Starting from zero: numbers, units, and orders of magnitude
This module assumes mathematics may be a distant memory. On Mars, a factor-of-one-thousand error can distort inventory, filtration, or power budgets, so the course begins by reading numbers together with their units.
The method is straightforward: identify the quantity, keep the unit, convert explicitly, estimate magnitude, then calculate. It turns the calculator into a checking instrument rather than an oracle.
1. Read the quantity before calculating
A number becomes useful only when you know what it measures. Three kilograms, three metres, and three kilowatts answer different questions. Keep the unit attached to the value. Decimal punctuation and thousands separators must never hide the physical scale.
0.1 is one tenth, 0.01 one hundredth, and 0.001 one thousandth. The gap between 0.003 and 3,000 is a factor of one million. Scale matters more than the number of digits shown.
Exercise 1
Order 0.003, 0.3, 3, 30, and 3,000, then calculate the ratio between the extremes.
Solution : 0.003 < 0.3 < 3 < 30 < 3,000; 3,000 / 0.003 = 1,000,000, six orders of magnitude.
Why this matters
A numerical value becomes useful only after the physical quantity, unit, and system boundary are known. Pressure, mass, flow, concentration, and energy cannot be added because they do not have the same dimension.
NASA’s 2026 limit of 0.1 mg/m³ for Martian particles <10 µm corresponds, in a perfectly mixed 100 m³ cabin, to 10 mg of airborne dust at that average. The calculation below is a consistency test, not a complete truth.
An average concentration does not describe a local airlock peak or the dose actually inhaled by one person.
2. SI units are a grammar
The metre measures length, kilogram mass, second time, ampere current, kelvin thermodynamic temperature, mole amount of substance, and candela luminous intensity. Derived units recombine them: N for force, J for energy, W for power, and Pa for pressure.
Units take part in reasoning. Distance divided by time gives m/s; force divided by area gives N/m² = Pa. An answer with dimensions that do not match the question is wrong before any decimal places are inspected.
| Quantity | Unit | Meaning |
|---|---|---|
| mass | kg | inertia |
| force | N = kg·m/s² | accelerates mass |
| energy | J | transferable quantity |
| power | W = J/s | energy rate |
| pressure | Pa = N/m² | force per area |
Workshop — keep a unit ledger
Imagine receiving three statements: a tank holds 2.4 m³ of water, a pump transfers 18 L/min, and an emergency plan requires 1,200 kg of usable water. Comparing 2.4, 18, and 1,200 directly is meaningless. Convert them into a common physical language. Using the teaching approximation 1,000 kg/m³ for water density, 2.4 m³ is about 2,400 kg. A flow of 18 L/min is 0.018 m³/min, about 18 kg/min. The tank therefore holds roughly twice the required mass, while transferring 1,200 kg would take about 66.7 minutes at constant flow. Every conclusion can be traced to a visible unit conversion.
A unit ledger records symbol, physical quantity, input unit, working unit, source, and precision for each value. This feels slow only until a project combines kilograms of water, kilograms per day of loss, electrical kilowatts, and kilowatt-hours of storage. Numbers become comparable after their meaning and units have been made explicit, not because they happen to be printed in the same table.
Physical reading
SI provides a shared grammar: metre for length, kilogram for mass, second for time, ampere for current, kelvin for thermodynamic temperature, mole for amount of substance, and candela for luminous intensity. Derived units keep physical relationships visible.
A pressure difference of 70 kPa is 70,000 Pa, or 70,000 N/m². Across a 1.2 m² hatch, the ideal pressure force is 84 kN.
The calculation gives a resultant force, not detailed stress in hinges, seals, or reinforcements.
3. Prefixes and thousand-fold steps
milli means 10⁻³, kilo 10³, mega 10⁶, and giga 10⁹. Case matters: mW and MW differ by a factor of one billion. Convert using factors equal to one: 2.5 kg × (1,000 g / 1 kg) = 2,500 g.
The same method handles compound units and prevents memorized shortcuts from silently reversing a conversion.
Move into calculation
Prefixes prevent long strings of zeros but become dangerous when treated as typography. Milli means 10⁻³, micro 10⁻⁶, kilo 10³, and mega 10⁶. The multiplier applies to the whole unit.
0.1 mg/m³ equals 100 µg/m³. An 80 kWh battery stores 80,000 Wh; in joules, 80×3.6 MJ = 288 MJ. The teaching value is seeing units turn into a decision.
Confusing kW and kWh mixes a rate of energy transfer with an amount of energy and can mis-size a battery or power converter.
