Originally Posted by niznai

You qoute values such as 2.93N/mm. This means you need to apply a weight of 2.93/9.81=0.29 grams to achieve a deflection of 1mm. If your scale can not measure this accurately, your results may be off by Newtons (not fractions thereof). I had to use only two decimal places because you don't give four.

Your comment relates to obtaining the elastic constant of your spring, which on your graph is the slope of the line, but I am arguing that if your experimentally determined points have errors of a few grams, that line can vary in its slope hugely because of this indetermination. Try to plot each point with its error bars and then see how many lines can fit through these error intervals and you'll understand what I mean. Come to think of it, you really need the calibration curve of your scale to be able to claim you have determined the spring constant to that accuracy.

Not to say the companies selling us their spring do any of this. I just take their values as orientation at best.