Soil Mechanics I: Phase Relationships, Classification, Compaction and Seepage

Section 3 of the Civil Engineering paper rests on a fact the rest of the paper does not have to face: soil is the one material a civil engineer does not get to specify, and everything in this section follows from that. It arrives as three phases at once — solid grains, water and air — in proportions that change with the weather, and almost every quantity in geotechnical engineering is a ratio between those proportions rather than a property of the mineral. So the first job is bookkeeping, and it is worth doing properly because the whole subject is built on it: void ratio e is voids over solids, porosity n is voids over total, water content w is water weight over solid weight, and degree of saturation S is water volume over void volume. One identity ties them together and does more work than any other line in the subject: e S = w G. Learn it and every unit-weight formula becomes derivable rather than memorised. Classification then sorts soils by the properties that predict behaviour — grain size for coarse soils, plasticity for fine ones — and the reason the plasticity chart works is that a fine soil’s engineering behaviour tracks how much water it can hold while staying plastic, not what mineral it is. Compaction is the one process on site that changes a soil’s properties deliberately, and its curve has a peak because water is doing two opposite things at once: lubricating grains into a denser packing, and then occupying the space the grains would have filled. Finally permeability and seepage, where a fact worth carrying is that the two extreme cases of layering — flow along the layers and flow across them — give equivalent permeabilities an order of magnitude apart, and it is the least permeable layer that governs the crossing.

The Three-Phase System, and the Identity That Does the Work

Draw the phase diagram every time: a column split into solids, water and air, with volumes on one side and weights on the other. Then the definitions are readings off that sketch. Void ratio e = V_v/V_s uses the solids as its base, which is why it can exceed 1 and why it is the quantity that appears in compressibility relations — the solids do not change volume, so a change in e is a change in the soil. Porosity n = V_v/V uses the total, so it is always below 1, and the two convert as n = e/(1+e). Water content w = W_w/W_s is a weight ratio and routinely exceeds 100 per cent in soft clays. Degree of saturation S = V_w/V_v closes the system.

🧠 e S = w G, and why it replaces a page of formulae
Write the water volume two ways. From saturation, V_w = S V_v = S e V_s. From weight, V_w = W_w/γ_w = w W_s/γ_w = w (G γ_w V_s)/γ_w = w G V_s. Equate them and V_s cancels: e S = w G. Everything else follows. A fully saturated soil has S = 1, so e = wG directly — with G = 2.65 and w = 20%, e = 0.53 and there is no formula to remember. From e the unit weights come out of the phase diagram in one step each: γ_d = Gγ_w/(1+e) = 16.99 kN/m³, γ_sat = (G+e)γ_w/(1+e) = 20.39 kN/m³, and γ_sub = γ_sat − γ_w = 10.58 kN/m³. Then check the arithmetic for free: γ_sat must also equal γ_d(1+w) = 16.99 × 1.2 = 20.39 ✓, because adding water adds weight without changing volume in a saturated soil. That cross-check catches the commonest error in the topic, which is confusing dry and bulk unit weight — and note that γ_d is not a lower bound on anything physical, it is the weight of the same skeleton with the water taken out, not a different soil. Two habits are worth forming. Always ask what base a ratio is measured against, because e and n differ only in that and are constantly interchanged in error. And treat γ_sub as the fundamental one for anything below the water table, since it is what drives effective stress in the next chapter.
One soil, four unit weights, from G = 2.65 and w = 20% at full saturation
QuantityFromValue (kN/m³)
Void ratio ewG/S0.530 (dimensionless)
Porosity ne/(1+e)0.346 (34.6%)
Dry unit weightGγ_w/(1+e)16.99
Saturated unit weight(G+e)γ_w/(1+e)20.39
Submerged unit weightγ_sat − γ_w10.58
Cross-checkγ_d(1+w)20.39 ✓

Index Properties, Classification, and the Peak of the Compaction Curve

Classification exists to predict behaviour from cheap tests, and it splits at a grain size because the physics splits there. For coarse-grained soils — sands and gravels — behaviour is governed by the packing of the grains, so the useful descriptors are grading (C_u, C_c) and how densely the deposit is packed. That density is reported as relative density D_r = (e_max − e)/(e_max − e_min), which places the field void ratio between the loosest and densest states the same soil can reach: with e_max = 0.85, e_min = 0.35 and a field e = 0.55, D_r = 60%, a medium-dense sand. Note what D_r is not — it is not e and not a unit weight; it is a position between two extremes of the same soil, which is why two sands at the same void ratio can differ completely in relative density. For fine-grained soils, packing matters far less than water, so classification runs on the plasticity indices instead: the liquid and plastic limits and the range between them, I_p = w_L − w_P. The plasticity chart’s A-line separates clays from silts, and it works because a fine soil’s strength, compressibility and swelling all track how much water it can hold while still behaving plastically.

