PROBLEMS
Grade 00 This set is kept in the standards room and is used for inspection/calibration of high precision only. It is also used to check the accuracy of the workshop and grade 1 slip gauges
4.10 NUMERICAL EXAMPLES
Example 4.1 In a vernier calliper, the main scale reads in millimetres with a least count of 0.1 mm.
Ten divisions on the vernier correspond to nine divisions of the main scale. Determine the least count of the calliper.
Bubble scale
Base plate
Brook level
Standard gauge
Test gauge
Standard gauge
Fig. 4.46 Brook-level comparator
Solution
1 MSD = 0.1 mm 10 VSD = 9 MSD
Therefore, least count = 1 MSD − 1 VSD = 1 MSD − 9/10 MSD = 1/10 (0.1) mm = 0.01 mm Example 4.2 The main scale in a vernier instrument is graduated in millimetres, with the smallest
division being 1 mm. Ten divisions on the vernier scale correspond to nine divisions on the main scale. Answer the following questions:
(a) Is the vernier scale a forward vernier or a backward vernier?
(b) What is the least count of the instrument?
(c) If the main scale reads 13 mm and the fifth division on the vernier scale coincides with a division on the main scale, what is the value of the dimension being measured?
Solution
(a) Since the length of one MSD is higher than the length of one VSD, the scale is a forward vernier scale.
(b) Least count = 1 MSD − 1 VSD = (1 − 9/10) MSD = 0.1 mm (c) The dimension is = 13 + 5 ´ 0.1 = 13.5 mm
Example 4.3 The barrel scale of a vernier micrometer has graduations in millimetres, with each division measuring 1 mm. The thimble scale has 50 equal divisions, and one complete rotation of the thimble moves it by one division over the barrel scale. If five divisions on the vernier scale attached to the micrometer correspond to six divisions on the thimble, calculate the least count of the instrument.
Solution
If one complete rotation of the thimble moves it by 1 mm on the barrel scale, the least count of the micrometer scale is 1/50 = 0.02 mm.
Since five divisions on the vernier scale correspond to six divisions on the thimble, the least count of the vernier scale is equal to 0.02/5 = 0.001 mm.
Example 4.4 Slip gauges have to be built up to a height of 26.125 mm using the 103-gauge set. Give the selection of slip gauges if wear blocks of 1.5 mm thickness are to be used at the bottom and top of the stack.
Solution
We need to deduct 3 mm from 26.125 mm, that is, 23.125 mm, and select the combination for this height. The selection of gauges in this example will be 1.005, 1.12, 21, and the two wear gauges.
Example 4.5 It is required to build a 0.95 mm height for some calibration purpose. This dimension is less than the least gauge in the 103-gauge set, that is, the
1.005 mm gauge. How can you meet the requirement?
Solution
In such cases, it is necessary to use a combination that is slightly greater than 0.95 mm and subtract a second combination to get the required result.
Accordingly, we can first choose the 1.005 mm gauge, which is in excess of 0.95 mm by 0.055 mm. Addition of 0.055 to 1.005 mm gives a value of 1.06 mm. Now,
0.95
1.005 1.06
Surface plate
Fig. 4.47 Indirect combination of slip gauges
LINEAR MEASUREMENT 115
for the second combination, select the 1.06 mm gauge. If the two gauges are kept on a surface plate, their difference will give the required value of 0.95 mm, as shown in Fig. 4.47. This can be measured by using a suitable comparator of high accuracy.
A QUICK OVERVIEW
• The foundation for all dimensional measurements is the ‘datum plane’, the most important one being the surface plate. A surface plate is a hard, solid, and horizontal flat plate, which is used as the reference plane for precision inspection, marking out, and precision tooling set-up.
V-blocks are extensively used for inspection of jobs with circular cross sections.
• Most of the basic linear measurements are derived from the simple steel rule. However, scaled instruments provide the most convenient means for linear measurement in contrast to steel rules, which are limited by accuracy, resolution, and mechanism to ensure ease of measurement.
• Callipers are the original transfer instrument to transfer such measurements onto a rule. They ensure physical duplication of the separation of reference point and measured point with a high degree of accuracy.
• The principle of vernier measurement is basic to metrology and the use of vernier instruments comprising vernier calliper, vernier depth gauge, vernier height gauge, vernier micrometers, etc., is still widespread in the industry. However, digital read-out instruments and dial callipers are rapidly replacing vernier instruments. The
advantages of vernier instruments are their long measuring range and convenience. Their chief disadvantages are reliance on the ‘feel’ of the user and susceptibility to misalignment.
• A micrometer can provide better least counts and accuracy than a vernier calliper. Better accuracy results from the fact that the line of measurement is in line with the axis of the instrument, unlike in a vernier calliper, which does not conform to this condition. This fact is best explained by Abbe’s principle, which states that maximum accuracy may be obtained only when the standard is in line with the axis of the part being measured.
• A floating carriage micrometer, sometimes referred to as an effective diameter-measuring micrometer, is an instrument that is used for accurate measurement of ‘thread plug gauges’.
Gauge dimensions such as outside diameter, pitch diameter, and root diameter are measured with the help of this instrument.
• Slip gauges provide the most practical standard for linear measurement. They are readily available and a small number of them, in a standard set, can be used to build thousands of dimensions. The errors involved with them are known and therefore predictable.
