§1The claim, and what was read
The mechanism, as its makers describe it: wire-reinforced belts arranged as a block and tackle, rolling contact instead of sliding, driven by a high-torque electric motor under torque control. The patent release (U.S. 11,255,416, 2022-12-22) covers a “high reduction belt-driven linear actuator” and quotes belt–pulley interface pressures above 1,400 psi, which is 9.65 MPa. A trade-press piece dated 2026-02-10 by a partner company quotes more than 90% efficiency from the motor shaft and roughly 85% system efficiency, and attributes to Oak Ridge National Laboratory a 21.1% average for mobile hydraulics.
What was and was not read
The vendor’s own page could not be fetched (robots.txt refused the request; it was not worked around). The trade article and the patent release were read. The ORNL/NFPA material that could be read states an average of 22% across fluid power, with a range from under 9% to 60% depending on application, and no mobile-only figure. The 21.1% mobile figure is therefore carried as attributed and UNCONFIRMED. No independent third-party test of the device is cited in anything read. “Proven faster, stronger and more accurate than hydraulics” is the vendor’s framing here, not a measured result.
§2What the two efficiencies imply
For the same useful work $W$, the energy drawn is $E=W/\eta$. The fraction saved by replacing a hydraulic system of efficiency $\eta_h$ with a belt system of efficiency $\eta_b$ is
$$s \;=\; 1-\frac{\eta_h}{\eta_b}, \qquad \eta_h \;=\; \eta_b\,(1-s).$$
| hydraulic ηh | ηb = 0.85 | ηb = 0.90 |
|---|---|---|
| 0.090 (bottom of ORNL range) | 89.4% | 90.0% |
| 0.211 (as attributed, mobile) | 75.2% | 76.6% |
| 0.220 (ORNL/NFPA average) | 74.1% | 75.6% |
| 0.300 | 64.7% | 66.7% |
| 0.600 (top of ORNL range) | 29.4% | 33.3% |
Finding 1 · computed
The vendor and laboratory averages, taken at face value, give 75.2% less energy (against 21.1%) or 74.1% (against 22%). The patent release’s “up to 90% less” requires the hydraulic comparator to sit at or below 8.5% efficiency (belt at 85%) or 9.0% (belt at 90%): the bottom edge of the range the same laboratory reports. Against the top of that range (60%) the saving is 29–33%. So “up to 90%” is a best case, and 74–75% is what the averages support. This is a statement about the claims, not about the device.
Finding 1b · computed
A reported gain of “10×–20×” cannot be an energy-efficiency ratio against these averages: the ratio $\eta_b/\eta_h\le 1/\eta_h$ is 3.86 against 22%. Ten times needs a hydraulic comparator at or below 8.5% (belt at 85%), and twenty times needs one at or below 4.25%, below the whole reported range (under 9% to 60%). If a 10×–20× figure is published, it is about another quantity (force or power density, service life, maintenance interval, cost per unit work) and the quantity, the baseline and the test conditions must be named before it is cited. “Same basic deployment cost” is likewise a claim with no source read here.
§3Stronger and faster: the identity behind both words
Let $N$ strands share the load and $T$ be the tension in each. Then the output force is $F=NT$ and the output speed is $v=v_{\text{belt}}/N$, so $F\cdot v = T\,v_{\text{belt}}=\eta P$ for every $N$. The reduction ratio moves a design along a curve of constant power; it cannot make it both stronger and faster.
Finding 2 · computed
A belt actuator beats a hydraulic cylinder on both force and speed at the same load point only if $\eta_b P_b > \eta_h P_h$. At equal input power the ratio of output power is $\eta_b/\eta_h$: 3.86 against 22%, 1.42 against 60%. Any claim of “stronger and faster” therefore has to be read as a claim about efficiency, motor power, or duty cycle, and each must be reported separately.
§4Why 85% against 22% is not yet a comparison
An average efficiency is energy-weighted, $\bar\eta=\sum E_{\text{out}}/\sum E_{\text{in}}$, over a duty cycle. A valve-throttled fixed-pressure hydraulic supply loses the difference between supply and load pressure as heat, so $\eta\approx\eta_{\text{peak}}\,p_{\text{load}}/p_{\text{supply}}$.