🎯 Why the compaction curve has a peak, and what the peak is not
Compact the same soil at increasing water contents with the same energy and the dry density rises, peaks, and falls. Both halves have the same cause read in opposite directions. Dry of optimum, adding water lubricates the grains, letting them slide past one another into a denser packing, so γ_d rises. Wet of optimum, the voids are nearly full, water cannot be squeezed out quickly enough during the blow, and it now occupies volume the grains would have filled — so γ_d falls. The peak is the optimum moisture content, and it is a property of the soil and the compactive effort together, not of the soil alone: raise the energy and the curve shifts up and to the left, giving a higher maximum at a lower OMC. That fact is worth holding, because it means an OMC quoted without its test is incomplete. Now the useful check. The zero-air-voids line, γ_d = Gγ_w/(1 + wG), is the density the soil would reach at that water content with no air left at all — a theoretical ceiling that compaction can approach and never touch. At w = 15% and G = 2.65 it is 18.60 kN/m³. So a measured peak of 17.5 kN/m³ at an OMC of 15% sits at 94.1 per cent of the ceiling, and back-solving the phase relations says exactly what is left over: e = Gγ_w/γ_d − 1 = 0.4855, hence S = wG/e = 81.9%, and the air still in the soil is n(1 − S) = 5.9% by volume. Compaction never expels the last of the air, and the 5 to 8 per cent typically remaining at the peak is why a field density of 95 per cent of the laboratory maximum is a normal specification rather than a lax one.

Permeability, and What Layering Does to It

Darcy’s law, v = k i, says the discharge velocity is proportional to the hydraulic gradient, with k carrying the units of velocity. Two cautions attach to it. The velocity it gives is a discharge velocity over the whole cross-section, not the speed of a water particle — that is the seepage velocity v_s = v/n, larger by a factor of three or so. And the law is linear only while the flow is laminar, which it is in sands and silts and is not in a coarse gravel or a fissured rock. k is measured by a constant-head test for permeable soils, where the discharge is large enough to time, and a falling-head test for fine ones, where it is not. What no laboratory test escapes is that k for a natural deposit spans ten orders of magnitude across soil types and varies by a factor of ten within one site — which is why the layering calculation below matters more than the precision of any single measurement.

ℹ️ Three layers, two answers, and a factor of fourteen
Take three layers, each 1 m thick, with k = 10⁻³, 10⁻⁴ and 10⁻⁵ cm/s. Flow parallel to the layers is like resistors in parallel: each layer carries what it can under the same gradient, and the discharges add, so k_h = Σ(k_i h_i)/Σh_i = (10⁻³ + 10⁻⁴ + 10⁻⁵)/3 = 3.70 × 10⁻⁴ cm/s. Flow across the layers is like resistors in series: the same discharge must pass through every layer, so the head losses add and k_v = Σh_i/Σ(h_i/k_i) = 3/(1000 + 10 000 + 100 000) = 2.70 × 10⁻⁵ cm/s. The two differ by a factor of 13.69 in a deposit whose layers are the same thickness and whose individual permeabilities span only two orders of magnitude. Look at which layer dominates each answer. k_h sits at 0.37 of the most permeable layer — the good layer carries almost everything, and the poor layers barely matter. k_v sits within a factor of 2.7 of the least permeable layer — the poor layer is a bottleneck the good ones cannot bypass. That asymmetry is the whole practical content: an aquifer is fed along its layers by whichever seam is best, while a clay blanket only has to be continuous to stop vertical flow almost completely. It also explains why natural deposits, which are almost always layered by the way they were laid down, are markedly anisotropic with k_h > k_v, and why a permeability measured on a small undisturbed sample can be an order of magnitude away from the value that governs flow at the scale of a site.

