapsed time experienced by the ship.
• X: Distance travelled (rectangular coordinate).
• v: Speed, instantaneous.
• a: Acceleration, derivative of b by T.
• v(avg): Average speed, distance over time.

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 intercepted	8.02e-5 Kg
Energy at 0.85c	7.22e12 J
Radiation dose (1.5 m e-folding)	4.81e12 gray
Assumed metallicity	2%
Dust intercepted	1.61e-6 Kg
Dust grain mass	3.16e-16 Kg
Kinetic energy at 0.85c	25.5 J (mc2(1/(1-(v/c)2)0.5-1))
Density of grains (approx)	4.8e-8 #/m3
Quantity of grains hit	6.12 #/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

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)

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 satellite	2.00e4 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 velocity	7850 m/s
Drag force (average)	2.5 Nt
Drag force (peak)	10.2 Nt
Air density	1.65e-8 Kg/m3
Perigee altitude	125 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 power	3e3 W
Heat sink temperature	70 C = 343 K
Radiance at that temp.	785 W/m2
Radiator area	3.8 m2
Square flat sheet, 2 sides	1.38 m edge length
Power per chip	30 W
Size of case	2 cm approximately square
Number of chips	100 (plus 100 partners)
Radiator area per chip	.038 m2
Radius of circular patch	11 cm
Pitch of square grid	19 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 drop	15 K
Conductivity of copper	394 W/mK
Thickness (for 30 W)	1.9 mm
Mass of copper	57 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)

* = 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.

CH4	0.25
H2O	0.25
CO2	0.10
CO	0.10
NH3	0.20
O (Total)	0.35 (Sum of O content of above items)
Al	0.061
Fe	0.038
Na	0.021
K	0.020
Ti	0.0033
Cu	5.3e-5

More authoritative cosmic abundances (by count of atoms):

H	1 - Y - Z (0.95?)
He	5e-2
O	6.7e-4
C	3.5e-4
N	1.1e-4
Si	3.5e-5
Mg	3.4e-5
Ne	2.8e-5
S	6.0e-6
Ca	2.0e-6
Fe	?

Chang Bush Growth (Chapter 4)

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

	Days	Kg per	Total	Seeds
	+arriv	plant	Kg	Kg/day
Most of planting done	30	0.01	500	0 (planted 50 Kg)
Start thinning weak plants	90	0.04	2000	0 (immature seeds start)
First harvest	120	0.08	2000	2.0
Mature plants	330	10.0	2000	4.0

Actual Schedule

	Days	Kg per	Total	Seeds
	+arriv	plant	Kg	Kg/day
Initial test sprouts	5	0.01	0.08	0
Most of planting done	30	0.01	40	0 (planted 4 Kg)
First harvest	120	0.08	320	0.32
Full production	200	0.5	2000	2.0
Mature plants	330	10.0	2000	4.0

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

Infants	Protein	2g/Kg.day
Adult male	Protein	56 g/day = 0.8 g/Kg.day (70 Kg male)
Adult female	Protein	46 g/day
Lumberjack	Energy	3700 kcal/day = 1.55e7 J/day
Household Worker	Energy	2100 kcal/day = 8.78e6 J/day
Geneva Convention	Energy	1500 kcal/day = 6.27e6 J/day
Florence on Diet	Energy	1000 kcal/day = 4.18e6 J/day
Killer Whale	Energy	55 kcal/Kg.day = 2.3e5 J/Kg.day

Composition of Foods

	Protein	Fat	CH2O	Calories	Joules
	Kg/Kg	Kg/Kg	Kg/Kg	Mcal/Kg	J/Kg x1e6
Millet (dry)	.103	.029	.655	3.28	13.7
Rice (dry)	.075	.020	.780	3.60	15.0
Wheat (seeds, dry)	.137	.028	.629	3.30	13.8
Soybean flour, no fat	.470	.010	.032	3.26	13.6
Soybean flour, full fat	.371	.200	.229	4.66	19.5
Chicken, meat only	.233	--	--	1.35	5.7
Sea otter, homogenized	--	--	--	1.81	7.6
Pork lard	--	1.0	--	8.85	37.0
Fat (on the hoof)	--	.925	--	8.19	34.0

