测量时，一个测量值通常写成：measurement=(best estimate ± absolute uncertainty) units，例如（20.1±0.1）cm，20.1cm是best estimate，指的是测量人员觉得该测量值非常接近20.1cm。Absolute uncertainty是0.1cm，指的是真实值可能在20.0cm和20.2cm之间出现。这里的absolute uncertainty只是uncertainty中的一种。
Uncertainty（不确定度）一共有三种，分别是absolute uncertainty，fractional uncertainty和percentage uncertainty。
通常说uncertainty指的是absolute uncertainty。而fractional uncertainty和percentage uncertainty实际上是同种uncertainty的两种不同表示方法，只是fractional uncertainty最终结果写成分数或小数，percentage uncertainty的最终结果写成百分数。所以在讨论uncertainty的计算问题时，主要掌握absolute uncertainty和percentage uncertainty的计算。本文只讨论absolute uncertainty的计算。
Part A. 测量一次absolute Uncertainly的计算
Absolute uncertainty简称uncertainty，计算时分两种情况，一种是只测量一次，一种是测量多次。
测量一次，uncertainty通常指的是测量仪器的最小刻度，即resolution（分度值）。也有些考试根据仪器的种类不同，把uncertainty的计算方法分成两种：
Analog instrument : ½ of the smallest increment (resolution)
Digital instrument : the smallest scale division
大部分的A-level考试都取分度值为uncertainty。
Example 1
The uncertainty in the value of the momentum of a trolley passing between two points X and Y varies with the choice of measuring devices.
Measurements for the same trolley made by different instruments were recorded.
1 distance between X and Y using a metre rule with cm divisions = 0.55m
2 distance between X and Y using a metre rule with mm divisions = 0.547m
3 timings using a wristwatch measuring to the nearest 0.5s at X = 0.0 s and at Y = 4.5s
4 timings using light gates measuring to the nearest 0.1s at X = 0.0s and at Y = 4.3s
5 mass of trolley using a balance measuring to the nearest g = 6.4 × 10–2 kg
6 mass of trolley using a balance measuring to the nearest 10g = 6 × 10–2 kg
Which measurements, one for each quantity measured, lead to the least uncertainty in the value of the momentum of the trolley?
A 1, 3 and 6 B 1, 4 and 6 C 2, 3 and 6 D 2, 4 and 5
| 解析：本题考查了三类物理量的uncertainly的比较，第1,2组是metre rule测量长度，第2组是mm比第1组cm小。第3,4组是测时间，第4组精度0.1s比第3组0.5s小。第5,6组测量质量，第5组精确到g比第6组精确到10g小，所以选D。 |
Part B. 多次测量absolute Uncertainly的计算
如果是多次测量某个物理量，uncertainty=half range=1/2(max-min)，也有些考试uncertainty=range，我们这里用的是第一种公式计算多次测量的uncertainty。
Example 2
The student recorded the following measurements.
Measurement |
Reading |
length / mm |
75.8 75.9 75.7 75.8 |
thickness / mm |
1.01 1.02 0.98 0.99 1.00 |
Use these measurements to estimate the uncertainty in the readings for length and thickness.
| 解析：对于length，uncertainly=half range=1/2(75.9-75.7)=0.01mm；对于thickness，uncertainly=half range=1/2(1.02-0.98)=0.02mm |
Part C. absolute Uncertainly的数乘和加减运算
如果是多次测量某个物理量，uncertainty=half range=1/2(max-min)，
除了掌握absolute uncertainty的基本计算，还要掌握absolute uncertainty的两种数学运算，即数乘和加减运算。
如果两个物理量x，y满足y=mx，m为常数，则absolute uncertainty of y=m×absolute uncertainty of x，即
；如果一个物理量z由两个物理量x，y相加或相减得到，即
Example 3
The long side of a rectangular piece of paper is measured to be (30±2) mm, and the short side is measured to be (20±3) mm. What is the perimeter of this piece of paper, together with its uncertainty?
a. (50±5) mm
b. (100±3) mm
c. (100±5) mm
d. (100±10) mm
e. (600±6) mm
| 解析：本题给出了长方形的长和宽，要求周长和周长的uncertainly，周长等于四条边相加，所以周长的uncertainly需要把四条边的uncertainly相加，即10mm。 |
Example 4
Four identical rods have a square cross-section. The rods are placed side by side and their total width is measured with vernier calipers, as shown.
The measurement is (8.4 ± 0.1)mm and the zero
reading on the calipers is (0.0 ± 0.1)mm.
What is the width of one rod?
A (2.10 ± 0.025)mm
B (2.10 ± 0.05)mm
C (2.1 ± 0.1)mm
D (2.1 ± 0.2)mm
| 解析：本题不仅考察了uncertainly的加减法，还考了数乘。首先four square rods的厚度为两个读数相减，则它们厚度的uncertainly=0.1+0.1=0.2mm，然后要求一个rod的厚度，需要除以4，即乘以1/4，则最终的uncertainly=0.2/4=0.05mm。 |
还有一些物理量，它的数值由几个物理量相乘或相除得到，对于这种求absolute uncertainty的问题，它的absolute uncertainty得先求出这个物理量的percentage uncertainty，然后根据percentage uncertainty的定义式算出absolute uncertainty，这个在percentage uncertainty的计算板块再讲。
