Dowel Pin and Hole Calculator

The actual interference or clearance range from the pin’s band and the hole’s, with the contact pressure, the insertion force and the hoop stress in the boss — every ISO 286 deviation derived from the IT grades rather than copied, and checked against the printed ISO 8734 pin limits. Plus the reason two round pins do not locate a plate and the diamond pin does.

Last updated 24 Sep 2026

Dowel pin and hole

Pin class and hole class → the real fit

Pin nominal diameterISO 8734 and ISO 2338 cover Ø1 to Ø50. This page starts at Ø4 because the 0–3 mm row of the ISO 286 interference hole bands could not be verified from a second source — see the references.

Pin tolerance classAn ISO 8734 hardened dowel pin is m6 — that is not a choice, it is what the standard specifies. ISO 2338 comes in m6 (type A) and h8. The hole is where the decision is.

Hole tolerance classThis is the whole answer. The pin’s band is fixed by the standard; the fit is whatever you ream the hole to.

Engaged length in this part (mm)How far the pin goes into the part you are pressing it into. The insertion force is proportional to it.

Outside diameter of the material round the hole (mm)How much material surrounds the hole. A hole near an edge or in a thin boss is a thin cylinder and carries far less pressure for the same interference — put the real figure in, because the difference is large.

Coefficient of friction, pin to hole0.10 to 0.15 for clean dry hardened steel in steel, lower with oil, higher if either surface is rough or galls. This is the least certain number on the page and the insertion force is directly proportional to it.

Young’s modulus of both parts (MPa)210,000 MPa for steel, 70,000 for aluminium, 100,000 for grey cast iron. The closed form used here assumes the SAME material both sides, which makes the two Poisson terms cancel; for a steel pin in an aluminium boss it will overstate the pressure.

Not a circuit: the tolerance bands as a number line, in MICROMETRES, which is the only scale they can honestly be drawn at — a 12 µm interference on a 6 mm pin is a five-thousandth of the diameter, so a drawing of the pin in the hole would show nothing at all. The vertical line is the nominal diameter; the scale runs from 40 µm under it to 60 µm over, with a tick every 10 µm. The top bar is the PIN's m6 band, which the standard fixes and you cannot change. The three below it are the H7, N7 and P7 hole bands, and under each one the hatched strip is the resulting interference range — from pin minimum minus hole maximum at one end to pin maximum minus hole minimum at the other. Read the H7 strip: it crosses zero, which is the whole point. An m6 pin in an H7 hole can be loose. N7 and P7 do not cross it. Notice also how much of every strip's width is the pin's own 8 µm band rather than the hole's: you cannot buy a tighter dowel pin, so the hole is the only decision.

0.0120mmExample

A Ø6 mm ISO 8734 m6 dowel pin in an H7 hole, 12 mm engaged, 24 mm of material round the hole, μ = 0.12, steel both sides

Two tolerance bands, then Lamé

maximum interference = pin_max − hole_min · minimum interference = pin_min − hole_max · contact pressure p = E·δ·(D₀² − d²) / (2·d·D₀²) · insertion force = μ·p·π·d·L · hoop stress at the bore = p·(D₀² + d²)/(D₀² − d²)

δ: the DIAMETRAL interference. Getting radial and diametral confused is a factor-of-two error and it is the commonest mistake in this calculation
d: nominal pin diameter
D₀: the outside diameter of the material around the hole. This is not a detail: a boss 1.2 d across carries less than 40% of the pressure an effectively infinite one does, for the same interference. A hole near an edge is a thin cylinder
E: Young’s modulus. The closed form assumes the SAME material both sides, which is what makes the two Poisson terms cancel and leaves this simple expression. For a steel pin in an aluminium boss it overstates the pressure
μ: coefficient of friction between pin and hole. 0.10 to 0.15 for clean dry steel on steel. The insertion force is directly proportional to it, and this is the least certain number on the page
the assumptions: plane stress, elastic throughout, smooth surfaces, no lead-in chamfer effect, the pin fully entered, and no thermal difference. It is a first-pass number: the real insertion force depends on the chamfer, the surface finish and whether the press is aligned

