Sawmill formula sheet: volume, recovery, kerf, moisture, shrinkage, density and conversions

This sheet collects the formulas used in the portal's sawmill articles and starter course in one place. Each entry gives the formula, the symbols and units, a worked example with numbers, and the caveat or regional note. Every example was recomputed in Python before it was written down. Examples marked "illustrative" use invented round numbers and are not benchmarks. Where a formula is textbook mathematics and not printed in a portal article, the entry says so. Formulas are numbered F1 onwards so that you can cite them.

Log volume

F1. Huber (mid-diameter) volume

• Formula: V = (pi/4) x d^2 x L. With d in cm and L in m: V (m³) = 0.7854 x d^2 x L / 10,000.
• Symbols: V solid volume of the log (m³); d diameter at the middle of the log (cm, or m if you drop the 10,000); L length (m).
• Example: d = 30 cm, L = 5.00 m gives 0.7854 x 900 x 5 / 10,000 = 0.3534 m³, reported as 0.35 m³.
• Error sensitivity: volume goes with d squared. If the caliper reads 31 cm instead of 30 cm, the volume error is (31² - 30²) / 30² = 6.78 %.
• Caveat (Germany): the German framework agreement for round timber trade (RVR, 5th edition 2023) measures mid-diameter without bark, rounded down to whole centimetres (one reading up to 19 cm, two perpendicular readings from 20 cm), and gives volume as solid cubic metres without bark. It is a voluntary agreement that applies only if the contract says so. Austria (Ö-NORM L1021) and Sweden (Biometria) use their own rules, so name the rule and edition in the contract.

F2. Smalian volume (two end areas)

• Formula: V = (A_butt + A_top) / 2 x L, with A = (pi/4) x d^2.
• Symbols: A_butt and A_top cross-section areas at the two ends (m²); L length (m).
• Example: butt d = 34 cm, top d = 26 cm, L = 5 m. A_butt = 0.09079 m², A_top = 0.05309 m², so V = (0.09079 + 0.05309) / 2 x 5 = 0.3597 m³. Huber with the mid-diameter of 30 cm gives 0.3534 m³, so Smalian is 1.8 % higher on this straight-taper example.
• Caveat: the portal's yield article says only that Kilborn (2002, Alaska) measured log volume by Smalian's formula; the formula itself is textbook mathematics and is not printed in the articles. The two methods give different volumes for the same log, so a yield figure is only comparable if the volume formula is the same.
• Newton's formula is not given in any of the portal's articles and is deliberately left out of this sheet.

F3. Bark deduction function (Sweden)

• Formula: Y = a + b x X, where X is the top diameter over bark (mm) and Y the double bark thickness (mm). The under-bark diameter is X - Y.
• Example: the Biometria regulation (in force 1 April 2025) gives y = 2.00 + 0.0153x for thin-bark pine in Norrbotten. For X = 250 mm, Y = 2.00 + 0.0153 x 250 = 5.83 mm, so the under-bark diameter is 244.2 mm.
• Caveat: the coefficients depend on species and area (the same regulation gives y = 3.10 + 0.0496x for spruce in "Area 1"). Do not carry them to another region. The same regulation separates m3to (a cylinder with the log length and the under-bark top diameter measured 10 cm from the top) from m3sub (the form-adjusted solid volume). Volume from top diameter and conversion factors may not be used as a basis for payment in that regulation.

F4. Stack volume (Sweden) and stacked-to-solid factors (Germany)

• Formula (Sweden): stack volume (m³sub) = length x width x height x wood-volume percentage.
• Example: the regulation's own stack 6.00 x 3.01 x 1.08 m = 19.505 m³t; at 55 % wood volume this is 10.73 m³sub.
• Base percentages in the regulation are 70 % for spruce and 68 % for pine, then adjusted for bark, stacking, crookedness, snow and ice and other factors.
• Formula (Germany): Fm o.R. = Rm m.R. x factor. The RVR gives the average factors 0.70 for 1 m wood, 0.65 for 2 m and 0.60 for 3 m (Rm m.R. is stacked cubic metre with bark, Fm o.R. is solid cubic metre without bark). Example: 100 Rm of 1 m spruce is about 70 Fm o.R.
• Weight: the RVR says 1 t atro (oven-dry) with bark equals 2.2 Fm o.R. for spruce and fir, 2.1 for pine and Douglas fir, 1.8 for larch, 1.5 for beech and 1.4 for oak. Example: 10 Fm o.R. of spruce is 10 / 2.2 = 4.55 t atro.
• Caveat: the RVR states that its factors are averages for internal and statistical use and must not serve as a sales measure. The Swedish figure is built from the real stack and adjusted; the German figure is one flat average. Do not use one country's factor for the other's wood.

