Hertzian Contact Stress Calculator

Contact pressure, patch size and approach for spheres, cylinders and crossed cylinders — with a concave surface entering as a negative radius, and the subsurface shear peak found where it actually is.

Last updated 24 Sep 2026

Hertzian contact stress

Radii, load and materials → pressure, patch and subsurface shear

Contact geometryPoint contact gives an elliptical patch and a pressure that rises as the CUBE ROOT of load; line contact gives a rectangular strip and a pressure that rises as the square root. That difference — and not the contact area — is why a roller bearing carries so much more than a ball bearing of the same size.

Normal load F (N)The force pressing the two bodies together, along the common normal. Not a tangential or traction force: this page is frictionless normal contact, and adding traction moves the subsurface shear peak towards the surface, which is a different and worse problem.

Body 1 radius R₁ (mm)The ball, roller or sphere — the convex one. For a ball this is half the diameter, so a 12 mm ball is 6 here. Body 1 is always taken as convex; put any concave surface in as body 2 with a negative radius.

Body 2 radius R₂ (mm) — 0 for a FLAT, NEGATIVE for concaveThe single most important field on this page. Zero means a flat surface. A positive number is another convex body. A NEGATIVE number is a concave one — a socket, a groove, a bearing outer raceway — and it enters the effective radius as a negative curvature, which makes the effective radius larger and the pressure lower. Getting the sign wrong here is the commonest mistake in contact stress work and it errs in the unsafe direction.

Body 2’s radius ACROSS the rolling direction (mm) — 0 to make it sphericalLeave at zero and body 2 has the same radius both ways, which is a sphere or a flat. Enter a second radius and the contact becomes elliptical. This is how a ball in a bearing groove is handled: the raceway is convex along the rolling direction and concave across it, so you would type the raceway radius above and the groove radius, negative, here. A groove cut at 0.52 of the ball diameter gives an effective transverse radius of 26 times the ball radius.

Contact length L (mm) — line contact onlyThe length of the roller actually in contact, which is the effective length and not the overall one: a crowned roller touches over less than its full length, and a straight roller in a misaligned housing touches over much less. Halving the effective length multiplies the pressure by the square root of two, not by two.

Body 1 materialOnly two numbers from it are used, E and Poisson’s ratio, and they combine with body 2’s into one effective modulus. Body 1’s Poisson’s ratio also sets the subsurface stress field reported below, because that field is the one INSIDE body 1.

Body 2 materialContact pressure depends on both materials through 1/E* = (1−ν₁²)/E₁ + (1−ν₂²)/E₂, so the SOFTER body dominates: steel on acetal sees a fraction of the pressure of steel on steel, which is why plastic gears and plain bearings survive contact loads that would brinell a hardened race.

Not a circuit: a contact drawn in section on the left and the shear stress below it on the right, both to the same depth scale, so the marked line is at the same place in each. The lobe hanging below the surface is the pressure over the patch — a semi-ellipsoid, which is why the peak is exactly 3/2 of the mean for a point contact and 4/pi for a line one. The right-hand plot runs DOWNWARDS into the material, with the shear stress measured across it. Read where the curve bulges. The shear is small at the surface, because the material there is squeezed almost equally in every direction and hydrostatic pressure cannot shear anything; it peaks about half a contact half-width down at 31 per cent of the peak pressure for a sphere, or 0.79 of a half-width down at 30 per cent for a line contact, and the marked line shows which of the two you have. Both axes are normalised — stress by p_max and depth by the contact half-width b — so this is the same drawing for every contact, and only the geometry you chose changes it. The figures underneath are yours, and the depth is in millimetres.

3,253.3MPaExample

A 12 mm hardened steel ball pressed onto a flat hardened steel plate with 500 N — about the weight of a person standing on a ball bearing

Two effective quantities, and everything follows

1/E* = (1−ν₁²)/E₁ + (1−ν₂²)/E₂ · 1/R = 1/R₁ + 1/R₂ · a = ∛(3FR/4E*) · p_max = 3F/2πa² · p_max(line) = √(F′E*/πR) · τ_max ≈ 0.31 p_max at z ≈ 0.48 b