Worked example

A Ø6 mm ISO 8734 m6 dowel pin in an H7 hole, 12 mm engaged, 24 mm of material round the hole, μ = 0.12, steel both sides
The pin is m6: at Ø6 that is +4 to +12 µm, so 6.004 to 6.012 mm. Fuller Fasteners' printed ISO 8734 table says 6.004 to 6.012, which is the check on the whole derivation
The hole is H7: 0 to +12 µm, so 6.000 to 6.012 mm
Maximum interference = pin max − hole min = 6.012 − 6.000 = 0.0120 mm, or 12 µm. Minimum interference = pin min − hole max = 6.004 − 6.012 = -0.0080 mm — which is NEGATIVE, an 8 µm clearance. So an m6 pin in an H7 hole is a TRANSITION fit: it may need a press and it may drop in, depending where in their bands the two parts landed
At the tight end the contact pressure is E·δ·(D₀² − d²)/(2·d·D₀²) = 210,000 × 0.012 × (576 − 36) / (2 × 6 × 576) = 196.9 MPa
Insertion force = μ·p·π·d·L = 0.12 × 196.9 × π × 6 × 12 = 5,344 N, about 5.3 kN — an arbor press, not a hammer. At the loose end of the same fit it is zero, because the pin is a clearance there
Change nothing but the hole class to N7 (−4 to −16 µm) and the picture changes completely: interference from 8 to 28 µm, never loose, and up to 459 MPa of contact pressure needing 12.5 kN to press. That is the choice this page is for: the pin's band is fixed by the standard, and the hole is the decision

An m6 pin at Ø6 mm against every hole class — the decision, in one table

The pin’s band is fixed: ISO 8734 says m6 and that is the end of it, +4 to +12 µm at Ø6. Everything in this table is the HOLE’s doing. H7 spans 0 to +12, so an m6 pin in it can be anything from 8 µm loose to 12 µm tight — a transition fit, which is why an H7 dowel hole sometimes needs a press and sometimes drops in, and why H7 is right for a joint you intend to take apart. N7 is −4 to −16, so the fit is 8 to 28 µm tight and never loose. P7 is −8 to −20 and always tighter still. Note how much of the uncertainty is the pin’s own 8 µm band: you cannot buy a tighter dowel pin, you can only choose a different hole. None of these deviations was copied from a table — they are derived from the IT grades and the shaft fundamental deviations, and checked against the published bands AND against the requirement that N7/h6 give the same fit limits as H7/n6. 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.
Hole class	Hole upper deviation (µm)	Hole lower deviation (µm)	Maximum interference (µm)	Minimum interference (µm)	What kind of fit that is	Contact pressure at the tight end (MPa)	Insertion force over 12 mm (N)
H7	12	0	12	-8	could be either	197	5,344
N7	-4	-16	28	8	always an interference	459	12,469
P7	-8	-20	32	12	always an interference	525	14,250
K7	3	-9	21	1	always an interference	345	9,352
M7	0	-12	24	4	always an interference	394	10,688
G7	16	4	8	-12	could be either	131	3,563
F7	22	10	2	-18	could be either	33	891
H8	18	0	12	-14	could be either	197	5,344

