Propagation of uncertainties

Formulas and graphs

Volume of a cylinder

1 2 3 4 5 6

D = (2.9 ± 0.16) cm

D = 2.9 cm D = 0.05 cm+ 0.01 cm+ 0.1 cm

12

34

56

h = 1.5 cm h = 0.05 cm+ 0.01 cm+ 0.05 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 %)

• 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?

Volume of a cylinder

Uncertainties and functions

dVV D

dD

V = ¼ p D2 h

d

V

D

V

D+DD-D

D

V

0

limd

V V dVV D

D D dD

Uncertainty in volume arising from uncertainty in diameter:

Propagation of uncertainty

V VV D h

D h

V = ¼ p D2 h

Uncertainty in V = contribution from D + contribution from h

21 1

2 4V hD D D 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%