Lumber volume and board foot

F5. Sawn volume from dimensions (metric)

• Formula: V = t x w x L.
• Symbols: thickness t, width w, length L, all in metres, result in m³.
• Example: 50 x 150 mm x 4.8 m = 0.050 x 0.150 x 4.8 = 0.036 m³ per piece.
• Caveat: state whether the dimensions are nominal, rough green, dried or planed. The same piece has a different volume in each state.

F6. Board foot

• Definition: 1 board foot = 1 in x 12 in x 12 in = 144 in³. Using 1 in = 0.0254 m exactly, 144 x 0.0254³ = 0.002359737 m³.
• Formula: bf = t_in x w_in x L_ft / 12, where thickness and width are in inches and length in feet (nominal sizes).
• Example: a nominal 2 x 4 in x 8 ft piece is 2 x 4 x 8 / 12 = 5.33 bf (nominal), or 0.01259 m³ by the definition above. The actual piece is smaller than the nominal size.
• Caveat: the portal's economics article notes that board-foot recovery uses nominal dimensions and differs from cubic recovery (Steele 1984).

F7. Thousand board feet (MBF) to cubic metres

• By the definition above, 1 MBF = 1000 x 0.002359737 = 2.3597 m³ nominal.
• The portal's energy and carbon article reports that its sources use 1.623, 1.59 and about 2.36 m³ per MBF in different documents and that these could not be reconciled. Our own check is only that 2.36 matches the nominal definition; the reasons for the other two factors are not stated in the sources.
• Example of the effect: the West Virginia average of 100 kWh/MBF becomes 100 / 1.623 = 61.6 kWh/m³ with one factor, and the California figure 194 kWh/MBF becomes 194 / 2.36 = 82.2 kWh/m³ with another. Convert only with the factor named in the source document, and say which one you used.

Recovery and conversion factors

F8. Volume recovery (yield) and conversion factor

• Formula: R (%) = V_sawn / V_log x 100; CF = V_log / V_sawn = 100 / R.
• Symbols: V_sawn sawn product volume (m³); V_log log volume (m³); CF cubic metres of logs per cubic metre of sawn product.
• Example: 52 m³ of boards from 100 m³ of logs gives R = 52 % and CF = 1/0.52 = 1.92 m³ per m³.
• National factors (FAO, ITTO and UNECE 2020): Germany 1.59 for coniferous sawnwood gives 100/1.59 = 62.9 %; the United States 2.04 gives 100/2.04 = 49.0 %. The same publication's illustrative construction softwood example gives rough green 1.57 (63.7 %, quoted as 64), rough dry 1.63 (61.3 %, quoted as 62) and sawnwood shipped 2.08 (48.1 %, quoted as 48).
• Caveat: the bases of the national factors were not verified, so 63 versus 49 percent is not a performance ranking. The 2.08 value was read from an extract with scrambled columns. Lumber state (green, dried, planed) and log measurement (over or under bark, formula or scanner) change the figure without any change in mill performance. A published yield is not a target: the portal's yield article finds no single typical figure for spruce or pine.

F9. Taper effect on main-product yield

• Formula: the same R = V_sawn / V_log; for a fixed product the numerator stays fixed while the denominator grows with log size.
• Example (Blinzer and Treml 2022, Austria, glulam lamella 43 x 146 x 4,000 mm, three-stem pattern): the sawn volume is 0.075 m³ in each case. Taper 5 mm gives 0.075 / 0.152 = 49.3 %; 15 mm gives 0.075 / 0.181 = 41.4 %; 25 mm gives 0.075 / 0.212 = 35.4 %. The source table rounds these to 49, 41 and 35 %.
• Caveat: main product only, one pattern, one conference presentation.

