Measurements : Appendix I

Measurements : Appendix IThe VERNIER Scale A Vernier scale is a small, moveable scale placed next to the main scale of a measuring instrument. It is named after its inventor, Pierre Vernier (1580 - 1637). It allows us to make measurements to a precision of a small fraction of the smallest division on the main scale of the instrument. (In the first example below the "small fraction" is one tenth.) Vernier scales are found on many instruments, for example, spectroscopes, supports for astronomical telescopes etc. One specific example, the Vernier calliper, is considered below.

Using a Vernier ScaleFigure 1 shows a Vernier scale reading zero. Notice that 10 divisions of the Vernier scale have the same length as 9 divisions of the main scale.

figure 1

In the following examples we will assume that the smallest division on the main scale is 1mm so the divisions on the Vernier scale are 09mm each. The position of the zero of the Vernier scale tells us the number of cm and mm in our measurement. For example, in figure 2, the reading is a little over 12cm.

figure 2

To find a more precise reading, consider figure 3 (which is a magnified view of part of figure 2).

figure 3

We are, in effect, trying to find the distance, x.

To find x, find the mark on the Vernier scale which most nearly coincides with a mark on the main scale. In figure 3 it is obviously the third mark.

Now, it is clear that ............x = d - d

Remembering that each division on the main scale is 1mm and that each division on the Vernier scale is 09mm, we have:

x = 3mm - 3(09)mm = 3(01)mm

Therefore, the reading in the example is: 123cm

Similarly, if it had been, for example, the seventh mark on the Vernier scale which had been exactly opposite a mark on the main scale, the reading would be: 127cm

Hence, the level of precision of an instrument which has a Vernier scale depends on the difference between the size of the smallest division on the main scale and the size of the smallest division on the Vernier scale.

In the example above, this difference is 01mm so measurements made using this instrument should be stated as: reading 01mm.

Another instrument might have a scale like the one shown in figure 4.

figure 4

Therefore, the precision is: 1mm - (49/50)mm = (1/50)mm = 002mm.

Results of measurements made using this instrument should therefore be stated as: reading 002mm.

This principle is used in the Vernier calliper shown below.

The diagrams below illustrate how to use a Vernier calliper to measure:

A. the internal diameter of a hollow cylinderB. the external dimensions of an objectC. the depth of a hole in a piece of metal.

A.B.

C.

Links to other pages about measurements:

Introduction

How many Decimal Places?

How does an uncertainty in a measurement affect the FINAL result?

Using the vernier calipers and micrometer screw gaugeThe precision of length measurements may be increased by using a device that uses a sliding vernier scale. Two such instruments that are based on a vernier scale which you will use in the laboratory to measure lengths of objects are the vernier callipers and the micrometer screw gauge. These instruments have a main scale (in millimetres) and a sliding or rotating vernier scale. In figure 1 below, the vernier scale (below) is divided into 10 equal divisions and thus the least count of the instrument is 0.1 mm. Both the main scale and the vernier scale readings are taken into account while making a measurement. The main scale reading is the first reading on the main scale immediately to the left of the zero of the vernier scale (3 mm), while the vernier scale reading is the mark on the vernier scale which exactly coincides with a mark on the main scale (0.7 mm). The reading is therefore 3.7 mm.

Figure 1 : The reading here is 3.7 mm.

Figure 1 : The reading here is 15.8 mm.

This Java applet will help you to understand how to read a vernier scale.

The vernier calipers

The vernier calipers found in the laboratory incorporates a main scale and a sliding vernier scale which allows readings to the nearest 0.02 mm. This instrument may be used to measure outer dimensions of objects (using the main jaws), inside dimensions (using the smaller jaws at the top), and depths (using the stem).

Figure 3: The vernier calipers

To measure outer dimensions of an object, the object is placed between the jaws, which are then moved together until they secure the object. The screw clamp may then be tightened to ensure that the reading does not change while the scale is being read.Watch this short movie to see how to do this.

Here is a nice vernier calipers applet.

The first significant figures are read immediately to the left of the zero of the vernier scale and the remaining digits are taken as the vernier scale division that lines up with any main scale division.

Some examples: Note that the important region of the vernier scale is enlarged in the upper right hand corner of each figure.

Figure 4: The reading is 37.46 mm.

In figure 4 above, the first significant figures are taken as the main scale reading to the left of the vernier zero, i.e. 37 mm. The remaining two digits are taken from the vernier scale reading that lines up with any main scale reading, i.e. 46 on the vernier scale. Thus the reading is 37.46 mm.

Figure 5: The reading is 34.60 mm.

