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Aug 23, 2010
05:47 PM

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Aug 23, 2010
05:47 PM

Segment of a circle (Pipeline in radius of curvature)

I have a very unique problem. In the attached worksheet I am trying to calculate the height of an arc at any point along the curve, not just the center. The arc is actually replicating a pipeline which has been placed in a radius of curvature.

The length of the chord and arc radius are know from a AutoCAD drawing, the first graph is correct and has been checked against my AutoCAD drawing, the problem is the sag bend which should be the opposite to the hog bend shown on the graph.

More explanation with in the worksheet.

**Mike**

60 REPLIES 60

Aug 24, 2010
12:13 AM

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Aug 24, 2010
12:13 AM

If you don't get replies, "saves as" 11 or lower versions,

or simply a diagram of the project.

Jean

Aug 24, 2010
02:32 AM

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Aug 24, 2010
01:15 PM

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Aug 24, 2010
01:15 PM

The very title "Segment of a circle" does not apply yet.

A very similar project was done in the former Mathcad collab [can't retrieve].

What is AutoCad doing in there ?

Here is my understanding, from total absence of analytical support in your work sheet.

jmG

Aug 24, 2010
02:04 PM

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Aug 24, 2010
02:04 PM

jean Giraud wrote:

What is AutoCad doing in there ?

AutoCad was stated because the pipeline has initially been drawn in the software and could be of aid to any members trying to solve my problem.

jean Giraud wrote:

Here is my understanding, from total absence of analytical support in your work sheet.

jmG

Your image in similiar to what I'm trying to achieve, but the pipeline is only placed through 1 hog and sag curve. On your image this would be represented by the section on the graph which lies between 2.47 & 3.25.

I'm not sure what you mean by "The very title "Segment of a circle" does not apply yet." either????

Mike

Aug 27, 2010
12:17 PM

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Aug 27, 2010
12:17 PM

>Your image in similiar to what I'm trying to achieve, but the pipeline is only placed through 1 hog and sag curve. On your image this would be represented by the section on the graph which lies between 2.47 & 3.25.

I'm not sure what you mean by "The very title "Segment of a circle" does not apply yet." either????

Mike

<<<<<<<<<<<<<<<<<<<<<<

It could be that I don't understand enough elementary physics. How can a pipe exist if not supported at both ends ? On a short run and technically feasible, I won't care and would install an intermediate support. Put yourself in the shoes of the designer, he must provide an installation drawing, usually in the 3 ISO views and most often isometric view. That piece of information is missing in your visit. If you have that ready "for construction" in Autocad, capture the screen image(s), let's say the visible project. Your drawing shows a curved line, how did it curve itself ? From an equation ? Putting the finger on he top of the screen ? Your boss did ? You understand my point: your drawing shows like "free end support", then the shape has an equation in that case. Your equation is a pure parabola, that can't be immediately, it starts "catenary equation". A suspended bridge is parabola, but a pure shape to start with is not parabola, but "chaînette".

Jean

Aug 27, 2010
12:40 PM

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Aug 27, 2010
12:40 PM

jean Giraud wrote:

It could be that I don't understand enough elementary physics. How can a pipe exist if not supported at both ends ? On a short run and technically feasible, I won't care and would install an intermediate support. Put yourself in the shoes of the designer, he must provide an installation drawing, usually in the 3 ISO views and most often isometric view. That piece of information is missing in your visit. If you have that ready "for construction" in Autocad, capture the screen image(s), let's say the visible project. Your drawing shows a curved line, how did it curve itself ? From an equation ? Putting the finger on he top of the screen ? Your boss did ? You understand my point: your drawing shows like "free end support", then the shape has an equation in that case. Your equation is a pure parabola, that can't be immediately, it starts "catenary equation". A suspended bridge is parabola, but a pure shape to start with is not parabola, but "chaînette".

Jean

Jean,

The pipeline will be subject to a radius of curvature which was drawn in Autocad. The self weight of the pipe will be enough to overcome the bending strain energy generated by bending the pipeline. No equation was needed. The radius was chosen after checks where carried out to ensure the pipeline would not fail under buckling.

The *.pdf I attached only shows the section of the pipeline which is passing over the proposed radius.