Seepage, Uplift, and the Gradient at Which Soil Stops Being Soil

Two-dimensional seepage is solved by a flow net: two families of curves, flow lines and equipotentials, crossing at right angles and dividing the region into approximately square figures. Once drawn, the discharge is q = k H (N_f/N_d) per unit length, where N_f is the number of flow channels and N_d the number of potential drops. For k = 5 × 10⁻⁵ m/s under a head of 4 m with N_f = 4 and N_d = 12, q = 6.67 × 10⁻⁵ m³/s per metre run — about 5760 litres per day for every metre of the structure. The flow net gives more than a discharge, though. The head remaining at any point tells you the pore pressure there, and integrating it under a dam gives the uplift that must be counted in the stability check as a force trying to float the structure.

ℹ️ Capillarity: water held above the water table, and the finer the soil the higher
Not all the water in the ground is below the water table. Surface tension draws water up through the pores against gravity, and equating the surface-tension force round a pore to the weight of the column it supports gives h_c = 4T cos θ/(γ_w d), with T ≈ 0.0728 N/m for water and d the effective pore diameter. The inverse dependence on d is the whole of it, and the numbers are startling: a pore of 1 mm lifts water 0.030 m, 0.1 mm lifts it 0.297 m, 0.01 mm lifts it 2.97 m, and 0.001 mm — a fine silt — lifts it 29.7 m. So a coarse sand is dry a few centimetres above the water table while a silt can be saturated tens of metres above it, which is why a capillary fringe is a real design consideration in fine soils and an irrelevance in gravel. Two consequences follow. The water in that fringe is under negative pore pressure, so by σ′ = σ − u the effective stress there is higher than the total — capillary tension is genuinely holding the grains together, and it is the entire explanation of why damp sand can be moulded into a castle and dry sand cannot. And it is a temporary strength: let the soil dry out or flood it, and the apparent cohesion vanishes with the menisci, which is why a trench that stood open in damp silt collapses after rain.
⚠️ The critical gradient, and why it is a property of the soil and not the dam
Upward seepage drags the grains with it. The drag per unit volume is the seepage force i γ_w — at a gradient of 0.4 that is 3.92 kN/m³, and it acts upward, directly against the submerged weight of the soil, γ_sub. When the two are equal the grains are carrying no contact stress at all: the effective stress is zero, the soil has no shear strength, and it boils. Setting i γ_w = γ_sub gives the critical hydraulic gradient, i_c = γ_sub/γ_w = (G − 1)/(1 + e). For G = 2.65 and e = 0.53 that is 1.65/1.53 = 1.078. Three things follow that are each worth a mark. First, i_c depends only on the soil — its specific gravity and how loosely it is packed — and not on the structure, the head or the geometry; a loose soil at e = 0.9 has i_c = 0.87 and boils under a gentler gradient than a dense one. Second, the factor of safety against piping is i_c/i_exit: at an exit gradient of 0.4 that is 2.70, at 0.8 it is 1.35, and at 1.078 it is exactly 1.00. Third, and this is where candidates go wrong, i_c is close to 1 for every ordinary soil, so the memorised statement "quicksand occurs at unit gradient" is a good approximation and a bad formula — it happens to work because G − 1 ≈ 1 + e for common values, and it fails for a loose or a heavy-mineral soil. The engineering response is not to raise i_c, which cannot be done, but to lower the exit gradient: lengthen the seepage path with a cut-off wall or a sheet pile, or load the exit area with a graded filter heavy enough to supply the missing effective stress while still letting the water out.

Key takeaways

  • e S = w G replaces a page of unit-weight formulae. Saturated soil at G = 2.65, w = 20% has e = 0.53, γ_d = 16.99, γ_sat = 20.39 and γ_sub = 10.58 kN/m³ — and γ_d(1+w) must return γ_sat, which is a free check.
  • Coarse soils are classified by grading and density, fine soils by plasticity. Relative density is a position between the loosest and densest states of the same soil, not a void ratio and not a unit weight.
  • The compaction curve peaks because water first lubricates and then occupies volume. OMC belongs to the soil and the effort together. The zero-air-voids line is a ceiling: a 17.5 kN/m³ peak at 15% sits at 94% of it, with 5.9% air still trapped.
  • Layering makes k anisotropic: parallel flow gives Σk_ih_i/Σh_i and is dominated by the best layer; crossing flow gives Σh_i/Σ(h_i/k_i) and is dominated by the worst. Three layers spanning two orders of magnitude differ by a factor of 13.7.
  • i_c = (G−1)/(1+e) = γ_sub/γ_w is a property of the soil, not the structure — 1.078 here. It is near 1 for ordinary soils, which is why "boiling at unit gradient" is a decent approximation and a poor formula.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A fully saturated soil has a water content of 20% and a specific gravity of solids of 2.65. Its void ratio is ______

    Numerical answer — type the value.

