Terminal Ballistics, From First Principles
Newton’s laws, momentum and calculus. Nothing else.

Terminal Ballistics,
From First Principles

Why sectional density, not velocity, not caliber, governs how a hunting bullet actually kills: derived, not asserted.

1The master variable2Why cartridges converge3The optimization4Why the debates never end
0 4 8 12 16 20 24 penetration into 10% ordnance gel, inches expansion: a front-end event coasting to a stop, velocity → 0 at rest ≈ 21″ impact 2,700 ft/s intact opening fully mushroomed
Part 1 of 4

The master variable

Build the penetration law from Newton’s second law and one integral, and watch frontal area quietly take over.

The puzzle

The same cartridge gets opposite reputations.

Modern .308 Winchester ammunition drives a 150-grain bullet as fast as a .270 Winchester. People hunt moose with the .270 and call the .308 “marginal.” Both can’t be true.

The gun world explains this with a folk taxonomy: this cartridge is a “deep penetrator,” that one is “explosive,” this one has “knockdown power.”

Those aren’t properties of cartridges. They’re one physical mechanism with the labels filed off.
“knockdown power” “explosive” “deep penetrator” “flat shooting = deadly” “magnum = better” one quantity: expanded sectional density
Five reputations, one governing number
Ground rules

Only Newton. Every equation here is force equals the rate of change of momentum.

the only law usedF = dpdt = m dvdt

Every quantity in this deck is a mass, a length, a time, or something built from them by that law: momentum p = mv, force, pressure, density. Nothing else is invoked, and no quantity is ever squared for its own sake. Where a v² appears, it is a momentum flux and you will see it derived.

Units: grains (1 gr = 64.8 mg), inches, feet per second, and SI where the arithmetic is cleaner. The reference load throughout is a .308″ 180-grain bonded bullet arriving at 2,700 ft/s.

m, m₀retained mass, launch mass
v, v₀velocity along the track, impact velocity
A, A₀expanded frontal area, caliber area
σ = m/Asectional density (sporting tables use m/d², the same thing up to π/4)
ρ, Rdensity and crush strength of the medium (force per area)
κnose factor: the fraction of swept momentum a face actually delivers
f, kretained-mass fraction, expansion ratio (expanded ÷ caliber diameter)
X, upenetration depth, dimensionless speed κρv²/R
Behind the equation

A soldier who rebuilt mathematics in a prison camp.

A lieutenant of engineers in Napoleon’s 1812 march on Russia, Jean-Victor Poncelet was left for dead at Krasnoi, taken prisoner, and held two years at Saratov on the Volga. With no books, he rebuilt geometry from memory and founded projective geometry in the process.

Home in Metz he turned to mechanics and military engineering and one brutally practical question: how deep does a projectile bury itself in earth, masonry, or flesh? His answer is the resistance law F = a + b·v². Everything below derives it from Newton and then integrates it.

engineer officer prisoner, Saratov Metz: mechanics & fortification Paris: professor, then general 1788 born in Metz 1812 left for dead at Krasnoi, POW 1822 founds projective geometry 1825–40 the penetration law F = a + b·v² 1867 dies a general

Poncelet (1788–1867) is one of the 72 engineers and scientists whose names are engraved around the first stage of the Eiffel Tower.

Step 1 · the retarding force

The medium pushes back for two reasons. Derive both from momentum.