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/m².day or 2.5e4 J/m².day. 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:

Joules/day	7e6	5e6
Percent full power	100%	71%
Kcal/day	1700	1200 (rounded)
Kg/day Chang seeds	0.507 (500g)	0.362 (360g)
Nbr of mature Chang plants	25	18
Floor area (mature plants)	279 m2	199 m2
Floor area (immature plants)	558 m2	398 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)

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

From	To	Distance (m)	Time (eday)
Kuiper Cloud	Njord	4.5e12	28.3
Njord	Wotan	8.60e10	3.9	Quadrature
Wotan	Thor	4.60e8	0.29
Thor	Freyja	2.24e11	6.3	Opposition
Freyja	Loki	1.03e11	4.3	Conjunction
Loki	Njord	1.38e11	5.0	Quadrature
Kuiper Cloud	Wotan	4.5e12	15.5

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.

Event	Day	B	N	Correlations
Arrival	0	50.4	50.5	Start chapter 4
Get oriented, arrive comet	10	50.4	50.5
Finish comet activities	30	50.5	50.8
Arrive at Njord	60	50.6	51.2	Start chapter 5
Finish Njord	90	50.6	51.6
Arrive at Wotan	94	50.7	51.7
Finish Wotan and Thor	125	50.7	52.1	harvest starts,160 g/day
Arrive at Freyja	130	50.8	52.2
Finish Freyja	140	50.8	52.3
Arrive at Loki	145	50.8	52.4	harvest 320 g/day at 150 days
Finish Loki	165	50.9	52.7
Arrive at Njord again	170	50.9	52.7	end of chapter 5
	180			(harvest 640 g/day)
	200			(1000 g/day, full production)

Ground Imaging Telescope (Chapter 5)

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 Eri	1.09e26 W
Power density at Kuiper cloud	0.708 W/m2
Stefan-Boltzmann const.	5.67e-8 W/m2K4
Blackbody temperature	42 K (P/4 = s T4)
Peak of normalized Planck distr.	hf/kT = 2.821
Wavelength of maximum power	1.21e-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)

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 senator@feinstein.senate.gov 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

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)

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:

• Enter decimal number.
• inv e^x (take the log).
• 7 inv e^x (get ln(7)).
• Divide. Integer part is the H exponent. Subtract the exponent. (Remember to convert exponent to heptal if it's 7 or over.)
• 7 x:y y^x (7 to power of fractional part).
• Loop: Integer part is a digit of the mantissa. Subtract it. (The point goes after the first digit obtained here.)
• Multiply by 7. Repeat Loop.

Nuclear Decay (Chapter 8 and 18)

(All energies in KeV)

Cobalt-60
	Halflife	5.26 eyr
	Decay energy	2819
	Beta rays	315 (99.87%), a few at 663 and 1488
	Gamma rays	1173 (99.88%)
		1332 (100%)
		2158 (0.001%)
Strontium-90
	Halflife	28.1 eyr
	Decay energy	546
	Beta rays	546 (100%)
	Gamma rays	None
Cesium-137
	Halflife	30.23 eyr
	Decay energy	1176
	Beta rays	511 (94%)
		1176 (6%)
	Gamma rays	662 (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 ocean	3.61e14 m2 (71% of surface) (For Terra)
Mean depth of ocean	3794 m
Maximum depth	10430 m
Volume of ocean	1.37e18 m3 (area * mean depth)
	1.66e18 m3 (mass of hydrosphere)
Water buried in lithosphere	2e18 ton approx.
Water in mantle as OH	2e20 ton approx.
Mass of atmosphere	5.14e18 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.