F10. Chain recovery from stem to sawn timber

• Formula: R_total = R_1 x R_2 x ... (multiply the surviving shares).
• Example: Brandstetter et al. (2020, Romania, silver fir, eight stems) report 11 % lost in crosscutting, 7 % in grading and 6 % in debarking, leaving 100 - 11 - 7 - 6 = 76 % as debarked sawlogs. A 20 m³ stem gives 15.2 m³ of sawlog. If the mill then recovers 53 % (the level of the Finnish spruce simulation), sawn timber is 15.2 x 0.53 = 8.06 m³, which is 40.3 % of the stem volume.
• Caveat: small sample, no sawing in the fir study, and the Finnish 53 % is a simulation, not mill data. Agree where yield is counted (standing tree, roadside, yard gate, saw infeed) before comparing.

F11. Log cost per cubic metre of sawn output, and value yield

• Formula: log cost per m³ sawn = log price / R (R as a fraction); value yield = (lumber revenue + by-product revenue) / V_log.
• Example (illustrative): log price 100 per m³ and R = 0.50 give 200 per m³ of sawn timber. A rise of 5 in the log price gives 105 / 0.50 = 210, so the cost rises by 10, twice the log increase.
• Value yield example: revenue of 30,000 from boards and 2,000 from by-products on 100 m³ of logs is (30,000 + 2,000) / 100 = 320 per m³ of log.
• Caveat: volume yield and value yield can move in opposite directions.

F12. North American cubic measures (definitions only)

• Lumber recovery factor LRF = board feet of lumber tally / cubic feet of log volume; cubic lumber recovery CLR = green lumber volume / log volume. A CLR of 0.50 means 50 percent in green-lumber terms.
• Lumber overrun is the lumber tally minus the log-rule volume, divided by the log-rule volume. Keegan et al. (2010) call it imprecise because log rules do not reflect solid-wood volume.
• Caveat: LRF values (for example 6.72 to 8.67 for western US mills) depend on log-rule conventions and are not comparable with a European volume percentage. Board-foot log rules such as Scribner, Doyle and International disagree most on small logs; the portal's log supply article shows Doyle 20 percent below International for a 16 in. log, and the two converge at 30 in.

Kerf and cutting-width arithmetic

F13. Kerf

• Formula: kerf = plate thickness + set (per side) x 2, if the set is counted on both sides.
• Example (Wieruszewski et al. 2023, Poland, pine): frame saw 2.2 + 2 x 0.9 = 4.0 mm; band saw 1.0 + 2 x 0.75 = 2.5 mm.
• Caveat: reading "set per side" as both sides is the portal's interpretation, and the paper is internally inconsistent about the yields (72.6 and 69.7 percent in the text, other values in the abstract and tables). With six logs it cannot demonstrate a kerf effect.

F14. Kerf share per cut

• Formula: kerf share = kerf / (board thickness + kerf).
• Example 1 (1 in = 25.4 mm board, kerf values from a USDA article): circular headsaw 0.300 in = 7.6 mm gives 7.6 / 33.0 = 23.0 %; band saw 0.140 in = 3.6 mm gives 3.6 / 29.0 = 12.4 %.
• Example 2 (26.5 mm board): frame saw 4.0 / 30.5 = 13.1 %; band saw 2.5 / 29.0 = 8.6 %. Note that 26.5 + 4.0 = 30.5 and 26.5 + 2.5 = 29.0.
• Caveat: ignores planing allowance and drying shrinkage, so it shows direction and not mill results.

F15. Boards from a cant height (illustrative)

• Formula: n = floor((H + k) / (t + k)), because n boards need only n - 1 kerfs between them.
• Symbols: H available height (mm); t board thickness (mm); k kerf (mm).
• Example (own illustration): H = 195 mm, t = 25 mm. With k = 4.0 mm, n = floor(199 / 29) = 6. With k = 2.5 mm, n = floor(197.5 / 27.5) = 7. For H = 200 mm both kerfs give 7 boards.
• Caveat: this shows why Steele (1984) finds kerf savings discontinuous: they pay only when they tip the log into a better pattern. It ignores sawing variation and planing allowance.

