A Vernier allows a precise reading of some value. In the figure to the right, the Vernier moves up and down to measure a position on the Scale. This could be part of a barometer which reads atmospheric pressure.
The "pointer" is the line on the vernier labelled "0". Thus the measured position is almost exactly 756 in whatever units the scale is calibrated in.
If you look closely you will see that the distance between the divisions on the vernier are not the same as the divisions on the scale. The 0 line on the vernier lines up at 756 on the scale, but the 10 line on the vernier lines up at 765 on the scale. Thus the distance between the divisions on the vernier are 90% of the distance between the divisions on the scale.
If we do another reading with the vernier at a different position, the pointer, the line marked 0, may not line up exactly with one of the lines on the scale. Here the "pointer" lines up at approximately 756.5 on the scale.
If you look you will see that only one line on the vernier lines up exactly with one of the lines on the scale, the 5 line. This means that our first guess was correct: the reading is 756.5.
Here is a final example, with the vernier at yet another position. The pointer points to a value that is obviously greater than 756.5 and also less than 757.0. Looking for divisions on the vernier that match a division on the scale, the 7 line matches fairly closely. So the reading is about 756.7.
In fact, the 7 line on the vernier appears to be a little bit above the corresponding line on the scale. The 8 line on the vernier is clearly somewhat below the corresponding line of the scale. So with sharp eyes one might report this reading as 756.73 ± 0.02.
This "reading error" of ± 0.02 is probably the correct error of precision to specify for all measurements done with this apparatus.
Now we shall use a simulation of a Vernier Caliper. A caliper measures a length, and in the following figure we show a caliper being used to measure the length of an Object. The Object will be placed between the "jaws" of the caliper. The Object is almost exactly 75 mm (2.95 in) long.
source :http://www.upscale.utoronto.ca/PVB/Harrison/Vernier/Vernier.html
Sabtu, 10 Januari 2009
Reading a Vernier
Vernier Caliper
Instructions on use
* The Vernier caliper is an extremely precise measuring instrument; the reading error is 1/20 mm = 0.05 mm.
* Close the jaws lightly on the object to be measured.
* If you are measuring something with a round cross section, make sure that the axis of the object is perpendicular to the caliper. This is necessary to ensure that you are measuring the full diameter and not merely a chord.
* Ignore the top scale, which is calibrated in inches.
* Use the bottom scale, which is in metric units.
* Notice that there is a fixed scale and a sliding scale.
* The boldface numbers on the fixed scale are centimeters.
* The tick marks on the fixed scale between the boldface numbers are millimeters.
* There are ten tick marks on the sliding scale. The left-most tick mark on the sliding scale will let you read from the fixed scale the number of whole millimeters that the jaws are opened.
* In the example above, the leftmost tick mark on the sliding scale is between 21 mm and 22 mm, so the number of whole millimeters is 21.
* Next we find the tenths of millimeters. Notice that the ten tick marks on the sliding scale are the same width as nine ticks marks on the fixed scale. This means that at most one of the tick marks on the sliding scale will align with a tick mark on the fixed scale; the others will miss.
* The number of the aligned tick mark on the sliding scale tells you the number of tenths of millimeters. In the example above, the 3rd tick mark on the sliding scale is in coincidence with the one above it, so the caliper reading is (21.30 ± 0.05) mm.
* If two adjacent tick marks on the sliding scale look equally aligned with their
counterparts on the fixed scale, then the reading is half way between the two marks. In the example above, if the 3rd and 4th tick marks on the sliding scale looked to be equally aligned, then the reading would be (21.35 ± 0.05) mm.
* On those rare occasions when the reading just happens to be a "nice" number like 2 cm, don't forget to include the zero decimal places showing the precision of the measurement and the reading error. So not 2 cm, but rather (2.000 ± 0.005) cm or (20.00 ± 0.05) mm.
source : http://www.physics.smu.edu/~scalise/apparatus/caliper/