E*: the effective, or reduced, modulus. Both bodies deflect, so both compliances add — and because they add as compliances, the SOFTER material dominates. Steel on acetal is barely stiffer than acetal on acetal. The (1−ν²) is there because the material under a contact patch is in plane strain, not uniaxial: it cannot spread sideways, so it behaves stiffer than its Young’s modulus alone would suggest
1/R: curvatures add, radii do not. A convex surface has positive curvature and a CONCAVE one negative, so a ball in a socket has 1/R = 1/R_ball − 1/R_socket and an effective radius much larger than either. This is the sign that gets fumbled, and fumbling it towards a smaller effective radius over-predicts the pressure while fumbling it the other way under-predicts it
the cube root: the patch grows as F^(1/3) and its area as F^(2/3), so the pressure grows only as F^(1/3). Contact is self-limiting in a way that bending is not: the harder you press, the more area you make. Eight times the load doubles the pressure on a ball; four times doubles it on a roller
the semi-ellipse: the pressure over the patch is not uniform and not parabolic — it is a semi-ellipsoid, p(r) = p_max√(1−r²/a²), which is what makes the peak exactly 3/2 of the mean for a point contact and 4/π for a line one. Those two ratios are a free check on any contact calculation
τ at depth: the number this page leads with. At the surface the material is in near-hydrostatic compression and the shear is only about a tenth of the pressure. The sideways stresses fall away faster than the vertical one, so the shear PEAKS below the surface — at 0.31 p_max about half a half-width down for a circular contact, and at 0.30 p_max at 0.79 of a half-width for a line contact. Metal fails in shear, so that is where the damage starts

Worked example

A 12 mm hardened steel ball pressed onto a flat hardened steel plate with 500 N — about the weight of a person standing on a ball bearing
THE TWO EFFECTIVE QUANTITIES FIRST. Both bodies are steel, so 1/E* = 2×(1−0.3²)/206,000 and E* = 113,187 MPa. One body is flat, so its curvature is zero and 1/R = 1/6 + 0, giving R = 6 mm. Note that a flat is not an infinite radius you have to type — it is a zero curvature, which is why this page asks for zero
THE PATCH. a = ∛(3 × 500 × 6 / (4 × 113,187)) = 0.2709 mm. That is a circle about half a millimetre across, and its area is 0.2305 mm² — smaller than the head of a pin
THE PRESSURE. p_max = 3F/(2πa²) = 3,253 MPa. Check it against the mean: 500 N over 0.2305 mm² is 2,169 MPa, and the peak is exactly 1.5 times that, as a semi-ellipsoidal distribution must be
AND NOW THE PART THAT SURPRISES PEOPLE. That is 3,253 megapascals from a 50-kilogram load. It is three times the tensile strength of most structural steel and it has not damaged anything, because the steel under the patch is squeezed from every side at once and cannot flow. Yielding starts when the von Mises stress reaches the uniaxial yield strength, and the maximum von Mises stress here is 0.6200×p_max = 2,017 MPa. So this contact is elastic in a steel with a yield strength above 2,017 MPa — true of hardened bearing steel, and not remotely true of mild steel, which would take a permanent dent
WHERE THE DAMAGE WOULD START. Not at the surface. The maximum shear is 0.3100×p_max = 1,009 MPa, and it sits 0.1303 mm below the surface — 0.48 of the contact radius down. At the surface itself the shear is only 325 MPa, three times less. That is why rolling contacts fail by a subsurface crack that grows to the surface and lifts a flake out, rather than by wearing away
WHAT IT WOULD TAKE TO OVERLOAD IT. Pressure grows as the cube root of load, so reaching ISO 76's 4,200 MPa — the stress at which a ball bearing takes a permanent dent of one ten-thousandth of the ball diameter — needs 500 × (4,200/3,253)³ = 1,076 N. Barely twice the load, for a 29 per cent rise in pressure. Contact stress hides overload well, which cuts both ways
AND THE SAME LOAD ON A ROLLER. Swap the ball for a 10 mm diameter roller 10 mm long at the same 500 N and the geometry becomes line contact: b = 0.0530 mm and p_max = 600 MPa, less than a fifth of the ball's figure. That is the whole reason roller bearings exist, and it is also why they are so much less tolerant of misalignment — the advantage is entirely in the length, and a roller that touches over half its length gives back 41 per cent of it

