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    Make New Entry, Make Followup Entry

    User name chench

    Log entry time 20:39:49 on October29,2008

    Entry number 245412

    Followups:

    keyword=Beam Position Scan

    A. Horizontal Scan (Beam Current~4.5uA)
    1.BPMA(x,y)=(-2.017,0.368) BPMB(x,y)=(-2.485,-0.099) T1=2.914e+03, T2=1.249e+01, T3=1.409e+04
    2.BPMA(x,y)=(-3.913,0.356) BPMB(x,y)=(-4.591,0.048) T1=1.635e+05, T2=7.071e+02, T3=3.622e+04
    3.BPMA(x,y)=(-2.943,0.361), BPMB(x,y)=(-3.988,-0.057) T1=8.2e+04, T2=2.91e+02, T3=2.75e+04
    4.BPMA(x,y)=(-0.042,0.396),BPMB(x,y)=(-0.516,0.112) T1=2.067e+04 T2=4.123e+01 T3=1483e+04
    5.BPMA(x,y)=(-1.005,0.372),BPMB(x,y)=(-1.561,-0.096) T1=1.925e+04 T2=1.249e+01 T3=1.460e+04
    For that case we find the centeral position of x is: BPMAX=-2.0 BPMB=-2.5

    B. Vertical Scan