From/To	Infinity	Njord	Wotan	Thor	Freyja
Loki	4.296e+08	-1.306e+09	-2.028e+09	-1.251e+09	-5.722e+08
Freyja	1.002e+09	-7.339e+08	-1.456e+09	-6.789e+08
Thor	1.681e+09	-5.503e+07	-7.772e+08
Wotan	2.458e+09	7.221e+08
Njord	1.736e+09

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 R^3/2 = 6.261e-10 R^3/2(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)

2R	Average rad. of Kuiper cloud	3.5e12 m (must divide by 2)
	1/2 Period (infall time)	7.25e8 sec = 22.96 eyr
	Velocity at Wotan's orbit	4.84e4 m/s (from potential energy)
T	Push time demanded by plot	3.15e8 sec = 10 eyr
m	Required comet mass	9.25e17 Kg (neglecting composition)
K	Mass of Kuiper cloud	5.983e24 Kg (1 x Terra)
F	Force required	1.40e10 Nt
	Number of large chips (3000 Nt)	4.66e6 chips
	Radius of equivalent sphere	9.03e4 m (density 0.3 g/cc)
	Number of comets, 1 Km radius	3.08e6 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 snowball.pl, 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):

Element	Rel. to H	Relative
O	6.7e-4	12
C	3.5e-4	6
N	1.1e-4	2
H		30 (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.

Compound	Rel.	By Mass	Kcal/Mole	J/Mole
NH3	2	11%	-3.9	-1.63e4
CH4	3	15%	-12.1	-5.06e4	(half the carbon)
CO2	3	41%	-94.3	-3.95e5
H2O	6	34%	-56.7	-2.37e5	(liquid)
CO	(small)		-32.8	-1.37e5
CH2O	--		-38.6	-1.62e5	(biomass, eth glycol)
C2H6			-20.2	-8.38e4
C3H8			-24.8	-1.03e5
CH2			-4.93	-2.04e4	(per additional unit)
C6H6			+19.8	+8.29e4

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.

Compound	Qty Kg	Qty Moles
H2O	1.66e24	9.22e25
O2	1.08e18	3.37e19
N2	4.06e18	1.45e19
CO2	A 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:

• CO₂: 4.61e26 Kg, 1.844e27 moles
• H₂O: -9.22e26Kg; not enough H₂O for reaction

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

• 1.22e5 J/M endothermic (per 2C) to biomass;
• 7.50e4 J/M exothermic (per 1C) to coal and water;
• 2.8e4 J/M exothermic (per 2C) overall.

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

• 4n - 4m = 2(n+m+1) for a saturated hydrocarbon, h = 2(c+1)
• n = 3m + 1
• delta H = 1.13e4 + 0.88e4 m J/mole endothermic.

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

• 2.76e4 J/M endothermic per formula unit

Reaction: C + 2 H₂O -> CO₂ + 2 H₂

• 7.99e4 J/M endothermic.

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.

Compound	Kg	Moles
NH3	2.61e23	1.54e25
CH4	3.69e23	2.30e25
CO2	1.02e24	2.30e25
H2O	8.30e23	4.61e25

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.

Compound	Kg	Moles
N2 + O2	5.14e18	1.78e20	Total atmosphere when finished
N2	3.94e18	1.41e20
O2	1.20e18	3.74e19
NH3	4.79e18	2.82e20	Composition of comet material
CH4	6.77e18	4.23e20
CO2	1.86e19	4.23e20
H2O	1.52e19	8.46e20

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 oxygen	1.76e25 J
Solar power	6.37e18 W (incl 50% atmos loss only)
Time to produce oxygen (50% eff)	2.76e6 sec = 31.9 days :-)
Mass of oxygen per area	2.36e3 Kg/m2
Biomass per area	2.21e3 Kg/m2
Tree production rate	10 ton/hectare-eyr = 1 Kg/m2.eyr
Time to produce oxygen (biomass)	2.21e3 eyr
Disposing of reduced species:
CH4 + H2O -> CH2O + 2 H2	30.2 Kcal/M, 1.26e5 J/M endothermic
Total energy to dispose of CH4	5.33e25 J
Mass of CH4 per area	1.33e4 Kg/m2
Biomass per area	2.50e4 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?