4. Powers of ten and orders of magnitude
Scientific notation writes a number as a × 10ⁿ, with |a| between 1 and 10. 45,000 = 4.5 × 10⁴ and 0.00045 = 4.5 × 10⁻⁴. Multiplication adds the exponents of powers of ten.
Before using a calculator, replace 198 × 51 with 200 × 50 ≈ 10,000. A screen result near 100 or ten million should trigger review.
Exercise 4
Write 6,500,000 and 0.000072 in scientific notation.
Solution : 6.5 × 10⁶ and 7.2 × 10⁻⁵.
Workshop — estimate the scale before the digits
Suppose a construction system must move 287,000 kg of regolith and actually processes 243 kg/h. Before exact division, round to 300,000 kg and 250 kg/h: 300,000 ÷ 250 ≈ 1,200 h. Exact division gives about 1,181 h. Both answers tell the same engineering story: the job is of order one thousand hours, not ten hours or one hundred thousand hours. An entry such as 118.1 h or 11,810 h should therefore trigger immediate review.
Order-of-magnitude thinking also decides which uncertainties deserve attention. If uncertainty in bulk density moves the mass estimate by ±20%, debating the third decimal place of throughput is wasted precision. Start with the parameters that can materially move the answer. Estimation is not permission to be vague; it is a way to spend measurement effort where it changes decisions.
Control point
Scientific notation separates the significant number from its scale. It makes 3×10² kg, 3×10⁵ kg, and 3×10⁸ kg easy to compare without losing the three orders of magnitude between each step.
Three hundred tonnes of regolith are 3×10⁵ kg. At 250 kg/h, an ideal system needs 1.2×10³ h, or about 50 continuous days. An engineer keeps numerical result, margin, and validity domain separate.
Order-of-magnitude reasoning does not replace detailed calculation; it catches impossible results before they are trusted.
5. Fractions, ratios, and percentages
A percentage is a fraction per hundred. 98% = 0.98, leaving 1 − 0.98 = 0.02 unrecovered. That small residual often drives makeup logistics.
Always state what a ratio compares. A language word-count ratio, a mass fraction, and an efficiency are not interchangeable.
Interpretation
A fraction describes part of a whole, a ratio compares quantities, and a percentage expresses a fraction per hundred. Before using a proportion, check that the quantity really scales linearly.
With 20 kg of process water per person per day, 98% overall recovery leaves 0.4 kg/day makeup per person; at 94%, the loss is 1.2 kg/day. For 100 people, the annual difference is 29.2 t in this teaching example. The formula is useful because it makes a system dependency visible.
The 20 kg/day figure is a teaching assumption, not a NASA requirement; mixing assumptions with institutional data changes the status of the result.
6. Dimensional analysis
A safe conversion is a chain of factors that cancels unwanted units. 72 km/h × 1,000 m/km × 1 h/3,600 s = 20 m/s. Each removed unit appears once above and once below the fraction line.
The same discipline converts flow into annual inventory, concentration into total mass, or power into energy.
Exercise 6
Convert 40 kg/day into tonnes over 365 days.
Solution : 40 × 365 = 14,600 kg = 14.6 t/year.
Field reasoning
Dimensional analysis tests an equation before any numbers are inserted. If the left side is energy, the right side must reduce to joules; if the unknown is mass, the final units must reduce to kilograms.
A flow of 40 kg/day for 365 days gives 14,600 kg because days cancel. Multiplying 40 kg/day by 365 kg is physically meaningless even though a calculator will display a number.
An equation can be dimensionally correct and still be physically wrong; the test eliminates some errors, not all.
7. Calculator discipline
A calculator executes syntax; it does not know whether your model is absurd. Predict sign, scale, and unit before each result. EXP or EE keys commonly enter powers of ten: 6.02 EE 23 represents 6.02 × 10²³.
Recalculate through a different route when the result matters. Convert back to the starting unit, add recovered plus lost fractions, or divide energy by time to recover the expected power.
Why this matters
Calculator discipline starts with a mental estimate. Write the units, anticipate the scale, then enter the numbers. A calculator checks arithmetic; it does not choose the model or assumptions.
Forty kg/day for a year should be near 40×400 ≈16,000 kg. The exact 14,600 kg is plausible. A display of 146 kg or 1.46 million kg would immediately reveal a factor-of-one-hundred error. The calculation below is a consistency test, not a complete truth.
Copying ten decimal places creates an illusion of precision when the inputs are known only to a few percent.
8. Uncertainty and significant digits
A measurement is not infinitely precise. NIST uncertainty guidance treats uncertainty as information about the dispersion reasonably attributable to a measured quantity. Extra decimal places do not reduce uncertain assumptions.
