Screw Thread

When does a screw hold its load without input?

Worked example — Screw Thread

Lead angle	2.73°
Ideal mechanical advantage	20.94
Friction angle	8.53°
Efficiency	24.0%
Input effort	199.2 N

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The holding condition contains no force 🖖

Unroll one turn of the thread and you have a ramp. It rises by the pitch while travelling once round the shaft, so its slope is tan λ = p / 2πr. Friction has a slope of its own: tilt a plank until a block on it starts to go, and the angle you measured is φ = arctan μ. The thread holds whenever λ ≤ φ.

Now look at what is missing from that comparison. The weight of the load is on neither side of it. Gravity pulls the load down the thread and presses it into the flanks in the same proportion, so the two cancel exactly, and a bolt that self-locks at 200 N still self-locks at 20 kN. Drag the load slider from 100 N all the way to 5,000 N and the verdict never moves.

Fifty percent is a ceiling, and nothing reaches it 🖖

Over the whole range in which a screw holds, efficiency rises with the lead angle, and holding is what caps the lead angle at φ. So the most efficient screw that still holds is the one sitting exactly on the boundary, where λ = φ turns η = tan λ / tan(λ + φ) into tan φ / tan 2φ. That simplifies to (1 − μ²)/2.

One half is what it approaches as friction vanishes, and a frictionless thread holds nothing, so no screw ever arrives there. At μ = 0.15 the ceiling is 48.9%. Load the 5 mm lead screw and push the pitch to 6.0 mm: it is still holding, at 49.2%. One more step, to 6.5 mm, and it back-drives at 51.1%. Across all 116,280 combinations of radius, pitch and friction the sliders reach, the best any self-locking screw manages is 49.86%.

A coarser pitch is always harder to turn 🖖

Each turn advances the nut by exactly one pitch, so the ideal advantage is the circumference over the pitch and nothing else. Double the M10's pitch from 1.5 mm to 3 mm and the advantage halves, from 20.94 to 10.47.

Efficiency moves the other way, from 24.0% to 38.3%, and that does not rescue the effort. What your hand pushes against is the product of the two, and the product simplifies to cot(λ + φ), which only ever shrinks as the pitch grows. The effort goes from 199.2 N to 249.1 N, and across all 116,280 slider settings there is not one where a coarser pitch is easier to turn. Fewer turns is the whole of what you get for it.

Problem solved in full

A thousand newtons held by a bolt you can tighten with two fingers 6 steps
Load the M10 coarse preset: a 1.5 mm pitch on a shaft of 5 mm radius, μ = 0.15, holding 1,000 N. The panel says 199.2 N of effort at 24.0% efficiency. Work out where the 199.2 N comes from, turn it into the pull your hand actually feels on a spanner, and then decide whether 24.0% is a number to mind.
1. Start by unrolling one turn of the thread. It rises by the pitch and travels once round the shaft, so it is a ramp whose slope is the pitch over the circumference. Take the arctangent of that slope and you have the lead angle.
  
  $tan λ = \frac{p}{2 π r} = \frac{1.5}{2 π \times 5} = 0.04775 ⟹ λ = 2.73 4^{\circ}$
2. Friction gets the same treatment, and that is the move which makes everything after it easy. μ = 0.15 is the tangent of the angle at which a block on a tilted plank starts to go, so friction is 8.531° of slope. The thread is shallower than that, which is why the bolt is still tight in the morning.
  
  $ϕ = tan^{- 1} μ = tan^{- 1} 0.15 = 8.53 1^{\circ} > 2.73 4^{\circ} = λ$
3. Pushing a load up a rough slope takes its weight times the tangent of the two angles added together: the ramp you are climbing, plus the friction you are dragging. A screw is that ramp wrapped round a shaft, so the same expression gives the effort, measured as a force acting at the thread radius.
  
  $E = W tan (λ + ϕ) = 1000 tan 11.26 4^{\circ} = 199.2 N$
4. The panel arrives from the other direction, dividing the load by the ideal advantage and then by the efficiency, and lands on the same 199.2 N. They agree for a reason worth keeping: the ideal advantage is 1/tan λ and the efficiency carries tan λ on top, so the lead angle cancels and any screw's real advantage is simply cot(λ + φ).
  
  $\frac{W}{M A _{i} η} = \frac{1000}{20.94 \times 0.2397} = 199.2 N, M A_{i} η = \frac{1}{tan λ} \cdot \frac{tan λ}{tan ( λ + ϕ )} = cot (λ + ϕ)$
5. Now the step the page does not take for you. That 199.2 N acts 5 mm from the axis, so what you have to supply is a torque, and it is a small one.
  
  $T = E r = 199.2 \times 0.005 = 0.996 N m$
6. A spanner on an M10 head is about 150 mm long and your hand is at the far end of it. The spanner is a second machine in series with the thread, and it is doing most of the lifting.
  
  $F = \frac{T}{L} = \frac{0.996}{0.15} = 6.64 N, \frac{W}{F} = \frac{1000}{6.64} = 151$
Answer

199.2 N at the thread, 0.996 N m of torque, and 6.64 N at the end of the spanner. Under 700 grams of pull, holding a hundred kilograms in the air. The thread's ideal advantage is 20.94 and the spanner multiplies it by another 30, for 628 in a world without friction; take 24.0% of that and you get the 151 the arithmetic above actually delivered. That is where the efficiency finally shows up as something you can feel.

So, is 24.0% worth minding? Count what it costs here. One full turn puts 6.26 J into the thread and raises the load 1.5 mm, which is 1.5 J of work; the other 4.76 J goes into the flanks as heat. Six and two-thirds turns raise the load 10 mm and lose 32 J, less than a second of a person's arm, and nobody in the history of bolts has noticed it.

Put the identical thread under a motor driving a feed screw eight hours a day and the same 76% sets the size of the motor, the running temperature of the nut and how fast both wear out. One number, two verdicts, and what decides between them is how often the machine runs rather than anything about the screw.

One thing this page leaves out. A real bolt also drags its head face against the joint, and on plain steel that term takes roughly as much torque again. 0.996 N m is the thread's share, not a tightening spec.

$T = W r tan (λ + ϕ)$

Learning path

Simple machines: trading force for distance

By the 42Math team · Updated 2026-09-04

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

M10 coarse - A 1.5 mm pitch on a 5 mm radius cuts a 2.73° helix, against a friction angle of 8.53°, so the thread holds its 1,000 N with the spanner off it. That margin is what the efficiency pays for: 24.0%, and 199.2 N of push at the thread to raise the load at all.
5 mm lead screw - Pitch of 5 mm on an 8 mm radius: the helix reaches 5.68° and efficiency nearly doubles, to 44.8%. It still holds, but with barely a degree of margin where the M10 had almost six, which is why a lead screw can be left standing under load and a ball screw needs a brake.
fast pitch - 12 mm of pitch on a 5 mm radius puts the helix at 20.91°, far past the 4.57° friction angle, so the load turns the screw back down the moment you let go. Efficiency is 80.2%, more than three times the M10's, and the push is still 285.9 N, because the ideal advantage has collapsed to 2.62.