The Screw Gauge

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a screw that moves forward by just half a millimetre in one complete rotation. Now you want to measure a thin wire, correct to one hundredth of a millimetre. How can a simple screw possibly do that? And here is a second surprise. The same kind of instrument can show one point two eight millimetres for a sheet that is really one point three one millimetres thick. Questions on this instrument come up often in JEE, so let's begin.

0:35Pitch and least count

So, this instrument is the screw gauge, also called a micrometer. The main scale, also called the pitch scale, runs along the screw, and the circular scale goes around it and turns with the screw, okay? Now, pitch is the distance the screw advances along the pitch scale in one complete rotation. So pitch equals distance moved, over number of rotations, right? Now let me ask you this. In one full turn, the circular scale goes through all its divisions, and the screw moves by one pitch. So if you turn by just one division, how far does the screw move? Yes, exactly, it moves by the pitch shared equally among all those divisions. That tiny distance is the least count, written L C on screen. So least count equals pitch, over the number of divisions on the circular scale. Got it? Two values are common: pitch one millimetre, written m m on the screen, with one hundred divisions, or pitch zero point five millimetres with fifty divisions. Both give a least count of zero point zero one millimetres. So the first puzzle is solved: half a millimetre shared among fifty divisions is one hundredth of a millimetre, clear?

1:51Taking a reading

Now look at the figure. On the left is the pitch scale in millimetres, and on the right is the circular scale, with fifty divisions. Beside them, the pitch is zero point five millimetres, and the least count, zero point five over fifty, is zero point zero one millimetres. The reading is worked out there too, so let's go through it. To read any screw gauge, use this one line. Reading equals the pitch scale reading, written PSR, plus the circular scale reading, written CSR, times the least count. Okay? First, the pitch scale reading. The red arrow points to the last visible mark, five point five millimetres. Be careful, and do not stop at the numbered five. The marks below the long horizontal line are the half millimetres, right? Next, the circular scale reading. That long horizontal line is called the reference line. The circular scale division lying on it is twenty eight, with twenty five below and thirty above, okay? Now put it together. We get five point five, plus twenty eight times zero point zero one, which is five point seven eight millimetres. Always multiply by the least count, and never add the bare twenty eight. With me?

3:08Zero error: studs in contact

Now comes a part that needs real care. Bring the two studs, the faces that grip the object, into contact. The reading should be zero, right? So the zero of the circular scale should sit exactly on the reference line. If not, the gauge has a zero error. Suppose the zero of the circular scale is below the reference line. In the figure, the numbers on this scale increase as you go up. So tell me, is this zero error positive or negative? It is positive, yes, because the gauge already reads a little with nothing between the studs. If the zero is small n divisions below the line, the zero error is plus small n times the least count, okay? And if the zero is small n divisions above the line, the error is negative, minus small n times the least count, right? Now here is the catch. With the zero above the line, the division you see on the line is not small n at all. It is capital N minus small n, where capital N is the total number of divisions on the circular scale. Keep this in mind for the second example. Clear? The correction rule is always the same: corrected reading equals the observed reading minus the zero error. Got it? Take the sign along with it, so a positive error gets subtracted, and a negative error gets added. Mistakes often happen right here, so watch it.

4:36Backlash error and the ratchet

Now two practical points, and the first is backlash error. Play between the screw and the nut makes the screw lag when the direction of turning is reversed, okay? So tell me, how would you keep this lag out of your reading? The answer is really simple: always approach the final position from the same side. Then the lag never enters your reading, right? Here is the second point. When a reading is taken, the screw is turned by its ratchet, and not by the thimble, which carries the circular scale. Why do you think that is? That is because the ratchet slips when the studs just grip the object, so the pressure is the same for every reading. Clear?

5:22Example 1: reading with pitch 1 mm and 50 divisions

Now let us try one. The gauge has a pitch of one millimetre and fifty circular divisions. The pitch scale reads two millimetres and the circular scale thirty five. Find the reading. Pause the video and try it first. Are you done? Okay, let's check it together. We have seen zero point zero one millimetres so often. So the least count is zero point zero one here too, right? No, no, that is the trap, so be careful. Least count is pitch over divisions, so one millimetre over fifty gives zero point zero two millimetres, okay? Now use the reading formula. We get two, plus thirty five times zero point zero two, and that product is zero point seven zero. So the reading is two point seven zero millimetres. Always work out the least count first, clear?

6:17Example 2: sheet thickness with negative zero error

Now a harder one, and this is the second puzzle from the start. With the studs in contact, the zero of the circular scale is three divisions above the reference line. The scale has one hundred divisions and least count zero point zero one millimetres, so the ninety seventh division is on the line. Then a sheet reads one point two eight millimetres. Find its thickness. Pause the video and try it first. Are you done? Okay, let's check it together. Ninety seven is on the line, so the zero error is ninety seven times zero point zero one, which is zero point nine seven millimetres, right? Oh no, careful, that is the trap we spoke about. The zero is only three divisions above the line, and one hundred minus three gives ninety seven, which is exactly capital N minus small n here. So small n is three, and above the line means a negative error, okay? So the zero error is minus three times zero point zero one, which is minus zero point zero three millimetres. Clear? Now, the sheet reads one point two eight millimetres. Should we subtract zero point zero three, or add it? We add it, yes, because corrected reading is observed minus zero error, and subtracting a negative means adding. So one point two eight minus, in brackets, minus zero point zero three, gives one point three one millimetres. And that solves the second puzzle: the gauge showed less than the true thickness because the zero error was negative. With me?

7:52Quick revision: least count, reading, zero error sign

So, quick revision. Least count is pitch over the number of circular divisions. Reading is the pitch scale reading, plus the circular scale reading times the least count. And corrected reading is observed minus zero error, with its sign. Now go and try the practice problems on this page. You have done the hard part already, so they will feel simple. All the best!