# weighing big box by carrying the box and stepping on the weighing machine yourself …?

I have seen people counter the problem of big box and small weighing machine by stepping on the machine while carrying the box in arms and subtract the body weight. Do you think it'll give accurate result (not counting negligible difference of course)?

It seems like a DUH kind of question, but I wanted to ask it anyway.

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""Do you think it'll give accurate result (not counting negligible difference of course)? "" What is the meaning of "accurate" and "negligible"? – Georg Nov 3 '11 at 11:01
@Georg I meant practically accurate.. like maybe less than 0.5 pound difference? – progtick Nov 3 '11 at 11:59
Depends on quality of Your "weighing machine" (ie scale or balance?) Old purely mechanic balances sold 20 or more years ago: no advice possible. Today person scales are much better, half a pond is no problem. – Georg Nov 3 '11 at 16:53

Provided your scales are linear enough then the logic is fine.

Edit: by linear I mean the scales read as accurately for large weights as small weights. This isn't guaranteed but probably true for home scales.

The logic of (man + box) - man = box is true.

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What do you mean? – progtick Nov 3 '11 at 13:04

A (man+box) measurement will be less accurate than a simple (box) due to the propagation of measurement errors. In the (man+box) case you are doing two measurements compared to a single measurement in the (box) case.

For example:

$Man = 200 lbs$

$Box = 50 lbs$

$Scale = \pm0.5 lbs$

$Box Measured Weight = (250 \pm 0.5) - (200 \pm 0.5) = 50 \pm 0.7$

In this example the box weight has more measurement error than if we just measured the box weight directly. Depending on the scale error and the box weight relative to the man, the final error can be large, for example:

$Box = 5 lbs$

$Scale = \pm 5\%$ (a really bad scale, probably not realistic)

$BoxMeasuredWeight = (205 \pm 5\%) - (200 \pm 5\%) = 5 \pm 14 lbs$

If we just measured the box directly the error would have only been 0.25 lbs.

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