Question about Pi

[AxelOliver]

So, using the area formula for a circle:

A = πr²

and plugging in A = 78.5, solving for r gives:

r = √(A/π) = √(78.5/π)

Using π ≈ 3.1416, that turns into:

r = √(78.5/3.1416) ≈ √25 = 5

Looks right, but if pi isn’t rounded too early, does it change the final answer much? And in real-world stuff like construction or engineering, how much does rounding pi affect accuracy?

[swansont]

√(78.500/3.1416) = 24.987 so you're rounding the answer. Let's say you wanted to change it so that the result was affected, to say, 5.1. 5.1^2 ≈ 26. You'd have to use π = 3 to get there

The issue here is "significant digits" which is the precision of the numbers you use. Each digit you add increases the precision of the result, but that increase in precision gets smaller with each digit you add.

Let's say you were using a constant that was 4/9 = 4.4444... 

4.4 is 10% bigger than 4 - that's a big change

4.44 is 0.9% bigger than 4.4

4.444 is 0.09% bigger than 4.44 (and so on)

each digit has a smaller impact on how close you are to the correct value

The impact on the answer is an equation depends on how many digits the other numbers in the equation have and whether you are adding or multiplying. For multiplying, if the answer only needs 1 digit, and you're rounding, only 2 digits matter (the second digit tells you if you round up or down) so you don't need more than 2 digits in your constant, since a 0.9% change in the answer won't affect the result.

You want a 3-digit answer? Then use 4 digits in the constant. etc. etc.

For adding, it's just where that last significant digit is. If the result is going to be rounded to the nearest 10, you only ned the "ones" digit, i.e. you use 4.  If the result is to the nearest 0.01, you need to use 4.444 (anything beyond that is going to be rounded off)

[studiot]

1 hour ago, AxelOliver said:

So, using the area formula for a circle:

A = πr²

and plugging in A = 78.5, solving for r gives:

r = √(A/π) = √(78.5/π)

Using π ≈ 3.1416, that turns into:

r = √(78.5/3.1416) ≈ √25 = 5

Looks right, but if pi isn’t rounded too early, does it change the final answer much? And in real-world stuff like construction or engineering, how much does rounding pi affect accuracy?

Good Morning Axel, I see you are new here so welcome.

😀

 

You are correct in thinking that the value used for π affects the end result in calculations.

There is indeed considerable theory which is used in Science and Engineering to select the appropriate number of decimal places to use.

In order to explain this further perhaps you would like to tell us a little about you maths background since calculus is needed for this.

Do you know any calculus or anything about 'significant figures' ?

 

In the meantime a couple of rules of thumb.

Firstly it is often good enough to adopt the value that π  =  √10

Which can make a calculation easier as this will often cancel.

[math]r = \sqrt {\frac{{78.5}}{\pi }}  \simeq \sqrt {\frac{{78.5}}{{\sqrt {10} }}}  = \sqrt {\frac{{7.85*10}}{{\sqrt {10} }}}  = \sqrt {7.85*\sqrt {10} }  = 4.98[/math]

 

This example is trivial but shows the principle.

 

Secondly it is common practice to emplou one or two 'guard digits'[

So if you are expecting the result to be accurate to 3 significant figures (3SF) you would work to 4 or 5 sf and round  at the end.

Better theory would give you a better answer.

 

[AxelOliver]

18 hours ago, swansont said:

√(78.500/3.1416) = 24.987 so you're rounding the answer. Let's say you wanted to change it so that the result was affected, to say, 5.1. 5.1^2 ≈ 26. You'd have to use π = 3 to get there

The issue here is "significant digits" which is the precision of the numbers you use. Each digit you add increases the precision of the result, but that increase in precision gets smaller with each digit you add.

Let's say you were using a constant that was 4/9 = 4.4444... 

4.4 is 10% bigger than 4 - that's a big change

4.44 is 0.9% bigger than 4.4

4.444 is 0.09% bigger than 4.44 (and so on)

each digit has a smaller impact on how close you are to the correct value

The impact on the answer is an equation depends on how many digits the other numbers in the equation have and whether you are adding or multiplying. For multiplying, if the answer only needs 1 digit, and you're rounding, only 2 digits matter (the second digit tells you if you round up or down) so you don't need more than 2 digits in your constant, since a 0.9% change in the answer won't affect the result.

You want a 3-digit answer? Then use 4 digits in the constant. etc. etc.

For adding, it's just where that last significant digit is. If the result is going to be rounded to the nearest 10, you only ned the "ones" digit, i.e. you use 4.  If the result is to the nearest 0.01, you need to use 4.444 (anything beyond that is going to be rounded off)

Thanks for the breakdown! So in real world cases, how do you decide the cutoff for precision? Is it just industry standards or do people actually calculate how much error is acceptable?

[swansont]

7 hours ago, AxelOliver said:

Thanks for the breakdown! So in real world cases, how do you decide the cutoff for precision? Is it just industry standards or do people actually calculate how much error is acceptable?

It depends on the precision of numbers you know, usually limited by your measuring instrument. If you’re measuring a distance and can only do so to the nearest meter, that would tell you . e.g. you measure 11 meters, so you have 2 significant digits. If you could measure down to the cm, e.g. 11.27 m, then you have 4 significant digits.

[studiot]

1 hour ago, AxelOliver said:

Thanks for the breakdown! So in real world cases, how do you decide the cutoff for precision? Is it just industry standards or do people actually calculate how much error is acceptable?

Calculating percentage 'accuracies' can be very misleading which is why it is a pity you did not answer my post.

The questions I asked were designed to help us to help you since you have asked the same question twice now, this is the one I was am still concentrating on.

The answer to this is that industry standards have grown up on the basis of calculating the error.

But the analysis depends upon the circumstances and the formulae you are using which I why I asked about calculus.

You really need a good understanding of the meaning of the words, error, significant figures, decimal places, accuracy, precision.

 

Your example of the area of a circle is more difficult and less versatile than one using the circumference since it only contains multiplication.

If you would like to work through the following question, which also contains addition, and its implications please let us know.

It is a much more useful question to start with.

 

A man who is 2 metres tall walks all the way round the Earth at the equator.

Given that the radius of the Earth is 6378137.3472 metres, how much further does his head travel than his feet ?

This question is about what is known as 'the difference of two large numbers' and the loss of accuracy that can arise if it is handled in the wrong way.

The answer is also quite suprising.
 

If you want to proceed further, please let us know.