    Show answer

    Answer: 0.53

    0.53. Straight from the identity e S = w G. Full saturation means S = 1, so e = wG = 0.20 × 2.65 = 0.53. Deriving the identity once is worth more than remembering it: the water volume is S e V_s from the definition of saturation and w G V_s from the weights, and equating those two expressions cancels V_s. Everything else about this soil now follows in one line each. n = e/(1+e) = 0.346; γ_d = Gγ_w/(1+e) = 2.65(9.81)/1.53 = 16.99 kN/m³; γ_sat = (G+e)γ_w/(1+e) = 3.18(9.81)/1.53 = 20.39 kN/m³; γ_sub = 20.39 − 9.81 = 10.58 kN/m³. Then spend a second on the free cross-check: in a saturated soil, adding the water adds weight without changing volume, so γ_sat must equal γ_d(1+w) = 16.99 × 1.20 = 20.39 ✓. That check catches the single commonest error in this topic, which is using the bulk unit weight where the dry one belongs or the reverse. Two traps in the wording. If the soil were not fully saturated the answer would be larger — e = wG/S, so S = 0.8 would give e = 0.6625 — which is why the phrase "fully saturated" is doing real work and not decoration. And w = 20% must go into the formula as 0.20; leaving it as 20 gives 53, which is not a void ratio any soil has.

  2. For the same soil compacted with a greater compactive effort, the compaction curve shifts such that

    1. the maximum dry density increases and the optimum moisture content decreases
    2. both the maximum dry density and the optimum moisture content increase
    3. the maximum dry density is unchanged, since it is a property of the soil
    4. the curve crosses the zero-air-voids line at the peak
    Show answer

    Answer: A — the maximum dry density increases and the optimum moisture content decreases

    Maximum dry density up, optimum moisture content down — the curve shifts up and to the left. More energy packs the grains closer, so a higher density is reachable; and because the extra energy does more of the work that water was doing by lubrication, less water is needed to reach the peak. Option C is the trap and it is worth being precise about why it fails: γ_d,max and OMC are not properties of the soil alone but of the soil and the compactive effort together, which is exactly why a laboratory result must be quoted with its test — light or heavy compaction, IS 2720 Part 7 or Part 8 — and why a field specification refers to a percentage of a named laboratory maximum rather than to an absolute density. Option B gets the density right and the moisture wrong, and is the answer that comes from thinking of water as simply helpful. Option D is impossible, and understanding why is the useful part: the zero-air-voids line is the density at which the soil contains no air at all, a theoretical ceiling that compaction can approach and never reach or cross, because no amount of rolling expels the last of the trapped air. At w = 15% and G = 2.65 that ceiling is 18.60 kN/m³, and a real peak of 17.5 sits at 94 per cent of it with 5.9% air still in the voids. Compaction curves for different efforts are therefore a family of curves all lying below one line, converging towards it on the wet side where the air has least room to hide.

  3. A deposit has three horizontal layers, each 1 m thick, with coefficients of permeability 10⁻³, 10⁻⁴ and 10⁻⁵ cm/s. The equivalent permeability for flow perpendicular to the layers, in cm/s, is ______ (give the answer as a multiple of 10⁻⁵, to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 2.70

    2.70 × 10⁻⁵ cm/s. Flow across the layers is a series problem: the same discharge must pass through every layer, so the head losses add and the equivalent permeability is the harmonic-type average k_v = Σh_i/Σ(h_i/k_i) = 3/(1/10⁻³ + 1/10⁻⁴ + 1/10⁻⁵) = 3/(1000 + 10 000 + 100 000) = 3/111 000 = 2.70 × 10⁻⁵ cm/s. Look at where that answer sits: within a factor of 2.7 of the least permeable layer, and nowhere near the arithmetic mean. The least permeable layer is a bottleneck the good layers cannot bypass, because the water has no choice but to go through it. Compare the other direction. Flow parallel to the layers is a parallel problem — each layer carries what it can under the same gradient and the discharges add — giving k_h = Σ(k_ih_i)/Σh_i = (10⁻³ + 10⁻⁴ + 10⁻⁵)/3 = 3.70 × 10⁻⁴ cm/s, which sits at 0.37 of the most permeable layer. The two differ by a factor of 13.69 in a deposit whose layers are equally thick and whose permeabilities span only two orders of magnitude. The wrong answer to expect here is the arithmetic mean, 3.70 × 10⁻⁴, which is the parallel result applied to the perpendicular case, and it overstates the vertical permeability by more than an order of magnitude — a mistake that would turn an adequate clay liner into a leaking one on paper. Two practical readings: natural deposits are laid down in layers and are therefore markedly anisotropic with k_h > k_v, and a clay blanket only needs to be continuous to be effective, because continuity is what makes it the bottleneck.