The three geometries, and how the pressure grows with load

The last column is the one to remember. Doubling the load on a BALL raises the peak pressure by the cube root of two, 26 per cent; doubling it on a ROLLER raises it by the square root of two, 41 per cent. Turned round: it takes eight times the load to double a point contact’s pressure and only four times to double a line contact’s. That is why contact stress is so forgiving of overload compared with bending, and why a small increase in pressure implies a very large increase in load. The approach row is not an omission: for two long cylinders the elastic displacement integral diverges logarithmically, so how far they move together depends on the overall dimensions of the bodies and not on the contact alone. Johnson’s *Contact Mechanics* states it plainly and no closed form for it exists. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Geometry	Patch size	Peak pressure	Approach	Pressure grows as
Point contact, circular patch (sphere on flat, two spheres, ball in a socket)	a = ∛(3FR/4E*)	p_max = 3F/(2πa²)	δ = a²/R	F^1/3
Point contact, elliptical patch (ball in a groove, crossed cylinders of unequal radius)	Hamrock and Brewe’s closed form in R_y/R_x	p_max = 3F/(2πab)	from the same fit	F^1/3
Line contact (cylinder on flat, two parallel cylinders)	b = √(4F′R/πE*)	p_max = √(F′E*/πR)	not determinate from the contact alone	F^1/2

The subsurface stress field, as a percentage of the peak contact pressure

Poisson’s ratio 0.30, stresses on the load axis, computed from the closed-form field rather than read off anybody’s graph. Two rows are marked because they are the answers: 0.4809 is where the shear peaks under a sphere, at 31.00 per cent of the peak pressure, and 0.7862 is where it peaks under a line contact, at 30.03 per cent. Read the top row to see why. At the surface the sphere case has σ_z at 100 per cent and σ_r at 80 per cent, so the shear is only 10 per cent — the material is in nearly hydrostatic compression and hydrostatic pressure cannot yield anything. Going down, σ_r falls much faster than σ_z, the gap opens, and the shear peaks three times higher than it was at the surface. That is the entire mechanism of subsurface-initiated spalling, and it is why a pitting bearing has a bright, unworn-looking raceway right up until a flake lifts out of it. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Depth (× b)	Sphere: σ_z	σ_r	τ	Line: σ_z	σ_x	τ
0.0000	100.0	80.0	10.00	100.0	100.0	0.00
0.1000	99.0	61.4	18.82	99.5	81.5	9.00
0.2000	96.2	46.2	24.97	98.1	65.9	16.08
0.3000	91.7	34.2	28.75	95.8	53.0	21.38
0.4000	86.2	25.0	30.60	92.8	42.6	25.14
0.4809	81.2	19.2	31.00	90.1	35.6	27.25
0.6000	73.5	12.9	30.33	85.7	27.5	29.13
0.7862	61.8	6.7	27.57	78.6	18.6	30.03
1.0000	50.0	2.9	23.55	70.7	12.1	29.29
1.5000	30.8	0.0	15.41	55.5	5.1	25.19
2.0000	20.0	0.5	10.27	44.7	2.5	21.11
3.0000	10.0	0.5	5.24	31.6	0.8	15.40

Conformity: what a groove does, for a 12 mm ball on a 60 mm raceway at 500 N

One negative number does all of this. The groove is concave, so its curvature is negative, and 1/R_y = 1/R_ball + 1/R_groove is a DIFFERENCE rather than a sum. At the 0.52 of ball diameter that bearing makers actually cut, the two curvatures nearly cancel and the effective transverse radius comes out at 26 times the ball radius: the contact stretches into a long narrow ellipse nine times longer than it is wide, and the pressure falls accordingly. Open the groove to 0.60 and the pressure rises 40 per cent; make the race flat across and it rises further still. Tighten the groove below about 0.515 and the pressure keeps falling, but the sliding within the contact and the friction torque rise sharply, which is why nobody does it. That trade — contact stress against spin friction — is the single most important number in a ball bearing’s internal design, and it is one negative radius. This is a first-pass calculation on an idealised part — uniform section, static load, no stress raisers beyond those stated, and room-temperature material properties. Real parts have fillets, keyways, surface finish and duty cycles that a closed-form answer cannot see.
Groove radius ÷ ball DIAMETER	… as a negative radius (mm)	Effective transverse radius R_y (mm)	… as a multiple of the ball radius	Ellipticity a/b	Peak pressure (MPa)	Relative to the tightest groove
0.515	-6.180	206.0	34.3	10.67	1,430	100.0
0.520	-6.240	156.0	26.0	8.94	1,520	106.3
0.540	-6.480	81.0	13.5	5.89	1,761	123.1
0.560	-6.720	56.0	9.3	4.66	1,918	134.1
0.600	-7.200	36.0	6.0	3.51	2,132	149.1
0.700	-8.400	21.0	3.5	2.49	2,443	170.8
1.000	-12.000	12.0	2.0	1.75	2,842	198.7