• P = pressure;
• R = gas constant, 8.31e0 J/mole-degK;
• T = temperature;
• v = specific volume (volume per mole);
• r = density (Kg/M³);
• m[j] = molecular weight of species j (Kg/mole);
• n[j] = atoms per molecule of j;
• f[j] = fraction of j;
• g = acceleration of gravity;
• z = altitude above (or below) reference level z0

The dependence of C_v and C_p 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 = K	Constant 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)z	Valid 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.

Cv	gamma	Mol wt	Scale	Press	Press	Temp	Composition
J/MK		Kg/M	ht, M	ratio	Nt/m2	K	percent Earth normal
16.62	1.50	34.00	-1.99e+4	3.38	1.03e+4	420	Ar 1.5 N2 1.5
15.25	1.55	36.00	-1.78e+4	3.54	1.61e+4	437	Ar 3.0 N2 1.5
16.62	1.50	34.00	-1.99e+4	3.38	2.06e+4	420	Ar 3.0 N2 3.0
18.00	1.46	32.00	-2.24e+4	3.22	2.94e+4	405	Ar 3.0 N2 6.0
25.35	1.33	28.99	-3.16e+4	3.05	5.53e+4	369	Ar 3.0 NH3 3.7 CH4,CO2 5.6
20.74	1.40	28.95	-2.73e+4	2.98	3.01e+5	383	O2 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

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.

Proto	Bfact	Description
2	.5	Small pusher chip, 100 Nt (includes partner)
10	2.5	Large pusher chip, 3000 Nt (includes partner)
2	.5	Video camera
2	.5	W32 CPU chip
2	.5	Memory 64 Mb total
15		APX detector
15		X-ray detector
15		Interplanetary bee prototype (4 sections) (not including partner)
15	0.45	Production bee (chiller, or pusher 10000 Nt)
15	0.45	Power nexus for set of four full-size bees

• General purpose lander (the first one they make, in chapter 5). Rock pockets hold 10 Kg. Let's say 10 Kg structure. Pushers: minimum 200 Nt. Needs 3 small pusher chips. Video camera. CPU and memory. Total chips: 12 days. On all these spacecraft the chassis can be assembled in parallel with chip production.
• Mapping orbiter. Similar. (Chapter 5)
• Water miner (chapter 9): The original miner can handle 100 kilos per day, getting 2 kilos of dust from it. The water miner is similar, yielding 33 kilos per day of water. It can bring in 200 kilos per load (6 days on station). Three small pusher chips would accelerate it at 0.3 G; three large pushers would move it at 9 G's (impractical). We'll use three segments of the bee prototype pusher (in standard encapsulation), needing 30 days for the pushers and their power nexus (partner chip). (18 small pushers would also do, but take 36 days.)
• Personnel lander. Needs 900 or 1000 Nt with nine chips: small pushers can be pushed to do this. 18 days including partners.

Concrete (Chapter 10)

From Encyclopedia Brittanica, ``Concrete'', the proportions to make it are:

Aggregate (rocks and sand)	9 units (by mass)
Cement	2 units
Water	1 unit
Total	12 units
Plus carbon dioxide	2.5 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):

• CaO + H₂O -> Ca(OH)₂
• Ca(OH)₂ + CO₂ -> CaCO₃ + H₂O

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.

Day	Wake	Events	Sleep	Events
1	Sunrise	Eclipse: wake+8	Sunrise+16	Sunset: sleep+2
2	Sunset+6	Sunrise: wake+12	Sunrise+4	Eclipse: sleep+4
3	Sunrise+12	Sunset: wake+6	Sunset+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.