F16. Oversize and sawing variation

• Formula: oversize (%) = (mean thickness - target thickness) / target x 100.
• Example (Kumasi, Ghana, band saw, wawa): mean 42.42 mm against a target of 41.00 mm gives 1.42 mm, or 3.46 %. (The thesis converted this to 3.46 m³ lost per 100 m³ of lumber.)
• Rule of thumb from Kilborn (2002, Alaska): roughly 2.5 % more recovery for each 0.030 in. removed from product thickness. 0.030 in x 25.4 = 0.762 mm.
• Caveat: US softwood data; the effect on a modern high-speed European line may differ.

Moisture content, density and green density

F17. Moisture content, dry basis

• Formula: MC = (m_wet - m_dry) / m_dry x 100, with m_dry the oven-dry mass.
• Example: 1.45 kg as received and 1.00 kg after oven-drying gives (1.45 - 1.00) / 1.00 x 100 = 45 %.
• Caveat: the dry basis is the standard for lumber and kilns, can exceed 100 %, and the oven-dry method (101 to 105 °C to constant mass) is the reference. Electrical meters are not reliable above about 30 %.

F18. Moisture content, wet basis, and conversion

• Formula: MC_wet = (m_wet - m_dry) / m_wet x 100; MC_wet = MC_dry / (100 + MC_dry) x 100; MC_dry = MC_wet / (100 - MC_wet) x 100.
• Example: for the sample in F17, the wet basis is 0.45 / 1.45 = 31.0 %. 25 % dry basis is 25 / 125 = 20 % wet; 100 % dry is 50 % wet; 150 % dry is 60 % wet. In reverse, 20 % wet is 25 % dry, and 50 % wet is 100 % dry.
• Caveat: wood fuels are often quoted on the wet basis. Always write "MC, dry basis" or "wet basis" in specifications, kiln logs and biomass contracts.

F19. Maximum moisture content

• Formula: MC_max = (1 / Gb - 1 / G_wall) x 100, with G_wall the specific gravity of the cell wall.
• Example: with G_wall = 1.5, Gb = 0.30 gives 266.7 % and Gb = 0.90 gives 44.4 %, matching the Wood Handbook values of about 267 and 44 %. For Gb = 0.40 the result is 183 %.
• Caveat: the Handbook states a cell-wall value of 1.54, which would give 268 and 46 %. The portal's article could not resolve the small difference, and the 183 % figure is our arithmetic, not a published value.

F20. Density at a given moisture content

• Formula: rho_12 = G12 x 1000 x 1.12 (kg/m³), so about 1120 x G12.
• Symbols: G12 specific gravity based on oven-dry mass and volume at 12 % MC.
• Example: white ash G12 = 0.605 gives 0.605 x 1000 x 1.12 = 677.6 kg/m³, the Handbook's 678.
• Definitions: basic specific gravity Gb is oven-dry mass over green volume (the steadiest reference, common in forestry); G12 is oven-dry mass over the volume at 12 %; G0 is oven-dry mass over oven-dry volume. For one wood, Gb is below G12, which is below G0.
• Caveat: a density without a moisture state is unusable. Czech measurements at 12 % give about 410 to 516 kg/m³ for Norway spruce (near pith to outer mature wood), and the Handbook gives about 10 % coefficient of variation within a species.

F21. Green density and water in green wood

• Formula: rho_green = Gb x 1000 x (1 + MC/100); water per m³ green = Gb x 1000 x MC/100 (kg per m³ of green volume).
• Example: spruce with Gb = 0.40 and sapwood at 120 % MC gives 0.40 x 1000 x 2.20 = 880 kg/m³, of which 0.40 x 1000 x 1.20 = 480 kg is water.
• Caveat: this is the number for truck loading and for the water a yard or kiln must remove. Sapwood and heartwood differ a lot (for example Sitka spruce 142 % sapwood against 41 % heartwood, North America; Swedish Scots pine sapwood 107 %, Norway spruce sapwood 95 % and heartwood 35 %).

