NIBEND

[nnaka]

Dear Michael,

I have a question about NIBEND.
We would like to fix the total deflection angle of a sector magnet with the fringe fields to 45 degrees in this element.
The bending radius is 1m for the hard-edge model without the fringe fields, but it should be changed more or less to keep
the same deflection angle. It is possible in this element?
In order to obtain the path length difference, we used the following descriptions (RUN1 & RUN2):

Point1:WATCH, filename=Point1.wat
Point2:WATCH, filename=Point2.wat
L600: DRIFT, L=0.6
BEND1 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0
BEND2 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0, FINT=0.36, HGAP=0.03, Model=linear, FUDGE_PATH_LENGTH=0

RUN1: LINE=(L600, BEND1, L600, Point1)
RUN2: LINE=(L600, BEND2, L600, Point2)

We calculated Ct values at Point 1 and 2 by elegant, obtained the Ct difference between the two points and compared them
with a simple analytical estimation. They were not consistent (analytical:4mm, elegant:0.5-0.6mm), so I think that the deflection
angle was changed in this element. It is true ? How about the bending radius ?
I would be grateful to you if you could answer the questions.

Best regards,
Norio Nakamura

[michael_borland]

Norio,

By default, NIBEND does not change the deflection angle. It uses one of two strategies to make sure it is correct:
1. Symmetrically adjust the locations in z of the edge of the field in order to get the required bending angle. This is controlled by the ADJUST_BOUNDARY parameter, which is 1 by default.
2. Adjust the overall field strength in order to get the required bending angle. This is controlled by the ADJUST_FIELD parameter, which is 0 by default.

If you set both of these to 0, the angle would be wrong. I get similar results regardless of which of these I choose to set to 1.

--Michael

[nnaka]

Michael,

I have understood two strategies (ADJUST_BOUNDARY and ADUST_FIELD) you suggested.
I can obtain almost the same result for these two strategies.
But the problem is that the difference of the two path lengths without and with fringe fields calculated
with elegant is not in agreement with the result of an easy estimation I attached to this mail.
The easy(rough) estimation is 3-4 mm for our bending magnet (bending radius=1m, angle=45 degrees).
This result is consistent with the result calculated from 3-D field analysis code OPERA (about 4 mm).
But elegant result is only 0.5 - 0.6 mm. I obtained the path length difference value from the difference of Ct-values
between the two Points (Point1 and Point2) described below.

Lattice:
BEND1 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0
BEND2 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0, FINT=0.36, HGAP=0.03, Model=linear, FUDGE_PATH_LENGTH=0, ADUST_FIELD=1

RUN1: LINE=(BEND1, Point1)
RUN2: LINE=(BEND2, Point2)

Result:
Ct at Point 1 = 2.620080e-009 sec
Ct at Point 2 = 2.618307e-009 sec

velocity=2.997611346x10^-8 at pCentral=69.17140 (about 35MeV).

I cannot understand why the two results of the path length difference are much different when the bending angle is fixed.
I would be grateful to you if you could answer my question or make some comment.

Best regards,
Norio Nakamura

[michael_borland]

Norio,

Sorry for the delayed reply. The forum email notification feature is not alerting me to new activity for some reason.

I think the explanation for this discrepancy is that NIBEND assumes that the fringe field is centered on the nominal exit plane of the magnet, whereas you've assumed that the fringe field is entirely outside the magnet. Neither of these assumptions is entirely correct, according to my experience. It depends on the design of the magnet end, e.g., whether there is a field clamp or not.

--Michael

[nnaka]

Michael,

If NIBEND assumes that the fringe field is centered on the nominal exit plane of the magnet as you said,
can we accurately calculate the path length or Ct-value of by NIBEND or elegant for a magnetic field distribution
I attached to this mail ? In this distribution, the fringe field is entirely outside the magnet and linearly
decreased. If we can, please tell me what values I should input into the NIBEND parameters for this
distribution. I already tried the followings:

(0) BEND1 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0
(1) BEND2 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0, FINT=1/6, HGAP=1/6, Model=linear, FUDGE_PATH_LENGTH=0, ADUST_FIELD=1
(2) BEND2 :NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=E2=0, FINT=1/6, HGAP=1/6, Model=linear, FUDGE_PATH_LENGTH=0
(3) BEND2 :NIBEND, L="1 pi * 4 / 0.06 +", ANGLE="pi 4 /", E1=E2=0, FINT=1/6, HGAP=0.03, Model=linear, FUDGE_PATH_LENGTH=0, ADUST_FIELD=1
(4) BEND2 :NIBEND, L="1 pi * 4 / 0.06 +", ANGLE="pi 4 /", E1=E2=0, FINT=1/6, HGAP=0.03, Model=linear, FUDGE_PATH_LENGTH=0

For Cases (1) and (2), the path lengths become shorter by 0.116 and 0.123 mm than that of Case (0) (hard-edge model) because the fringe field is centered
on the exit plane of the magnet as you said. For Cases (3) and (4), we input the effective length into L. For these cases the path lengths become larger
by 59.892 and 60.000 mm than that of Case (0). This path length differences are too large. If 60 mm is subtracted from these path length differences,
the new differences are -0.108 and 0.000 mm. Anyway, these are much different from the path length difference (3-4mm) we estimated in the previous slide.
Thank you in advance for your reply.

Best regards,
Norio Nakamura

[michael_borland]

Norio,

I've looked into this in more detail and come to the conclusion that elegant is giving the correct answer. I'll try to write it up in the next week. However, one point is that simple estimates of the path-length change can be misleading.

I found an approximate semi-analytical solution for two cases: (1) fringe centered on the reference plane and (2) fringe entirely outside the reference plane. I also modified elegant to allow offseting the fringe from the centered position so that case (2) can be simulated. For case (1), using the bending magnets defined in your last message, I got ds=-0.64mm analytically and -0.53mm from elegant. For case (2), I got -7.3 mm analytically and -7.2 mm from elegant.

Incidentally, if the fringe is entirely inside the reference plane, the result (from elegant), is ds=+6.1mm.

As I said, I'll try to write this up in the next week and also release the new version that allows controlling the fringe positioning.

--Michael

[nnaka]

Michael,

Thank you very much for your time and effort to answering my question.
I would like to check whether your calculation result of ds=-0.53mm is for the case of FINT=0.36
and not the case of FINT=1/6(=0.167).
Anyway I am looking forward to the new version.

Best regards,
Norio Nakamura

[michael_borland]

Norio,

I used
NIBEND, L="1 pi * 4 /", ANGLE="pi 4 /", E1=0,E2=0, FINT=0.36, HGAP=0.03, ...

So the length of the fringe region is 12*FINT*HGAP=0.1296 m

--Michael

[michael_borland]

Norio,

Here's a summary of my analysis and tests related to the path-length issue in the NIBEND element. My conclusion is that there is good agreement between the semi-analytical results and elegant.

--Michael