1 · The medium has strength

R · A same push, any speed

Crushing a hole takes a stress R (force per area) whatever the speed. Over the face area A that is a constant force. Creep or sprint, it is paid the same.

a = R · A
independent of speed

2 · The medium has mass

κ·ρ·A·v² double the speed → 4× the momentum flux

In time dt the face sweeps a volume Av dt, a mass ρAv dt, and throws it aside at a speed κv. Momentum handed over per second: ρAv × κv. By the third law that is the force back on the bullet.

b·v² = κ · ρ · A · v²
a momentum flux: mass per second × speed given to it
add themF = a + b·v² = RA + κρAv²
Both terms carry the frontal area A. The v² is not a new quantity: it is mass flux times velocity, and it is what stops the bullet.
Step 1 · how that force behaves

A bullet in tissue obeys the Poncelet equation.

retarding forceF(v) = a + b·v²
In plain words

Two tolls. A steady tearing toll (a) to rip tissue apart, the same whether the bullet is racing or crawling, and a plowing toll (b·v²) for shoving mass aside, which grows with the square of speed. At impact the plowing dominates by a factor of thousands; only in the last inch is the tearing toll all that is left.

a = R·A the strength floor
b = κ·ρ·A the momentum-flux factor
R ≈ 64 kPa in 10% gelatin (fitted)
ρ ≈ 1,030 kg/m³ (water-like)
κ ≈ 0.5 for a blunt mushroom
A expanded frontal area
0″ 7″ 14″ 21″ distance along the wound track, inches retarding force b·v² shoving gel aside a — tearing tissue, constant to the end impact, v high rest, v = 0 F(x) = R·A + κρA·v(x)²
Both terms scale with area. Along the track, the plowing toll collapses onto the constant floor.
Both terms scale with area. Frontal area is the master variable.
Step 2 · solve for total penetration

Along the wound track, integrate Newton to a full stop.

1Newton along the trackm dv/dt = −(a + bv²)
2chain rule, dv/dt = v·dv/dxmv dv/dx = −(a + bv²)
3separate the variablesdx = −mv dv / (a + bv²)
4substitute w = a + bv², dw = 2bv dvdx = −(m/2b) dw/w
5integrate from impact (w = a + bv₀²) to rest (w = a)X = (m/2b) ln[(a + bv₀²)/a]
penetration depthX = m2b ln(1 + bav₀²) = m2κρA ln(1 + u),  u = κρv₀²R

Note what dropped out: b/a = κρ/R. The argument of the logarithm contains neither mass nor area. Mass multiplies depth; area divides it; velocity only enters through ln(1 + u), and u ≈ 5,452 at 2,700 ft/s. The strength term is small at any hunting speed, but it is the floor of the logarithm: without it the bullet would never stop.

typical hunting impact band 1,800 2,200 2,600 3,000 3,400 10 20 30 impact velocity, ft/s penetration, inches if depth ∝ velocity X(v), the logarithm +50% velocity → +10% depth (19.5″ → 21.5″)
Computed from the line above. The dashed line is the intuition the mathematics overturns.
A magnum’s extra speed buys almost no extra depth. The logarithm is why.
Step 2 · the same integral, read the other way

Where the momentum goes: most of it in the first hand-span.

Solve step 4 for w as a function of depth instead of stopping at the end point, and the whole track falls out in closed form:

vvelocity profilev(x)² = [(a + bv₀²)·e−2bx/m − a] / b
pmomentum still in the bulletp(x) = m·v(x)
Fmomentum delivered per secondF(x) = (a + bv₀²)·e−2bx/m
ttime to slow from v₀ to vt = (m/√(ab))·[arctan(v₀√(b/a)) − arctan(v√(b/a))]

The force decays exponentially with a length scale m/2b ≈ 2.4″. Half of the bullet’s momentum has been handed to the medium by 3.5″, ninety percent by 11″; the first foot takes about 1.6 ms. The violence is front-loaded, which is exactly where gel shows the stretch cavity.

0″ 4″ 8″ 12″ 16″ 20″ 0.25 0.50 0.75 1.00 depth along the wound track, inches fraction of impact value half the momentum delivered by 3.5″ 90% delivered by 11.2″ p(x) = m·v(x): momentum still in the bullet F(x) = a + b·v(x)² momentum handed to the medium per second
Reference load. Momentum and force along the track from the closed forms at left.
Penetration is what is left after the momentum is spent. The medium, not the animal, receives almost all of it.
The headline result

Penetration is expanded sectional density, times a logarithm.

