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Sine bar principle, setting and error

Short answer

Sine bar sets an angle with stack height h = L sinθ. Limitations: inaccurate above 45°. For L=200 mm, θ=28°30', h=95.432 mm. With 0.002 mm height error, angular error is 2.35 s at 28.5°, 4.13 s at 60°; larger error at steeper angles.

(a) Explain the principle of the sine bar and state its limitations. (b) Calculate the height of slip gauge stack required to set a 200 mm centre-distance sine bar to an angle of 28∘30′28^\circ 30'. (c) If the stack has an error of 0.002 mm, find the resulting angular error at 28∘30′28^\circ 30' and at 60∘60^\circ, and comment.

(a) Principle

A sine bar is a hardened steel bar with two equal precision rollers fixed at a known centre distance LL (100, 200 or 300 mm). When one roller rests on slip gauges of height hh and the other on the surface plate, the bar is inclined at angle θ\theta to the plate:

        ________  <- sine bar
      /   L    /(roller)
    /  theta  /|
   /_________/_| h  (slip gauges)
   surface plate
sin⁡θ=hL\sin\theta = \frac{h}{L}

Used to set a component at a known angle, or to measure an unknown angle by adjusting hh until the work surface is level (checked by a dial indicator).

Limitations

  • Accuracy falls at large angles (above 45∘45^\circ) because cos⁡θ\cos\theta becomes small (see (c)); not suitable above 60∘60^\circ.
  • Needs a surface plate and a good set of slip gauges; cannot measure steep tapers directly.
  • Errors in centre distance, roller diameters and parallelism of rollers directly give angle error.
  • Only conical or flat surfaces that can be set against the bar can be checked; tedious for large work.

(b) Stack height

h=Lsin⁡θ=200sin⁡(28.5∘)=200×0.47716=95.432 mmh = L\sin\theta = 200\sin(28.5^\circ) = 200 \times 0.47716 = 95.432\ \text{mm}

Answer: slip gauge height = 95.432 mm.

(c) Error

Differentiating h=Lsin⁡θh = L\sin\theta gives δh=Lcos⁡θ δθ\delta h = L\cos\theta\,\delta\theta, so

δθ=δhLcos⁡θ\delta\theta = \frac{\delta h}{L\cos\theta}
Anglecos⁡θ\cos\thetaδθ\delta\theta (rad)δθ\delta\theta
28.5∘28.5^\circ0.87880.002200(0.8788)=1.138×10−5\dfrac{0.002}{200(0.8788)} = 1.138\times10^{-5}2.35 s
60∘60^\circ0.50.002200(0.5)=2.0×10−5\dfrac{0.002}{200(0.5)} = 2.0\times10^{-5}4.13 s

The same stack error causes a larger angular error at steeper angles. Hence the sine bar is best used below 45∘45^\circ, and for large angles the complement is measured.

✓ Answer checked by Claude Opus 5.5 (high effort)

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