# What does it means to increase the accuracy of a screw gauge? And if so how to increase it?

This question is related to the following question from NCERT:

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

Now when I searched for the answer all websites say something like this:

"It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only."

So I have the following questions:

1. What does it mean to increase accuracy (for eg. of a screw gauge)?

2. If possible how to increase the accuracy (say of a screw gauge)?

3. Why in this case it isn't possible to increase the accuracy (as per the answers on the websites)?

Increasing the number of divisions on a measuring device increases the precision, but does not change the accuracy. The precision is dependent on how closely you can read or replicate a particular setting. The accuracy, as explained in Farcher's answer, depends on the correct calibration of the mechanics of the measuring device.

Accuracy refers to the closeness of a measured value to a standard or known value.

If the pitch of screw gauge thread was equal to $$1.000 \,\rm mm$$ as measured by the National Institute of Standards and Technology then a reading of $$73$$ divisions on a scale divided into $$200$$ divisions is equal to $$\frac{73}{200} \times 1.000 = 0.365 \,\rm mm$$.

However, if due to some error in manufacture and unknown to you the pitch of the screw gauge thread was equal to $$1.010 \,\rm mm$$ if measured by NIST and not $$1.000\,\rm mm$$ as you have assumed then $$73$$ divisions would correspond to a reading of $$\frac{73}{200} \times 1.010 = 0.36865 \,\rm mm$$ and thus your reading would be in error by $$1\%$$.

Using a scale subdivided into $$400$$ divisions a reading of $$147$$ divisions gives values of $$\frac{147}{400} \times 1.000 = 0.3675 \,\rm mm$$ if the pitch was accurately measured and $$\frac{147}{400} \times 1.010 = 0.371175 \,\rm mm$$ if the thread had not be accurately made.
The difference between these two readings is again $$1\%$$ and so there is no improvement in the accuracy if the number of scale divisions has been increased.

The accuracy of a screw gauge with a stated pitch of one millimetre depends on how close the pitch is to one millimetre as measured at NIST.

If you have access to a number of wooden metre rules it might be an interesting exercise to place them side by side to see if they are all the same length and then finding that they are not all exactly the same length, asking the question "which of these metre rules will give me an accurate measurement of length?".

• The most vital part to the accuracy of a screw gauge is the machining of the screw. Generally they strive to maintain accuracy within .00005 inch, also have you read en.wikipedia.org/wiki/… Aug 6, 2019 at 14:52

theoretically speaking least count decreases on increasing the number of divisions on the circular scale. Hence, accuracy would increase. Practically it may not be possible to take the reading precisely due to low resolution of human eye.