Allowable contact pressure: the named figures, and what they mean

The first two columns are ISO 76 and nothing else: the basic static load rating C₀ is *defined* as the load producing those contact stresses at the centre of the most heavily loaded contact, at which the total permanent deformation of rolling element and raceway is about 0.0001 of the rolling element diameter. They are not allowable stresses in the ordinary sense and they are not derived here; they are quoted. The third column is computed: divide by Green’s coefficient 1.6132 at ν = 0.30 and you get the uniaxial yield strength at which the material would FIRST yield under that pressure. The fourth divides by Tabor’s factor of three, which is the pressure at which the plastic zone has spread to the whole contact and indentation is fully plastic — the same three that connects hardness to yield strength. The gap between the columns is the point: a contact can carry roughly 1.6 times the uniaxial yield before it yields at all, and roughly three times before it dents properly, because the material under the patch is held in triaxial compression by the material around it. These dimensions come from a published standard’s table, not from a formula. The standard itself is cited below and the printed values are attributed to the catalogue they were taken from; a different publisher may round differently in the last digit.
ISO 76:2006, clause 3.1	Contact stress at C₀ (MPa)	Uniaxial yield that implies (MPa)	Indentation hardness that implies (MPa)
Self-aligning ball bearings	4,600	2,852	1,533
All other radial ball bearings	4,200	2,604	1,400
All radial roller bearings	4,000	2,480	1,333

The damage starts below the surface, and the pressure is not a stress you compare with yield

The maximum shear stress under a contact is not at the surface. That single fact explains why rolling contacts fail the way they do. Right at the middle of a contact patch the material is squeezed almost equally in all three directions — σ_z is the full contact pressure and the two sideways stresses are about 80 per cent of it — and near-hydrostatic pressure cannot shear anything. Go down and the sideways stresses fall away much faster than the vertical one, the gap between them opens, and the shear reaches a maximum of about 31 per cent of the peak pressure roughly half a contact half-width below the surface. Metal yields and fatigues in shear, so the most damaged material in a bearing or a gear flank is a thin buried layer, typically a tenth to a few tenths of a millimetre down. A crack starts there, runs parallel to the surface, turns up and lifts a flake out. That is a spall, and it is why a bearing that is pitting has a bright, unworn raceway right up until the flake comes away.

Contact pressure is not a stress you compare with yield. Five hundred newtons — about fifty kilograms — through a 12 mm ball onto a flat plate makes 3,253 MPa, several times the tensile strength of any structural steel — and does no damage at all provided the steel is hard enough. The reason is the triaxiality: the material under the patch is held from spreading by the material around it, so its deviatoric stress is far smaller than its hydrostatic stress. Yielding starts when the von Mises stress reaches the uniaxial yield strength, and the maximum von Mises stress is only about 0.62 of the peak pressure for a circular contact. In other words a contact can carry about 1.61 times the material’s uniaxial yield strength before it yields anywhere, and roughly three times it before the plastic zone breaks out to the surface and it dents properly — which is the same factor of three that connects indentation hardness to yield strength. Comparing p_max with a yield strength directly will tell you that everything fails.

A concave surface is a negative radius, and that is the whole of conformity. Curvatures add, so 1/R = 1/R₁ + 1/R₂ becomes a difference when one surface is concave, and the effective radius grows. A 12 mm ball in a groove cut at 0.52 of the ball diameter has an effective transverse radius of 26 times the ball radius: the contact stretches into a long thin ellipse and the pressure falls by more than half against the same ball on a flat. That one number — the groove conformity — is the most important dimension inside a ball bearing, and it is a trade: tighter grooves give lower contact stress and more spin friction, looser grooves the reverse. Get the sign of the radius wrong and you will compute a ball on a hill instead of a ball in a valley, which errs in the unsafe direction.