	Mithrim	Gondolin
Longitude	0 deg	+50 deg (East)
Sunrise (type 1)	0 hrs	19 hrs (= -5 hrs)
Eclipse (center)	9 hrs	9 hrs
(Range)	7 - 11 hrs	7 - 11 hrs
Eclipse (after sunrise)	9 hrs	14 hrs
(Range)	7 - 11 hrs	12 - 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 day	92423 approx.
Hours (Thor) per day	24
Hours (Earth) per day	25.673
Thor rotations per day	2/3
Seconds per (Wotan) year	1.810e7
Days per year	195.83
Days per month	28 (1.016 x average Earth month)
Months per year	7 (minus 1 day on unleap years).
Seconds per Earth year	3.156e7
Earth years per Wotan year	0.574 = 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

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)

Timeline in Chapter 11

• 1 eyr for Willie to design the implanter machine.
• 2.5 eyr of production with 50% of Willie's time devoted to making implanters at 5e5 sec each (2000 productive hours per eyr, no charge for ribbon smelters).
• 3.5 eyr total between the end of chapter 10 and the beginning of chapter 11.
• 1.49e8 chips required (actually twice this number) in 3.08e8 secs

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.

Nova.	H.sap	Event
0	-9	Implantation
7	0	Birth
9	3	Ready to implant next set of kittens
16	--	First handsigns
19	15-20	Toilet training (goes as needed, and presses own buttons)
20	13	First words (voice)
30	24	Useful conversation
40	?	Can pay attention to stories
50	42	Reading on their own

Age of				Event
Lions	Otters	'Uomi	Jaguars
0				Lions implanted (start of chapter 11)
7				Lions born (end of ch 11)
8				Finish dome 2
9	0			Otters implanted
14	5			Iris finds a jacinth crystal (ch 12)
15	6			Lions ejected from pockets
16	7			Otters born
16	7			Selen plays pattycake with Lucent (ch 12)
19	10	1		Finish dome 3
25	16	7		'Uomi born
25	16	7		Swimming lessons (ch 12)
27	18	9	0	Jaguars implanted; work on other life
29	20	11	2	Finish dome 4
34	25	16	7	Jaguars born
39	30	21	13	Simba finishes air plant; finish dome 5
42	33	24	15	Jaguars ejected from pockets
43	34	25	16	Reading lesson (ch 12)
49	40	31	22	Finish 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.)

Age	Mass	Total	Food	Area	Domes	Kitten
eyr	Kg/ktn	Kg	Kg/day	m2		Multiplier
0	0	260	1.88	1036	3.5
1	6	452	3.27	1800	6	2x
5	20	900	6.52	3600	12	2x
10	35	1380	10.0	5500	18	2x
15	55	1580	11.4	6300	21	1.5x
20	70	1380	10.0	5500	18	1x

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):

1A	Chang bushes
2A	Chang bushes
3A	Vegetables
4A	Nonfoods: 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 2^M combinations of sexes. To accommodate every one requires 2^M domes. In N domes the number of M-species combinations that can't be realized is 2^M - 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.

