Significant digits rule when determining radius from diameter

In summary, the significant digits rule should not be applied to Math calculations when dealing with measurements.
  • #1
vcsharp2003
897
176
Homework Statement
If ##r## and ##d## are radius and diameter of a given circle, then to determine radius from diameter we use the formula ##r = d \div 2##. Suppose, ##d=5## then ##r=5 \div 2 = 2.5##. I have a question regarding significant digits in this calculation. We know the significant digits rule for dividing two numbers is that the resulting value must have as many significant digits as the minimum number of significant digits of dividend and divisor. In this case the minimum number of significant digits is 1 for dividend or the divisor, and therefore why we don't apply this rule and express the answer as a number up to 1 significant digit?
Relevant Equations
##r = d \div 2##
Probably, to satisfy the significant digits rule for division, we should consider ##r = 5.0 \div 2.0##. But I'm unable to come up with a reason why significant digits rule should not apply to ##r= d \div 2##. Also, if we apply significant digits rule to this calculation then we loose accuracy and so we should ignore the significant digits rule.
 
Last edited:
Physics news on Phys.org
  • #2
vcsharp2003 said:
the significant digits rule for dividing two numbers is that the resulting value must have as many significant digits as the minimum number of significant digits of dividend and divisor.
It is not a precise science. Common sense must be applied.
vcsharp2003 said:
if we apply significant digits rule to this calculation then we loose accuracy and so we should ignore the significant digits rule.
Quite.
 
  • Like
Likes vcsharp2003
  • #3
Hi,

The golden rule for error propagation is $$\Bigl (\Delta f(x,y)\Bigr )^2 = \left({\partial f\over\partial x}\right )^2(\Delta x)^2 + \left({\partial f\over\partial y}\right )^2(\Delta y)^2\ \ .$$
In the case ##\ r = d/2\ ## there is no uncertainty in the factor 2, so we have $$\Delta r = {\Delta d\over 2}$$the first example formula here.

##\ ##
 
  • Like
Likes vcsharp2003
  • #4
BvU said:
In the case r=d/2 there is no uncertainty in the factor 2
How does this fact imply that significant digits rule should not be applied to given scenario?
 
  • #5
haruspex said:
It is not a precise science. Common sense must be applied.
We can say that accuracy of r is same as accuracy of d since 2 is an exact number. Now, if we assume that ##d## is accurate then ##d \div 2## must also be accurate and we don't need to apply significant digits rule to accurate numbers. Only if there is some uncertainty in values of dividend and divisor, do we need to apply significant digits rule. Does this reasoning sound correct?
 
  • #6
vcsharp2003 said:
We can say that accuracy of r is same as accuracy of d since 2 is an exact number. Now, if we assume that ##d## is accurate then ##d \div 2## must also be accurate and we don't need to apply significant digits rule to accurate numbers. Only if there is some uncertainty in values of dividend and divisor, do we need to apply significant digits rule. Does this reasoning sound correct?
It depends on how the value of d is known. If it is given as 5cm in a made-up question you can take it to be exact. If someone measured it and stated the measurement as 5cm then you have a problem. Were they really not able to measure it more precisely than that, or are they just being lazy in not specifying it as 5.0cm, or whatever.

Taking it at face value, we are being told it is between 4.5cm and 5.5cm, so r is between 2.25cm and 2.75cm. If we try to express that by using an appropriate number of digits we have the choice of 2.5cm, implying it is between 2.45cm and 2.55cm, or 3cm, implying it is between 2.5cm and 3.5cm. Neither is satisfactory.
The only solution here is to state it as ##2.50\pm 0.25cm##.

Implying the precision from the number of digits works better when there are more of them to play with. If we start with d=1.005cm then it might not matter if we take r to be 0.503cm.
 
  • Like
Likes vcsharp2003
  • #7
I think the rules of significant digits should only be applied when dealing with measurements, otherwise they have no relevance and should not be applied.

For example, if I ask someone to tell the result of ##1 \div 3## then there is absolutely no need to apply significant digits since its just a Math calculation and not related to measurements. As a student reading the chapter on significant digits, I was always curious when these rules should be used and when they should not.
 
Last edited:

What is the significant digits rule?

The significant digits rule, also known as the significant figures rule, is a method used to determine the level of precision or accuracy in a measurement. It helps to determine the number of digits that are considered reliable in a measurement or calculation.

Why is the significant digits rule important?

The significant digits rule is important because it helps to ensure the accuracy and precision of a measurement. It also helps to prevent misleading or incorrect results in calculations.

How does the significant digits rule apply to determining radius from diameter?

When using the significant digits rule to determine the radius from diameter, you must consider the number of significant digits in both the diameter and the radius. The rule states that the result should have the same number of significant digits as the least precise measurement, which in this case is the diameter.

What are the steps for applying the significant digits rule to determine radius from diameter?

The steps for applying the significant digits rule to determine radius from diameter are as follows:

  • Count the number of significant digits in the diameter measurement.
  • Round the diameter measurement to the same number of significant digits as the least precise measurement.
  • Divide the rounded diameter by 2 to get the radius measurement.

Can the significant digits rule be applied to any measurement?

Yes, the significant digits rule can be applied to any measurement, whether it is length, mass, time, or any other physical quantity. It is a universal rule for determining the level of precision and accuracy in a measurement.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
6K
Replies
8
Views
3K
  • Other Physics Topics
Replies
13
Views
3K
  • Classical Physics
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
351
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • General Math
Replies
4
Views
7K
  • Beyond the Standard Models
Replies
19
Views
5K
Back
Top