I want to investigate the effect of x-, y-, and z-displacements of my quadrupole magnets on several quantities (for example the envelope in x-direction and R56).
I thought I can solve my problems with the quantities DX, DY, and DZ in the QUAD-element, but at this point I do not understand how these quantities are working: I expected that for example DZ=1 leads to a shift of the quadrupole magnet in z-direction by 1 meter, but it doesn’t work. For example:
My idea was, to convert the original lattice
Code: Select all
D1a: DRIFT,L=2
Q1: QUAD,L=2, K1=1
D1a: DRIFT,L=2
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D1a: DRIFT,L=3
Q1: QUAD,L=2, K1=1
D1b: DRIFT,L=1
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D1a: DRIFT,L=2
Q1: QUAD,L=2, K1=1 , DZ=1
D1b: DRIFT,L=2
Furthermore I thought DX and DY work like shifting in the transverse positions, and if so, then I expected that the displacement of the quadrupole magnet in transverse direction leads to a “new” element which works like the superposition of a dipole magnet and a quadrupole magnet, or to say it in other words, B(x)=B_0*x becomes B(x)=B_0*x+B_1. If this is the case, R56 should only change in the range of this new element (because it is like a bending magnet), but in my example it changes even in the drift area behind the quadrupole magnet. I don’t understand this fact, because R56 is defined by $\int R16/\rho ds$ and the absolute value of $\rho$ should be infinity in the drift area, so R56 should not change. Besides this, R56 is zero in the quadrupole magnet (range 2 to 4 meters), but R16 is not zero in this range, so with regard to the aforementioned formula, R56 should start changing at postion s=2 meters, if the new element works like described before.
This is the result of R16 and R56, if I use the 4th test-beamline, using DX=1:
Can you please explain how DX, DY, and DZ are working, and if I can solve my problem with the already existing elegant elements.
Thanks in advance and best regards,
Felix