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Chen
Apr18-04, 10:45 AM
In our labs we often need to find the error percent between the expected value and the measured value. So if we expected to get a value of 9.8 m/s2 for g but got 9.9 m/s2 instead, we say that the error was (9.9 - 9.8)/9.8 % = 1.02%. But what happens if we expect a value of 0 for something, and measure 1cm instead?

jdavel
Apr18-04, 11:04 AM
Chen,

Are you really measuring something whose correct length is zero, or are you measuring two non-zero lengths that are supposed to be the same?

Chen
Apr18-04, 12:34 PM
It is an experiment about conservation of momentum and energy in an elastic collision. The original momentum in the Y axis is 0, so the total momentum in that axis after the collision must also equal 0. Dropping the units for a second, let's say that the momentum in the Y axis after collision is (6) + (-5.9) = 0.1. What's the error percent then? :smile:

matt grime
Apr18-04, 12:42 PM
Undefined, that's what.

jdavel
Apr18-04, 01:21 PM
It is an experiment about conservation of momentum and energy in an elastic collision. The original momentum in the Y axis is 0, so the total momentum in that axis after the collision must also equal 0. Dropping the units for a second, let's say that the momentum in the Y axis after collision is (6) + (-5.9) = 0.1. What's the error percent then? :smile:

How about .1/6 or .1/5.9 depending on whether 6 or 5.9 is more likely to be correct?

Integral
Apr18-04, 02:07 PM
Compute the error in each of your measured values (.6) (.59) Then compute either a room mean square error or just use the Max error for the error of the final computation. It may be that your error will be larger then your final result. In which case you can claim to have a correct measuement within your error.

\mbox{rms} = \sqrt {{ \Delta x_1 }^2 + {\Delta x_2}^2}