To get the round trip Gouy phase of the cavity to 1/3, the total round trip length should be 2.4 m.
Current mirror locations: (X,Y) = (10.655 in, 1.1033 in)
Current total round trip length = 4*X+4*sqrt(X2+Y2) = 85.47 in = 2.17 m
Need 2.4 m, so need to add 0.23 m = 9 in
This means every mirror needs to go 9 in / 8 = 1.125 in back
8 comes from 4 mirrors and the round trip
New mirror locations: (X,Y) = (11.779 in, 1.163 in)
New total round trip length = 4*X+4*sqrt(X2+Y2) = 94.46 in = 2.40 m
Thus, I plan on buying sixteen 1.125 in spacers for the every mirror.
Central vessel:
Treating as a cylinder
11.9 inch radius
11 inch height
V = pi*r^2*h = 4900 in^3 = 0.080 m^3
SA = 2*pi*r*h + 2*pi*r^2 = 1,700 in^2 = 1.1 m^2
End cube:
Treating as a cube, even though the inside is more spherical
L = 10 in
V = L^3 = 1,000 in^3 = 0.016 m^3
SA = 6*L^2 = 600 in^2 = 0.39 m^2
Tubes:
4 inch radius
275 in height (two 10 ft tubes + 1 ft T = 6.4 m, round up to 7 m including gate valve, reducer, etc)
V = pi*r^2*h = 13,800 in^3 = 0.23 m^3
SA = 2*pi*r*h = 6,900 in^2 = 4.5 m^2
Total:
Vessel + 5 Cubes + 2.1 Tubes (power recylcer ~0.7m):
V = 0.643 m^3 = 643 L
SA = 12.5 m^2
Assumed desorption: 1e-8 mbar*l/(s*cm²)
Assumed leak rate: 3e-9 mbar*l/s
Can then figure out what pumps and how many to use using the calculator
BTgeom_gouyph_1C.pdf Shows the round-trip Gouy phase (fractional) of the cavity with the 1.6m optic as a function of the total round trip length. We want a Gouy phase of 0.33333 to filter, so the total round trip length should be 2.4m.
Entire mouser order is in the uline cabinet. Using this google sheet to track what goes where. Anyone with this link + a caltech address should be able to edit it.
Using the calculated \( H_\infty \) limited controllers in the attached plot I show that as \( \gamma \) decreases the phase margin and the gain margin both increase. It also shows that the gain and phase margin are strongly correlated. In the plot the gamma is represented by the colors in the color bar and the current LIGO controller is shown as a blue triangle. The different lines of points are diffrent F1 gains and show that the relation between the gain margin and the phase margin is not always linear. However, generally speaking, this shows that gain margin and phase margin can be used somewhat interchangeably in this case. I'm not sure if this extends to all admissible controllers.
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]
The source for this is:
When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are \[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]
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