Seems like this problem won't be solved.

Richard gave a few tips, but I haven't had chance to look into his suggestions, yet.

Mike

Aug 27, 2010
01:34 PM

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Aug 27, 2010
01:34 PM

Jean,

The pipeline will be subject to a radius of curvature which was drawn in Autocad. The self weight of the pipe will be enough to overcome the bending strain energy generated by bending the pipeline. No equation was needed. The radius was chosen after checks where carried out to ensure the pipeline would not fail under buckling.

The *.pdf I attached only shows the section of the pipeline which is passing over the proposed radius.

Seems like this problem won't be solved.

Richard gave a few tips, but I haven't had chance to look into his suggestions, yet.

Mike

___________________________________

That's what I'm saying: drawn from what ? Magic Autocad ?

Your equation is a parabola, but you are saying that because of this, because of that it's not. Then what is it ? The other alternative about your project is simple: give me the certified drawing and I will digitize the trace and repost an approximation, from which approximation you will calculate what you want. Simple, is it ! You understand my point: many collabs had that attitude of pushing for an approval from here. Personally I can only treat what id defendable in some ways, in some ways mostly traceable. My offer will be traceable to your certified design.

Jean

Aug 27, 2010
01:41 PM

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Aug 27, 2010
01:41 PM

jean Giraud wrote:

The other alternative about your project is simple: give me the certified drawing and I will digitize the trace and repost an approximation, from which approximation you will calculate what you want. Simple, is it ! You understand my point: many collabs had that attitude of pushing for an approval from here. Personally I can only treat what id defendable in some ways, in some ways mostly traceable. My offer will be traceable to your certified design.

Jean

Jean,

I would be happy to supply the drawing.

What format do you want it in?

Mike

Aug 27, 2010
02:04 PM

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Aug 27, 2010
02:04 PM

The pipeline will be subject to a radius of curvature which was drawn in Autocad. The self weight of the pipe will be enough to overcome the bending strain energy generated by bending the pipeline. No equation was needed. The radius was chosen after checks where carried out to ensure the pipeline would not fail under buckling.

I believe what Jean is trying to point out is that a pipe that is supported at both ends and sagging due to gravity will not form a circular arc. I think it will form a catenary, but I'm not 100% sure of that.

I haven't had time to do anything with your drawing.

Aug 27, 2010
02:09 PM

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Aug 27, 2010
02:09 PM

Richard Jackson wrote:

I believe what Jean is trying to point out is that a pipe that is supported at both ends and sagging due to gravity will not form a circular arc. I think it will form a catenary, but I'm not 100% sure of that.

Richard,

The pipeline will be fully supported throughout the launch on conveyor rollers. The rollers which be leveled and positioned to ensure the pipeline follows the proposed radius.

I haven't had time to do anything with your drawing.

No problem, I understand collabs are busy and grateful for any help received.

Mike

Aug 27, 2010
02:25 PM

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Aug 27, 2010
02:25 PM

The pipeline will be fully supported throughout the launch on conveyor rollers. The rollers which be leveled and positioned to ensure the pipeline follows the proposed radius.

So why not just make it flat. Then you don't need Autocad or Mathcad. You can draw the pipe with a ruler and a pencil, and h(x)=0

Aug 27, 2010
02:33 PM

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Aug 27, 2010
02:33 PM

Richard Jackson wrote:

The pipeline will be fully supported throughout the launch on conveyor rollers. The rollers which be leveled and positioned to ensure the pipeline follows the proposed radius.

So why not just make it flat. Then you don't need Autocad or Mathcad. You can draw the pipe with a ruler and a pencil, and h(x)=0

Exactly. What couldn't we make it simple???

My company doesn't seem to do simple!!! If I started to explain my reasons for putting vertical bends in our pipeline lauchline, I would probably put you to sleep.

I am currently working on the problem and will post if resolved.

Mike

Aug 24, 2010
10:41 AM

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Aug 24, 2010
12:36 PM

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Aug 24, 2010
12:36 PM

Richard Jackson wrote:

Is this what you mean?

Richard,

As usual you are very close to finding the solution. Have a look at the attached worksheet - easier to explain.

I would like to post an image of the problem, but due to the scale in Autocad it's hard to see when plotted.