MODEL · mechanism only, assumed numbers
With an assumed $\eta_{\text{peak}}=0.88$, a throttled system at a load-to-supply pressure ratio of 0.25 has $\eta=0.220$, exactly the reported average, at 0.50 it has 0.44, at 1.00 it has 0.88. The point is qualitative: a fleet average reflects part-load duty, while a torque-commanded electric drive does not throttle. An 85% figure quoted without duty cycle, and a 22% average taken over many machines, are not a matched test.
Chain accounting shows what the 85% must contain. Total efficiency is the product of stage efficiencies. If the quoted 85% system figure includes motor and inverter at an assumed 93%, the belt and pulley stage is 91.4%. The assumed 93% is not from a source; the sources do not break the 85% into stages.
§5What a matched test would hold constant
Proposal (not a result)
Same load profile and stroke, same input-energy measurement at the electrical or engine shaft, energy per unit useful work over a stated duty cycle including standby, the hydraulic comparator specified (pump type, load-sensing or not, accumulator or not), position accuracy measured as repeatability under load, and temperature at steady state. The result to report is energy per unit work with its duty cycle, not a single efficiency.
§6The author’s hypothesis (not yet shown)
HYPOTHESIS · untested · idea not yet disclosed to this page
The author intends improvements to the belt-actuator architecture and expects a gain of 10×–20×. That figure is the author’s projection, not a vendor claim and not a measurement, and the improvements are not described here, so nothing on this page tests them. To be a result it needs four things stated before any test: the quantity that gains (Finding 1b shows it cannot be energy efficiency against these averages), the baseline it gains over, the load and duty cycle, and the outcome that would count as failure. A robot has many joints, so the natural unit is the joint: per-joint torque density, per-joint efficiency over a stated duty cycle, per-joint mass and cost, with the system total being the sum over joints plus the extra structure needed to carry the mass of the distal joints. The improvement that matters is the one that survives that accounting.
§7Compounding along a load path
Suppose each actuator’s mass is a fixed fraction $\alpha$ of the load it must carry, and that load is the payload $P$ plus every actuator distal to it on the same load path. Then $L_{k-1}=(1+\alpha)L_k$, and a chain of depth $n$ has total actuator mass $M=P\left[(1+\alpha)^n-1\right]$. If a design lowers $\alpha$ by a factor $g$, the mass ratio is
$$R(n)=\frac{(1+\alpha)^n-1}{(1+\alpha/g)^n-1}\;\sim\;\exp\!\Big(n\big[\ln(1+\alpha)-\ln(1+\alpha/g)\big]\Big).$$
MODEL · assumed α = 0.05, g = 2 · computed by joint-compounding-verify.py
Depth 7 gives R = 2.16; 30 gives 3.03; 50 gives 4.30; 100 gives 12.07; 150 gives 38.05; 300 gives about 1,380. R first reaches 10 at depth 92 and 20 at depth 123 (with g = 3: depths 59 and 85). So a per-joint gain compounds exponentially in load-path depth, and a 10×–20× system gain from a 2× per-joint gain needs a depth of order 100. Link mass and lever arms are ignored here; both make the real compounding stronger.
Finding 3 · computed
Two limits on the compounding. First, the exponent is the depth of one load path, not the number of joints: a humanoid has tens of joints but a torso-to-fingertip path of order ten (approximate, unsourced), where R stays near 2; only serial structures (snake, continuum, cable-driven chains) have depth equal to joint count. Second, energy and power do not compound across joints: they add. Three hundred joints each 1.5× more efficient give a system energy ratio of 1.5, not 1.5³⁰⁰. Efficiencies multiply only where power passes in series through stages of one drive. The compounding claim is therefore true for mass (and for accumulated positioning error) along a load path, and not for energy.
