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# Propagation of uncertainties PowerPoint PPT Presentation

Propagation of uncertainties. Formulas and graphs. 3. 3. 4. 4. 5. 5. 6. 6. 1. 1. 2. 2. Volume of a cylinder. D = 2.9 cm D = 0.05 cm+ 0.01 cm+ 0.1 cm. h = 1.5 cm h = 0.05 cm+ 0.01 cm+ 0.05 cm. D = (2.9 ± 0.16) cm. h = (1.5 ± 0.11) cm. Volume of a cylinder.

Propagation of uncertainties

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## Propagation of uncertainties

Formulas and graphs

3

3

4

4

5

5

6

6

1

1

2

2

### Volume of a cylinder

D = 2.9 cm

D = 0.05 cm+ 0.01 cm+ 0.1 cm

h = 1.5 cm

h = 0.05 cm+ 0.01 cm+ 0.05 cm

D = (2.9 ± 0.16) cm

h = (1.5 ± 0.11) cm

### Volume of a cylinder

D = (2.9 ± 0.16) cm

h = (1.5 ± 0.11) cm

Result will have 2 significant figures

V = ¼ p D2 h

V = ¼ p (2.9 cm)2 1.5cm

V = 9.907789463 cm3

V = 9.9 cm3

How sure can we be about the result?

Lowest end: D=2.74 cm, h= 1.39cm V = 8.2 cm3 (-17%)

Highest end: D=3.06 cm, h= 1.61cm V = 11.8 cm3(+19 %)

Volume of a cylinder

• Using physical quantities with uncertainty in a formula leads to calculation results with an uncertainty.

• How much uncertainty?

• How does the formula influence this uncertainty?

• Is there a way to predict this?

V

d

### Uncertainties and functions

V = ¼ p D2 h

V

V

Uncertainty in volume

arising from uncertainty

in diameter:

D-D

D+D

D

D

### Propagation of uncertainty

V = ¼ p D2 h

Uncertainty in V = contribution from D + contribution from h

Every regular equation has an error equation.

Every error equation has one term for each measured quantity.

### Volume of a cylinder

D = (2.9 ± 0.16) cm

h = (1.5 ± 0.11) cm

V = 9.9 cm3

V = (9.9 ± 1.9) cm3

Relative error: V/V 100% = 1.9/9.9  100% = 19%