• 3 domes, 2 species:
  • Domes: ff fm mm
  • Failed (1.00): mf
• 4 domes, 2 species:
  • Domes: ff mf fm mm
  • All sexes fit
• 4 domes, 3 species:
  • Domes: mff fmf ffm mmm
  • Failed (4.00): fmm mfm mmf fff
• 4 domes, 4 species:
  • Domes: mfff fmff ffmm mmmm
  • Failed (30): --fm --mf fmm- mfm- mmf- fff- fm-m mf-m mm-f etc.
• 5 domes, 2 species:
  • Domes: ff mf fm fm mm
  • All sexes fit
• 5 domes, 3 species:
  • Domes: mff fmf ffm fmm mmm
  • Failed (3.07): mfm mmf fff
• 5 domes, 4 species:
  • Domes: mmff ffmf fffm mfmm fmmm
  • Failed (23): mmm- fmf- mff- mm-m fm-f mf-f m-fm m-mf f-ff etc.
• 6 domes, 2 species:
  • Domes: ff ff mf fm mm mm
  • All sexes fit
• 6 domes, 3 species:
  • Domes: fff mff fmf mfm fmm mmm
  • Failed (2.00): ffm mmf
• 6 domes, 4 species:
  • Domes: ffff mmff mfmf fmfm ffmm mmmm
  • Failed (18): fmm- mff- mf-m fm-f m-fm f-mf -ffm -mmf fmmm etc.

(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.

Dome	Lions	Otters	'Uomi	Jaguars
1	(Adults)
2	Male	Male	Female	Female
3	Female	Female	Male(H)	Female(H)
4	Female	Female	Female(H)	Male(H)
5	Male(H)	Female(H)	Male	Male
6	Female(H)	Male(H)	Male	Male

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

Species		mm	mf	fm	ff
Lion	Otter	2	5	6	3,4
Lion	'Uomi	5	2	3,6	4
Lion	Jaguar	5	2	4,6	3
Otter	'Uomi	6	2	3,5	4
Otter	Jaguar	6	2	4,5	3
'Uomi	Jaguar	5,6	3	4	2

Dome assignments during emergency (chapter 14):

Dome	Lions	Otters	'Uomi	Jaguars
1	Male(H)	Male(H)	Female	Female	Adults(H)
2	Male	Male	Female(H)	Female(H)
3	Female(H)	Female(H)	Male	Male
4	Female	Female	Male(H)	Male(H)
5	Burned
6	No Cover

Dome assignments after mate selection (chapter 17 and 18): * indicates moved.

Dome	Occupants
1	Adults
2	Open space
3	Quin and Valeria*
	Wolf* and Xena
4	Oso* and Petra
	Orion and Rose*
5	Iris and Lucent*
	Selen* and Titania
6	Ken* and Jacinth
	Mica and Night*

Timeline for Chapter 13 and 14

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.

Grade	H.sapiens	N.leo	Skill
K-1	6	4-5	Reading
1	7	6	1 or 2 digit add
2	8	7	Simple multiply
3	9	6	Writing (unstructured)
5	11	7-8	Formal grammar
5	11	9	Fractions
6	12	9	Decimals
6	12	10	Long division
7	13	11	Syntactic analysis
7	13	11	Writing paragraphs
8	14	12	Writing essays
9	15	13	Algebra
10	16	14	Geometry

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:

• p₀ = p(t₀) and p₁ = p(t₁).
• p₁/p₀ = exp((t₁-t₀)/g)
• g = (t₁-t₀)/ln(p₁/p₀)
• t₁ = t₀ + g ln(p₁/p₀)

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,

• t₀ = 0 (age of parents)
• t₁ = 22 eyr (average age at birth of kittens)
• p₀ = 2
• p₁ = 8
• g = 15.87 eyr

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.

Date	Population
54.7	16 (kittens)
77	80
99	326
121	1350
157	1.3e4
194	1.3e5
230	1.3e6
266	1.3e7
303	1.3e8

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)

Ages are in eyr.

Results (program snowball.pl):

• Scenario 1: up to date 100 with 2.5 FTE (no population growth) yields 5.2e14 Kg water. After the E factories are activated it takes only 7 years to fill the oceans, not to mention the atmosphere: 3.5e20 Kg water.
• Scenario 2: No E factories, but yes population growth, 16% of the population (FTE) working on factories. At date 180 (126 eyr) they fill the atmosphere with 1.9e18 Kg of material, using 2.5e3 FTE. At date 267 (213 eyr) they fill the oceans with 5.0e20 Kg water using 6.0e5 FTE.

