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TheLuteceSibling t1_jdg3msf wrote

It shouldn't. Scales function on compression between two surfaces. Whether the surface under the scale is a little squishy or not shouldn't make a difference.

Does your scale have a two-piece construction? Like a box that fits in a box that compresses when you stand on it? If so, then the outer shell might be touching the carpet when you stand on it, so not all of your weight is on the scale... it's on the carpet.

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chiuchebaba OP t1_jdg6p1i wrote

Yes even I felt it shouldn’t happen.

And the scale isn’t really a box inside a box. But it has tiny footrests kind of thing at the bottom that are just around 1 cm tall. My guess is that the compression on the footrests gets translated into the measured weight but if I keep the scale on a soft surface then the body of the scale may come in contact with that surface instead of just the footrests. Thanks.

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pichael289 t1_jdggvvh wrote

You might be putting more weight to the side, not evenly distributed. That could throw off the mechanism inside, introduce friction or something

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veemondumps t1_jdgj3mu wrote

A scale works by compressing something and measuring how much that thing is compressed. In an analogue scale, that's usually a spring that turns a dial when compressed. In a digital scale, its usually a wire with a current being run through it - when the wire is compressed, the wire's resistance increases so the amount of current flowing decreases.

When you put a scale on a hard surface, there is nothing under the scale to compress, so 100% of the compression is focused on whatever inside the scale is measuring compression.

When you put a scale on a soft surface, like a carpet, the carpet functions as a continuation of the measuring surface inside of the scale. IE, some of the compressive force being applied to the scale is instead applied to the carpet, reducing the amount of force on the internal components of the scale.

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xepci0 t1_jdhbkci wrote

The scale tilts a bit and some of the force is applied "from the side" and not paralel with the spring.

That's the eli5 part.

To be more specific, it will only measure a portion of the straight down force vector.

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