X = σexp · ln(1 + u)2κρ,  σexp = mretainedAexpanded

Write the expanded area as a disk of diameter D = k·d and the retained mass as f·m₀. Expanded sectional density factors into the three things a bullet engineer controls:

the three dialsσexp = f · σ0k²
f fraction of mass retained
σ₀ launch sectional density
k expansion ratio, D ÷ d
u depends on speed and medium only
In plain words

Surviving mass and a lean-and-heavy start add depth; fattening steals it, and because a face is a disk, fattening counts twice. Double the mushroom’s width and you quarter the depth.

1.0× 1.5× 2.0× 2.5× 20 40 60 80 expansion ratio k = expanded diameter ÷ caliber penetration, inches unexpanded slug ≈ 76″ bonded, 1.8–2.0× → ≈ 21″ X ∝ 1 / k²
Depth falls as 1/k². The squared term swamps everything else.
Differentiate the answer

What a 50% increase in each input actually buys.

Take the logarithm of the closed form and differentiate. The elasticities, the percentage change in depth per percentage change in each input, are exact:

mmass∂ ln X / ∂ ln m = +1
Afrontal area∂ ln X / ∂ ln A = −1
vvelocity∂ ln X / ∂ ln v = 2u / [(1 + u) ln(1 + u)] ≡ ε(u)

For u ≫ 1, ε ≈ 2/ln u: it is 0.25 at 2,000 ft/s, 0.23 at 2,700 and 0.21 at 4,000, and it keeps falling. A unit of recoil momentum spent on bullet mass buys about 4 times the depth that the same unit spent on velocity buys.

Mass is linear. Area is inverse. Velocity is a logarithm with an elasticity near one quarter. Those three numbers settle most arguments.
change in depth when the input rises by 50% retained mass ∂ ln X / ∂ ln m = +1 +50% expanded frontal area ∂ ln X / ∂ ln A = −1 -33% impact velocity ∂ ln X / ∂ ln v = ε(v) ≈ 0.23 +10% -50% +0% +50% reference: .308, 180 gr retained 95%, k = 1.9, 2,700 ft/s → 21.0″; velocity bar 2,000 → 3,000 ft/s
Recomputed from the closed form, not linearized: the velocity bar includes the full logarithm.
Part 2 of 4

Why cartridges converge

Once retention and expansion are pinned by design, wildly different loads stop at the same depth.

The convergence

Why wildly different cartridges stop at the same depth.

Bonded bullets pin two of the three dials by design, and makers cluster the third:

retention  f ≈ 0.93–0.97, nearly constantexpansion  k ≈ 1.8–2.0, by designhunting  σ₀ ≈ 0.24–0.30, by convention
bullet momentum m·v, N·s computed penetration X, inches (f = 0.95, k = 1.9) 6.5 Creedmoor 140 gr · σ 0.287 · 2,500 ft/s 6.91 21.8″ .270 Winchester 150 gr · σ 0.279 · 2,650 ft/s 7.85 21.5″ 7 mm Remington Magnum 160 gr · σ 0.283 · 2,800 ft/s 8.85 22.1″ .308 Winchester 180 gr · σ 0.271 · 2,500 ft/s 8.89 20.6″ .30-06 Springfield 180 gr · σ 0.271 · 2,550 ft/s 9.07 20.7″ .300 Winchester Magnum 180 gr · σ 0.271 · 2,800 ft/s 9.95 21.2″ 0 5 10 0″ 7″ 14″ 21″ 28″ top to bottom: +44% all within ±4%
Six bonded loads at 100-yard impact velocities, run through the closed form with f = 0.95 and k = 1.9. Momentum spans +44%; depth spans 20.6″ to 22.1″.
Constant inputs → constant output. Depth lands near 21″ regardless of caliber: the cartridge never enters the equation, only σ₀, f, k and ln(1 + u) do.
Where “12–18 inches” comes from

The benchmark everyone quotes was written in Miami, 1986.

