# How would you determine the measurement error in the following example?

I'm at a bit of a loss as to how to determine the error in measurement in a project I'm working on. The project involves taking a picture of an object, and then using the image to determine the width of the object.

So, a sample image would be like the one below:

Things we know:

• Distance from the object (via a sensor)
• Field of view of camera (measured at 48.8 degrees)

So I came up with the following to calculate the width of the frame of view:

The example is based on the object taking up the entire field of view, but that doesn't have to be the case. The object could be smaller, say 1/2 of the frame of view; to find the width you would simply take half of the field of view.

Now my question is, how do you determine the measurement error?

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Wouldn't there be some distortion due to the lens? – J. M. Apr 28 '11 at 4:39

## 1 Answer

You take each piece of data that goes into your calculation, figure out the error on that piece of data, and see how that error changes the output value. So for your case you have measured the distance and the angle of view of the camera and your equation is $width=2distance\cdot \tan \theta$. As the errors are small, you can use the derivatives: $\frac{d(width)}{d(distance)}=2\tan \theta \approx 0.91$ So an error in the distance measurement of 1 inch becomes an error in the width of 0.91 inches. If the object doesn't fill the frame, the ratio will be different. You can also assess the impact of an error in $\theta$ the same way and add them together to get the total possible error.

Sometimes the errors in the data are fractional-a mass measurement might be $\pm 1 \%$, for example. Then if you take the logarithm of the equation before the derivative you get things like $\frac{dm}{m}$ which is the $1 \%$

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Would % error = abs((estimate - actual) / actual * 100) work? And if so, would it be less accurate than your approach? – Ryan Apr 29 '11 at 5:05
It works fine. In fact it is not dependent on approximate linearity of the error. Sometimes my approach is easier if you want to find how small the measurement error has to be to keep the final calculation below a certain amount, or you have a lot of error terms and want to assess how important each one is. – Ross Millikan Apr 29 '11 at 12:54
Ah, ok. Thank you! – Ryan Apr 30 '11 at 0:46