Calculating the uncertainty in an A4 page

Smythe

I want to calculate the uncertainty in the area of an A4 page using a 30cm ruler with mm divisions.

Page is measured as:
Height H = 297mm
Width W = 210mm
H x W = 62370 mm^2

I'm using this formula to calculate the area uncertainty:
(dA/A)^2 = (dH/H)^2 + (dW/W)^2

(0.5/297)^2 + (0.5/210)^2 = (dA/62370)^2
(0.000002834 + 0.000005669)^0.5 = dA/62370
(0.000008503)^0.5 = dA/62370
0.002916 = dA/62370
dA = 181.87

So the area is:
62370 +/- 180 mm^2

Have I done this correctly?

Thank you.

krd

Smythe wrote: »

I'm using this formula to calculate the area uncertainty:
(dA/A)^2 = (dH/H)^2 + (dW/W)^2

I'm not sure. But I think that formula is Pythagoras theorem.

I think the formula should be:

(dA/A)= (dH/H)(dW/W)


18cm^2 for an error measuring an A4 page sounds a little huge.

Delphi91

Your formula is correct, in that if
[LATEX]x=a \times b[/LATEX]
then
[latex]\frac{\triangle x}{x}=\sqrt{(\frac{\triangle a}{a})^2 + (\frac{\triangle b}{b})^2}[/latex]
or, as you put it,
[latex](\frac{\triangle x}{x})^2=(\frac{\triangle a}{a})^2 + (\frac{\triangle b}{b})^2[/latex]
Your figures look right - haven't double checked them, but they look ok. Errors in H and W would be 0.5mm

18cm^2 for an error measuring an A4 page sounds a little huge.

180mm^2 is 1.8cm^2

krd

Delphi91 wrote: »

180mm^2 is 1.8cm^2

Oooops.......it's so long since I've had to do "maths homework", I make really silly boo boos. I know I need to sit down with a few school books and practice the exercises, but my inner child keeps screaming himself hoarse crying NO!!!!!

There is another method for calculating the error - which I can't remember either - that's simpler for calculating the error in longer equations. You just take the error and then use a simple set of rules to get your total error.

Smythe

Thanks guys, was able to work that out.