What this page owns, and what it does not. It owns the general contact problem: any two elastic bodies, any combination of radii, any pair of materials. It does NOT own the gear mesh — the gear tooth bending stress page already computes the Hertz pressure at a gear pitch point from the flank radii (d/2)sin α, which is this same line-contact formula written in gear terms, and that page is the right place for it because the gear geometry is what is hard about it. It does not own bearing life: the L10 page does that, and this page is the explanation of why its exponent is 3 rather than something gentler. It is not an AGMA or ISO pitting rating, which carry geometry, load distribution, lubrication, surface finish and life factors that no closed form can see. And it says nothing about traction: adding a tangential force drags the shear peak towards the surface, and once the traction coefficient passes about 0.3 the maximum arrives AT the surface and the failure mode changes from subsurface spalling to surface-initiated micropitting.

What the model leaves out. Hertz assumes both bodies are elastic half-spaces, that the patch is small compared with both radii, that the surfaces are frictionless and perfectly smooth, and that nothing is moving. Real surfaces are rough, and at light loads the real contact happens on asperities at pressures far above the nominal one. Real contacts are lubricated, and an elastohydrodynamic film changes the pressure distribution near the exit of the contact with a spike that Hertz has no way to produce. Real rolling contacts have traction, spin and slip. And residual stresses from hardening, which are compressive at the surface in a case-hardened part, shift the effective stress field by an amount comparable with the numbers on this page. Every one of those is a reason the answer here is a first-pass figure.

Frequently asked questions

Where is the maximum shear stress in a Hertzian contact?

Below the surface, not at it. For a circular point contact at Poisson’s ratio 0.3 the maximum shear is 0.310 of the peak contact pressure at a depth of 0.481 contact radii; for a line contact it is 0.300 of the peak pressure at 0.786 half-widths, and that line-contact figure does not move with Poisson’s ratio at all. At the surface itself the shear is only about a tenth of the pressure, because the material there is in nearly hydrostatic compression. This is why rolling contacts fail by subsurface crack initiation and spalling rather than by surface wear, and why case depth has to exceed the depth of the shear peak.

Can I compare contact pressure with the yield strength?

No, not directly, and doing so will tell you that every bearing in the world has failed. The material under a contact patch is in triaxial compression and yields on the von Mises stress, which for a circular contact peaks at only about 0.62 of the contact pressure. So first yield happens at a contact pressure of roughly 1.61 times the uniaxial yield strength, and the contact does not dent properly until about three times it. If you want a named allowable rather than a derived one, ISO 76 uses 4,200 MPa for ball bearings and 4,000 MPa for roller bearings as the stress at which the permanent deformation reaches 0.0001 of the rolling element diameter.

How do I handle a ball in a groove or a socket?

Enter the concave radius as a NEGATIVE number. Curvatures add, so a concave surface subtracts and the effective radius grows. A ball in a bearing groove needs two radii for the raceway: the raceway radius along the rolling direction, which is positive for an inner ring, and the groove radius across it, which is negative. At the 0.52 of ball diameter that bearing makers actually use, the transverse effective radius comes out at 26 times the ball radius and the contact becomes a long thin ellipse. Getting that sign wrong is the commonest error in contact stress work.

Why does a roller carry so much more load than a ball?

Because the pressure grows more slowly with load only in one sense and the area grows in the other. A point contact makes a patch whose area grows as F^(2/3), so pressure grows as F^(1/3); a line contact makes a strip whose area grows as F^(1/2), so pressure grows as F^(1/2). At first glance the point contact looks better. But the line contact starts from a vastly larger area at any useful load, because the patch runs the whole length of the roller. A 12 mm ball on a flat at 500 N sees 3,253 MPa; a 10 mm roller 10 mm long at the same load sees 600 MPa. That is the whole argument, and the catch is that it depends entirely on the roller touching over its full length, which misalignment destroys.

What is the effective modulus E* in Hertz contact?

1/E* = (1−ν₁²)/E₁ + (1−ν₂²)/E₂. Two things about it. The compliances add, so the softer material dominates — steel on acetal behaves very nearly like acetal on acetal. And the (1−ν²) is there because the material beneath a contact is in plane strain and cannot spread sideways, so it behaves stiffer than its Young’s modulus alone. Some texts use E′ = 2E*, which doubles every appearance of it; check which convention a formula is written in before mixing two sources.