If a teaching scenario assumes 20 kg/person/day, reporting 14.600000 t/year creates false precision. Use 14.6 t/year and keep the assumption visible.
Workshop — separate precision, uncertainty, and margin
A measurement of 10.0 ± 0.2 kg is different from a requirement of at least 10 kg. The first describes an estimate and uncertainty; the second sets an acceptance threshold. Margin is different again: if calculated need is 10 kg and inventory is 12 kg, nominal margin is 2 kg or 20% of need. Treating uncertainty and margin as interchangeable hides risk.
Consider a 500 L tank with a gauge uncertainty of ±2% and an operational minimum of 430 L. Two percent of 500 L is 10 L. If the display reads 440 L, a simplified 430–450 L interval touches the operational minimum. The displayed ten-litre surplus is therefore not automatically comfortable. This simple example prepares the logic needed for engineering margins, alarm thresholds, redundant sensing, and acceptance testing.
Physical reading
A measurement is not just a value; it carries uncertainty from repeatability, calibration, the method, and sometimes the model that turns signal into a physical quantity. NIST distinguishes Type A and Type B evaluations among others.
If a flow is 10.0±0.2 kg/h, the simple relative uncertainty is 2%. Over 24 h the nominal flow is 240 kg, but a systematic 2% uncertainty does not disappear merely because the value is multiplied by 24.
Random scatter and common bias must be separated: averaging many readings may reduce one without correcting the other.
9. Mars laboratory: dust, water, energy
NASA published a preliminary 2026 requirement for Martian particles under 10 µm: 0.1 mg/m³ as a 24-hour time-weighted average for specified exposure scenarios up to 30 days. In 100 m³ of air that concentration corresponds to 10 mg suspended at the average.
For water, use a teaching assumption of 20 kg/person/day gross flow. At 100 people and 98% total recovery, theoretical makeup is 40 kg/day, or 14.6 t/year. The 20 kg value is not a NASA requirement.
Exercise 9
A 2 kW load operates for 18 h. How much energy is consumed?
Solution : 2 × 18 = 36 kWh. kW is power; kWh is energy.
Move into calculation
A good Mars exercise combines several quantities without mixing them. The goal is not to accumulate formulas but to build a dimensionally coherent chain from the observed phenomenon to an operational decision.
A 25 m³ airlock measured at 0.08 mg/m³ contains an idealized 2 mg of airborne dust. An airflow of 100 m³/h does not automatically mean 8 mg/h is removed: filter efficiency and actual mixing are still required. The teaching value is seeing units turn into a decision.
A material balance requires a boundary: airlock air, filter capture, surface deposits, and resuspended dust are different inventories.
10. Five checks before believing a result
Ask whether the final unit answers the question, the sign is physically possible, the magnitude matches an estimate, the result moves in the right direction when an input changes, and the dominant uncertain assumption is visible.
An engineering calculation is a short argument: data, symbols, units, operation, result, check, and physical meaning. Speed comes later; reliable reasoning is the goal.
Final workshop — audit a complete balance
A small installation draws 4.8 kW for 20 hours of a working sol and 1.2 kW for the remaining 4 hours. Daily energy is 4.8 × 20 + 1.2 × 4 = 100.8 kWh. If usable electrical storage is 320 kWh, that is about 3.17 days at the same average consumption, assuming no generation and deliberately ignoring conversion losses. Naming that simplification is part of the answer, because it separates a teaching model from real system behaviour.
Now assume an 85% round-trip storage efficiency and interpret 320 kWh as energy stored before those losses. Deliverable energy becomes 272 kWh and autonomy falls to about 2.70 days. The answer changes because one explicit assumption changed. Check the result: the final unit is days, 272/100 must be a little below three, and lower efficiency should reduce autonomy. This is the complete discipline: data → units → operation → assumptions → scale check → physical meaning.
Integrated case — the Mars shift-control sheet
You take an operations shift and receive four statements: usable water inventory is shown as 8.4 t, average net loop loss is estimated at 27 kg/day, a 3.2 kW non-critical load can be shed, and the dust sensor reads 0.07 mg/m³ over a monitoring period. The first task is not fast arithmetic but putting each datum into a form that answers a question. The water inventory is 8,400 kg. At 27 kg/day with no other input, 8,400/27 is about 311 days. That is theoretical inventory endurance against this one loss stream, not total base autonomy.
Now suppose net loss may vary by ±20%. The high case is 32.4 kg/day and endurance falls to about 259 days. The low case, 21.6 kg/day, gives about 389 days. Reporting only 311 days would hide the dominant sensitivity. An operational reserve changes the question again: if sixty days must remain untouched, usable inventory under the conservative high-loss case is 8,400 − 60×32.4 = 6,456 kg. The calculation changes because the decision criterion changed.
