half inch condit

[kwired]

Your fill is off. Look at the one small circle in large circle. You can fit about six more smaller circles around the existing one. The one circle is far from 53% of the larger. At 53%, the smaller circle should have a diameter 73% of the larger. What you have for diameter is more likely the radius.

was radius on that one think others are correct. Corrected version:

 

[K8MHZ]

was radius on that one think others are correct. Corrected version:

I say no.

Gimme a minute to do some math.......but it looks like the other examples are off, too.

 

[K8MHZ]

I say no.

Gimme a minute to do some math.......but it looks like the other examples are off, too.

For two conductors to have 31 percent fill they need to be .3937 inches in diameter each.

And three conductors having 40 percent fill would be a diameter of .367 inches each.

 

[kwired]

For two conductors to have 31 percent fill they need to be .3937 inches in diameter each.

And three conductors having 40 percent fill would be a diameter of .367 inches each.

Not doing very good - I had radius in all of the originals and not diameter- but there is enough consistancy that maybe the idea may be understood - will try to correct things and repost but don't know how soon I will get to it.

 

[K8MHZ]

Not doing very good - I had radius in all of the originals and not diameter- but there is enough consistancy that maybe the idea may be understood - will try to correct things and repost but don't know how soon I will get to it.

Somewhere on the net there is a graphic depicting exactly what you are trying to depict and in the same format. If I can find it I will post it or a link.

 

[K8MHZ]

Here is the best I can do:

Notice that the smallest percent fill, on the table is for two conductors (no more than 31% of the conduit?s interior volume). The reason for this is that two conductors, of the same size, collectively form an oval shape. One, or any number of cables greater than two, will tend to form a circular shape (see figure ZZ).

View attachment 6592

http://www.lanshack.com/DesigningConduitRuns.aspx

Not quite what I was looking for, but does a nice job explaining the reasoning.

 

[kwired]

Hopefully things are correct in this version. Don't do enough geometric calculations to be really sharp on it anymore apparently.

They are drawn to scale and results look more like what I expected this time around.

 

[K8MHZ]

kwired:

Much better! :thumbsup:

 

[kwired]

Here is the best I can do:



View attachment 6592

http://www.lanshack.com/DesigningConduitRuns.aspx

Not quite what I was looking for, but does a nice job explaining the reasoning.

In my drawing (if I can ever get it right) I am attempting to show to scale a circle filled with 2 circles that take up 31% of the area, and 3 circles that take up 40% of the available area, as well as one that takes up 53% of available area.

Just any two or three sizes will not necessarily demonstrate the 31 and 40% fill.

 

[K8MHZ]

In my drawing (if I can ever get it right) I am attempting to show to scale a circle filled with 2 circles that take up 31% of the area, and 3 circles that take up 40% of the available area, as well as one that takes up 53% of available area.

I think your corrected version does just that. I went over your figures and I don't see anything wrong with them. Nice job. I didn't check your figures for the distance to the sides, though.

I'm taking it for granted they are correct. Plus I don't know where you are measuring from.

 

[Smart $]

In my drawing (if I can ever get it right) I am attempting to show to scale a circle filled with 2 circles that take up 31% of the area, and 3 circles that take up 40% of the available area, as well as one that takes up 53% of available area.

Just any two or three sizes will not necessarily demonstrate the 31 and 40% fill.

Looks right. Here's mine...

 

[Dennis Alwon]

So no matter how you look at it 31% is still less fill then 40%. This seems to lead us to think it all has to do with the pulling of the wires not just the amount of free space

 

[kwired]

I think your corrected version does just that. I went over your figures and I don't see anything wrong with them. Nice job. I didn't check your figures for the distance to the sides, though.

I'm taking it for granted they are correct. Plus I don't know where you are measuring from.

I took the diameters and (if you but three together) you can draw a triangle with the sides being 2 times diameter. I then figured if the center of the triangle is also the center of the larger circle - what distance each corner of triangle should be from circumference of the larger circle. Think I have it correct.

 

[acrwc10]

So no matter how you look at it 31% is still less fill then 40%. This seems to lead us to think it all has to do with the pulling of the wires not just the amount of free space

I think the existence of MC cable proves your point.

 

[kwired]

So no matter how you look at it 31% is still less fill then 40%. This seems to lead us to think it all has to do with the pulling of the wires not just the amount of free space

Take a close look at the illustrations. 2 conductors that use up 31% of the area use more of the diameter of the raceway than 3 conductors that use 40 percent of area of same raceway.

You wouldn't need to increase the size of conductors very much and they will be really tight as far as diameter goes but there would still be a lot of unused area.

 

[JDB3]

Just wondering if the test question, means the ampacity of the conduit itself? 1 square inch of copper 1,000 amps,
1 square inch of aluminum 750 amps, 1/2" EMT equals ___amps?