F22. Oven-dry mass and water to remove

• Formula: m_od = rho_MC / (1 + MC/100); water removed = m_od x (MC_in - MC_out) / 100.
• Example: the Swedish Wood EPD product is 489 kg/m³ at 16 %. m_od = 489 / 1.16 = 421.6 kg. Dried from an assumed 60 % to 16 %, water removed is 421.6 x 0.44 = 185.5 kg per m³; from an assumed 100 % it is 421.6 x 0.84 = 354.1 kg.
• Second example: per 100 kg of oven-dry wood dried to 12 %, sapwood at 140 % must lose 128 kg of water, heartwood at 40 % only 28 kg.
• Caveat: the entry moistures here are assumptions for illustration, not data.

F23. Basic to 12 % specific gravity

• Formula: S_x = S_0 x (1 - x/30) for x below 30; G_x = Gb / (1 - S_x/100).
• Symbols: S_0 total green-to-oven-dry volumetric shrinkage (%); x the target MC (%); 30 is the working fibre saturation point.
• Example: white ash, Gb = 0.55, S_0 = 13.3 %. At 12 %, S_x = 13.3 x (1 - 0.4) = 7.98 %, so G12 = 0.55 / 0.9202 = 0.598, close to the Handbook's 0.60.
• Caveat: good to a few hundredths; the linear shrinkage and the 30 % fibre saturation point are conventions (see the shrinkage section).

Shrinkage and swelling estimates

F24. Linear shrinkage from the fibre saturation point

• Formula: S (%) = S_0 x (30 - MC) / 30 for MC below 30; final dimension = green dimension x (1 - S/100).
• Symbols: S_0 green-to-oven-dry shrinkage in the direction concerned (% of green dimension); 30 is the fibre saturation point (22 for the few species the Handbook flags).
• Example: bur oak, tangential S_0 = 8.8 %, green width 200 mm, dried to 8 %. S = 8.8 x 22 / 30 = 6.45 %, a loss of 12.9 mm, final width 187.1 mm. A quarter-sawn board with the radial S_0 = 4.4 % gives S = 3.23 %, a loss of 6.5 mm, final width 193.5 mm.
• Use tangential values for the width of flat-sawn boards and when the ring orientation is unknown; radial for quarter-sawn. Commercial lumber is rarely perfectly flat- or quarter-sawn, so this probably overestimates width shrinkage and underestimates thickness shrinkage.
• Caveat: Bergman says the linear equation tends to underestimate the real change by about 5 percent. Individual pieces scatter (coefficient of variation about 15 % for transverse and volumetric shrinkage). The S_0 values in the portal articles are North American species; no authoritative table for Norway spruce, Scots pine, beech or European oak was found.

F25. Dimensional change coefficient (service range)

• Formula: change = D x c x (MC_2 - MC_1).
• Symbols: D initial dimension (mm); c change coefficient per percentage point of MC; valid for the service range of about 6 to 14 % MC.
• Example (Bergman, white fir, tangential c = 0.00245): a 232 mm board going from 8 % to 11 % grows by 232 x 0.00245 x 3 = 1.71 mm. With the radial c = 0.00112 the same change is 0.78 mm.
• Caveat: North American species; this method was developed for the 6 to 14 % range and does not carry the 5 % underestimate of F24.

F26. Tangential to radial (T/R) ratio

• Formula: T/R = S_tangential / S_radial.
• Example (Wood Handbook values as quoted in the portal): bur oak 8.8 / 4.4 = 2.0; sugar maple 9.9 / 4.8 = 2.06; black walnut 7.8 / 5.5 = 1.42; loblolly pine 7.4 / 4.8 = 1.54; Sitka spruce 7.5 / 4.3 = 1.74; eastern cottonwood 9.2 / 3.9 = 2.36. The coefficients above give 0.00245 / 0.00112 = 2.19.
• Caveat: "about twice" is a rule of thumb, not a precision claim. The selection spans 1.4 to 2.4 and a cited review gives 1.5 to 2.5. Longitudinal shrinkage is only 0.1 to 0.2 % for most wood but about 2 % in reaction and juvenile wood, which causes bow, crook and twist.