Mike

Aug 24, 2010
12:37 PM

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Aug 24, 2010
12:37 PM

Just got the "JIVE" error screen - made up now

Aug 24, 2010
02:14 PM

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Aug 24, 2010
02:14 PM

I will leave it there as it stands, i.e: not traceable R, l, s .

Underdesigned Font family and size

Lambda can't be made other than default size 10.

jmG

Aug 24, 2010
03:55 PM

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Aug 24, 2010
03:55 PM

Jean,

What is the root function calculating?

Mike

Aug 24, 2010
08:45 PM

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Aug 24, 2010
08:45 PM

MIke Armstrong wrote:

Jean,

What is the root function calculating?

Mike

Some useful stuff [eventually].

Jean

Aug 26, 2010
06:17 PM

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Aug 26, 2010
06:17 PM

jean Giraud wrote:

Some useful stuff [eventually].

Jean

Jean,

While I agree that images are useful some of the time, but the image must shown all defined variables. In the above image it's not clear how xxx has been defined, therefore not so clear to follow. A worksheet in this came would be much more useful.

Mike

Message was edited by: MIke Armstrong

Nice worksheet though.

Aug 26, 2010
09:09 PM

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Aug 26, 2010
09:09 PM

Jean,

While I agree that images are useful some of the time, but the image must shown all defined variables. In the above image it's not clear how xxx has been defined, therefore not so clear to follow. A worksheet in this came would be much more useful.

Mike

Message was edited by: MIke Armstrong

Nice worksheet though.

__________________________________

What I could do quick, did in a blank sheet then made an image. Images can be reposted to suit visits, not work sheets.

>In the above image it's not clear how xxx has been defined, therefore not so clear to follow<

The variable xxx comes from nowhere. Mathcad 11 and earlier versions take the scalar argument as dummy, for it to be assigned properly at the point it will be used [either plot or else calculations]. For instance, in erf(x) ... x is dummy and had served to code the construct of the numerical algorithm that defines erf. . So, if erf is erf, the coding is from the Mathcad built-in library, but if erf is whatever defined by the coding as expressed in the RHS of := ,,, from there, anytime later in the work sheet the dummy (x) argument can be infinitely assigned and ranged for various uses ,,, for various end use like for plotting several segments of different colors, of different colors signifying also and most probably different use. Operators like their process evolution to change colors when too high or too low before an alarm rings.

In two months and 3 days, I have posted 109 images, reflecting mostly collaboration, collaboration from quick help done in the collab work sheet. Do you think it would make sense I had resaved in file and plug my box with most stuff so badly dressed and that I have already.

My mother language is not english or worse "American", and I had no problem earning good money in both languages but I find damn insulting that a spell check for professionals does not suggest the correct spelling.

Jean

Jean

Aug 27, 2010
04:40 AM

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Aug 27, 2010
04:40 AM

jean Giraud wrote:

>In the above image it's not clear how xxx has been defined, therefore not so clear to follow<

The variable xxx comes from nowhere. Mathcad 11 and earlier versions take the scalar argument as dummy, for it to be assigned properly at the point it will be used.

Jean,

Ok my error, wasn't aware that was the case with M11. I am all for the posting of images as long as they show all the defined variables, so the image can be re-produced.

In two months and 3 days, I have posted 109 images, reflecting mostly collaboration, collaboration from quick help done in the collab work sheet. Do you think it would make sense I had resaved in file and plug my box with most stuff so badly dressed and that I have already. With Akiva, easy to repost the work sheet but with the PTC "facebook", that is not possible unless spending time and effort and accumulating useless stuff. Why should I make my collaboration more miserable than PTC has decided to make the Mathcad "Facebook" already so miserable.

Agree. Point taken.

Mike

Aug 27, 2010
08:56 AM

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Aug 27, 2010
08:56 AM

PTC does not want collabs to correct the work sheet and repost, that's why they have invented the *.zip. What I could do quick, did in a blank sheet then made an image. Images can be reposted to suit visits, not work sheets.

Just extract it from the zip, edit it, save it, repost it. The only difference to the Collab is the need to extract it from the zip file first. That one extra step is slightly irritating, but it harldy prevents you from reposting an edited worksheet.