On 11 April 1986, eight agents of the Federal Bureau of Investigation cornered two heavily armed bank robbers on a Miami street. Early in the fight a 9 mm round hit Michael Platt and stopped just short of his heart. A wound that should have ended the fight didn’t: Platt kept shooting, and two agents were killed and five wounded.

The post-mortem verdict wasn’t “wrong caliber.” It was too little penetration; the bullet never reached the vitals. So the Bureau stopped arguing calibers and built one repeatable test: a duty bullet fired into calibrated 10% ordnance gelatin must stop between 12 and 18 inches. That window is the yardstick behind every gel number in this deck.

0 6 12 18 24 Bureau window 12–18″ 14–16″ bullseye hunting loads ≈ 21″ penetration into 10% gel, inches …and again through each of five barriers: heavy clothing wallboard plywood sheet steel auto glass
12″ reaches the vitals through a raised arm, an odd angle or a barrier; 18″ stops before the bullet exits and endangers someone behind.

It is a defensive standard for stopping a human; big game often wants more, which is why bonded hunting loads already sit past it near 21″. Gel is a consistent yardstick, not a 1:1 model of tissue.

The counterintuitive part

More speed does not mean more penetration.

Inside the expansion window, two effects fight to a draw:

  • faster → deeper, but only through ln(1 + u): elasticity ≈ 0.23
  • faster → wider mushroom → more area → less depth, as 1/k²

They cancel. An expanding bullet is self-regulating. The chart runs the closed form with an empirical expansion ratio k(v) and retention f(v); the flat band is not drawn in, it is computed.

A magnum’s velocity isn’t buying penetration. It’s buying a wider crush channel, a bigger stretch cavity and a flatter trajectory. Not a deeper hole.
won’t open deep & narrow self-regulating the flat band sheds its core depth collapses 1,000 2,000 3,000 4,000 1× 2× 3× impact velocity, ft/s penetration ÷ reference (21″) ≈ 1.0× X(v) with k(v) and f(v) k: 1.0 → 1.9 across 1,500–1,950, then +0.2 per 1,000 f: 0.95, falling past 3,000
Push velocity to the extremes and the self-regulation breaks: deep and narrow below the window, shallow and erratic above it.
A clean test of the area term

Copper “pencils” because it cheats area.

lead mushroom A = the full disk gap gap copper petals A ≈ half the span same span 0 10 20 30 lead ≈ 21″ copper 30″+ same weight, same speed → depth
Same span, far less solid frontal area. X ∝ m/A cashes the difference as depth.

A lead mushroom is a continuous disk: big A, wide channel, shallower.

Copper opens into stiff petals with gaps between them. The medium passes through the gaps instead of being swept, so the effective A in κρAv² collapses, and since X ∝ m/A, depth shoots up.

Add near-100% retention, and if the petals shear off you’re left with a near-caliber slug that drives 30″ or more.

“Pencil-through” is the right image. Deeper isn’t tougher. Deeper is narrower.
What bullet weight really does

At fixed caliber, extra weight is length, not width.

Mushrooming is a front-end event, set by construction and velocity. A heavier same-caliber bullet doesn’t open wider; it trails a longer intact shank. Weight acts on the mass term, not the area term, and ∂ ln X/∂ ln m = 1: depth is exactly linear in mass.