Chinese Vocabulary in Chapter 18

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

1	High even
2	Rising
3	Down-Up
4	Falling

• Major Tsun2 Lan2 Ying1

  Tsun2

  Family name (one of the 100 names)

  Lan2

  Orchid; strong and beautiful

  Ying1

  Hero; strong and brave

• Colonel General Liu2 Kai3

  Liu2

  Family name (one of the 100 names)

  Kai3

  (?)

• Derogatory Descriptions

  Ju1 Ba1 Jie2

  Eighth sister in a family of pigs; very ugly in a story.

  General Fam4 Pi4

  General Fart Gas.

  Sse4 Lang2

  ``Color wolf''; color is a metaphor for a female you chase; wolf is for a man avid for females (like English). In the story it's called ``colored skirt wolf''.

  Gong1 ji1

  Male chicken (rooster).

  Go3 ro4

  Dog meat. Usage in the story is non-authentic, as this is not derogatory in Chinese: it means meat of a dog to be eaten by people, not meat to be eaten by a dog.

  Wang2 ba1 dan4

  King's number 8 egg. Derogative predicate applied to the listener. Origin of metaphor unknown.

  Ta1 ma1 de*

  His mother's... Derogative predicate applied to a situation or a third party. Normally pronounced with exaggerated tones: ta4 ma (low constant tone) de (attenuated).

  Ro4 dan4

  Meat bomb. Slang for tit. Dan4 literally means egg, and is used metaphorically for bomb.

Available Power (Chapter 19)

(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:

Unit	Quantity	Fam/vil	People
Villages	50	20	6000
Homesteads	100	1	600
Capital	1	300	1800
Total			8400

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.

Unit	Quantity	Fam/vil	People
Villages	8	10	480
Homesteads	5	1	30
Capital	1	12	70
Total			580

Character Names in Chapter 21

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.

Day	Type	Sunrise	Eclipse	Sunset	Hr	Lt	Events
1	1	0	8	18	1	d	Kittens run away from home
					0?	d	Tiger starts hiking
2	2	12	20	--	6	n	Tiger meets kittens
					7	n	Rescue Cinder from hole
					8	n	Back to Mithrim
					10	n	Pick up Simba from trail
					11	n	Confront Talisman; he runs
					19	d	Talisman collects supplies
3	3	--	None	6	0	d	Talisman at Mithrim, hides
					1	d	Talisman tries to rape Tiger
					4	d	Death 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.)

Name	Age eyr	Mass Kg
Talisman	24	70
Joy	24	70
Ardent	6.7	20	(Kittens count double mass)
Blaze	5.5	18	(Deceased, not counted)
Cinder	4.3	15
Diablo	3.2	10
Ember	2.0	6
Flame	10mo	3	(Age after begetting, not birth)
Total mass		248 Kg
Food needed		1.80 Kg/day
Starvation		1.28 Kg/day
Plants		90	mature healthy plants (minimum)
Floor area		1000 m2	for 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)

N-B:

5.4 eyr (when she reads messages from her friends, they are 5.4 eyr older than she is)

R-B:

27.0 eyr (Friends reading her reports are 27 eyr older than Tiger when she sent them)

S-B:

16.2 eyr (In a simultaneous reference frame, friends differ in age by the journey time)

A-B:

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.

.0	Mar	.5	Sep
.1	Apr	.6	Oct
.2	May	.7	Nov
.3	Jun	.8	Jan
.4	Aug	.9	Feb

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.

