Percentage errors

lavster · Sep 17, 2012

How can you get positive and negative errors to be different in magnitude?

for example -

when calculating the error of the area of an annulus from two circular areas (ie subtracting one from the other, why is the positive error greater than the negative error

Thanks

Naty1 · Sep 17, 2012

because the denominators are different.....

like, say, you have $100 invested and you lose $20...that's a $20 loss....
Now you have $80...What percentage gain do you need to get your $20 back.....
20/80 is 25%.

figures don't lie, but liars figure!

lavster · Sep 17, 2012

i understand the money analogy, but not when talking about the areas - sorry! :S we are only subtracting once

mathman · Sep 17, 2012

lavster said: ↑

i understand the money analogy, but not when talking about the areas - sorry! :S we are only subtracting once

Could you illustrate by an example what you are concerned about?

mfb · Sep 17, 2012

Imagine a square where both sides are known with a precision of 10% - they might be 10% shorter or 10% longer, but not more. What is the maximal deviation?

Larger area: Both sides 10% longer, total area 1.1^2 = 1.21 of the original area (21% more).
Smaller area: Both sides 10% shorter, total area 0.9^2 = 0.81 of the original area (19% less).
Do you see the difference?

mathman · Sep 18, 2012

mfb said: ↑

Imagine a square where both sides are known with a precision of 10% - they might be 10% shorter or 10% longer, but not more. What is the maximal deviation?

Larger area: Both sides 10% longer, total area 1.1^2 = 1.21 of the original area (21% more).
Smaller area: Both sides 10% shorter, total area 0.9^2 = 0.81 of the original area (19% less).
Do you see the difference?

I see the difference, but why do you see this as a problem?

Khashishi · Sep 18, 2012

The error depends on how the measurement was taken and what analysis was done. When you do math on the measurement, the errors will change shape. If you measure a circle's radius with a ruler with a symmetric error in the length, then the error in the area is not symmetric, because area goes as length squared.

mfb · Sep 19, 2012

mathman said: ↑

I see the difference, but why do you see this as a problem?

Where did I say that it is a problem?
I just said that you can get asymmetric uncertainties in this way.

lavster · Sep 21, 2012

ah great :) thanks :)

Log in or Sign up

Percentage errors

lavster

Naty1

lavster

mathman

mfb

Staff: Mentor

mathman

Khashishi

mfb

Staff: Mentor

lavster

Percentage of Uncertainty (Replies: 5)

Errors, errors (Replies: 1)

Light, photons, reflection and percentages (Replies: 1)

Percentage excitation (Replies: 1)