EMC and drying time scaling

F27. Equilibrium moisture content (table lookup)

• Wood in air of constant relative humidity (RH) and temperature approaches an equilibrium moisture content (EMC). The Wood Handbook table for 21.1 °C (derived mainly for Sitka spruce, said to apply to any species for most practical purposes) gives:

• Example: a shop at 21 °C and 50 % RH has an EMC of 9.2 %. Stock delivered at 7.0 % will gain 2.2 points; stock delivered at 12 % will lose 2.8 points.
• Caveat: EMC is more sensitive to RH than to temperature (at 65 % RH the Handbook gives 12.1 % at 15.6 °C and 12.0 % at 21.1 °C). The table is North American; the portal did not open European tables. No interpolation formula is given in the articles, so read the nearest tabulated values for other humidities.

F28. Hysteresis ratio

• Formula: EMC_adsorption = about 0.8 x EMC_desorption.
• Example: drying wood settles at 16.0 % in air at 80 % RH; the same wood re-wetted in that air settles at about 0.8 x 16.0 = 12.8 %.
• Caveat: 0.8 is an average near room temperature; individual species and temperatures vary.

F29. Air-drying time and thickness

• Formula: t_2 / t_1 = (thickness_2 / thickness_1)^1.5.
• Source: FPL-GTR-117 states that drying time rises roughly with thickness to the 1.5 power. McMillen and Wengert (1978) say air-dried 2-inch stock takes three to four times as long as 1-inch stock.
• Example (illustrative): if 25 mm stock needs 60 days to reach 20 %, 50 mm stock needs about 60 x 2^1.5 = 60 x 2.83 = 170 days.
• Caveat: this is a rough scaling from US hardwood work. Species, site and season dominate the absolute time (US air-drying of 25 mm lumber to 20 to 25 % runs from about 15 to 30 days to 200 to 300 days).
• Not supported: the articles do not give a thickness-squared law, so none is used here. Squaring would give exactly 4 for a doubling of thickness, which sits at the top of the sources' 2.8 to 4 range but is not what they state.

F30. Kiln drying time and thickness

• Rule (USDA Dry Kiln Operator's Manual, ch. 7): doubling the thickness raises the kiln drying time by a factor of about 3 to 3.5.
• Example (illustrative): a charge that dries in 10 days at one thickness needs about 30 to 35 days at double thickness.
• Own conversion: a factor of 3 to 3.5 for a doubling corresponds to an exponent of log2(3) = 1.58 to log2(3.5) = 1.81 (own calculation). This is not stated in the source.
• Caveat: North American hardwood schedule guidance; actual cycles also depend on species, sapwood share, initial moisture and kiln type.

Economics

All numbers in this section come from the portal's economics article and course, which use invented round numbers. They are illustrative, not data from any mill.

F31. Contribution per m³ and result

• Formula: contribution = revenue per m³ - variable cost per m³; result = contribution - fixed cost per m³.
• Example (illustrative mill, 100,000 m³ of logs a year, recovery 50 %, 50,000 m³ sawn): revenue 260 (sawn) + 24 (residues) = 284; variable cost 200 (logs, 100 / 0.50) + 15 + 6 + 8 = 229; contribution 55. Fixed cost 20 + 10 + 8 + 12 + 3 = 53. Result 2 per m³, or 100,000 a year.

F32. Break-even volume and utilisation

• Formula: break-even volume = fixed cost per period / contribution per m³; break-even utilisation = break-even volume / capacity.
• Example: fixed cost 53 x 50,000 = 2,650,000; 2,650,000 / 55 = 48,182 m³ sawn. Capacity is 62,500 m³ sawn, so break-even utilisation is 77.1 %. After a 5 rise in the log price, contribution falls to 45 and break-even is 2,650,000 / 45 = 58,889 m³, or 94.2 % of capacity.
• Caveat: fixed and variable are not clean categories (labour is semi-fixed, maintenance partly variable).

F33. Value of a recovery gain

• Formula: gain = extra m³ sawn x (sawn price - extra variable cost - residue revenue displaced).
• Example: two more points of recovery on 100,000 m³ of logs give 2,000 m³ more sawn timber. Each extra m³ sells for 260, adds about 29 of energy, consumables and logistics, and displaces 24 of residue revenue: 2,000 x (260 - 29 - 24) = 414,000 a year, about four times the base profit.

F34. Operating leverage

• Formula: fixed cost per m³ = fixed cost / volume.
• Example: 53 per m³ at a given volume falls to 53 / 1.1 = 48.2 when volume rises 10 %, a fall of 4.8 or 9.1 %.