B	N	Event
0	0	Tiger assembled (in March). (Birth 195 Earth days later.)
16.0	16.0	Tiger and Simba marry. (Simba is 1 month younger.)
17.5	17.5	Tiger starts college
21.3	21.3	Attila begotten (other kittens 1 year apart). Finish college.
28.3	28.3	Surya begotten (zero age for him).
42.3	42.3	Anansi begotten (grandkitten: Alex x Attila)
44.6	44.6	Begin ch 1: discover CQMT. Surya and Holly get into MIT.
45.0	45.0	Ch 2: A Rescue in Space. Stock market collapse.
45.5	45.5	Surya starts college (MIT) (his age 17.2 eyr)
47.8	47.8	Ch 3: UNESCO Conference, starship announced.
49.3	49.3	Ch 3: Test flight; test freezing process.
49.5	49.5	Ch 3: Wolf starts college (Wooly 1 year later).
50.4	50.4	End of Ch 3: Launch. 16.2 years pass frozen.
50.4	50.4*	Ch 4: Arrive at epsilon Eridani.
50.46	50.7*	Election results (not a presidential election) (ref. in ch 8)
50.6	51.2*	End ch 4: grow plants, find comet, mine it, go to Njord, ch 5.
50.6	51.3*	Wolf finishes college
50.7	51.4*	Chinese launch starship.
50.7	52.0*	Charlie nominated for SEC, not confirmed.
50.8	52.3*	Wooly finishes college
50.8	52.4*	Chinese lose contact with ship.
50.86	52.7*	Election results (presidential). Faraldo out, Martin in.
50.9	52.9*	Ch 6, 7, 8, 9. Heptapi discovered, destroyed.
51.0	53.3*	Surya finishes grad school (4+4 eyr), thesis published
51.0	53.4*	Charlie appointed to SEC, confirmed.
51.0	53.4*	Revolution in China, Maoists in power.
51.0	53.4*	End of ch 9.
51.2	54.4*	End of ch 10: start work on Gondolin.
54.7	60.1	Ch 11: implant lion kittens.
55.1	60.5	Anansi starts college. (Not mentioned in story.)
55.2	60.6	Lion kittens born (end of ch 11, start ch 12). Dome 2 done.
55.4	60.8	Implant otters.
56.2	61.6	Implant uomi.
56.3	61.7	Dome 3 done (takes 1.1 yr due to kitten care).
57.0	62.4	Implant jaguars.
57.1	62.5	Dome 4 done. (Not mentioned in story.)
57.6	63.0	Jaguars born.
57.9	63.3	Dome 5 done. (Not mentioned in story.)
58.4	63.8	Jaguars out of pockets. Kittens learn to read.
58.4	63.8	Last date for return to Terra.
58.6	64.0	Charlie appointed chairman of the FRB, confirmed.
58.7	64.1	Dome 6 done. Food crisis, begin building greenhouse domes.
59.7	65.1	Four ag domes built. Iris asks for reading lesson. End ch. 12.
62.0	67.4	First comet impact.
66.0	71.4	Big comet impact. Chapter 13, 14, 15. (Lions 11.3 eyr)
66.7	72.1	Night suggests putting snowballs in orbit. Chapter 16.
67.3	72.7	Chinese launch second starship.
71.2	76.6	Iris sexually mature. Ken delayed until 71.9.
71.9	77.3	Lions and otters get married. Chapter 17.
71.9	77.3	Revolution in China, Maoists out of power.
72.0	77.4	Tiger sees Terran reaction to arrival at e Eri. (Not in story.)
72.5	77.9	Terran reaction to Heptapus debacle. Obliquely mentioned in 17.
72.7	78.1	Chinese starship arrives. Chapter 18. Lan Ying's age 31 eyr.
72.7	78.1	Uomi and jaguars get married. Chapter 19.
76.3	81.7	Terran reaction to colony formation orders. (Not in story.)
77.5	82.9	Wilma dies. Described (flashback) in chapter 21.
77.8	83.2	Willie dies.
100.0	105.4	E factory capable of making A factory and smelter.
106.0	111.4	Full atmosphere mass of 5.14e18 Kg. Chip production restrained.
111.7	117.1	Chapter 21, 22, 23.
126.0	131.4	Full ocean mass of 1.66e21 Kg.