MODEL · lever arms · computed by joint-compounding-verify.py
Let the torque at a joint be the sum of every distal mass times its distance, and let each actuator’s mass be $\beta$ times its torque. Successive joint masses then have a constant ratio $r=1/x$ where $x$ solves $(1-x)^2=\beta x$, so the growth is exponential, not factorial, with base close to $1+\sqrt{\beta}$ instead of $1+\beta$. At $\beta=0.05$ the ratio is exactly 1.25 per joint; at $\beta=0.02$ it is 1.1518. Halving $\beta$ from 0.05 to 0.025 lowers the base from 1.2500 to 1.1711, a factor 1.0674 per joint, about 678× over a depth of 100 (asymptotic). Lever arms make the compounding stronger than in the simple model; the growth stays exponential because the ratio does not itself grow with depth. In this model factorials do not appear.
Finding 4 · computed · the series with 1/k!
The compounding is itself a series with factorials. Expanding, $(1+\alpha)^n=\sum_k \binom{n}{k}\alpha^k$, and $\binom{n}{k}=n(n-1)\cdots(n-k+1)/k!$. Term $k$ counts the ways to choose $k$ of the $n$ joints, each contributing a factor $\alpha$. For small $\alpha$ and large $n$ the terms tend to $x^k/k!$ with $x=n\alpha$, and the sum to $e^x$. The robot enters through one number, $x=n\alpha$ (load-path depth times actuator mass fraction), and the gain from lowering $\alpha$ by a factor $g$ is, to first order,
$$R\approx\frac{e^{x}-1}{e^{x/g}-1}\;\longrightarrow\;e^{x(1-1/g)}\ \ (x\ \text{large}).$$
At $n=100$, $\alpha=0.05$ ($x=5$), $g=2$ the exact $R$ is 12.07 and this form gives 13.18; the largest series term is $k=4$ of value 24.5. The form replaces $\ln(1+\alpha)$ by $\alpha$, so its error is of order $n\alpha^2$: 3% at $n=30$, 9% at $n=100$, 31% at $n=300$. Reaching $R=10$ or $20$ needs, in the large-$x$ limit, $x\ge 4.61$ or $5.99$ for $g=2$ ($3.45$ or $4.49$ for $g=3$). Alpha and $g$ are assumed, not data.
MODEL · where to improve · computed by joint-compounding-verify.py
Improving joints is a sensitivity ranking: the mass saved by halving one joint’s coefficient, at the robot’s own numbers. In the simple model every joint is worth exactly the same (a proximal joint has the bigger mass and fewer joints upstream to benefit; a distal joint the opposite, and the product $\prod(1+\alpha_k)$ is symmetric). In the lever-arm model, with 12 equal joints at $\beta=0.05$, the saving from halving one joint runs from 0.805 at the base to 0.163 at the tip: proximal joints are worth about five times more, because their own mass is far larger. The tip joint’s saving does propagate to every joint upstream, but it does not outweigh that. The ranking depends on geometry, masses and torques and is not a general law.
Finding 5 · computed · the limit of infinitely many joints
With equal fractions the mass $P[(1+\alpha)^n-1]$ grows without bound as $n\to\infty$: that infinity is the model saying that an unbounded serial chain with no taper is not buildable. With per-joint fractions $\alpha_k$ the total is $P\big[\prod_k(1+\alpha_k)-1\big]$, and the infinite product is finite exactly when $\sum_k\alpha_k$ converges. For $\alpha_k=c/k^2$ the product has a closed form,
$$\prod_{k\ge1}\Big(1+\frac{c}{k^2}\Big)=\frac{\sinh(\pi\sqrt c)}{\pi\sqrt c}.$$
Checked numerically to $k=200{,}000$ at $c=0.05,\,0.5,\,1$ (agreement to 1e-5 or better). At $c=0.5$ an infinite chain carries 1.051 times the payload in actuator mass; dividing every fraction by $g=2$ gives 0.465, a gain $R=2.26$, and by $g=3$ gives 0.298, $R=3.53$. So once the taper of the chain is known, the gain of an infinite chain is a finite number. The real limits are physical (material strength, joint size, control bandwidth), and the equal-fraction model holds only while they do.
