Dave restarted the gateway and the channels have reconnected.

I had forgotten to mention the Pcal LASERs.

Forgot to mention: all plots are calibrated in nm/nrad.

More clue on the TMSX measurement showing a different TF than before the vent (cf log below 19246). I looked at the response from pitch and vertical drive to the individual osems (LF, RT) and it looks like something is funny with the RT osem which should have the same response as the LF one, cf light red curves below 

EDIT : Actually, LF and RT osem response are superposed on the graphs (green LF lies under red RT), so their response is the same.

EY measurements

ETMY ant TMSY transfer functions were measured on thursday after the doors went on, with chamber at atmospheric pressure. They did not show signs of rubbing, cf the first two pdf attached showing good agreement with previous measurements.

Today, I remeasured the vertical dof, after the suspension sagged ~120um from the pressure drop, and it still looks fine, cf figures below.

 

 

 

EX measurements

ETMX and TMSX were measured today when pressure in chamber was about 3 Torrs, after the QUAD sagged by about 130um. 

 

 

The second pdf attachment on the log above was supposed to show TMSY transfer functions. Attached is the correct pdf. 

Here are photos from EY TMS work today:  https://ligoimages.mit.edu/?c=1616

This is the picture of the stainless steel strain relief in resource space.

The certificate for ligoimages.mit.edu has expired, so this site is currently not accessible.

I'm told this was intentional. I wasn't informed.

This was actually completely accidental.  Peter and I went out to see if we could pull a log off of the NPRO UPS as part of our investigation into this morning's NPRO trip.  Turns out that when you plug in a DB9 cable to the back of the UPS, the UPS shuts off.  We did not know this.  Now we do.  Note to self...

In other news we were not able to establish communication between the laptop and the UPS, not sure why.  Will continue to investigate.

That make me feel much better about not having gone out there alone to try it!

I used the channel H1:GRD-ISC_LOCK_STATE_N to identify locklosses in to make the pie chart of locklosses here, specifcally I looked for times when the state was lockloss or lockloss_drmi.  However, this is a 16 Hz channel and we can move through the lockloss state faster than 1/16th of a second, so doing this I missed some of the locklosses.  I've added 0.2 second pauses to the lockloss states to make sure they will be recorded by this 16 Hz cahnnel in the future.  This could be a bad thing since we should move to DOWN quickly to avoid ringing up suspension modes, but we can try it for now.  

A version of the lockloss pie chart that spans the end of ER7 is attached.  

I'm bothered that you found instances of the LOCKLOSS state not being recorded.  Guardian should never pass through a state without registering it, so I'm considering this a bug.

Another way you should be able to get around this in the LOCKLOSS state is by just removing the "return True" from LOCKLOSS.main().  If main returns True the state will complete immediately, after only the first cycle, which apparently can happen in less than one CAS cycle.  If main does not return True, then LOCKLOSS.run() will be executed, which defaults to returning True if not specified.  That will give the state one extra cycle, which will bump it's total execution time to just above one 16th of a second, therefore ensuring that the STATE channels will be set at least once.

reported as Guardian issue 881

Note that the corrected pie chart includes times that I interprerted as locklosses that in fact were times when the operators made requests that sent the IFO to down.  So, the message is that you can imagine the true picture of locklosses is somewhere intermediate between the firrst and the second pie charts. 

I realized this new mistake because Dave asked me for an example of a gps time when a lockloss was not recorded by the channel I grabbed from nds2, H1:GRD-ISC_LOCK_STATE_N.  An example is

1117959175 

I got rid of the return True from the main and added run states that just return true, so hopefully next time around the channel that is saved will record all locklosses. 

Looking at the initial power up, we can see that an increase of a factor of ~10 causes ~0.7 µrad of pitch misalignment. During the accelerated drift in the last 3-5 minutes before the lock loss another 0.4 µrad of pitch misalignment was acquired with only ~10% of power increase. One might wonder, if we see a geometrically induced wire heating run away.

I modeled how much the two front wires have to heat up to casue a bottom mass pitch of 1 microradian. A very small temperature increase is needed to predict this.

* Assuming a constant temperature profile along the wire length (I'm sure this is not the case, but it is easy to calculate), it is

0.003 [C]

* Assuming a linear temperature profile where with the max temperature is in the middle, and the ends of the wire have no temperature increase

0.006 [C]

So we can say an order of magnitude estimate is greater than 1 mC / urad and less than 10 mC / urad.

