toolfoundry Mechanical Engineering

Mechanical Engineering

Shaft Power, Torque & Speed Calculator

P = T·ω solved either way — plus efficiency, service factor, torsional stress, twist and minimum shaft diameter.

Enter the other two; the field you are solving for is ignored

Mechanical power transmitted by the shaft (output side)

Gearbox, belt, coupling and bearing losses between the prime mover and this shaft. A single helical gear stage is ~97–98%, a worm set can be 50–90%

Multiplier on torque for load character and duty — roughly 1.0 uniform, 1.25–1.5 moderate shock, 1.75–2.5 heavy shock or reversing. Use the value from your drive standard

Used only when the section is hollow; ignored for a solid shaft

Length of shaft over which the torque acts, for the angle-of-twist result

≈79 300 MPa (79.3 GPa) for steel, ≈26 000 MPa for aluminium alloy, ≈41 000 MPa for grey cast iron

Your design value, not a material limit — commercial steel shafting with a keyway is often held near 40 MPa; take the figure from the code you are designing to

Results
Shaft power — kW
Shaft torque — N·m
Rotational speed — rpm
Shaft power — hp
Input power required at this efficiency — kW
Drive rating (input power × service factor) — kW
Design torque (torque × service factor) — N·m
Torsional shear stress at design torque — MPa
Angle of twist over the length — °
Minimum diameter for the allowable stress — mm

Method reviewed 2026-08-09

Method

Last reviewed

What this calculator does

Given any two of power, torque and speed, it returns the third — then it keeps going and tells you whether the shaft can actually carry it. Mechanical efficiency turns the shaft power into the input power the prime mover has to supply; a service factor turns the arithmetic torque into the design torque you size against. From that it works out the torsional shear stress in a solid or hollow circular section, the angle of twist over the length you give it, and the minimum diameter your allowable shear stress permits.

The three-way solve is the part everyone comes for. The stress, twist and diameter results are the part that decides whether the drive works.

The formula

Power is torque times angular velocity:

P = T · ω          ω = 2πN/60

P in watts, T in N·m, ω in rad/s, N in rev/min. Substitute ω, divide by 1000 for kilowatts:

P(kW) = T(N·m) · N(rpm) · 2π / (60 × 1000) = T(N·m) · N(rpm) / 9549.30

because 60 000 / 2π = 9549.2966.

Where 5252 comes from. Exactly the same substitution in imperial units. One mechanical horsepower is defined as 33 000 lbf·ft per minute. A shaft turning at N rpm under torque T lbf·ft does T · 2πN lbf·ft of work per minute, so:

P(hp) = T(lbf·ft) · 2πN / 33 000 = T(lbf·ft) · N(rpm) / 5252.11

because 33 000 / 2π = 5252.11. There is nothing empirical in 5252 or 9549 — both are 2π folded into a unit conversion. It also explains the famous crossover on a dyno plot: at 5252 rpm the horsepower and lbf·ft curves must intersect, because at that speed the two numbers are equal by construction.

Torsion of a circular shaft:

τ = T · r / J
J = π d⁴ / 32                 solid
J = π (D⁴ − d_i⁴) / 32        hollow
θ = T · L / (G · J)           angle of twist, radians

τ shear stress at radius r, maximum at the outer surface; J polar second moment of area; G shear modulus; L twisted length. Inverting the solid-shaft form gives the sizing equation:

τ = 16 T / (π d³)   →   d_min = ∛( 16 T / (π τ_allow) )

and for a hollow shaft at a fixed bore ratio k = d_i/D, d_min = ∛( 16T / (π τ_allow (1 − k⁴)) ).

Reading the result

Input power vs shaft power. The shaft power is what the shaft delivers. The input power is what the motor must produce to get it there after gearbox, belt and bearing losses — always larger. The drive rating multiplies that by the service factor, and that is the number you take to a motor catalogue.

