Error And Measurement

Tom

2008-09-21 7:11 am

In an experiment, the external diameter D and internal diameter d of a metal tube wew found to be 64±2mm and 47±1mm respectively. What is the meximum percentage error in the cross-sectional area of the metal tube?

My solution:
A=pi/4(D^2-d^2)
(〥A)/A=(2〥D)/D+(2〥d)/d
(〥A)/Ax100%=[2(2/64)+2(1/47)]x100%
So, the %error=10.5%

But the answer:
A=pi/4(D^2-d^2)=pi/4(D-d)(D+d)
(〥A)/Ax100%=±[〥(D+d)/(D+d)+〥(D-d)(D-d)]x100%
(〥A)/Ax100%=±[3/(64+47)+3/(64-47)]x100%
(〥A)/Ax100%=±20%

What's wrong I have made?

更新1:

D^2-d^2 =(D+d)(D-d) =D^2+Dd-Dd+d^2 From D^2-d^2 (2〥D)/D+2(〥d)/d =2[(〥D)/D+(〥d)/d] From D^2+Dd-Dd+d^2, (2〥D)/D+(〥D)/d+(〥d)/d+(〥D)/d+(〥d)/d+(2〥d)/d =4[(〥D)/D+(〥d)/d] Why are't they equal?

回答 (1)

用戶222683

2008-09-21 7:50 am

✔ 最佳答案

對不起容許我用中文作答。
有關計算誤差的公式應為
當兩物理量相加或減時
X = Y ± Z
δX = δY + δZ

若是乘或除是
X = YZ 或 X = Y/Z
δX/X = δY/Y + δZ/Z

而
A = π(D^2 – d^2)/4
是兩項相減，不是乘除所以
δA/A ≠ 2(δD/D) + 2(δd/d)

因此計法便是第二種方法
A=π(D^2-d^2)=pi/4(D-d)(D+d)/4
(δA)/Ax100%=±[δ(D+d)/(D+d)+ δ(D-d)(D-d)]x100%
(δA)/Ax100%=±[3/(64+47)+3/(64-47)]x100%
(δA)/Ax100%=±20%
不是第一種方法

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