§8Case: a leg exoskeleton for mobility support in space (first pass)
Quoted inputs: NASA’s X1 exoskeleton has 10 degrees of freedom, four motorized (hips and knees) and six passive, and weighs “under 60 lbs” (under 27.2 kg); its stated uses include walking on extraterrestrial surfaces and keeping leg strength in microgravity. A 2020 review recommends maximum joint moments for daily life of 2.4 Nm/kg (hip), 1.5 (knee) and 1.9 (ankle). For an assumed 80 kg wearer that is 192, 120 and 152 Nm as capability ceilings, not walking averages. To first order the gravitational part of a joint torque scales with local gravity: 0.378 of Earth on Mars, 0.165 on the Moon, and none in microgravity, where the torque left is inertial or from exercise and suit loads.
Finding 6 · computed
Two results for the exoskeleton, both from the script exoskeleton-verify.py. First, per-joint compounding is negligible at this depth. The powered load path of an X1 leg has depth 2, and at $\alpha=0.05$, $g=2$ the mass gain is $R=2.025$, essentially $g$; even depth 7 gives 2.158. The “hundreds of joints” compounding belongs to deep or serial structures (hands, snakes, cable chains), not a leg. Second, a different loop does compound at shallow depth: carried mass feeds itself (a heavier system needs more battery and stronger actuators, which weigh more). If $\kappa$ is the fraction of total mass added only to carry the mass, then $M=M_0/(1-\kappa)$, and dividing $\kappa$ by $g$ gives a mass ratio $(1-\kappa/g)/(1-\kappa)$. That is 1.5 at $\kappa=0.5$, $g=2$; 5.5 at $\kappa=0.9$; and 10.5 at $\kappa=0.95$. A 10× gain from a 2× efficiency gain therefore needs a baseline within 5% of runaway, which is a nearly self-defeating design. $\kappa$ is assumed here; it must be measured for a real design.
What this case still needs
The wearer’s mass and the exoskeleton’s actuator masses and torques per joint, the mission (surface walking or resistive exercise), the energy budget, and the battery specific energy, which together give a measured $\kappa$. In microgravity the leg is not used for walking, so the use case there is resistive exercise and rehabilitation, where the load is set by the exercise prescription, not by body weight. The 2.4/1.5/1.9 Nm/kg figures are from a review that itself notes such data are rarely gathered in one place; the X1 page gives no torque values. Sources: NASA Technology Transfer, X1 Exoskeleton (MSC-TOPS-71); Frontiers in Robotics and AI, 2020. Not read: the primary gait data (Winter, 1987).
§9Case: the suit that walks its wearer back
The scenario, from the author’s reading of a spacewalk: an astronaut tires quickly because every movement works against the pressurized suit, and if the astronaut loses consciousness the suit itself should be able to walk the person back to base. NASA states a related requirement for the lunar south pole: a single crew member should transport a suited, fully incapacitated crew member of 343 kg up to 2 km and a slope of up to 20 degrees without a rover. NASA’s suit requirements also state that resources shall be provided to rescue an incapacitated suited crew member, that EMU joints are designed near neutral stability at 4.3 psid (29.6 kPa), and give metabolic rates of 490–2,607 kJ/h for lunar surface EVA and 1,675–3,167 kJ/h for an advanced walkback test. A study of EVA injuries reports hand injuries as the most common, and fatigue in the quadriceps, thighs, gluteal muscles and lower back on a 10 km lunar walkback test.
