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## Re: [Opal] the initial phase and energy gain of double-gap RF Cavity in opal-cycl

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**From**: "Snuverink Jochem (PSI)" <jochem.snuverink AT psi.ch>**To**: Dou Guoliang <douglas610322 AT 163.com>**Cc**: opal <opal AT lists.psi.ch>**Subject**: Re: [Opal] the initial phase and energy gain of double-gap RF Cavity in opal-cycl**Date**: Thu, 4 Jun 2020 13:23:54 +0000**Accept-language**: en-US, de-CH

Hi Dou,

I can't quite follow the calculation, why do you take the minus the voltage for one the cavities? (and there is a typo, the first sin should be a cos I guess, otherwise the second line is wrong).

I don't think I understand exactly how your double gap cavity behaves. I can try to clarify the Injector 2 example:

I hope you already got a reply, if not apologies for the late reply. Some answers between the text.

I am using opal-cycl to simulate our compact-cyclotron these days. I started with single particle mode. But I always didn't get right initial phase （PHI0 in the element of 'RF cavity') of our double-gap RF cavities. So, I have some questions about the parameter "PHI0".

At first, I found "InjectII" in opal user-guide , which is an example with double-gap RF cavity.But there is no B-filed and cavity E-field file for running it (maybe you should add them and I will get answer).So I reference its computing method of PHI0(I think I get right method) and set it for our RF cavities . But I got wrong energy gain(dE) in the 'inputfile.out' . Maybe I got wrong calculation results,so I will explain my calculation process below.

The supplementary files are indeed missing from the manual for the injector-2 example, thanks for trying and the suggestion. Tomorrow these should be present on http://amas.web.psi.ch/opal/Documentation/master/OPAL_Manual.html#sec.tutorial.examplesbeamlines

Our cavites are two double-gap cavities with 33 deg opening angle in the azimuth position of 16.5，-16.5，163.5，-163.5 deg (I added structure diagram in attachment). I know that the opal-cycl use 2 single-gap replacing the double-gap cavity . We set the initial azimuth(PHIINIT in element of 'cyclotron') of particle equal to 90 deg . When we pay no attention with PHIINIT, I calculated the accelerate phase , which are -156 and -124 deg with the function of cos(wf*t+phi) RF volt(peak volt=0.065MV) for the first gap and second gap. The dE of two gaps are -0.065*26*cos(-156) and 0.065*26*cos(-24) . But 'inputfile.out' shows different results in dE and phase . My detailed calculation process had added in the attachment .

I don't know whether I get right understanding. When we use 2 single-gaps to replace the double-gap , their phases has no relation with opening angle of double-gap cavity. Their acceleration phase should set to 0. And for dE ,do we need to modify the peak volt to V*cos(phi)?

With OPAL, you need to treat a double gap as 2 single gaps, but then you need to input the single gap cavities how you would expect them as if they were actually single gap cavities. OPAL has no knowledge
that the two cavities are actually a double gap and treats them independently. So yes, you might want to change the peak voltage (with VOLT) to whatever peak voltage you effectively have. Note that the value 1.69 that you got is simply the peak voltage 0.065*26.

By the way,because of phase shift, the proper initial phase is difficult to find . Can you give me some advice to choose proper initial phase in OPAL ?

INJECTOR2: CYCLOTRON, TYPE="INJECTOR2", CYHARMON=10, PHIINIT=30.0, PRINIT=-.00550008,

RINIT=392.0,SYMMETRY=1.0, RFFREQ=50.6370, BSCALE=1, FMAPFN="ZYKL9Z.NAR";

RFM11: RFCAVITY, VOLT=.303619, FMAPFN="Cav1.dat", TYPE="SINGLEGAP", FREQ=50.6370,

RMIN=350.0, RMAX=3150.0, ANGLE=35.0, PDIS=0.0,

GAPWIDTH=0.0, PHI0=50.0;

RFM12: RFCAVITY, VOLT=.303619, FMAPFN="Cav1.dat", TYPE="SINGLEGAP", FREQ=50.6370,

RMIN=350.0, RMAX=3150.0, ANGLE=55.0, PDIS=0.0,

GAPWIDTH=0.0, PHI0=((50.0)+200);

GAPWIDTH=0.0, PHI0=50.0;

RFM12: RFCAVITY, VOLT=.303619, FMAPFN="Cav1.dat", TYPE="SINGLEGAP", FREQ=50.6370,

RMIN=350.0, RMAX=3150.0, ANGLE=55.0, PDIS=0.0,

GAPWIDTH=0.0, PHI0=((50.0)+200);

The cyclotron is a 10 harmonic. The injection angle (PHIINIT) is at 30 degrees.

The first cavity (first part of double gap) is at 35 degrees. To calculate the PHI0: (35-30) * 10 = 50.

The second cavity (second part of double gap) is at 55 degrees. PHI0: (55-30) * 10 = 250, (or 50 + 20*10 = 50 + 200)

This ensures that the reference particle sees the maximum voltage.

Hope that helps,

Jochem

**Re: [Opal] the initial phase and energy gain of double-gap RF Cavity in opal-cycl**,*Snuverink Jochem (PSI), 06/04/2020*

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