Aug 27, 2010
11:23 AM

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Aug 27, 2010
11:23 AM

Richard Jackson wrote:

PTC does not want collabs to correct the work sheet and repost, that's why they have invented the *.zip. What I could do quick, did in a blank sheet then made an image. Images can be reposted to suit visits, not work sheets.

Just extract it from the zip, edit it, save it, repost it. The only difference to the Collab is the need to extract it from the zip file first. That one extra step is slightly irritating, but it harldy prevents you from reposting an edited worksheet.

They seem to have improved the zip: click left on the file, on the click it is uzipped, double click on the work sheet and it takes then few minutes. That's with Google Chrome. Can't save corrections made in the work sheet, you have to save as a new file in a new filing folder. All that BS because of the *.zip that does not reduce the original file size.

Aug 24, 2010
02:24 PM

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Aug 24, 2010
02:24 PM

I would like to post an image of the problem, but due to the scale in Autocad it's hard to see when plotted.

Sorry, but I'm lost. Can't you plot from Autocad to a pdf and post that?

Aug 24, 2010
03:38 PM

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Aug 24, 2010
03:38 PM

Richard Jackson wrote:

I would like to post an image of the problem, but due to the scale in Autocad it's hard to see when plotted.Sorry, but I'm lost. Can't you plot from Autocad to a pdf and post that?

Sorry, sometimes my explanations are a little poor.

I have attached a pdf of the Autocad sketch. The hog bends intercepts the sag bend at the tangent point.

Hope this helps.

Cheers

Mike

Aug 25, 2010
10:58 AM

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Aug 25, 2010
10:58 AM

Are you trying to calculate this in Mathcad (i.e. solve for the tangent point), or just plot it?

To plot it you need to know the location of the tangent point relative to the center of both arcs. You could get this from Autocad (I would need the length of the overlap of the two arcs). Then the equation of an arc centered at x0,y0, with radius R, is

where y1 is the positve half of the circle (hog bend) and y2 is the negative part of the circle (sag bend).

To make the second arc meet the first one at the correct point you need to caclulate x0 and y0 for the second arc. To do that you need to know the location of the tangent point.

Aug 25, 2010
02:38 PM

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Aug 25, 2010
02:38 PM

Richard Jackson wrote:

Are you trying to calculate this in Mathcad (i.e. solve for the tangent point), or just plot it?

To plot it you need to know the location of the tangent point relative to the center of both arcs. You could get this from Autocad (I would need the length of the overlap of the two arcs). Then the equation of an arc centered at x0,y0, with radius R, is

where y1 is the positve half of the circle (hog bend) and y2 is the negative part of the circle (sag bend).

To make the second arc meet the first one at the correct point you need to caclulate x0 and y0 for the second arc. To do that you need to know the location of the tangent point.

Richard have you got Autocad? I could post the drawing.

Basically, I am trying to find the "h" value at any length along the arc. I am not that concerned about the tangent point, just the plot.

If you haven't got Autocad, below is information you require.

tangent distance from center of hog bend = 10.447m

tangent distance from center of sag bend = 10.418m

Cheers for taking your time to look at it - been bugging me for a whie now.

Mike

Aug 25, 2010
06:18 PM

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Aug 25, 2010
06:18 PM

>Cheers for taking your time to look at it - been bugging me for a whie now.<

_________________________________

If your pipeline is zillions km long, it must be supported and the shape responds to a mathematical equation or approximation. If that piece of maths is not entered in AutoCad, then the graph is erroneous and the project stops there, with the missing analytical justification. Consult expertise near you. There must be some standard formulation. I don't have the piping design handbook [rats had breakfast with many of my best stuff except Mathcad !]. I think Mike you are pedalling in butter.

Jean

Aug 26, 2010
06:11 PM

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Aug 26, 2010
06:11 PM

jean Giraud wrote:

If that piece of maths is not entered in AutoCad, then the graph is erroneous and the project stops there, with the missing analytical justification.

Jean

Jean,

The pipeline radius is represented in Autocad as an arc with can be drawn at any desired radius. No maths is entered into Autocad.

I think Mike you are pedalling in butter.

Pedaling in butter? - Care to elaborate?

Mike