Calculations:

From gwinc, the thermal coefficient of expansion for C70 steel wire is

alpha = 12e-6 [1/C].

From the HLTS model at ../SusSVN/sus/trunk/Common/MatlabTools/TripleModel_Production/hltsopt_wire.m

wire length L = 0.255 [m]

front-back wire spacing s = 0.01 [m]

The change in wire length for pitch = 1 urad is then

dL = s * pitch = 0.01 * 1e-6 = 1e-8 [m]

* For uniform wire heating of dT, this change comes from

dL = alpha * L * dT

So, solving for dT

dT = dL / (alpha * L) = 1e-8 / ( 12e-6 * 0.255 ) = 0.0033 [C]

* For a linear temperature increase profile (max at middle, 0 at ends), I break the wire into many constant temperature segments of length Lsegment.

The temperature increase profile is a vector defined by

dT = dTmax * TempPrile

where TempProfile is a vector of the normalized shape of the temperature prodile. It is triangular, 0 at the ends and 1 at the peak in the middle. Each element of the vector corresponds to a constant temperature segment of the wire. dTmax is a scalar representing the maximum temeprature increase at the middle of the wire.

The change in wire length is then given by

dL = sum( alpha * Lsegment * TempProfile ) * dTmax

solving for dTmax

dTmax = dL / sum( alpha * Lsegment * TempProfile )

with 101 segments, this gives us

dTmax = 0.0063 [C]

about double the uniform heating case.

* I also considered that since the wire has significant stress due to the test mass weight, the Young's modulus's temperature dependence might cause a different effective thermal expansion coefficient alpha_effective. This appears to be a negligible effect.

From gwinc, the temperate dependence of the young's modulus E is

dE/dT = -2.5e-4 [1/C]

and young's modulus E is

E = 212e9 [Pa]

from https://alog.ligo-la.caltech.edu/aLOG/index.php?callRep=12581, we know that the change in spring length due to the modulus of eleasticity dependence is

dL = -dE/dT * dT * Tension / Stiffness

where Tension is the load in the wire and Stiffness is the vertical stiffness of the wire.

The Stiffness is given by

Stiffness = E * A / L = E * pi * r^2 / L

where A is the cross sectional area of the wire, and r is the radius.

So plugging this in above

dL = -dE/dT * dT * Tension * L / ( E * pi * r^2 )

We get the correction on alpha by dividing this by L and dT, which eliminates both from the equation. From the HLTS model, the bottom mass is 12.142 kg and the wire radius is 1.346e-4 m.

Tension = 12.142 * 9.81 / 4 = 29.8 [N]

The correction on alpha is then

-dE/dT * Tension / ( E * pi * r^2 ) = 2.5e-4 * 29.8 / (212e9 * pi * 1.346e-4^2) = 6.2e-7 [1/C]

This changes alpha from

12e-6 to 12.6e-6 [1/C]

Not enough to matter for the estimates above.

Localizing the heat source:

I made a calculation of the heat absorption by wires.

Based on Brett's temperature estimate, assuming the radiation as the only heat dissipation mechanism, the heat the front wires should be absorbing is about 1uW total per two wires when SR3 tilts by 1 urad regardless of the temperature distribution.

If you only look at the power, any ghost beam coming from PRC power (about 800W per 20W input assuming recycling gain of 40) can supply 1uW as each of these beams has O(10mW) or more.

I looked at BS AR reflection of X reflection, CP wedge AR both ways, and ITM AR both ways. I'm not sure about the first one, but the rest are mostly untouched by anything and falls on SR3 off centered.

The attachment depicts SR3 outline together with the position of CP wedge AR (green) and ITM AR (blue) reflections, assuming the perfect centering of the main beam and the SR3 baffle on SR3. Note that ITMX AR reflection of +X propagating beam falls roughly on the same position on SR3 as ITMY AR reflection of +Y propagating beam. Ditto for all ITM and CP AR reflections. The radius of these circles represent the beam radius. The power is simply 20W*G_rec(40)*(AR(X)+AR(Y))/4 (extra factor of 2 due to the fact that the AR beam goes through the BS) for ITM and CP, and 20W*40*AR/2 for BSAR of -X beam.