m6 pin limits and the three fits, size by size

The m6 limits here are DERIVED, not transcribed, and the check on that derivation is Fuller Fasteners’ printed ISO 8734 table: its minimum and maximum pin diameters from Ø1 to Ø20 reproduce these figures exactly at every size. Read the H7 columns down the page and notice what happens: the minimum interference is negative at every size, so an H7 hole never guarantees a press fit, and the band widens with diameter. That is why two H7 dowel holes in a plate are not a repeatable location on their own and why a location scheme needs the geometry described below rather than a tighter pin. 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.
Pin	Pin min (mm)	Pin max (mm)	H7 max interference (µm)	H7 min (µm)	N7 max (µm)	N7 min (µm)	P7 max (µm)	P7 min (µm)
Ø4	4.0040	4.0120	12	-8	28	8	32	12
Ø5	5.0040	5.0120	12	-8	28	8	32	12
Ø6	6.0040	6.0120	12	-8	28	8	32	12
Ø8	8.0060	8.0150	15	-9	34	10	39	15
Ø10	10.0060	10.0150	15	-9	34	10	39	15
Ø12	12.0070	12.0180	18	-11	41	12	47	18
Ø16	16.0070	16.0180	18	-11	41	12	47	18
Ø20	20.0080	20.0210	21	-13	49	15	56	22
Ø25	25.0080	25.0210	21	-13	49	15	56	22
Ø30	30.0080	30.0210	21	-13	49	15	56	22

Two pins locate a plate. Three over-constrain it. And the second one is not round

This is the paragraph worth the page, and it is geometry rather than arithmetic. The interference table above tells you how tight each pin is; this table tells you why the tightness is not the problem. Two round pins in two round holes fail not because the fit is wrong but because the scheme demands that two independent centre distances agree to within a few micrometres, and they do not. The diamond pin is the standard answer: it constrains one direction and releases the other, which is exactly one degree of freedom and exactly what is left to remove.
Scheme	What it does	What goes wrong
One round pin	Removes two degrees of freedom in the plane — the plate can still rotate about the pin	Nothing, except that it is not a complete location. It is the right answer when a second feature (a face, a slot) does the rotation.
Two ROUND pins in two round holes	Removes the rotation as well, in principle	Over-constrained in practice. The centre distance in the plate and the centre distance in the fixture each have a tolerance, and the two pins have to bridge whatever the difference turns out to be. With H7 holes and m6 pins there is at most a few micrometres of play per pin, so the plate either binds or the pins bend. The usual description is exactly that: two round pins of tight fit cause binding.
One ROUND pin and one DIAMOND pin	The round pin locates the point; the diamond pin kills the rotation and nothing else. Its flats are relieved, so it constrains the direction perpendicular to the line of centres and leaves the direction ALONG that line free	Nothing, and this is the standard practice for a reason. The centre-distance tolerance is absorbed along the free direction. Orient the diamond so its relieved flats face along the line between the two pins, not across it — turned 90° it does the opposite of what it is for.
Three or more pins	Nothing useful	A plate is fully located in the plane by two pins. A third pin must either be in a slot or be accepting whatever misalignment the tolerance stack hands it. It is the classic over-constraint, and what it costs is repeatability: the plate seats differently depending on the order the pins engage.

Blind holes, and getting the pin out again

None of this is exotic and all of it gets found out late. The vent and the extraction thread are decisions taken at the drawing stage for a few pence; discovered at the first strip-down they cost a scrapped part. This page sizes a part; it does not certify one. Where the answer carries a consequence — a load path, a lifting duty, a pressure boundary, a fastener holding something that can fall — confirm it against the design code that governs the application, and against the manufacturer’s own rating, before relying on it.
Problem	Why	The fix
A pin will not go fully into a blind hole	The air under it has nowhere to go. Press a close-fitting pin into a sealed hole and you are compressing the trapped air, which pushes back — and worse, any oil in the hole is incompressible and will simply stop the pin	Vent it. A small cross-drilling into the side of the hole, a flat ground along the pin, or a vent groove. Dowel pins are often supplied with a small chamfer and, on some forms, a flat for exactly this.
A pin in a blind hole cannot be driven out	There is nothing behind it to push against. This is the situation people discover during the first strip-down, not during design	An extractable pin: ISO 8735 (DIN 7979 D) is the parallel pin with an internal thread, so a puller or a bolt can pull it. The thread sizes go from M4 at Ø6 to M20 at Ø30. Design it in — retro-fitting it means drilling out the pin.
A pin in a THROUGH hole is hard to remove	Usually not a problem, but a press fit on a long engagement can still take real force, and the force to remove is not necessarily the force that went in	Drive it through from the far side. Note that the extraction force can exceed the insertion force, because the pin has burnished the hole on the way in and because any corrosion since has made things worse.
The boss splits when the pin goes in	The hoop stress at the bore of a thin boss. This page computes it, and it is larger than the contact pressure — for a boss twice the pin diameter across it is about 1.7 times the contact pressure at the bore	More material round the hole, less interference, or a shrink fit rather than a press. The outside diameter of the surrounding material is an input on this page precisely because it changes the answer by a factor of two or more.