Does the contact pressure depend on the hardness?

Not at all, as long as everything stays elastic. Hertz contains only Young’s modulus and Poisson’s ratio, and hardening a steel changes neither to any useful degree — a 60 HRC bearing steel and a soft mild steel have almost identical moduli. What hardness changes is whether the answer is allowed to happen: a hard steel stays elastic at 3,000 MPa and a soft one takes a permanent dent. So hardness sets the allowable, not the stress.

How deep should a case-hardened layer be?

Deeper than the shear peak, with margin, and this page gives you the depth. The maximum subsurface shear sits at about 0.48 of the contact half-width for a circular contact and 0.79 of it for a line contact, so a contact making a 0.5 mm half-width puts its peak roughly 0.24 to 0.4 mm down. If the case is thinner than that, the highest shear lands in the soft core, a crack starts under the case, and the whole case flakes off in sheets rather than in the small pits a properly cased part produces. The failure looks completely different and it is diagnostic: case crushing means the case was too thin, not that it was too soft.

Related calculators

References

Heinrich Hertz, Über die Berührung fester elastischer Körper, Journal für die reine und angewandte Mathematik 92 (1882), 156–171. The original. Everything on this page above the subsurface section is in it: the elliptical contact patch, the semi-ellipsoidal pressure distribution, and the cube-root dependence of the patch size on load that makes contact pressure rise so slowly with force.
K. L. Johnson, Contact Mechanics (Cambridge University Press, 1985), chapters 3 and 4. The standard modern account, and the source for two statements made here that Hertz’s paper does not contain: that the approach of two bodies in LINE contact is not determined by the contact solution alone, because the integral diverges and the answer depends on the overall dimensions of the bodies; and that the subsurface stress field under a line contact has its maximum shear below the surface. Cited by chapter, not reproduced.
Itzhak Green, Poisson ratio effects and critical values in spherical and cylindrical Hertzian contacts (Georgia Tech, itzhak.green.gatech.edu, read 29 September 2026). Used as the independent check on the one number this page leads with. Green gives the contact pressure at first yield as p₀/σ_Y = 1.30075 + 0.87825 ν + 0.54373 ν², which at ν = 0.30 is 1.6132. The reciprocal, 0.6199, is the maximum von Mises stress as a fraction of p_max — and the scan of the closed-form stress field done here returns 0.6200, from a completely different calculation. Half of it, 0.310, is the maximum shear.
ISO 76:2006, Rolling bearings — Static load ratings, clause 3.1 (read from the publisher’s preview on standards.iteh.ai, 29 September 2026). Cited by number and clause; three numbers are taken from it and nothing else. The basic static load rating C₀ is defined as the load giving a calculated contact stress at the centre of the most heavily loaded contact of 4 600 MPa for self-aligning ball bearings, 4 200 MPa for all other radial ball bearings and 4 000 MPa for all radial roller bearings — at which the total permanent deformation of rolling element and raceway is about 0.0001 of the rolling element diameter. It is the named allowable this page uses, in preference to deriving one.
D. E. Brewe and B. J. Hamrock, Simplified Solution for Elliptical-Contact Deformation Between Two Elastic Solids, Journal of Lubrication Technology 99(4) (1977), 485–487; and B. J. Hamrock and D. E. Brewe, Simplified Solution for Stresses and Deformations, Journal of Lubrication Technology 105(2) (1983), 171–177. The closed form this page uses for an elliptical contact, because the exact Hertz solution needs complete elliptic integrals and this calculator has no way to evaluate them. The accuracy quoted on this page is not theirs: it was measured here, by solving the exact elliptic-integral problem with the arithmetic-geometric mean and comparing, at radius ratios from 1 to 100.
Earle Buckingham, Analytical Mechanics of Gears (McGraw-Hill, 1949; Dover reprint 1988). Cited for the gear surface-durability tradition that this page deliberately stays out of. Buckingham’s surface stress formula is Hertz line contact with the flank radii written in gear terms, and the gear tooth bending stress page in this plugin already carries it. What is NOT in Buckingham, and is here, is the general elliptical case and the subsurface field.