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Chapter 5 · 3 hours

Angular and Taper Measurement

Practice questions

Practice questions and answers

3 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 8 marks

(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.

Answer

(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.

  • Practice · 8 marks

Explain with neat sketches the principle and working of an autocollimator. Describe the angle dekkor and state where each is used.

Answer

Autocollimator

An autocollimator is an optical instrument that measures very small angular tilts (down to 0.1 s) of a reflecting surface at a distance. It combines a collimator and a telescope.

Principle: A target (cross-wire) at the focus of an objective lens gives a parallel beam. If the beam strikes a plane mirror exactly normal to it, the beam returns on itself and the reflected image falls on the target. If the mirror tilts by an angle θ\theta, the reflected beam turns by 2θ2\theta and the image shifts by

x=2 θ fx = 2\,\theta\, f

where ff is the focal length of the objective.

 lamp--> [cross-wire] --|beam splitter|--> objective --->| mirror
                               |            (f)  parallel |
                           eyepiece  <---- reflected beam-|
                           micrometer

Light from the lamp illuminates the target, passes through the beam splitter and objective, goes to the mirror, returns and forms an image of the target in the eyepiece. The shift is read by a micrometer eyepiece graduated in seconds of arc.

Uses: checking straightness and flatness of machine beds and surface plates, squareness, parallelism of slideways, angle of prisms and polygons, and calibration of angle gauges and indexing tables. The mirror is fixed on a carriage moved along the bed.

Angle dekkor

It is also an autocollimator but with a fixed illuminated scale and a fixed graticule with a cross-line in the eyepiece. The reflected image of the scale is seen superposed on the fixed graticule, and the reading gives the tilt directly in minutes. Angle dekkor measures angles up to about ±1∘\pm 1^\circ with a least count of 1 minute (and 0.5 min estimated).

For measuring an angle of a component, slip gauges and a sine bar are used to set the working face to a known angle, and the dekkor reading shows the difference from the nominal angle (e.g. checking angle gauges, taper plugs, and bevel angles).

AutocollimatorAngle dekkor
TargetCross-wireScale and fixed graticule
RangeVery small anglesUp to about ±1∘\pm 1^\circ
Accuracy0.1 to 1 sAbout 1 minute
  • Practice · 6 marks

(a) What is a spirit level? Explain how its sensitivity is expressed. (b) A level has a vial whose divisions are 2 mm apart, and a division corresponds to 10 seconds of arc. Calculate the radius of curvature of the vial. (c) This level is placed on a 500 mm long base on a machine slide and the bubble moves 3 divisions. Find the difference in height between the ends of the base.

Answer

(a) Spirit level

A spirit (bubble) level is a sealed glass vial, slightly curved, partly filled with alcohol or ether so that a bubble remains. The bubble always rises to the highest point. If the base is horizontal the bubble is at the centre; if the base tilts by a small angle the bubble moves along the graduated scale. It is used to check the levelness of machine tools, surface plates and slideways.

Sensitivity is the tilt of the base per division of bubble movement. It is expressed as an angle (seconds of arc per division) or as slope (mm per metre per division). For a vial of radius RR and division length pp:

θ=pR  ⇒  R=pθ\theta = \frac{p}{R} \;\Rightarrow\; R = \frac{p}{\theta}

A larger RR gives greater sensitivity (small tilt moves the bubble far) but a longer settling time.

(b) Radius

θ=10′′=10×4.848×10−6=4.848×10−5\theta = 10'' = 10 \times 4.848\times10^{-6} = 4.848\times10^{-5} rad

R=2 mm4.848×10−5=41 253 mm≈41.3 mR = \frac{2\ \text{mm}}{4.848\times10^{-5}} = 41\,253\ \text{mm} \approx 41.3\ \text{m}

(c) Height difference

Total tilt =3×10′′=30′′=1.454×10−4= 3 \times 10'' = 30'' = 1.454\times10^{-4} rad.

Δh=500×1.454×10−4=0.0727 mm\Delta h = 500 \times 1.454\times10^{-4} = 0.0727\ \text{mm}

(a slope of 0.145 mm per metre).

Answer: R=41.3R = 41.3 m; end-to-end height difference = 0.073 mm over 500 mm.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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