Finding 7 · computed · exoskeleton-rescue-verify.py
Take the 343 kg load on the Moon: its weight is 555.7 N, 71% of the Earth weight of an unsuited 80 kg person. Energy is not the binding constraint. Climbing the whole 2 km at 20 degrees costs 380 kJ (105.6 Wh) of potential energy. Adding a flat-ground gait cost $E=c_t\,m g d$ for three assumed values $c_t=0.1,\,0.3,\,1.0$, and an assumed drive efficiency of 0.7, the total is 195, 283 or 592 Wh, which is 0.97, 1.42 or 2.96 kg of battery at an assumed 200 Wh/kg. Torque is the harder number. Scaling the quoted daily-life ceilings by weight gives about 136 Nm at the hip, 85 Nm at the knee and 108 Nm at the ankle at lunar weight, first order only (inertia in swing and the slope add to it). Both results rest on assumptions listed on the page, not on a measured design.
MODEL · the barrier · geometry assumed
A pressurized limb of diameter 0.05 m carries an end-cap pressure force of 58 N at 29.6 kPa, and one of 0.10 m carries 233 N (24 kgf). The restraint layer must react it, and a joint away from neutral stability adds a restoring torque. Because the EMU is designed near neutral stability, the resisting torque is only the residual, and the sources give no value for it. So the “barrier” the astronaut works against cannot be quantified from what was read; it has to be measured joint by joint.
What must be settled before this is more than arithmetic
Safety is the first question: the suit must not take control of a conscious wearer, so the detection of incapacitation, its false-positive rate, the override, and joint-limit and torque-limit protection for an unresponsive body are design requirements, not details. The cause of incapacitation matters as much as the walk: if it is a life-support failure, walking back does not help unless the consumables last. The gait cost, drive efficiency, battery energy density and limb geometry above are assumptions; the residual pressure-joint torque is unmeasured; the sources are NASA web pages and a NASA-related review, not a design study. Sources: NASA, South Pole Safety: Designing the NASA Lunar Rescue System; NASA-STD-3001 Technical Brief, Spacesuits (OCHMO-TB-050); Chappell et al., EVA operations, injuries and compromised performance (NASA JSC).
§10Case: a distribution warehouse on a moonbase
The author’s point, on reflection: a suit worn by a person has limited degrees of freedom, and moving someone with a spinal injury is not safe, so a rescue or remote-controlled suit has to be restricted in how it can move a body. That case does not carry the compounding argument. The case that does is a large warehouse for a giant distribution center on a moonbase: many actuators, deep load paths, no person inside the machine. The warehouse’s size is not yet known, so what follows is per tonne and per unit, not a design.
Finding 8 · computed · moonbase-warehouse-verify.py
The launch multiplier is the compounding that matters most. Sourced: trans-lunar injection takes 3.05–3.25 km/s (Apollo), and the Apollo lunar module’s powered descent took 5925 ft/s, which is 1.806 km/s. With an assumed lunar-orbit insertion of 0.9 km/s (not read), assumed specific impulses of 420 s (TLI stage) and 320 s (landing stages), and inert-mass fractions of 0 to 10% of propellant, each kilogram on the lunar surface costs 5.09 to 6.66 kg in Earth orbit. That is a product over stages of $e^{\Delta v/v_e}$, the rocket-equation form of compounding, and it does not change the per-joint ratio $R$; it multiplies every tonne saved. At $\alpha=0.05$ and $g=2$, a load path of depth 7 gives $R=2.16$ and saves 0.54 t per tonne, which is 2.7–3.6 t in Earth orbit; at depth 30, $R=3.03$ (3.4–4.5 t); at depth 100, $R=12.07$ and 4.7–6.1 t saved in orbit per tonne of surface actuator mass. Depth is the lever the warehouse designer controls. The Moon lowers static loads more than dynamic ones. Actuator force scales as $m(g_{\text{local}}+a)$, so the Moon-to-Earth force ratio is 0.165 for a static load but 0.206, 0.242 and 0.307 at accelerations of 0.5, 1 and 2 m/s²: fast throughput erodes the gravity advantage.
MODEL · thermal environment · sources give no material ratings
NASA gives the lunar equator at more than 121 °C by day and −133 °C by night, and permanently shadowed polar craters below −246 °C. That is a 254 K swing at the equator (394 K to 140 K), and 27 K in polar shadow. A fluid-free drive has no fluid to freeze or seal, which is a plausible advantage over hydraulics there, but that claim is an inference: no fluid datasheet was read. The same swing is a concrete risk for the belt, whose polyurethane rating was not read.
