Is Weighing Quarters Individually More Precise Than Together?

[brycenrg]

Homework Statement

What is a more precise way of finding the total mass? Weighing 5 individual quarters or weighing them all together?

2. Relevant equation
individual weights n1+n2+n3+n4+n5 = total weight
total weighed together N = total weight

The Attempt at a Solution

I think weighing individually is more precise, but we weigh things together because the loss error is
negligible.[/B]

 

[Buzz Bloom]

Hi brycenrg:

I think you need to specify something about the scale you are using to make the weight measurements. For example: is the precision of measurement plus/minus a fixed weight, or is it plus/minus a fixed fraction of the measurement? Or is it a variable fraction of the measurement depending on the weight?

Regards,
Buzz

 

[brycenrg]

Hello Buzz Bloom:

It would be an analytical balance with a ± 0.0005 g
Im not sure what the differences are for fixed weight or fixed fraction or the variable fraction.

Thank you.

 

[Rx7man]

If the scale has the capacity to weigh all parts together, I'm quite certain you'd have better accuracy.
Most scales are ±1 on the least significant digit (in some cases ±5 which sounds like yours), and ± x% total error.. by weighing all parts together you reduce the cumulative error on the significant digit part, though the percentage error would remain the same... The smaller your pieces are, the more accurate you'll be by weighing them all together.

For example.. if the real weight is.0010g for each piece it means the scale could show between .0005 and .0015g grams... weighing each piece individually means you only know the total weight will be between .0025 and .0075g.
Weighing them together you will get .005g ± .0005g
But how are you getting 5 quarters?

 

[brycenrg]

If the scale has the capacity to weigh all parts together, I'm quite certain you'd have better accuracy.
Most scales are ±1 on the least significant digit (in some cases ±5 which sounds like yours), and ± x% total error.. by weighing all parts together you reduce the cumulative error on the significant digit part, though the percentage error would remain the same... The smaller your pieces are, the more accurate you'll be by weighing them all together.

For example.. if the real weight is.0010g for each piece it means the scale could show between .0005 and .0015g grams... weighing each piece individually means you only know the total weight will be between .0025 and .0075g.
Weighing them together you will get .005g ± .0005gBut how are you getting 5 quarters?

Thank you, and to clarify i mean 25 cent quarters.
So i could just say (x ± .0005g) + (c ± .0005g)+ (a ± .0005g)+ (z ± .0005g)+ (d ± .0005g) = (x1+x2+x3+x4+x5) ± .0025
Which if i weighed them all at once on the scale it would be (X) ± .0005g?
Soo.. if that's the correct logic then it would be more precise weighing it all at once? It seems like individual would be better but I dono.

 

[Rx7man]

Unless they're new quarters, they're going to be a little worn, and that can shave some weight off them... seems that if you're trying to count quarters, the scale is far more accurate than you need... what are you trying to accomplish?

 

[ChrisVer]

how did you get the uncertainty of 0.0025?

 

[Buzz Bloom]

So i could just say (x ± .0005g) + (c ± .0005g)+ (a ± .0005g)+ (z ± .0005g)+ (d ± .0005g) = (x1+x2+x3+x4+x5) ± .0025

Hi brycenrg:

Your answer that measuring all five at the same time is better than measuring one at a time is correct. However, what the ± .0005g means is that the .0005g is the standard deviation of the scale's error distribution. To get the standard deviation from adding the five separate measurements, you have to calculate the square root of the sum of the squares. Thus the standard deviation is
√(5 × .0005²) g = .0005 × √5 g. Thus the answer from 5 separate weighings is
(x1+x2+x3+x4+x5) ± .0005 × √5 g ≈ (x1+x2+x3+x4+x5) ± .0011 g.

Regards,
Buzz

 

[brycenrg]

Hi brycenrg:

Your answer that measuring all five at the same time is better than measuring one at a time is correct. However, what the ± .0005g means is that the .0005g is the standard deviation of the scale's error distribution. To get the standard deviation from adding the five separate measurements, you have to calculate the square root of the sum of the squares. Thus the standard deviation is
√(5 × .0005²) g = .0005 × √5 g. Thus the answer from 5 separate weighings is
(x1+x2+x3+x4+x5) ± .0005 × √5 g ≈ (x1+x2+x3+x4+x5) ± .0011 g.

Regards,
Buzz

Thank you buzz