Design torque. Stress, twist and minimum diameter are all evaluated at torque × service factor, because that is the torque the shaft is actually sized for. With a service factor of 1.0 the design torque is the arithmetic torque and nothing changes.

Stress and twist together. Short shafts are usually strength-limited and long ones stiffness-limited. A common rule for transmission shafting is to hold twist to about 0.25° per metre; beyond that, torsional wind-up starts to affect positioning accuracy and pushes the torsional natural frequency down toward excitation orders. The tool flags it when you cross that line.

Hollow shafts. A bore at half the outside diameter removes 25% of the cross-sectional area but only 6.25% of the polar moment (1 − 0.5⁴ = 0.9375). That is the whole argument for hollow shafting, and the calculator will show it directly if you switch the section type.

What this does not cover. Torsion alone. Real shafts also see bending from gears, sprockets and overhung loads, axial thrust, stress concentration at keyways, splines, shoulders and cross-holes, and fatigue under reversed bending — any of which can govern. A keyway alone can cut the effective torsional strength by 25%. Treat the minimum diameter here as a floor, never as a design.

Typical values

Worked example

A 22 kW-class drive: measured shaft torque 140 N·m at 1470 rpm (a 4-pole motor at full-load slip), drive efficiency 95%, service factor 1.25, on a solid 60 mm steel shaft, 1.5 m between the driving and driven elements, G = 79 300 MPa, allowable shear 40 MPa.

P = 140 × 1470 / 9549.30 = 205 800 / 9549.30 = 21.55 kW  (28.90 hp)
P_input = 21.55 / 0.95                       = 22.69 kW
drive rating = 22.69 × 1.25                  = 28.36 kW
T_design = 140 × 1.25                        = 175 N·m

Torsion at the design torque:

J = π × 60⁴ / 32 = π × 405 000       = 1 272 345 mm⁴
τ = 16 × 175 000 / (π × 216 000)     = 4.13 MPa
θ = 175 000 × 1500 / (79 300 × 1 272 345) = 2.602 × 10⁻³ rad = 0.149°

That is 0.149° over 1.5 m, i.e. 0.099°/m — comfortably inside the 0.25°/m guideline. And the diameter torsion alone would require:

d_min = ∛(16 × 175 000 / (π × 40)) = ∛22 282 = 28.14 mm

So the 60 mm shaft is carrying about 10% of its allowable torsional stress. It is not sized by torque — it is sized by bending, by bearing bore, or by the coupling that has to fit on it. That is the normal outcome for a machine shaft, and it is worth knowing before you spend an afternoon optimising a torsional stress that was never the constraint.

FAQ

Where does 5252 come from? 33 000 / 2π. One horsepower is defined as 33 000 lbf·ft per minute, and a shaft does 2πN foot-pounds per minute per lbf·ft of torque. It is a unit conversion, not a material or empirical constant. Its metric counterpart is 60 000 / 2π = 9549.30 for kW, N·m and rpm.

Why do horsepower and torque curves always cross at 5252 rpm? Because hp = T·N/5252, so at N = 5252 the two numbers are numerically equal. Every dyno plot drawn with hp and lbf·ft on the same axis crosses there. It tells you nothing about the engine.

Should I use the shaft power or the input power to size the motor? Input power, then multiply by the service factor. The shaft power is what the load needs; the motor also has to cover the transmission losses and the duty severity.

Is the minimum diameter safe to build to? No. It is the diameter at which torsional shear alone reaches the allowable stress. Add bending, keyway and fillet stress concentration, fatigue and deflection limits before you commit to a size — every one of them makes the required diameter larger.

Does the hollow-shaft option handle thin walls? The torsion formulae are exact for any circular annulus, so the numbers are right. But a thin-walled tube can fail by local buckling well below its torsional yield stress, and that check is not in here.

Indicative figures for preliminary sizing and checking. Torsion is only one of several actions on a real shaft; verify against the design code, drive standard and fatigue criteria applicable to your equipment.

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