The pin’s band is fixed; the hole is the decision

A dowel pin’s diameter is not the decision, because the standard has already made it. ISO 8734 specifies m6 for a hardened dowel pin, and at Ø6 that is +4 to +12 µm — you cannot buy a tighter one. The hole is the whole decision, and the answer changes character completely across the classes. An H7 hole is 0 to +12 µm, so the fit ranges from 8 µm LOOSE to 12 µm tight: a transition fit, which may need a press and may drop in, and which you cannot predict from the drawing. N7 is −4 to −16, giving 8 to 28 µm of interference and never a clearance. P7 is −8 to −20 and tighter still. Notice how much of the spread is the pin’s own 8 µm band: at Ø6 the H7 fit is 20 µm wide and 8 of that is the pin.

Those bands are derived here rather than copied, and the derivation had to settle a contradiction. Every deviation on this page comes from the ISO 286 IT grades and the shaft fundamental deviations, with the Δ correction the standard applies to the K, M, N and P hole classes. Two published tables disagreed: one gave N7 at 10–18 mm as −4/−22 and another as −5/−23. The test that settles it is that a hole-basis fit and its shaft-basis twin must give identical limits — N7/h6 has to be the same fit as H7/n6, and P7/h6 the same as H7/p6 — and only one of the two tables satisfies that. The derived m6 band then gets an independent check for free: it reproduces Fuller Fasteners’ printed ISO 8734 pin limits exactly at every size from Ø1 to Ø20. One row is deliberately missing. The 0–3 mm row of the interference hole classes was printed by the source consulted as a copy of the 3–6 mm row, which cannot be right, and no second source was found — so this page starts at Ø4 rather than publish a band it could not verify.

The insertion force follows from the interference, and the assumptions are worth stating plainly because they are doing a lot of work. The contact pressure is the classical Lamé result for a solid pin in a hub, in plane stress, with the same material both sides: p = E·δ·(D₀² − d²)/(2·d·D₀²). This batch did not take it from a textbook — it asserted it against a finite-difference solve of Navier’s radial equation for the hub and the pin separately, with the pressure entering only through the boundary conditions, at five geometries, and showed the residual falling as the mesh was refined. What the formula assumes: elastic behaviour throughout, smooth surfaces, the pin fully entered, no thermal difference, and the same modulus both sides. For a steel pin in an aluminium boss it will overstate the pressure. The force itself is then μ·p·π·d·L, and μ is the least certain number here — 0.10 to 0.15 for clean dry steel on steel, and the force is directly proportional to it. Treat the answer as a size of press, not as a specification. And note what D₀ does: a boss 1.2 d across carries under 40% of the pressure an effectively infinite one does, so a hole near an edge is a different calculation from a hole in the middle of a plate.

Now the part that matters more than any of the arithmetic: two pins locate a plate and three over-constrain it, and the second pin is not round. One round pin removes two degrees of freedom and leaves the rotation. Two round pins in two round holes remove the rotation too — in principle. In practice the centre distance in the plate and the centre distance in the fixture each carry a tolerance, and the two pins have to bridge whatever the difference turns out to be. With m6 pins in H7 holes there are a few micrometres of play per pin and nothing else, so the plate binds or the pins bend. The standard answer is one round pin and one DIAMOND pin: the diamond’s flats are relieved, so it constrains the direction perpendicular to its flats and leaves the direction along them free, which is exactly one degree of freedom and exactly the one left to remove. Orient it so the relieved flats face along the line between the two pins — turned 90° it does the opposite of what it is for. A third pin is worse than useless: it must either sit in a slot or accept whatever the tolerance stack hands it, and what it costs is repeatability, because the plate then seats differently depending on which pin engages first.