The 3.2 kW load demonstrates a different trap. Shedding it for ten hours saves 32 kWh, not 32 kW. If a battery has 180 kWh of genuinely usable energy, the saving is about 17.8% of that energy. It still says nothing about whether required peak power can be delivered. A system may have plenty of stored energy yet lack instantaneous current capability, so power and energy remain separate columns in an engineering balance.
Finally, 0.07 mg/m³ does not mean '70 percent safe.' The preliminary Mars dust criterion discussed earlier is a 24-hour time-weighted average for a particle class and defined exposure scenarios. You need the sensor averaging window, method, uncertainty, and knowledge of post-EVA peaks. The correct check is whether quantity, unit, integration period, and criterion describe the same physical object. A correct unit attached to the wrong definition still produces a bad decision.
A useful shift sheet therefore ends each line with four fields: measured datum or assumption, conversion, result with unit, and validity limit. This discipline becomes the bridge to every later module. Mathematics and physics will become more sophisticated, but the rule remains: never let a number travel without its meaning.
| Question | Minimum calculation | What it does not prove |
|---|---|---|
| Water | 8,400/27 ≈ 311 d | total base autonomy |
| Sensitivity | 8,400/32.4 ≈ 259 d | probability of high loss |
| Energy | 3.2×10 = 32 kWh | available peak power |
| Dust | 0.07 mg/m³ | compliance without time window |
11. End-of-module check
- I distinguish mass, force, energy, power, and pressure.
- I convert prefixes safely.
- I read scientific notation.
- I turn percentages into decimal factors.
- I check dimensions and magnitude.
Final exercise
A habitat processes 1,200 kg water per day at 98% recovery. Find daily and yearly makeup and its order of magnitude.
Solution : 1,200 × 0.02 = 24 kg/day; ×365 = 8,760 kg = 8.76 t/year, roughly 10 t/year.
Corrected drill set — independent check
1. Convert 0.35 t to kilograms and then grams.
0.35 t × 1,000 = 350 kg; 350 kg × 1,000 = 350,000 g. Moving tonne→kilogram→gram multiplies by one thousand twice, which supplies an immediate scale check.
2. A loop loses 1.5% of 800 kg each day. What makeup is required?
1.5% = 0.015; 800×0.015 = 12 kg/day. Across thirty unchanged days, makeup associated with this one loss stream is 360 kg.
3. A 750 W device runs for 16 h. Give energy in kWh.
750 W = 0.75 kW, so energy is 0.75×16 = 12 kWh. Multiplying 750×16 gives 12,000 Wh, the same energy expressed in a different unit.
4. A measurement is 4.0 ±0.3. Is reporting 4.0000 justified?
A calculator can print the digits but the measurement does not support them. Uncertainty of 0.3 overwhelms ten-thousandth precision. Report a precision consistent with uncertainty and state what that uncertainty represents.
5. Why is 5 kg + 3 kW invalid?
Kilograms and kilowatts have different physical dimensions: mass and power. Only compatible quantities can be added. Dimensional analysis rejects the operation before arithmetic begins.
Deep practice workshop
Before calculating, name the quantity you are trying to find and convert the data into a coherent set of units. In this module, a trustworthy answer depends less on calculator speed than on catching a missing prefix, an impossible power of ten, or false precision. Estimate the order of magnitude first, then compare your reasoning with the solution.
1. Cross conversion
A pump is rated at 2.5 L/min. Convert to m³/s and estimate water mass moved in 8 h with ρ=1,000 kg/m³.
Reasoned solution : 2.5 L/min = 2.5×10⁻³ m³/min = 4.167×10⁻⁵ m³/s. Over 8 h the volume is 2.5×60×8=1,200 L=1.2 m³, about 1,200 kg. A few litres per minute for several hundred minutes should indeed yield roughly a thousand litres.
The two conversion routes are the check: volumetric flow and integrated mass should tell the same story. If density differed substantially from 1,000 kg/m³, the final mass step would have to be recomputed rather than copied.
2. Prefix error
A sensor reads 250 µg/m³. Express it in mg/m³ and compare with 0.1 mg/m³.
Reasoned solution : 250 µg/m³ = 0.250 mg/m³, or 2.5 times 0.1 mg/m³. This alone does not determine acceptability because the cited 2026 NASA requirement is a 24-hour time-weighted average for particles <10 µm; duration and size distribution still matter.