130 gr same width 200 gr same width longer shank = more mass travel →
Same mushroom, more length behind it.
140 160 180 200 10 20 30 bullet weight, grains (.308, 2,700 ft/s impact, k = 1.9, f = 0.95) penetration, inches 150 gr → 17.5″ 180 gr → 21.0″ 200 gr → 23.3″ X ∝ m (A fixed)
A straight line through the origin: the exact opposite of velocity’s flat logarithm. Letting the heavier bullets arrive slower, as they do from one case, moves these by under 1%.
.308 at 2,700 ft/s: 150 gr → 17.5″ · 180 gr → 21.0″ · 200 gr → 23.3″
The validity window

The linear rule holds until the bullet can’t hold its shape.

controlled-expansion band: linear 30 gr 60 gr 120 gr 220 gr 400 gr 600 gr bullet weight at fixed caliber (log scale) penetration → 30 gr wafer fragments, tumbles 600 gr rod, L/D ≈ 9 bends, yaws, erratic
Linear in the controlled-expansion band; off the chart at the extremes.

Too light: a 30-grain wafer

The stagnation stress on its face, ρv², exceeds the strength of lead. It fragments and tumbles, and depth collapses.

Too heavy: a 600-grain rod, length ≈ 9 diameters

It bends and yaws in fluid; lead is no rigid penetrator. Depth becomes erratic.

What actually wounds

Two cavities. Only one is reliable.

0 4 8 12 16 20 24 depth, inches (block drawn to scale) impact 2,700 ft/s neck temporary cavity: stretch springs back in elastic tissue permanent crush cavity: the hole actually punched at rest ≈ 21″

Permanent crush cavity: the hole actually punched

Frontal area × path length, A·X. Always wounds. This is the reliable kill.

Temporary cavity: stretch

The medium is thrown outward at a speed proportional to v, so the force profile F(x) sets its size. Tissue that cannot stretch that fast (liver, kidney) tears; lung and muscle spring back. Reliable only above roughly 2,000–2,600 ft/s.

“Momentum dump” isn’t a wounding mechanism at all; the next slide shows how little momentum there is to dump.

Momentum is conserved

The animal cannot be pushed harder than the shooter is.

The bullet’s momentum at impact is mv₀ = 9.6 N·s for the reference load, and that is the most it can ever deliver to the animal, gel, or anything else: momentum is conserved, and the medium gets it all when the bullet stops inside.

The rifle delivered that same bullet momentum, plus the powder gas, into the shooter’s shoulder: about 13 N·s for a .308 Winchester. By the third law the push on the animal is never larger than the push on the shooter.

the animal’s change in speedΔv = mv₀ / Manimal = 9.6 N·s / 90 kg ≈ 0.11 m/s
“Knockdown power” would have to knock the shooter down first. Animals fall because of what the crush cavity destroys, never because of what the bullet pushes.
into the shooter’s shoulder bullet + powder gas, .308 Winchester 180 gr 13.3 N·s into the animal the bullet’s whole momentum, if it stops inside 9.6 N·s a 90 kg deer gains 0.11 m/s ≈ 0.35 ft/s: a slow walk’s worth of speed, nothing more Newton’s third law: the two pushes are equal and opposite, and the shooter’s is the larger one
Momentum bookkeeping for a .308 Winchester, 180-grain load. Gas jet taken at 1,400 m/s.
One goal, three routes

Every great cartridge spends its budget differently.

.45-70 Government

slow · fat · heavy
0 8 16 24 32 ≈ 28″

diameter + mass: a wide, slow hammer. No stretch at all.

.300 magnum

fast · heavy bullet
0 8 16 24 32 ≈ 21″

wide channel + depth: velocity buys both, at the cost of recoil.

6.5×55 Swedish

slow · skinny · long
0 8 16 24 32 ≈ 27″

extreme sectional density: a deep, gentle dart.

Three philosophies, one physics: get a high-sectional-density bullet through the vitals. Everything else is how violently it tears on the way.
Part 3 of 4

The optimization

Run the model forward and it points at one specific bullet — one the market has already built.

