Originally Posted by
gameover
nah it's not a problem, the weight on the scale at 1mm compression is around 200-300g and to compress the spring to 5mm takes close to 1500g (in some cases more). 1-2 grams error doesn't matter in the slightest.
the absolute load on the spring is not critical, actually when i calculate the final rate i ignored the first reading at 0.5mm to remove the variation. It takes the same weight to compress the spring from 1.0mm - 1.5mm as it does from 4.5mm to 5.0mm, its only the relative difference that is important for these linear springs.
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.