What the warehouse case needs
Its size and layout, the payload per unit, the accelerations, which structures have deep load paths (tall stacker masts, telescoping booms, cable-suspended or modular lattice robots) and their depth, the number of units, and the thermal rating of the belt material. The lunar-orbit insertion delta-v, the specific impulses and the inert fractions are assumed. Sources: Trans-lunar injection (Wikipedia); MIT OCW STS.471, Lunar landing; NASA, Weather on the Moon.
§11Resources in place before arrival
The author’s requirement: the resources must be put in place before the crew arrives, not while they are there. So the warehouse is built and must survive unattended, and the question changes from “how efficient is the actuator” to “how many must be delivered so that enough still work when the crew lands.” The reliability figures below are assumed: no actuator failure data was read.
Finding 9 · computed · moonbase-preposition-verify.py
Without redundancy the warehouse does not arrive intact. With $N$ actuators of mean time between failures $\text{MTBF}$ over an unattended interval $T$, the probability of no failure is $e^{-NT/\text{MTBF}}$. For $N=1000$, $\text{MTBF}=10^5$ h and one year ($8{,}766$ h) the expected number of failures is $\mu=87.66$ and the probability of none is $8.5\times10^{-39}$. With spares, the count is a series with $1/k!$ terms. The probability that at most $s$ actuators fail is $e^{-\mu}\sum_{k\le s}\mu^k/k!$. Spares needed for that probability to reach 0.9, 0.99, 0.999 and 0.99999 are 100, 110, 118 and 130, that is $\mu$ plus a few $\sqrt{\mu}$: spares scale with the expected number of failures, not with $N$ alone. At 118 spares (11.8% of the fleet) the extra landed mass is 118 unit-masses, which is 601–786 unit-masses in Earth orbit at the launch multiplier of §10. Doubling the MTBF (if a simpler fluid-free actuator did that, which is unproven) cuts the spares to 66 and the orbit mass to 336–440 unit-masses: a mass gain from reliability, not from efficiency, and a different argument from the compounding in §7. Remote operation: the Earth–Moon light time is 1.282 s one way and 2.564 s round trip (standard constants, not read here), so anything faster than a 2.6 s loop needs on-site autonomy.
What this assumes and leaves out
Independent failures at a constant rate, which real hardware does not obey (wear-out, common-cause failures such as dust or a thermal event, infant mortality). A spare is useful only if a robot on site can swap it, and that swapping capability is itself mass that must be delivered and can fail. The order of landings (power, communications, construction robots, the warehouse, then people) and the amount of local resource production are not modelled here.
§12The hangar, printed from the site
The author’s reasoning: the hangar for the vehicles has to exist before they arrive, so a construction robot must land first and build it from lunar soil and rock. Quoted: heating 1 t of regolith to 1000 °C needs at least 233 kWh; compressive strengths are 2–3 MPa for regolith bagging, 12.6–37 for microwave sintering, 84.6–218.8 for direct sintering, and 206 for sintered basalt regolith; direct-sintered tensile or flexural strength is 7.2–8.2 MPa; the RASSOR 2.0 excavator has a dry mass of about 66 kg and carries 80 kg. The hangar size is not known, so an illustrative shell is used: a half-cylinder vault, 20 m span, 50 m long, 0.5 m thick, at an assumed 2000 kg/m³.
Finding 10 · computed · moonbase-hangar-verify.py
The shell is 1571 m² and 785 m³, about 1571 t. Launched instead of built locally it would be 7,995–10,462 t in Earth orbit; an assumed 2000 kg printer-and-power system is 10–13 t in orbit, a payback of 785 to 1 by mass. The cost moves from mass to energy and time. The 233 kWh per tonne is re-derived (1 t × 0.84 kJ/kg·K × 1000 K = 233.3 kWh) and is a ceiling for full heating; applied to the whole shell it is 367 MWh, 42 kW average over a year. Excavating 19,635 loads of 80 kg with 10 excavators, at an assumed 1 h per trip and daytime-only operation, takes 0.45 years (0.09 years with 50). Self-weight is not the constraint: the stress scale $\rho g r$ is 0.032 MPa on the Moon (0.196 on Earth), 62 to 2,600 times below the strengths above. The binding loads are the 254 K thermal swing, impacts, and tension. Density, geometry, trip time and printer mass are assumptions.
