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    Make New Entry, Make Followup Entry

    User name chench

    Log entry time 20:58:16 on October 29, 2008

    Entry number 245420

    This entry is a followup to: 245412

    keyword=Beam Position Scan

    1. BPMA(x,y)=(-2.173,-1.907) BPMB(x,y)=(-2.662,-2.286) T1=4.14e+05
    T2=3.892e+02 T3=2.522e+05
    2.BPMA(X,Y)=(-2.082,-1.001) BPMB(x,y)=(-2.551,-1.022) T1=2.25e+04
    T2=2.999e+01 T3=2.075e+04
    3.BPMA(x,y)=(-1.911,2.038) BPMB(x,y)=(-2.416,1.989) T1=2.333e+03,
    T2=1.249e+01, T3=1.305e+04
    4.BPMA(x,y)=(-1.866,2.984) BPMB(x,y)=(-2.332,3.014) T1=2.294e+03
    T2=1.229e+01 T3=1.329e+04
    5.BPMA(x,y)=(-1.838,4.038) BPMB(x,y)=(-2.323,3.968) T1=2.201e+03,
    T2=1.249e+01 T3=1.351e+04
    6.BPMA(x,y)=(-2.018,1.095) BPMB(x,y)=(-2.501,1.050) T1=2.409e+03
    T2=1.274e+01 T3=1.340e+04
    To summarize, the central position of y is BMPAy=0.3 and BPMB=0.0

    Therefore, the centeral position of the target is BMPA(x,y)=(-2.0,0.3)
    BPMB(x,y)=(-2.5,0.0)