A hidden result

Crush volume does not depend on caliber or expansion at all.

multiply depth by areaVcrush = A · X = A · m2κρA ln(1 + u) = m2κρ ln(1 + u)
k = 1.3 0.40″ wide 44.9″ deep volume 5.65 in³ k = 1.9 0.59″ wide 21.0″ deep volume 5.65 in³ k = 2.4 0.74″ wide 13.2″ deep volume 5.65 in³ height ∝ frontal area, length ∝ depth: the drawn area is the crush volume

The area cancels exactly. For a given retained mass and impact speed the bullet punches a fixed volume of hole; expansion only decides its shape, trading depth for width one-for-one. That is the trade the .45-70, the .300 magnum and the 6.5×55 were making on the previous slide.

Caliber is decided by the constraints volume ignores: reaching the vitals from any angle (depth), retained velocity downrange, and recoil.
The same law in air

Downrange, the same momentum flux governs retained velocity.

In air the strength term vanishes (air has no crush strength) and the momentum-flux term is all that is left, with a shape factor c that varies slowly with speed and the air density ρair:

1Newton in airmv dv/dx = −c·ρair·A₀·v²
2separate and integrateln(v/v₀) = −c·ρair·A₀·x/m
3retention lengthv(x) = v₀·e−x/L,  L = σ₀ / (c·ρair)

Sectional density sets the retention length, just as it set the penetration depth. The sporting “ballistic coefficient” is σ₀ divided by a form factor: the same quantity in different clothes. For the 7 mm bullet on the next slide, L ≈ 1,883 m.

One number, σ₀, buys depth in the animal and velocity at the animal. It is the only quantity worth maximizing.
0 100 200 300 400 500 2,000 2,400 2,800 range, yards retained velocity, ft/s below ≈ 1,900 ft/s a bonded bullet stops opening reliably σ = 0.301: 2,429 ft/s at 400 yd σ = 0.250, same shape: 2,335 ft/s at 400 yd v(x) = v₀·e^(−x/L), L = σ₀ / (c·ρ_air) ≈ 1,883 m for σ₀ = 0.301
Same shape, same muzzle velocity; only sectional density differs. The red band is where a bonded bullet stops opening reliably.
Now run it forward

Fix the sectional density and the caliber choice is a budget, not a debate.

Hold σ₀ at 0.29 (enough for 24″ with margin) and the impact velocity inside the expansion window. Then bullet mass is σ₀d², and both of the things caliber still controls scale with it, linearly and together:

Vcrush volume∝ m = σ₀d²
precoil momentum= mv₀ + mpowdervgas ∝ m
Xdepthunchanged: σ₀ is fixed

Every step up in caliber buys crush volume in exact proportion to recoil. There is no optimum in the mathematics; there is a budget. Most people place a 4 kg rifle well up to about 4.5 m/s of recoil velocity, and that budget spans .264 to .308 inches with 7 mm in the middle.

The plateau is real, but it is a recoil budget, not a property of any cartridge.
shaded band, 13–18 N·s: 3.3–4.5 m/s in a 4 kg rifle, the recoil most people place well from field positions .224 .243 .264 .284 .308 .338 .375 10 20 30 bullet diameter, inches (σ₀ = 0.29, 2,950 ft/s, powder 38% of bullet mass) recoil momentum, N·s 102 gr 120 gr 141 gr 164 gr 193 gr 232 gr 285 gr crush volume ∝ m ∝ d² rises on exactly this curve depth is the same at every point (σ fixed)
Recoil momentum for σ₀ = 0.29 bullets at 2,950 ft/s with a 38% powder charge and a 1,400 m/s gas jet. Crush volume rises on the same curve.
The number, then the match

The physics asks for .284″. The market already built it.

.284″ 7 mm ≈ 1.50″ long · 170 gr boattail base bonded lead core copper jacket polymer tip σ = 0.301 ballistic coefficient 0.646
diameter.284″ (7 mm)
weight165–175 gr · σ₀ ≈ 0.30
muzzle velocity≈ 2,950 ft/s
ballistic coefficient (G1 reference)≈ 0.65
retained at 300 yd≈ 2,550 ft/s
penetration, f = 0.95, k = 1.9≈ 23″ at 300 yd

This is the 7 mm Precision Rifle Cartridge firing a 170-grain Federal Terminal Ascent: a bonded bullet at about 2,950 ft/s, ballistic coefficient 0.646, sectional density 0.301. The made-up ideal and a real factory load are the same object.