Two practical things that get discovered late. A pin pressed into a BLIND hole is compressing the air under it, and any oil in the hole is incompressible and will simply stop it: the hole needs a vent, a cross-drilling, or a flat along the pin. And a pin in a blind hole cannot be driven out from behind, so if it is an interference fit it needs the ISO 8735 / DIN 7979 D extractable form with an internal thread — M4 at Ø6 up to M20 at Ø30 — which is a decision taken on the drawing for a few pence and a scrapped part if it is not. For the clearance hole that a bolt goes through rather than a pin, see the counterbore and countersink calculator, and remember that a bolt in a clearance hole locates nothing at all — that is what the pins are for. For axial restraint rather than location, the retaining ring groove calculator; for grip on a shaft, the set screw dimensions calculator.

Frequently asked questions

Is an m6 dowel pin in an H7 hole a press fit?

No — it is a transition fit, and that surprises people. At Ø6 the pin is +4 to +12 µm and the hole is 0 to +12 µm, so the fit runs from 8 µm loose to 12 µm tight. Whether a particular pin needs a press depends on where that pin and that hole happened to land in their bands, which the drawing does not tell you. H7 is the standard reamed LOCATION hole and it is the right choice for a joint you intend to take apart. For a fit that is always an interference, N7 gives 8 to 28 µm and P7 more again.

What hole tolerance should I use for a dowel pin?

H7 where the joint comes apart and something else retains the pin. N7 or P7 where the pin must stay put and never work loose. Published advice differs on this and the page does not hide it: one dowel-pin reference calls m6 in H7 “a light interference/transition fit: the pin stays put without a retainer” and recommends G7 or F7 for the removable side, which is a different scheme from the conventional H7-locates, N7-presses reading. Rather than pick, this page computes the actual interference range for every class so you can see what each one really gives you.

How much force does it take to press in a dowel pin?

For a Ø6 m6 pin at the tight end of an H7 hole, 12 mm engaged in steel with 24 mm of material round the hole and μ = 0.12, about 5.3 kN — an arbor press. At the loose end of the same fit, zero, because it is a clearance there. Change the hole to N7 and the tight end goes to about 12 kN. The calculation is Lamé for the pressure and μ·p·π·d·L for the force, and the friction coefficient is the least certain part: the answer is proportional to it. Treat it as the size of press you need, not as a number to design a tool against.

Why one round pin and one diamond pin rather than two round ones?

Because two round pins in two round holes over-constrain the plate. The centre distance in the plate and in the fixture each have a tolerance, and the pins have to absorb the difference — but a tight pin in a tight hole has a few micrometres of play and nothing more, so the plate binds or the pins bend. A diamond pin has its flats relieved, so it constrains one direction and leaves the other free: the round pin sets the position, the diamond kills the rotation, and the centre-distance tolerance is absorbed along the free direction. Orient the diamond with its relieved flats along the line between the two pins.

How many pins should I use to locate a plate?

Two, one of them a diamond. A plate is fully located in its plane by two features — the first removes two degrees of freedom, the second removes the rotation — and the third pin has nothing left to constrain. What a third pin actually does is take whatever misalignment the tolerance stack gives it, which costs you repeatability: the plate seats differently depending on which pin engages first. If you need more than two for load-sharing rather than location, put the extras in slots.

Do I need a vent in a blind dowel hole?