§13The minimum necessary is dominated by the night
Finding 11 · computed · g6-architecture-check-verify.py
The author observes that the minimum logistics for the mission is large. One reason is arithmetic: a lunar night at the equator lasts about 354 h (half a synodic month, a standard constant not read here). Storing energy for a continuous load at an assumed 200 Wh/kg takes 8.9 t of battery for 5 kW, 17.7 t for 10 kW and 35.4 t for 20 kW, which is 45–59, 90–118 and 180–236 t in Earth orbit at the launch multiplier of §10. The alternative is to work by day only and wait, trading time for battery mass. Neither number is a design: the load, the battery and the site are assumptions.
Finding 11b · computed · g6-architecture-check-verify.py · radiation and the day-only option
The author’s point is that working by day only means working when the radiation hits, and that there is no ozone layer. Both are right about the day: daytime adds the full solar ultraviolet spectrum (no ozone layer and no atmosphere to absorb it), the hottest temperatures (121 °C at the equator) and solar particle exposure. But waiting through the night does not remove the measured dose. Quoted from the Chang’e-4 lander (Zhang et al., Science Advances 2020, text supplied by the author): the charged-particle (galactic cosmic ray) dose equivalent is 57.1 ± 10.6 µSv/h, an average of 1,369 µSv/day, against 731 µSv/day inside the ISS of which 523 were galactic cosmic rays; 1,369/523 = 2.62 matches the quoted factor of 2.6, so the comparison is lunar cosmic-ray dose against ISS cosmic-ray dose. Neutral particles add 3.1 ± 0.5 µGy/h, 23% of the 13.2 ± 1 µGy/h total absorbed dose, and the paper gives no dose equivalent for them. The instrument measured only while the lander was awake (two lunar days in 2019); night was not measured, so the page cannot say night lowers or keeps the dose; the paper calls the cosmic-ray exposure chronic and solar particle events sporadic (none seen, Sun near minimum). An 8 h EVA at 57.1 µSv/h is about 457 µSv (cosmic rays only, ~1 g/cm² shielding). On shielding, the authors (per the Science news report) calculate that at least 50 cm of regolith suffices for a base and about 10 m of water for solar storms, with about 30 min of warning; the illustrative 0.5 m shell (100 g/cm²) matches that minimum, but the paper also says neutron flux rises with regolith depth (Apollo 17, up to ~150 g/cm²), so the shell is in the build-up regime and the neutron dose was not computed here. Ultraviolet and the missing ozone layer are not in the paper; the day adds UV and heat (121 °C at the equator) for machines and polymer parts, including belts, whose UV resistance was not read. So the trade of §13 (day only, wait) removes battery mass but not the cosmic-ray dose, and it adds UV and heat exposure for the machines and any polymer parts, including belts, whose UV resistance was not read. The regolith thickness needed to shield the hangar was not read either.
§14The dimensions of the books
The author asked to use the dimensions from the books. They are in Book 2, TOGT, chapter 5 “The G6 Crystal” and chapter 9 “Martian Colony Architecture” (PDF only; the chapter text is not in the repository HTML): a hexagonal tower with aspect ratio height/apothem $=66=33\cdot\tau$, $\tau=2$; an Earth height of about 15,080 m for an apothem of 229 m; a Mars height of about 39,808 m from $h_{\text{Mars}}=h_{\text{Earth}}\,\gamma^{-1}$; a proposed 1:1000 model test; and a lonsdaleite layer quoted at a Vickers hardness of 114 GPa. Each was checked.