I haven't done any more calculations and I don't intend to, but just by looking at the numbers (total power in green and blue beams in the figure is about 240mW, 5 orders of magnitude larger than the heat absorbed by wires), and considering that the centering on SR3 cannot be perfect, and that SR3 baffle is somewhat larger than SR3 itself, and that CP alignment is somewhat arbitrary, it could be that these blobs seeps through the space between the baffle and the SR3 and provide 1uW.

The red thing is where BSAR reflection of -X beam would be if it is not clipped by the SR2 scraper baffle. If everything is as designed, SR2 scraper baffle will cut off 90% of the power (SR2 edge is 5mm outside of the center of the beam with 8mm radius), and remaining 10% comes back to the left edge of the red circle.

Any ghost beam originating from SRC power is (almost) exhonerated, because the wire (0.0106"=0.27mm diameter) is much smaller than any of the known beams such that it's difficult for these beams to dump 1uW on wires. For example the SRC power hitting SRM is about 600mW per 20W input, SRM AR reflection is already about 22uW.

Details of heat absorption:

When the temperature on a section of wire rises, the stretching of that section is proportional to the length of that section itself and the rise in temperature. Due to this, the total wire stretch is  proportional to the temperature rise integrated over the wire length (which is equial to the mean temperature rise multiplied by the wire length) regardless of the temperature distribution as is shown in effect by Brett's calculation:

stretch prop int^L_0 t dL = mean(t) * L

where L is the length of the wire and t is the difference from the room temperature.

Likewise, the heat dissipation of a short wire section of the length dL at temperature T+t via radiation is

sigma*E*C*dL*[(T+t)^4-T^4] ~ 4*sigma*E*C*dL*T^3*t

where sigma is Stefan-Boltzmann constant, E the emmissivity, C the circumference of the wire, T the room temperature (about 300K). The heat dissipation for the entire length of wire is obtained by integrating this over the length, and the relevant integral is int^L_0 t dL, so again the heat dissipation via radiation is proportional to the temperature rise integrated over the wire length regardless of the temperature distribution:

P(radiation) ~ 4*sigma*E*T^3*(C*L)*mean(t).

I assume the emmissivity E of the steel wire surface to be O(0.1). These wires are drawn, couldn't find the emissivity but it's 0.07 for polished steel surface and 0.24 for rolled steel plate.

I used T=300K, t=3mK (Brett's calculation for both of the temperature distributions), C=pi*0.0106", L=0.255m*2 for two front wires, and obtained:

P(radiation) ~ 0.8uW ~ 1uW.

ITM AR:

ITM has a wedge of 0.08 deg, thick side down.

ITM AR reflection of the beam propagating toward ETM is deflected by 2*wedge in +Z direction. For the beam propagating toward BS, ITM AR reflects the beam, deflecting down, and this beam is reflected by ITM and comes back to BS. Deflection of this beam relative to the main bean is -(1+n)*wedge.

AR beam displacement at BS is +14mm for +Z-deflection and -17mm for -Z-deflection. Since the BS baffle hole "radius" seen from ITMs is 100+ mm, and since the beam radius is about 53mm, AR  beams are not blocked much by BS baffle and reaches SR3.

ITM AR reflectivity is about 300ppm.

CP AR:

Similar calculation  for CP except that they have horizontal wedge, thick part being -Y for CPX and -X for CPY.

CP wedge is about 0.07 degrees.

I only looked at the surface of CP that is opposite of the ITM, and assumed that the surface facing ITM is more or less parallel to ITM AR, within an accuracy of O(100urad).

I assumed that S1 is the surface close to the ITM, and took S2 AR numbers from galaxy web page (43.7ppm for X, 5ppm for Y).

BS AR propagation:

BS wedge is 0.076 degrees, with a reflectivity of 50ppm.

Deflection of BS AR reflection of -X beam relative to the main beam is NOT -2*wedge as BS is tilted by 45 degrees. With some calculation it turns out that it is about -0.27 degrees, with a displacement of +48mm (positive = +X).

This beam is not obstructed at all by the BS baffle, hits SR3 and makes it to SR2 baffle edge. What made it to the SR2 surface doesn't go to SRM and instead comes back to SR3 as SR2 is convex  and the beam is heavily off-centered.

If there's no SR2 baffle and if SR2 is much larger, the center of the reflected beam is going to be  50cm in -X direction from the center of SRM, which happens to be on SR3.

I don't know what happens to the edge scattering and the reflection from SR2, but both of these are highly dependent on SR2 centering.