Yes, in practice. A close-fitting pin pressed into a sealed hole is compressing the air underneath it, and if there is any oil in the hole — from machining, from handling — it is incompressible and the pin simply stops. A small cross-drilling into the side of the hole, a flat ground along the pin, or a vent groove all solve it. And plan the removal at the same time: a pin in a blind hole cannot be driven out from behind, so an interference fit there wants the ISO 8735 / DIN 7979 D form with an internal thread, M4 at Ø6 up to M20 at Ø30.

Where do these tolerance numbers come from?

They are derived, not copied. Every band on this page is built from the ISO 286 IT grades and the shaft fundamental deviations, with the Δ correction the standard applies to the K, M, N and P hole classes. Two checks back it up: the derived bands reproduce the published hole and shaft tables, and they satisfy the requirement that a hole-basis fit and its shaft-basis twin give identical limits — N7/h6 must equal H7/n6 — which is what settled a contradiction between two published sources. The m6 pin band also reproduces Fuller Fasteners’ printed ISO 8734 pin limits exactly at every size from Ø1 to Ø20. One row is deliberately absent: the 0–3 mm interference classes, because the source printed that row as a copy of the next one and no second source was found.

Related calculators

References

ISO 286-1 and ISO 286-2, Geometrical product specifications (GPS) — ISO code system for tolerances on linear sizes. Cited by number. The bands used here are DERIVED from the IT grades and the shaft fundamental deviations rather than copied, and checked two ways: against RoyMech’s published hole and shaft tables, and against the requirement that a hole-basis fit and its shaft-basis twin give identical limits (N7/h6 must equal H7/n6, P7/h6 must equal H7/p6). That second check settled a straight contradiction: a second published table gave N7 at 10–18 mm as −4/−22 where RoyMech gives −5/−23, and only RoyMech’s value satisfies the equivalence. The 0–3 mm row of the K, M, N and P hole bands is NOT used on this page: the source printed it as a copy of the 3–6 mm row, which cannot be right, and no second source for it was obtained.
ISO 8734, Parallel pins, of hardened steel and martensitic stainless steel (dowel pins), and ISO 2338 for the unhardened parallel pin. Dimensions from Fuller Fasteners’ ISO 8734 specification page, whose printed m6 limits from Ø1 to Ø20 reproduce the derived m6 band exactly at every size — which is the independent check on the whole ISO 286 derivation. ISO 8735 (DIN 7979 D) is the extractable form with an internal thread, needed for a blind hole.
EKINSUN. Dowel Pin Size Chart — ISO 8734 / DIN 6325 m6 Tolerances. Recommends H7 for the retaining side — “an m6 pin in an H7 hole provides a light interference/transition fit: the pin stays put without a retainer” — and G7 or F7 for the removable side. That is a DIFFERENT recommendation from the one that pairs H7 with a location fit and N7 or P7 with a press fit, and this page prints the computed fit for every class rather than choosing between them.
On the round-pin / diamond-pin pair: the standard reasoning is that two round pins in two round holes over-constrain the plate, so that “using two round pins of tight fit caus[es] binding” once the centre-distance tolerances stack, while a diamond pin “with its relieved flats, constrains the remaining needed degree perpendicular to the flat but allows slight movement along the flat”. Collected from jig-and-fixture teaching material; the geometry itself is computed on this page rather than asserted.
The press-fit relation used here is the classical Lamé thick-cylinder result for a solid pin in a hub, in PLANE STRESS, as machine-design texts state it: with the same material either side the contact pressure is p = E·δ·(D₀² − d²) / (2·d·D₀²) for a diametral interference δ. It is not taken from a table: this batch asserted it against a finite-difference solve of Navier’s radial equation for the hub and the pin separately, with the pressure entering only through the boundary conditions, at five geometries, and showed the residual falling with the mesh. Engineers Edge’s Pin / Shaft Press Fit Force Equation was consulted and is NOT used: it is an empirical tons-per-cubic-inch table keyed to a machine-steel shaft in a cast-iron hub with the hub diameter fixed at twice the pin’s, which is a narrower case than this page covers.