I also have Y and Z feedforward working on the SRC HAMs. I attach performance plots (taken at the same time as the plots from my main post). These have been running on HAM4 for a little while (sometime after ER7 ended), but I never got around to doing the alog and I was a little more organized when I installed them on HAM5 today.  First plot is Y, second is Z. Active measurement is with FF on, reference is FF off. Again, we really need a cavity to say if these are good enough, but I leave them running for now.

I've looked at X, RX and RZ, but RX and RZ show low coherence and X looks... messy, see last plot.

That was quick! Looking good.
Thanks.

J. Kissel, J. Warner

Some additional information and/or a "current status:"
ISIs HAM2 and HAM3 do not have any ST0 / HEPI L4C feed-forward running.
ISIs HAM4 and HAM5 have Y, Z, and RY ST0 L4C (not HEPI L4C) feed-forward running.
(HAM6 is currently vented and the ISI and HEPI are locked.)

The HAM4 and HAM5 filters, for Y, Z, and RY live in FMs 2, 3, and 4 respectively. The gain for all DOFs on both HAMs is 0.5.

Norna's designing / modelling how adding blades between the HSTS's lowest stages will improve performance in the SUS's vertical displacement. The input motion for the SUS's suspension point in vertical is composed of the ISI's center of mass moving in Z, RX, and RY (see T1100617). She noticed from the results Jim posted (T1500289), that at 25-30 [Hz], the input V motion was dominated by RX / RY of the table. So, among other ways to improve the performance at these frequencies (see them discussed in SWG aLOG 11327), Jim tried improving the RY DOFs today -- and won!

Nice work, as always, Jim!

There was a problem with HAM5 at the time I took this data. I've taken new measurements from HAM4, see alog 19343. Conclusions remains the same, I think, but the data is cleaner.

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.

I don't think the IFO stayed locked for the whole time. The summary page says it lost lock at 20:48:41, and a time series of DCPD_SUM (first plot) seems to confirm that. I did a few spectrograms, and Omega scanned each 10 seconds in the second series, and the only glitch I find is a very big one at 20:48:00.5. Here is an Omega scan. The fourth tap after this one caused the lock loss. Detchar will look closer at this time to see if there are any quiet glitches. First we'll need to regenerate the Omicron triggers... they're missing around this time probably due to having too many triggers caused by the lockloss.

I'm not sure why there's a discrepancy in the locked times with Robert's report.

Any chance this could be scattered light? at 1 m/s^2 and 1 kHz, it is a displacement of 25 nm, so you don't need fringe wrapping.

I think that you would expect any mechanism that does not require release of a particle to occur pretty much with every tap. These glitches dont happen with every tap. 

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.