Finding 12 · computed · g6-architecture-check-verify.py
What holds. $66=33\cdot2$. The rule that the maximum height scales as $1/g$ is the constant-stress limit $h_0=\sigma/(\rho g)$; on the Moon, sintered basalt regolith at 206 MPa gives 63.6 km (Earth 10.5, Mars 27.8). A tower or arm of constant stress has mass $P(e^x-1)$ with $x=\rho g h/\sigma$, checked numerically at three heights, which is the continuous limit of the joint model of §7: a tower and a robot arm obey the same law. What does not hold as written. (1) 229 × 66 is 15,114 m, not 15,080 (0.23% apart); 15,080/0.379 is 39,789 m, not 39,808. (2) The hexagon is not “the isoperimetric optimum among regular polygons”: perimeter²/area is 13.856 for the hexagon and 12.862 for the 12-gon. The true statement is the honeycomb theorem of Hales (1999), for tilings of equal-area cells. (3) A solid hexagonal prism 15,080 m tall at an assumed 2500 kg/m³ has a base stress of 370 MPa, above the 144–292 MPa basalt range, and its self-weight buckling limit (Greenhill, at an assumed E = 50 GPa) is 6,152 m, so the stated height is 2.45 times the limit; the Mars height is 2.45 times its limit too. A tapered or hollow design is not ruled out, but the chapter does not give one. (4) A 1:1000 model of the same material and shape is 15.08 m tall with a buckling limit of 61.5 m, so it is stable where the full tower is not; stress and the buckling limit do not scale with size the way the height does, and the model test cannot detect a full-scale structural failure. (5) Vickers hardness (114 GPa) is not compressive strength, so it cannot be compared with the 206 MPa regolith figure or support the “more than 10×” claim; the cited paper was not read.
Status: source patched 2026-09-30; PDFs not rebuilt
On the author’s instruction the LaTeX source
main-2-fixed-2.tex (the copy whose chapter 9 text matches the PDF; a backup .bak-2026-09-30 sits beside it) was corrected: heights 15,114 m and about 39,900 m, the polygon sentence rewritten to the honeycomb theorem, the 1:1000 test restated as a test of the coupling only, and the hardness claim reworded, with a labelled verification note where the tower structure and the lonsdaleite strength remain the author’s call. Compiled in scratch: same 249 pre-existing errors and 6 undefined citations before and after, none added. Not fixed: the published PDFs (Book 2 eBook, Principia Orthogona Complete v8 and the other editions) still carry the old text until rebuilt, and their build source is not the same file (the PDF has 122 pages, this source builds 114); the older copies main.tex, main_fixed.tex, main-2.tex, main-2-fixed.tex and book2_outline.tex still carry the old numbers or polygon sentence. The structure of the tower and the lonsdaleite strength remain open for the author. What can be used now for the hangar is the hexagonal module and the gravity-rescaling rule; the 66:1 aspect ratio describes a tower, and applied to a 20 m hangar it would give a 1,320 m height, so a hangar needs its own dimensions.§15Open
Left to the author
The factorials in the author’s argument are read here as the $1/k!$ series of Finding 4; if they enter elsewhere, that is not yet stated. The author’s own improvements and a real robot’s numbers are not yet supplied. The ORNL mobile-hydraulics figure (21.1%) is not confirmed from primary text; the full ORNL/NFPA report should be read before it is cited. The 85% and 90% figures have no stated test conditions. The unit of the “2-ton” cylinder is not stated (17.79 kN short ton, 19.61 kN metric). Whether this paper belongs in the engineering chapters, beside the HVEH hydraulics, or in a grant appendix is undecided. No claim on this page has been added to a volume index.
Sources read: Machine Design, 2026-02-10 (R. Grodzki, Turntide Technologies; a partner of the vendor) · BusinessWire, 2022-12-22 (U.S. 11,255,416) · ORNL/NFPA, Estimating the Impact (Energy, Emissions and Economics) of the US Fluid Power Industry, Dec 2012 · ORNL/NFPA 2010 study slides (L. Love), DOE Clean Cities. Not read: the vendor’s Beltdraulic page.