Comparison with a limit is meaningful only after the units match. The factor 2.5 is arithmetic; an exposure decision still depends on duration, particle size, sensor location, and measurement uncertainty.
3. Power versus energy
A load draws 3 kW for 20 min. Find energy in kWh and MJ.
Reasoned solution : 20 min = 1/3 h. E=3×1/3=1 kWh=3.6 MJ. Power is the rate; energy is what is transferred over the stated duration.
Converting minutes to hours is essential because kilowatts multiplied directly by minutes would not give the requested energy unit. The megajoule conversion provides an independent route and catches a missing factor of 3.6.
4. Loss percentage
A loop processes 5,000 kg/day and recovers 97.5%. What ideal makeup is required?
Reasoned solution : Loss is 2.5%, so 5,000×0.025=125 kg/day, or 3,750 kg in 30 days. Percentages should be translated into absolute mass before logistics are judged.
A percentage becomes an operational mass when applied to throughput. That translation matters for storage and makeup: a small relative loss can become tonnes over a long campaign.
5. Relative uncertainty
A scale reports 42.0±0.6 kg. What is relative uncertainty?
Reasoned solution : 0.6/42.0≈0.0143, or 1.43%. Stating ±0.6 kg alone makes comparisons across very different measurement scales difficult.
Relative uncertainty lets measurements at very different scales be compared. The percentage expresses spread relative to the reading, but it does not prove absence of bias or a valid calibration.
6. Dimensional check
Why does P×t/m have units of J/kg?
Reasoned solution : P is W=J/s. Multiplying by time in seconds gives joules and dividing by kilograms gives J/kg. The check confirms specific-energy dimensions but does not by itself establish the correct physical model.
7. Order of magnitude
An architecture processes 4×10⁶ kg of local material in 200 days. What average t/day is required?
Reasoned solution : 4×10⁶ kg=4,000 t. Dividing by 200 days gives 20 t/day. A machine processing 0.25 t/h for 16 h/day supplies only 4 t/day, so throughput, operating hours, or machine count must increase.
The average throughput does not describe machine availability. Comparing the required daily tonnage with one machine’s realistic output immediately shows whether more units, more operating hours, or a longer campaign are needed.
8. Capstone
Build a check sheet for a 30 m³ airlock at 0.06 mg/m³ dust, 95% filter efficiency, 180 m³/h recirculation for 15 min. Find initial airborne mass and discuss an ideal removal estimate.
Reasoned solution : M₀=0.06×30=1.8 mg. In 15 min the filter sees a nominal 45 m³, more than the airlock volume because air is recirculated. Therefore C×45×0.95 is not a valid one-pass mass removal without a dynamic mixing model; the same air can cross the filter repeatedly. The correct lesson is to recognize when a static balance must become a decay model.
The capstone separates airlock volume from recirculated air volume: a filter can process the same air more than once. Captured mass therefore depends on mixing, real efficiency, and repeated passes, not flow multiplied by time alone.
Additional advanced problems
1. Unit chain
A 1.8 kW pump runs 45 min. Energy in kWh and MJ?
Reasoned solution : 45 min=0.75 h. E=1.8×0.75=1.35 kWh=4.86 MJ. The same result is 1,800 J/s ×2,700 s=4.86 MJ.
2. Concentration margin
A limit is 0.1 mg/m³ and measurement is 0.072±0.010 mg/m³. Is the limit exceeded under a simple worst-case reading?
Reasoned solution : Upper reading=0.082 mg/m³, still below 0.1. This does not replace duration, particle-size, or sensor-bias analysis, but immediate numerical margin is 0.018 mg/m³.
3. Thousand factor
Why does 3 MW for 2 h not equal “6 MWh of power”?
Reasoned solution : MW is power and MWh is energy. Integration over two hours gives 6 MWh. Calling that “power” mixes dimensions.
4. Mini-project
Build a unit sheet for habitat water, air, energy, and dust.
Reasoned solution : A strong sheet lists variable name, symbol, SI unit, operational unit, expected range, precision, and sampling frequency. It prevents kg/day, L/min, kW, kWh, and mg/m³ from being mixed.
Sources and references
These references anchor SI units, uncertainty, and the two Mars examples used in the exercises. Teaching flow assumptions are identified as such when they do not come directly from a source.
- NIST — Definitions of SI Base Units — SI base units and exact defining constants.
- NIST Technical Note 1297 — Measurement Uncertainty — Framework for expressing measurement uncertainty.
- NASA — ISS 98% water recovery milestone — Demonstrated total water recovery milestone and prior 93–94% level.
- NASA — Martian dust exposure limits — 2026 preliminary exposure requirement for Martian particles under 10 µm.