In figure 5 above, the first significant figures are taken as the main scale reading to the left of the vernier zero, i.e. 34 mm. The remaining two digits are taken from the vernier scale reading that lines up with any main scale reading, i.e. 60 on the vernier scale. Note that the zero must be included because the scale can differentiate between fiftieths of a millimetre. Therefore the reading is 34.60 mm.

Figure 6: The reading is 40.00 mm.

In figure 6 the zero and the ten on the vernier scale both line up with main scale readings, therefore the reading is 40.00 cm.

Try the following for yourself.

Figure 7: Click here for the answer.

Figure 8: Click here for the answer.

Figure 9: Click here for the answer. The micrometer screw gauge

The micrometer screw gauge is used to measure even smaller dimensions than the vernier callipers. The micrometer screw gauge also uses an auxiliary scale (measuring hundredths of a millimetre) which is marked on a rotary thimble. Basically it is a screw with an accurately constant pitch (the amount by which the thimble moves forward or backward for one complete revolution). The micrometers in our laboratory have a pitch of 0.50 mm (two full turns are required to close the jaws by 1.00 mm). The rotating thimble is subdivided into 50 equal divisions. The thimble passes through a frame that carries a millimetre scale graduated to 0.5 mm. The jaws can be adjusted by rotating the thimble using the small ratchet knob. This includes a friction ?clutch? which prevents too much tension being applied. The thimble must be rotated through two revolutions to open the jaws by 1 mm.

Here is a useful applet to learn how to use the micrometer screwgauge.

Figure 10: The micrometer screw gauge

In order to measure an object, the object is placed between the jaws and the thimble is rotated using the ratchet until the object is secured. Note that the ratchet knob must be used to secure the object firmly between the jaws, otherwise the instrument could be damaged or give an inconsistent reading. The manufacturer recommends 3 clicks of the ratchet before taking the reading. The lock may be used to ensure that the thimble does not rotate while you take the reading. Watch this short movie to see how to do this.

The first significant figure is taken from the last graduation showing on the sleeve directly to the left of the revolving thimble. Note that an additional half scale division (0.5 mm) must be included if the mark below the main scale is visible between the thimble and the main scale division on the sleeve. The remaining two significant figures (hundredths of a millimetre) are taken directly from the thimble opposite the main scale.

Figure 11: The reading is 7.38 mm.

In figure 11 the last graduation visible to the left of the thimble is 7 mm and the thimble lines up with the main scale at 38 hundredths of a millimetre (0.38 mm); therefore the reading is 7.38 mm.

Figure 12: The reading is 7.72 mm.

In figure 12 the last graduation visible to the left of the thimble is 7.5 mm; therefore the reading is 7.5 mm plus the thimble reading of 0.22 mm, giving 7.72 mm.

Figure 13: The reading is 3.46 mm.

In figure 13 the main scale reading is 3 mm while the reading on the drum is 0.46 mm; therefore, the reading is 3.46 mm.

Figure 14: The reading is 3.56 mm.

In figure 14 the 0.5 mm division is visible below the main scale; therefore the reading is 3.5 mm + 0.06 mm = 3.56 mm.

Try the following for yourself.

Figure 15: Click here for the answer.

Figure 16: Click here for the answer.

Figure 17: Click here for the answer.

Taking a zero reading

Whenever you use a vernier calipers or a micrometer screw gauge you must always take a ?zero reading? i.e. a reading with the instrument closed. This is because when you close your calipers, you will see that very often (not always) it does not read zero. Only then open the jaws and place the object to be measured firmly between the jaws and take the ?open? reading. Your actual measurement will then be the difference between your ?open? reading and your ?zero? reading.

Recording the result of your vernier measurement

Let us say you take a reading with an object between the jaws of a vernier calipers and you see the following:

Say that you decide that the best estimate of the reading l 1 is 37.46 mm.

What about the standard uncertainty u(l1) in this reading?

Using a triangular probability density function, you might decide that you are 100% sure that the reading is not 37.42 mm and 100% sure that the reading is not 37.50 mm.

Then mm = 0.0163 mm

When you remove the object and read the vernier calipers with the jaws closed, you might decide that the best estimate of the "closed" reading l0 = 0.04 mm with standard uncertianty u(l 0) = 0.0204 mm

What should you then record as the best estmate of the length of the object you are measuring?

The best estimate of the length l = l 1 - l0 = 37.46 - 0.04 = 37.42 mm

with a standard uncertainty = = 0.0261 mm

Therefore l = 37.420 0.026 mm (65% level of confidence).

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