  4. A soil has G = 2.65 and void ratio 0.53. The critical hydraulic gradient at which quick condition occurs is ______ (round to three decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 1.078

    1.078. At the quick condition the upward seepage force per unit volume, i γ_w, exactly balances the submerged weight of the soil, γ_sub, so the effective stress falls to zero and the soil loses all shear strength. Setting i γ_w = γ_sub gives i_c = γ_sub/γ_w = (G − 1)/(1 + e) = 1.65/1.53 = 1.0784. Both forms are the same expression and it is worth seeing that they agree numerically: γ_sub = (G+e)γ_w/(1+e) − γ_w = 10.579 kN/m³, and 10.579/9.81 = 1.0784 ✓. Three consequences. First, i_c is a property of the soil alone — its specific gravity and how loosely it is packed — and has nothing to do with the head, the structure or the geometry driving the seepage; a loose soil at e = 0.9 boils at i_c = 0.868, a gentler gradient. Second, the factor of safety against piping is i_c/i_exit, so an exit gradient of 0.4 gives 2.70, one of 0.8 gives 1.35, and 1.078 gives exactly 1.00. Third, and this is the point most worth taking away, i_c comes out near 1 for almost every ordinary soil because G − 1 ≈ 1 + e for common values — so "quicksand at unit gradient" is a serviceable rule of thumb and a bad formula, and it will mislead on a loose deposit or a heavy-mineral sand. Since i_c cannot be raised, the engineering response is to lower i_exit: lengthen the seepage path with a cut-off or sheet pile, or place a graded filter over the exit area, heavy enough to restore the missing effective stress while still letting water escape.

  5. The seepage velocity in a soil is related to the discharge velocity given by Darcy’s law as

    1. v_s = v n, so it is smaller
    2. v_s = v/n, so it is larger
    3. they are equal, since the same water passes through
    4. v_s = v/e, using the void ratio
    Show answer

    Answer: B — `v_s = v/n`, so it is larger

    v_s = v/n, so the seepage velocity is larger. Darcy’s v = ki is a discharge velocity: the total flow divided by the whole cross-sectional area, solids included. But water can only travel through the voids, which occupy a fraction n of that area, so to carry the same discharge through a smaller area it must move faster. Continuity says v A = v_s (n A), hence v_s = v/n. For a typical porosity of about 0.35 the seepage velocity is nearly three times the discharge velocity — which matters whenever the question is about how long something takes rather than how much flows: the travel time of a contaminant, the arrival of a tracer, or the time for a pore pressure front to move. Option A inverts it and is the answer that comes from treating n as a reduction factor without asking what it reduces. Option C is the intuition to dislodge; the same water does pass through, which is exactly why it has to speed up when the channel narrows. Option D uses the wrong ratio: e is voids over solids and can exceed 1, so v/e would give a smaller velocity for a loose soil, which is backwards. The distinction is worth keeping straight because both quantities are legitimate and answer different questions — use v for discharge and quantity of flow, and v_s for anything to do with time or transport. Neither, incidentally, is the actual speed of a water molecule, which is higher still because the real path through the pores is tortuous rather than straight.

  6. A flow net beneath a sheet pile has 4 flow channels and 12 potential drops under a head difference of 4 m. If k = 5 × 10⁻⁵ m/s, the seepage discharge per metre run, in units of 10⁻⁵ m³/s, is ______ (round to two decimals)

    Numerical answer — type the value.