MULTIPLE-CHOICE QUESTIONS
1. The foundation for all dimensional measurements is the
(a) datum plane (c) datum point
(b) datum line (d) geometry of work piece 2. As per the practice devised by Sir Whitworth,
the surface plates are fabricated in sets of
(a) two (c) four
(b) three (d) ten
3. Granite surface plates are free from burrs or
protrusions because of
(a) lower coefficient of expansion (b) light weight
(c) very fine grain structure (d) none of these
4. The preferred instrument for measuring holes, grooves, and recesses is
(a) plain scale (c) slip gauge (b) vernier calliper (d) depth gauge
REVIEW QUESTIONS
1. Why are callipers and dividers called dimension transfer instruments?
2. List the important considerations for the design of linear measurement instruments.
3. When do you prefer cast iron surface plates over granite surface plates and vice versa?
4. What is the main purpose of a ‘V-block’? What is the basis for their classification?
5. How is a scale different from a rule?
6. What are the common errors associated with measurements using a steel rule?
7. How do you measure the depth of a hole or recess
using a depth gauge? What are the limitations of this instrument?
8. How do you employ a combination set to measure the following?
(a) Height
(b) Angle of a surface (c) Centre of a bar stock
9. Give a classification of callipers. Illustrate with sketches the different types of callipers.
10. Differentiate using sketches the measurements of inside and outside dimensions using a vernier calliper.
5. The centre head attachment on a combination set is used to
(a) measure angles
(b) measure height and depth (c) measure distance between centres
(d) locate the centre of a bar stock or a circular job
6. A scribing tool comprising one divider leg and one calliper leg is known by the name
(a) hermaphrodite calliper (b) hermasonic calliper (c) transfer calliper (d) firm joint calliper
7. vernier calliper has jaws on both sides for making measurements and the jaws have knife edge faces for marking purpose.
(a) Type A (c) Type C (b) Type B (d) Type D
8. The degree to which an instrument conforms to the law determines its inherent accuracy.
(a) Moore’s law (c) Johnson’s law (b) Abbe’s law (d) Mikelson’s law 9. Micrometer measuring faces are tipped with
to prevent rapid wear.
(a) aluminium (c) molybdenum oxide (b) chromium (d) tungsten carbide 10. Gauge blocks owe their existence to the
pioneering work done by
(a) Johansson (c) Taylor (b) Abbe (d) None of these 11. Which of the following best describes wringing
of slip gauges?
(a) Squeezing the oil out from between two gauges
(b) Causing blocks to adhere by molecular attraction
(c) Effect of atmospheric pressure (d) Process of removing minute burrs
12. Which of the following is preferred for selecting a combination of slip gauges?
(a) Selective halving
(b) Writing out all combinations and selecting the closest one
(c) Averaging
(d) Determining the fewest number of blocks 13. Which of the following statements with reference
to a micrometer is false?
(a) It is not as precise as a vernier calliper.
(b) It can be used for end measurement only.
(c) It has no parallax error.
(d) It has a shorter measuring range compared to a vernier calliper.
14. Which of the following part features is easiest to measure with a vernier calliper?
(a) Large distances between outside planes (b) Heights from a surface plate
(c) Cylindrical features (d) Concave features
15. Slip gauges marked with the letter ‘P’ are (a) wear gauges
(b) preferred gauges (c) discarded gauges (d) inspection grade gauges
LINEAR MEASUREMENT 117
PROBLEMS
1. In a vernier calliper, the main scale reads in millimetres with a least count of 0.5 mm.
Twenty divisions on the vernier correspond to 19 divisions on the main scale. Determine the least count of the calliper.
2. The main scale in a vernier instrument is graduated in millimetres, with the smallest division being 1 mm. Ten divisions on the vernier scale correspond to 11 divisions on the main scale. Answer the following questions:
(a) Is the vernier scale a forward vernier or backward vernier?
(b) What is the least count of the instrument?
(c) If the main scale reads 12 mm and the 10th division on the vernier scale coincides with a division on the main scale, what is the value of the dimension being measured?
3. The barrel scale of a vernier micrometer has graduations in millimetres, with each division measuring 1 mm. The thimble scale has 100 equal divisions, and one complete rotation of
the thimble moves it by one division over the barrel scale. If 10 divisions on the vernier scale attached to the micrometer corresponds to 11 divisions on the thimble, calculate the least count of the instrument.
4. A selection of slip gauges is required to build a height of 48.155 mm. Propose the best combination of gauges using the 103-gauge set.
5. It is required to set a height difference of 0.45 mm to set the sine bar to a known angle.
How will you select the combination of slip gauges to set the sine bar using a set of 56 slip gauges? (Hint: Since the required value is less than the dimension of the minimum available gauge, it is required to select combinations on either sides of the sine bar, so that the difference in height should give the required value.) 6. For the case described in problem 4, what will be
the combination of slip gauges if two protection gauges of 1.5 mm thickness need to be used?
11. Discuss the guidelines to be followed for the proper use of a vernier calliper.
12. Explain the working principle of a dial calliper.
13. What are the major benefits of an electronic digital calliper?
14. Explain how a micrometer conforms to Abbe’s law.
15. Write a note on the types of micrometers.
16. Differentiate between an inside micrometer and
an inside micrometer calliper.
17. What is the application of a floating carriage micrometer?
18. Why are slip gauges called ‘Johansson gauges’?
19. Explain the phenomenon involved in ‘wringing’
of slip gauges.
20. What is the significance of calibration of slip gauges?
ANSWERS
Multiple-choice Questions
1. (a) 2. (b) 3. (c) 4. (d) 5. (d) 6. (a) 7. (c) 8. (b) 9. (d) 10. (a) 11. (b) 12. (d) 13. (a) 14. (c) 15. (a)
Problems
1. 0.025 mm 2. (a) Backward vernier scale (b) 0.1 mm (c) 13 mm
3. 0.01 mm 4. 1.005 + 1.15 + 21 + 25
5. First combination: 1 + 1.1 Second combination: 1.05 + 1.5
6. First combination: 1.5 + 1 + 1.1 + 1.5 Second combination: 1.5 + 1.05 + 1.5 + 1.5