Error Analysis of wall thickness

AI Thread Summary
The discussion focuses on calculating the wall thickness of a hollow cylinder using measured diameters and determining the associated error. The calculated thickness is 0.18 cm with an error of ±0.01 cm, but sources suggest an error of ±0.02 cm. A statistical approach is proposed, where the error is divided by the square root of 2 to account for potential cancellation of errors. This method is deemed appropriate for engineering tolerances, emphasizing the importance of precision in manufacturing. The conversation highlights the balance between simple error calculations and more complex statistical methods depending on the context.
Nikhil Rajagopalan
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Homework Statement


The internal and external diameter of a hollow cylinder are measured with the help of a vernier calipers. Their values are (3.87 ± 0.01) cm and (4.23 ± 0.01) cm respectively. The thickness of the wall of the cylinder is ?

Homework Equations


Thickness of cylinder wall= 1/2 (Outer Diameter - Inner Diameter)
t = 1/2 (Do - Di)

The Attempt at a Solution



for value of t,

t = 1/2 (Do - Di)
t = 1/2(4.23 - 3.87)
t = 1/2(0.36)
t = 0.18

for finding the error in t,
by differentiating on both sides,

Δt = 1/2 (ΔDo - ΔDi)
Δt = 1/2 (0.01 + 0.01)
Δt = 0.01

thickness t ± Δt = 0.18 ± 0.01

most sources show the answer as 0.18 ± 0.02. Kindly help me to figure out the mistake that i made in calculating the error.
 
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Nikhil Rajagopalan said:
most sources show the answer as 0.18 ± 0.02
Sources you regard as reliable?
I agree with your answer.
Many would take a statistical approach. This allows that the two errors will often cancel out somewhat, and rarely be at opposite extremes. So they would divide the error by √2. But when you need to be sure that engineering tolerances are met, the simple method you used is appropriate.
 
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The formula for finding error in the case of ##t=au+bv## where a and b are constants is:
$$\sigma_t = \sqrt{(a\sigma_u)^2+(b\sigma_v)^2}$$
since a and b both equal 1/2 and both ##\sigma_u## and ##\sigma_v## equal 0.01, this nicely simplifies to:
$$\sigma_t = \sqrt{2(1/2(0.01))^2} = \frac{0.01}{\sqrt{2}}$$
So I would agree with @haruspex
 
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Jamison Lahman said:
So I would agree with @haruspex
As I wrote, the approach should depend on the purpose. In manufacturing, the engineer sets tolerances for each component. Each machinist works to those specifications. If the resulting components don't fit the engineer is in trouble.
 
haruspex said:
As I wrote, the approach should depend on the purpose. In manufacturing, the engineer sets tolerances for each component. Each machinist works to those specifications. If the resulting components don't fit the engineer is in trouble.
True, but if you'd like to take the statistical approach as you suggested many would in your first post, that's the rationale behind the root 2. The equation is directly out of Bevington and Robinson (2003).
 
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