F35. Overall equipment effectiveness (OEE)

• Formula: OEE = availability x performance x quality.
• Example (illustrative): 0.90 x 0.90 x 0.95 = 0.7695, or 77 %.
• Caveat: keep capacity utilisation, throughput at the true bottleneck, and uptime separate from OEE. The portal found no verified European utilisation or uptime benchmark.

F36. Inventory carrying cost and simple payback

• Formula: carrying cost = stock value x interest rate x days / 365; simple payback = investment / annual gain.
• Example: 60 days of finished stock at an invented 250 per m³ on 50,000 m³ a year is 12,500,000 x 60 / 365 = 2,054,795 of stock; at 6 % this costs 123,288 a year, more than the whole base profit of 100,000. Each 10 days of stock is 342,466 of capital or 20,548 a year. A 1.0 million scanner upgrade adding one point of recovery earns 1,000 x 207 = 207,000 a year, a payback of 4.8 years.
• Caveat: payback is a screening tool and ignores cash flows after the payback date and the time value of money; the portal recommends net present value and internal rate of return across a full price cycle for long-lived assets.

Energy and carbon

F37. Specific energy

• Formula: specific energy = energy used (kWh) / output (m³) over the same period and boundary; kWh = MJ / 3.6; MJ = 3.6 x kWh.
• Example: Norwegian mills' specific electricity fell from 66.8 to 53.1 kWh/m³ between two data points, a reduction of (66.8 - 53.1) / 66.8 = 20.5 %. Swedish drying heat: 4.9 TWh / 16.4 million m³ = 4.9 x 10^9 / 16.4 x 10^6 = about 299 kWh/m³ (own calculation, averaging dried and undried product).
• Caveat: reference volume (logs, rough green, dried, planed) and system boundary (electricity only, with or without kiln or planer) differ between studies. The portal found no verified open European softwood study giving both electricity and heat per m³.

F38. Kiln heat from water removed

• Formula: heat (kWh/m³) = water removed (kg/m³) x specific heat demand (kWh/kg).
• Example: using the reference of 1.6 kWh per kg of water and F22, 185.5 kg x 1.6 = 296.8 kWh/m³ (1,068 MJ) from 60 % to 16 %; 354.1 kg x 1.6 = 566.6 kWh/m³ from 100 %. A batch that evaporates 7,500 kg of water with a heat meter reading of 13,500 kWh uses 1.8 kWh/kg, which is 12.5 % above 1.6.
• Reported values: for oak, Câmpean and Lăzărescu state 2.7 to 4.0 kWh per kg of water for conventional drying; 2,500 kg of water therefore needs 4,000 kWh at 1.6 and 6,750 to 10,000 kWh at 2.7 to 4.0.
• Caveat: the 1.6 kWh/kg is a theoretical benchmark from a trade article (Garrahan, FPInnovations, 2011, Canada) and is flagged unverified; it is not a target that every kiln can reach. Species, thickness and boundaries differ between the oak figures and the softwood benchmark. Hungarian oak data (28 mm): 1,089 MJ/m³ in summer and 2,312 MJ/m³ in winter thermal energy, which is 302.5 and 642.2 kWh/m³.

F39. Yard saving per point of moisture

• Source figure: McMillen and Wengert (1978, USA) estimated that each percentage point of moisture removed outdoors saves 50 to 85 Btu per board foot of later kiln heat.
• Own conversion (1 Btu = 1055.06 J; 1 bf = 0.002359737 m³): 22.4 to 38.0 MJ/m³ per percentage point. Removing 30 points outdoors would save about 0.7 to 1.1 GJ/m³.
• Caveat: a 1978 hardwood figure; use it for order of magnitude only.

F40. Fan affinity and compressed-air rules

• Fan power follows the cube of flow: P / P_full = (Q / Q_full)^3. The Energy Trust of Oregon guide gives 51 % of full power at 80 % flow and about 13 % at 50 % flow, consistent with 0.8³ = 0.512 and 0.5³ = 0.125.
• Compressed air: each 2 psi reduction in pressure saves about 1 %. Example: a system using 500,000 kWh a year, reduced by 6 psi, saves about 6 / 2 x 1 % = 3 %, or 15,000 kWh.
• Caveat: trade guide and US case figures; the 1 % per 2 psi rule is approximate, and payback depends on your equipment.

