Significant Figures Question

[phiphika1453]

Ok, here is a question posed to me at work while a coworker was helping with his younger sisters homework. I think I explained it correctly but I could not find any corroboration to my explanation, so I turn to SMP.

Ok, she measured the diameter of a circle to be 1.20cm. She was asked to calculate the volume, which is given by pir^2L.

The radius comes to be 1.20cm/2 = 0.600cm, but here is the question as it was asked to me. 0.600cm is more precise than a metric ruler can measure, so (assume all the original measurements had 3 sig figs) how can it maintain the 3 significant figures that is required by the original measurement.


MY answer in white:

<font color="white">The original measurement of 1.20cm has 3 sig figs and you arrived at 0.600cm by doing a whole number calculation, therefore the answer of the calculation can be more precise than the original measurement. </font>

Is my answer correct and does anyone know where I can find info to support this?

[Drag]

That's wrong.

If we can measure diameter D with an uncertanity of dx, the calculation will go as follows:

S=Pi/4* (D+dx)^2=Pi/4*D^2+Pi/4*2*D*dx
(we omitted term porportional to dx^2)
dS=Pi/4*2*D*dx

dS/S=2*dx/D

So your uncertanity is 2 times higher, but of the same order of magnitude giving 3 significant figures.

Normally uncertanity increases with the calculations.

[Kaj]

Or more simply,

D = 1.20 +/- 0.01

R = D/2, so...

R = 0.600 +/- 0.005

If you want to carry through the volume calculation, use the largest and smallest values for R to measure your uncertainty of the final answer.

And no, you didn't explain it correctly to your coworker.

[Kaj]

Scratch that last statement. I think I misread your explanation. I think my calculation jives with what you were explaining in words.

[oe39]

[ QUOTE ]
That's wrong.

If we can measure diameter D with an uncertanity of dx, the calculation will go as follows:

S=Pi/4* (D+dx)^2=Pi/4*D^2+Pi/4*2*D*dx
(we omitted term porportional to dx^2)
dS=Pi/4*2*D*dx

dS/S=2*dx/D

So your uncertanity is 2 times higher, but of the same order of magnitude giving 3 significant figures.

Normally uncertanity increases with the calculations.

[/ QUOTE ]

do you really need all of these calculations? he is only dividing by two!

of course you can get more precise than the ruler would measure. how else would you be able to measure the thickness of a sheet of paper with a ruler?

[phiphika1453]

[ QUOTE ]
[ QUOTE ]
That's wrong.

If we can measure diameter D with an uncertanity of dx, the calculation will go as follows:

S=Pi/4* (D+dx)^2=Pi/4*D^2+Pi/4*2*D*dx
(we omitted term porportional to dx^2)
dS=Pi/4*2*D*dx

dS/S=2*dx/D

So your uncertanity is 2 times higher, but of the same order of magnitude giving 3 significant figures.

Normally uncertanity increases with the calculations.

[/ QUOTE ]

do you really need all of these calculations? he is only dividing by two!

of course you can get more precise than the ruler would measure. how else would you be able to measure the thickness of a sheet of paper with a ruler?

[/ QUOTE ]

we are limited by the instruments we have available. she didnt have a SEM sitting at home.

[m_the0ry]

[ QUOTE ]
we are limited by the instruments we have available. she didnt have a SEM sitting at home.

[/ QUOTE ]

Well the point is you can get accuracies beyond that of the measuring device in scenarios like this, by utilizing the symmetry of the problem. You get a more accurate measurement of the radius when you measure the diameter instead of the radius. Can you see why? Consider oe39's statement about measuring the thickness of a sheet of paper versus measuring the thickness of a stack of papers when the number of papers in the stack is known.

[Siegmund]

The essential point is that significant figures are a shortcut to give you a rough idea of the uncertainty of a final answer. If you need to know the exact uncertainty you'll have to do propagation of errors. (In general, science classes that don't have calculus as a prerequisite use sig figs, and classes that do have calculus as a prerequisite expect full-fledged propagation of error.)

In your case, you have a measurement of 1.20±.01 cm. Dividing by two (exactly two, not a measurement of 2 point something) results in 0.600±.005 cm, and if you were using the propagation of errors method, you'd carry that .005 forward with you for the rest of the calculation (error of 1 part in 120 in R = error of 1 part in 60 in R^2, which will be combined with some error in L to give a final answer.)

Using significant figures, you have no way to represent an uncertainty of .005. You are forced to use either .60±.01 or .600±.001. Using the former would be more honest.

[jay_shark]

[ QUOTE ]


Ok, she measured the diameter of a circle to be 1.20cm. She was asked to calculate the volume, which is given by pir^2L.



[/ QUOTE ]

You must either mean volume of a sphere or area of a circle .

[Kaj]

[ QUOTE ]
[ QUOTE ]


Ok, she measured the diameter of a circle to be 1.20cm. She was asked to calculate the volume, which is given by pir^2L.



[/ QUOTE ]

You must either mean volume of a sphere or area of a circle .

[/ QUOTE ]

It's the volume of a cylinder of height L.