High σ₀ for depth and retained velocity, adequate frontal area, and recoil you can actually place.
A corollary in the powder

Efficient cartridges plateau early. Overbore ones keep climbing.

14″ 18″ 22″ 26″ 30″ -400 -200 0 +200 barrel length, inches velocity vs. a 24″ barrel, ft/s .308: flat past ~20″ −20 to −25 ft/s for every inch cut below it 7 mm PRC: still climbing +30 to +37 ft/s per inch at 26″ 20″
Fast powder flattens early; slow magnum powder keeps working down the bore.

.308 Winchester: small case, fast powder, essentially all burned by ~20″. Past that the gain shrinks to a trickle, a few ft/s an inch, trending toward zero. Going the other way, every inch you cut costs ~20–25 ft/s; chop it to a 6″ stub and you’ve gutted it: unburned powder, a fireball, hundreds of ft/s gone.

7 mm Precision Rifle Cartridge, .300 magnums: overbore, slow powder still accelerating the bullet at the muzzle. Roughly 30–37 ft/s per inch and still climbing at 26″; they need that length just to realize the case. Their plateau is real too; it sits further down the bore.

Every cartridge plateaus, and with enough barrel eventually loses velocity to friction. The only question is where.
The capacity illusion

More case, barely more bullet.

A .30-06 Springfield burns about 17% more powder than a .308 Winchester. At the muzzle that buys only ~4% more velocity. Momentum shows exactly where the rest goes.

impulse is momentumm·v = A₀ · ∫ P(t) dt
  • peak pressure is capped by the brass: a bigger case can’t raise it, only stretch the tail
  • the bullet is already moving fast, so it crosses the late inches, where the extra powder burns, almost instantly and at falling pressure
  • +17% powder adds only ~4% to the area under the curve → ~4% more bullet momentum
Recoil = bullet momentum + gas-jet momentum. The surplus charge leaves as fast gas: the bullet gains ~4%, the gas jet ~18%. You pay near-full freight in recoil and blast and pocket a few percent in the bullet.
0″ 6″ 12″ 18″ 24″ bullet travel down the bore, inches chamber pressure peak capped by the brass, same for both .308 .30-06 the extra powder lives in this thin rim +17% powder → +4% velocity
Same capped peak. The bigger case only adds a thin rim of impulse.
Part 4 of 4

Why the debates never end

The internet argues along the flat axis and ignores the steep one.

The contaminated evidence

Most of the debate runs on bullets that came apart.

The convergence assumed one thing: retained mass barely moves, f ≈ 0.93–0.97. Cheap cup-and-core bullets shatter that premise.

  • rapid cup-and-core designs (the Hornady SST and dozens like it) can shed 35–50% of their mass
  • the lost mass becomes a snowstorm of lead: hundreds of fragments flung inches past the wound, through meat the bullet never touched
  • retained m collapses, so X ∝ m/A craters: the same launch weight now penetrates far less
When a bullet loses half its mass, depth stops telling you about the cartridge and starts telling you about the bullet. Most “stopping power” arguments are comparing construction and calling it caliber.
rapid cup-and-core · −45% mass · stops ≈ 12″ lead snowstorm 0 4 8 12 16 20 24 penetration into 10% gel, inches bonded or monolithic · −3% mass · one clean channel to ≈ 21″
Same caliber, same launch weight. The bullet that came apart is a different experiment.
The decision that actually matters

Caliber is the flat axis. Construction is the steep one.