F41. Carbon dioxide from carbon

• Formula: CO2 = C x 44/12 (about 3.667).
• Example: the Fiskarheden EPD declares 218 kg C/m³, which gives 218 x 3.667 = 799 kg CO2, matching its declared -800; the Swedish Wood EPD 211 kg C gives 774, matching -773. Wood is about 50 % carbon by oven-dry mass.
• Caveat: a large negative biogenic value does not mean the mill removed carbon from the air. It means the product contains carbon the tree absorbed. Whether that is a climate benefit depends on forest regrowth and the product's fate. Always state whether a footprint is fossil only, includes biogenic flows, or includes land-use change.

F42. Mill carbon estimate (electricity and fuel)

• Formula (the portal's own method for an internal figure): kg CO2e per m³ = (kWh x grid factor + litres of fuel x fuel factor) / m³ produced.
• Factors: take official emission factors for your grid and fuel and write them down. The articles give no generic factor. For orientation only, the SCA EPD models Swedish production-mix electricity at 35.3 g CO2e per kWh, so 60 kWh/m³ gives 60 x 0.0353 = 2.1 kg CO2e/m³ from electricity alone.
• Caveat: this is not an EPD and must not be marketed as one. A coal-heavy grid gives a very different result. EPDs report primary energy, while meters read final energy.

F43. Fuel energy per kilogram

• Source (NIBIO, Norway): lower calorific value 19.1 MJ/kg for softwood and 18.9 MJ/kg for hardwood, on dry and ash-free wood without bark.
• Conversion: 19.1 / 3.6 = 5.31 kWh/kg and 18.9 / 3.6 = 5.25 kWh/kg.
• Caveat: moisture reduces the useful heat per tonne, and energy per cubic metre of dry wood then follows density. The page gives no moisture figures.

Unit conversion table

• Example: a Wood Handbook modulus of rupture of 105 MPa (white oak, clear wood, 12 % MC) equals 105 N/mm², or 105 / 0.0980665 = 1,071 kgf/cm². In the other direction, 1,000 kgf/cm² is 98.07 MPa.
• Caveat: a "105 MPa oak" in a table is mean clear-wood strength. It does not mean structural oak beams are rated at 105 MPa; characteristic bending strengths in structural standards are lower (the portal's species course contrasts 87 MPa for small clear Swedish pine with 40.5 N/mm² for Norwegian structural planks). The metre-based and inch-based factors above are definitions. The kgf/cm² factor is not printed in the portal's articles and is stated here from the definition of the kilogram-force.

Limits of this sheet

• Not a standard. The sheet is a study aid. For billing and contracts, use the measurement rule named in the contract (for example the German RVR or the Biometria regulations) and its current edition.
• Regional. Shrinkage coefficients, EMC tables and dimensional change coefficients come from the US Wood Handbook and extension sources and describe North American species. The portal found no authoritative table for Norway spruce, Scots pine, beech or European oak, so no European coefficients are given.
• Missing on purpose. Newton's log volume formula, any thickness-squared drying law and any generic emission factor are not in the portal's articles and are not included. The Smalian formula and the kgf/cm² factor are standard definitions, not taken from a portal article.
• Not benchmarks. National conversion factors, simulation yields, the 1.6 kWh/kg kiln figure and every economics example illustrate method. Do not use them as targets, contract values or evidence of performance.
• Unresolved source conflicts. MBF to m³ factors (1.623, 1.59, about 2.36) disagree between sources; the cell-wall specific gravity is 1.54 in the Handbook against 1.5 reproduced here; the Polish pine yields differ between abstract, text and tables.
• Averages and approximations. The 30 % fibre saturation point, the linear shrinkage method, the 0.8 hysteresis ratio, the T/R ratio of about two and the stack factors are averages with scatter. The linear shrinkage method underestimates real change by about 5 percent.
• Arithmetic. Every worked example was recomputed, and rounding is shown to the precision used. Illustrative numbers are invented and say so.
• Not advice. This is general information, not an engineering design, tax, financial or carbon-accounting statement.