    Show answer

    Answer: 6.67

    6.67 × 10⁻⁵ m³/s per metre run. From the flow net, q = k H (N_f/N_d) = (5 × 10⁻⁵)(4)(4/12) = (5 × 10⁻⁵)(4)(0.3333) = 6.667 × 10⁻⁵ m³/s per metre. Put it in units an engineer can act on: 6.667 × 10⁻⁵ × 1000 × 86 400 = 5760 litres per day for every metre of the sheet pile, so a 50 m long cofferdam takes about 288 cubic metres a day and the pump has to be sized for it. The structure of the formula is what to hold on to. N_f and N_d are counted off the sketch — flow channels between adjacent flow lines, and potential drops between adjacent equipotentials — and the ratio N_f/N_d, called the shape factor, is a property of the geometry alone, so it does not change if k or H change. That means the same flow net answers the same problem for any permeability and any head, and it is why flow nets are drawn once for a given cross-section. The commonest error is inverting the ratio to N_d/N_f, which here would give 1.5 × 10⁻⁴, more than double the truth — the sanity check is that N_d is nearly always the larger count, since a well-drawn net has many more drops than channels, so the shape factor should come out less than 1. Two further readings the same net supplies for free: the head remaining at any point gives the pore pressure there, and hence the uplift force under a dam; and the steepest gradient at the downstream exit, Δh/Δl in the last square, is the i_exit to compare against i_c in the piping check.

  7. Two sands have the same void ratio in the field. It follows that they have

    1. the same relative density
    2. the same dry unit weight, whatever their specific gravities
    3. neither the same relative density nor necessarily the same dry unit weight
    4. the same permeability, since permeability depends on void ratio
    Show answer

    Answer: C — neither the same relative density nor necessarily the same dry unit weight

    Neither. Relative density is D_r = (e_max − e)/(e_max − e_min), so it measures where the field void ratio sits between the loosest and densest states that particular soil can reach — and e_max and e_min are properties of the grading and grain shape, so they differ between sands. Two sands both at e = 0.55 can have quite different D_r: one with e_max = 0.85 and e_min = 0.35 gives 60 per cent, while another with e_max = 0.65 and e_min = 0.45 gives 50 per cent and is looser relative to its own range. That is why D_r, not e, is the meaningful measure of how densely a sand is packed, and why a specification calls for a relative density rather than a void ratio. Option B fails for a separate reason: γ_d = Gγ_w/(1+e), so equal e gives equal γ_d only if G is also equal, and a sand with a heavy-mineral fraction has a higher G and therefore a higher dry unit weight at the same void ratio. Option D is the most tempting of the three, because permeability does correlate with void ratio — but it correlates far more strongly with the size of the pore channels, which is governed by the fine end of the grading. A uniform coarse sand and a well-graded sand at the same void ratio can differ in k by orders of magnitude, because the fines in the graded soil sit in the throats between the large grains and choke exactly the passages that carry the flow. The general caution the question is teaching: a single index is not a soil, and two soils agreeing on one number need not agree on any other.

  8. Which of the following statements about void ratio and porosity are correct? (One or more)

    1. Void ratio can exceed 1 but porosity cannot
    2. They are related by n = e/(1+e)
    3. Void ratio is preferred in compressibility relations because the solid volume is constant
    4. For a fully saturated soil, porosity equals the water content
    Show answer

    Answer: A — Void ratio can exceed 1 but porosity cannot; B — They are related by `n = e/(1+e)`; C — Void ratio is preferred in compressibility relations because the solid volume is constant

    A, B and C. All three come from the one difference between the definitions: e = V_v/V_s is measured against the solids, while n = V_v/V is measured against the total. A follows at once — the voids can exceed the solid volume, so e > 1 is common in soft clays and peats, but the voids can never exceed the whole, so n < 1 always. B is the conversion: substituting V = V_s + V_v gives n = e/(1+e), and inverting, e = n/(1−n). C is the reason e is the working variable in geotechnical engineering rather than n: under load, soil compresses by losing voids while the mineral grains keep their volume, so V_s is a constant base and a change in e is a clean measure of what happened to the soil. Measure the same compression as a change in n and the denominator moves too, which is why e–log p plots are linear over a useful range and n–log p plots are not. D is false, and it is a plausible confusion because both are around 0.3 to 0.4 for many soils — but they are not the same kind of quantity. Porosity is a volume ratio; water content is a weight ratio, W_w/W_s. For the saturated soil worked through this chapter, n = 0.346 while w = 0.20, and the correct relation between them for S = 1 is n = wG/(1 + wG), which is just e/(1+e) with e = wG substituted. The habit worth forming is to ask of every ratio in this subject what its base is, because almost every error in phase relationships is a base substituted for another.