Determining Diameter of Pipe with Drilled Hole

[Casio1]

A textbook shows a circle with a line drawn through the centre and advises that the line represents the diameter of the circle, ok with that, but what happens if the circle was a pipe and the pipe had a hole drilled into it, so the pipe now has an outer diameter of 10 but the hole inside the pipe is 5.

How do you define the diameter of the pipe now, is it at the outer edge of the circle or the inner circle diameter, where the hole was drilled through?

Thanks.

 

[Ackbach]

In applications, you call one the inner diameter, and the other the outer diameter. Common abbreviations are ID and OD.

 

[SuperSonic4]

A textbook shows a circle with a line drawn through the centre and advises that the line represents the diameter of the circle, ok with that, but what happens if the circle was a pipe and the pipe had a hole drilled into it, so the pipe now has an outer diameter of 10 but the hole inside the pipe is 5.

How do you define the diameter of the pipe now, is it at the outer edge of the circle or the inner circle diameter, where the hole was drilled through?

Thanks.

You have an outer diameter and an inner diameter as Ackbach said and any good textbook/exam paper/problem/application will make this clear. For a thin pipe you may be able to assume they are equal, especially if the gap in the middle is comparatively large.

Consider this picture of a washer (it's the same as a cross-section of your pipe but easier to visualise) - the red horizontal line is the outer diameter and the blue horizontal line is the inner diameter.

You usually come across these problems when you need to find the area of the shaded part (or, using the example above, the area of the washer)

 

[Casio1]

You have an outer diameter and an inner diameter as Ackbach said and any good textbook/exam paper/problem/application will make this clear. For a thin pipe you may be able to assume they are equal, especially if the gap in the middle is comparatively large.

Consider this picture of a washer (it's the same as a cross-section of your pipe but easier to visualise) - the red horizontal line is the outer diameter and the blue horizontal line is the inner diameter.

You usually come across these problems when you need to find the area of the shaded part (or, using the example above, the area of the washer)

I was OK with the above understanding but I did have a method for calculating the inside diameter as above, which I thought was something like;

OD^2 - (OD^2 - Id^2) / pi

The problem is that I can't find the original material I got the above out of so am at a loss if the above is correct or I have got something wrong?

The other problem is that I believe that in your washer example as is the same as my pipe, the annulus is the wall, and there is only one circular wall and some people are trying to tell me there are two walls when trying to calculate the thickness of the wall?

I had this lot in a book which I can't find now

 

[HallsofIvy]

I was OK with the above understanding but I did have a method for calculating the inside diameter as above, which I thought was something like;

OD^2 - (OD^2 - Id^2) / pi

I don't understand what you mean here. You said this was "a method for calculating the inside diameter" but the formula involves the inside diameter. You can't have a formula for inner diameter based only on the outer diameter because you can drill a hole in a solid pipe with diameter anywhere from 0 up to the outer diameter.

The problem is that I can't find the original material I got the above out of so am at a loss if the above is correct or I have got something wrong?

The other problem is that I believe that in your washer example as is the same as my pipe, the annulus is the wall, and there is only one circular wall and some people are trying to tell me there are two walls when trying to calculate the thickness of the wall?

The people, I suspect, are talking about the two sides of the 'wall'. The thickness of the 'wall'- actually of the pipe- is the difference between inner and outer radius- that is (OD- ID)/2, half the difference of outer diameter and inner diameter because 'radius' is half the 'diameter'.

I had this lot in a book which I can't find now