Prev: Concat string to cell array of strings
Next: ignore output arguments of a function
From: Nicholas on 30 Jun 2010 15:06
Hello,
I am plotting a polar chart of the edge of a pipe. I am plotting out the deformation from the pipe's original location. On the chart there are different shaped rings showing different deformation at different times during the test.

Is there a way to make the center of the polar chart be negative? I want to show how the pipe shifted from its zero position and the number of millimeters it moved from that zero location.

Any help would he appreciated
From: Walter Roberson on 2 Jul 2010 01:13
"Nicholas " <andreanc(a)mcmaster.ca> wrote in message <i0g4jb$786$1(a)fred.mathworks.com>...

> I am plotting a polar chart of the edge of a pipe. I am plotting out the deformation from the pipe's original location. On the chart there are different shaped rings showing different deformation at different times during the test.

> Is there a way to make the center of the polar chart be negative? I want to show how the pipe shifted from its zero position and the number of millimeters it moved from that zero location.

You can change the labels on the rings (I detailed the method in some earlier posts -- it isn't _hard_ but it isn't trivial).

On the other hand, it makes no sense for a polar plot to start from a negative radius. Consider where the origin (0) would have to be on such a plot: since the origin has no associated angle, the origin would have to be smeared over the entire ring of radius labeled 0.
From: Nicholas on 5 Jul 2010 09:39
"Walter Roberson" <roberson(a)ibd.nrc-cnrc.gc.ca> wrote in message <i0jsgu$3ma$1(a)fred.mathworks.com>...
> "Nicholas " <andreanc(a)mcmaster.ca> wrote in message <i0g4jb$786$1(a)fred.mathworks.com>...
>
> > I am plotting a polar chart of the edge of a pipe. I am plotting out the deformation from the pipe's original location. On the chart there are different shaped rings showing different deformation at different times during the test.
>
> > Is there a way to make the center of the polar chart be negative? I want to show how the pipe shifted from its zero position and the number of millimeters it moved from that zero location.
>
> You can change the labels on the rings (I detailed the method in some earlier posts -- it isn't _hard_ but it isn't trivial).
>
> On the other hand, it makes no sense for a polar plot to start from a negative radius. Consider where the origin (0) would have to be on such a plot: since the origin has no associated angle, the origin would have to be smeared over the entire ring of radius labeled 0.

Thanks Walter Robertson for your help. I am reading through the page where you described how to change the labels. It will take me some time to figure it out.

I know that a standard polar graph would not make sense in this situation, but leaving this as a standard polar graph in this situation would be confusing.

I have a pipe of 1.5m in diameter and I am looking at a deformation of 1mm. I took the values and subtracted off .995% of the whole pipe. Now I have a graph ranging from 0 to 4mm in radius. 0 has no relation to the radius of the pipe other than it is 3.5mm less than the average radii. All values are in mm.

This isn't matlab code, but it may help describe things.

%to find the range of values I want to use
Radius2 = radius-0.995*mean(radius)

%the original radius of the pipe
OriginalPipeRadius2 =OriginalPipeRadius-0.995*OriginalPipeRadius = 3.5

The edge of the original pipe is now at the value of 3.5. Now, if the pipe edge moves in by one mm from the original pipe radius, it lands in the range of 2.5. If the pipe edge moves out by 1mm, it lands in the range of 4.5.

The original values are confusing to describe and the zero radius value of that chart does not relate to the pipe radius. I would like the original radius of the pipe to be a ring on the graph with a value of 0 so I can say that the radius is deformation in mm. When the ring moves inwards by 1mm, I want there to be a ring inwards at -1. Outwards by 1mm, I want a 1. If the center of the graph is -3, so be it; the center of the polar graph is unimportant. Only the deformation from the original pipe radius is critical.
 | 
Pages: 1
Prev: Concat string to cell array of strings
Next: ignore output arguments of a function