Bureau window 12–18″ 6″ 12″ 18″ 24″ 30″ gel penetration, 10% gel 12–18″ 17″ 56 gr kept Fragmenting very-low-drag match · soft point keeps ~40–60% 18–27″ 24″ 66 gr kept Bonded AccuBond · Fusion · Partition keeps ~75–95% 22–32″ 32″ 127 gr kept Monolithic copper Barnes · CX · E-Tip keeps ~98–100% ringed dots: three 6.5 mm bullets, 127–142 gr, same impact speed only the construction changes, and depth nearly doubles small dots: published 10% / synthetic-gel tests · ringed dots: Berger 140 gr very-low-drag · Nosler 142 gr AccuBond Long Range · Barnes 127 gr Long-Range X

Hold the construction fixed and the caliber fight really is a plateau: the six bonded cartridges earlier all stopped near 21″. Pin one caliber and one bullet weight, change only how the bullet is built, and measured gel penetration runs from ~17″ to ~32″. It nearly doubles, exactly as f/k² predicts.

Representative figures from published 10% and synthetic-gel testing (Brass Fetcher, Field & Stream Bullet Lab, maker labs); exact depth shifts with medium, impact velocity and barrel.

What the internet fights over (6.5 vs .270 vs 7 mm vs .308) is the flat axis. What it almost never debates (cup-and-core vs bonded vs copper) is the steep one. Stop arguing calibers. Pick the bullet.
The bullet that should exist

Copper that opens like lead: the physics allows it. The shelf doesn’t sell it.

0 4 8 12 16 20 24 28 32 depth, inches · same retained mass, same impact speed, equal crush volume an 18″ chest behind the animal bonded lead f = 0.95 · k = 1.9 5% of the mass left in the meat 21″ · 86% inside petal copper, as sold f = 1.00 · face ≈ 74% of a disk narrow; the tail of the hole is in the dirt 30″ · 60% inside full-face copper, as the model asks f = 1.00 · k = 1.9 wide, no fragments, stops inside 22″ · 81% inside

Petal copper pencils because its petals leave gaps: the effective face is about three-quarters of a disk, so the same retained mass drives 30″ instead of 21″. The invariant says the crush volume is identical; on an 18″ chest, 40% of it is spent in the dirt behind the animal.

Nothing forbids a full-face copper mushroom. Annealed copper yields near 70 MPa; the stagnation stress on the nose at 2,700 ft/s is ρv² ≈ 700 MPa, ten times that. A soft copper front bonded to a hard copper shank would open like lead and hold together like copper: f = 1, k = 1.9, 81% of the hole inside the animal and no lead in the meat.

The petal is a manufacturing convenience, not a law of physics. A hunting bullet that spends 40% of its wound behind the animal is a slower kill, and a bullet that could avoid it, and isn’t made, is a choice the animal pays for.

Position of this deck, not a measured result. Published petal-copper depths run 22–32″; the 74% face is fitted to that range. The metallurgy argument is about yield stress, not about any product on sale.

What to carry out

The model that defuses the rest of the noise.

For any claim, ask: which variable does the work, which is merely correlated, and which is a logarithmic afterthought?

  • Penetration = retained mass ÷ expanded frontal area, times ln(1 + u). Velocity has an elasticity near one quarter.
  • Crush volume = m·ln(1 + u)/2κρ. Expansion trades depth for width at constant volume.
  • Momentum is conserved: the animal is pushed less than the shooter. Wounding is what the crush cavity destroys.
  • Sectional density buys depth and retained velocity together: the one quantity worth maximizing. Construction and placement dwarf the cartridge.
X = mretained 2κρ·Aexpanded · ln ( 1 + κρv² R ) retained mass ÷ expanded area, times a logarithm of speed ρ, R: density and crush strength of the medium · κ: nose factor · nothing else

“Knockdown power,” magnum-versus-standard, barrel-length wars, the numbers printed on the box: all of it gets clear the moment you ask which quantity is actually doing the work.

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