This was a continuation of the work covered in the previous paper. The aim was to see if the complete form of the LY and RY plots could be generated from radial-arm inclinometry. The conclusion was that they could be, but with residual errors of up to 4 arcseconds.

Data and tools

I used the radial-arm inclinometry from July 1997, and a standard inclinometry set from August 1997. The sets were 970712 RADY and 970819 LY and RY. I made extensive use of Excel and in particular the "solver" which I used to numerically optimise fits between observed and generated data.

Expected results

The radius of the track (and hence the length of the radial arm) is about 5m; a slope of 1 arsecond gives a movement at the end of about 25 microns. The length of the bottom of the A frame is about 7m, and so a movement of a wheel of 35 microns gives a slope of about 1 arcsec. If the track was a completely rigid surface, and the wheels each moved over it in the same way, I would expect the slope of the left-hand inclinometer (for example), to be given by

LY = (RADY Wheel 3 – RADY Wheel 4) * 5/7

Where "RADY Wheel 4" means the radial-inlcinometry result shifted in azimuth to the wheel 4 position.

There will be a certain amount of contamination of the RADY data because of the known rising and falling of the center bearing due to it’s known problems. The rise and fall is of order 20 microns and can thus be expected to "pollute" the answers at around the 1 arcsecond level. Sadly, because we do not know how many details of this movement, we cannot really say any more than that.

Generation of LY and RY from RADY

The radial arm inclinometry only gives results from wheel 3, at unevenly distributed azimuth angles and with an included slope of the telescope base. I removed the slope by fitted a sine curve, with minimised square residuals this had the form 20.8 * sin (Az + 76.6 deg) –3.22 arcsec. In order to generate track-joint interaction profiles for all four wheels I took a set of azimuths from 0 to 360 in steps of 0.25degrees. At each azimuth I simply looked up the value for next-largest azimuth for wheel 3. The data for RADY, thus processed, is shown in figure 1. For each other wheel I added the appropriate angular offset (W4: 84.9 deg; W1: 180 deg; W2: 264.9 deg) before the lookup. I only used the radial arm data from 0 to 360, and ignored the fifth quadrant. I did the same sin fitting and interpolation with the LY and RY data. The LY fitted sin curve was 4.55 * sin (Az + 22.7 deg) –0.70 arcsec. Note that the angle between the RADY and LY axes is about 50 deg; the difference between the fitted phase angles for RADY and LY of 53 degrees is thus as we would expect.

I made a generated LY profile in three ways, in each case recording the sum of the squares of the errors when compared to the actual LY result. In the first case I used a sum of the Wheel 3 and 4 profiles, each with a selected factor. (Recall that wheels 3 and 4 are on the left-hand a-frame.) See figure 1, which shows a fairly good fit between the generated profile and the actual data. The optimised factors gave a square difference sum of 3937 arcsec² and were

LYgen = W3* 1.22 – W4 * 0.67

Note that wheel 3 plays a much bigger role than wheel 4 – this can be seen in the forms of the wheel 3 and RADY data in figure 1.

In the second case, I included wheels 1 and 2 in the mix. This gave an interesting result because wheel 1 obviously plays a large role even though it is on the other side of the telescope. In this case the square difference sum was 3068, and the optimised factors were

LYgen = W3 * 1.34 – W4 * 0.64 + W1 * 0.39 + W2 * 0.08

In the third case, I also included quadratic terms. This did not have a dramatic effect; the square difference sum came down to 2736, and the quadratic factors were small. The optimised factors were

LYgen = W3 * 1.30 + W3² * 0.03 – W4 * 0.69 – W4² *0.03

+ W1 * 0.37 + W1² * 0.02 + W2 * 0.04 – W2² * 0.04

The difference plots for all three methods are shown in figure 3.

I carried out the same operations on the RY data; the results for both sets are given in table 1:
 

For RY the results were similar, but with two intriguing differences. The fits were noticeably worse – perhaps because wheel 3, on which the radial-arm inclinometer was set, is on the left. And the contribution from the wheels was much more even front-to-back, both for the direct contribution (wheels 1 and 2) and the indirect (3 and 4). The curves for the 4-variable case are given in figure 4, and the error term is added to figure 3. A final point to note (figure 3) is that the residuals for the left-hand side seem to mirror those for the right-hand side, at least to some extent. Returning to the LY data, I considered a possible explanation for wheel 3 being much more important than wheel 4, and for wheel 1 playing a role while wheel 2 didn’t. Perhaps wheels 1 as well as 3 influence the radial arm data, and so what we are really seeing is some sort of combined term in the wheel 3 and 1 factors. But of course the track/joint interactions are not at the same azimuths, so I don’t see how this theory could really hold water. I explore another similar thought in the next paragraph.

Investigation of RADY profile

From what I understand about the form of the track segments, each one is bowed up in the middle. This is borne out, I believe, by the Leica measurements. So I would expect RADY to have the form of consecutive "humps" with downwards-pointing spikes between them. This is by and large the case (figure 1) but in the region around 40<Az<60 there is an inverted hump. Interestingly the upwards spikes correspond more closely to wheel 1 intersections than to wheel 3 intersections, suggesting that the wheel 3 results are being affected by wheel 1 passing over the features 180 degrees away. This is credible in terms of the telescope structure. I made a model in which the observed result was given by the shape of the track as seen by wheel 3 plus some contribution from the part of the track wheel 1 is seeing, and worked back to get from the observed results to the "true" track profile. In order to completely get rid of the "inverted hump" I had to assume a large cross-coupling (about 0.8) which also implied that the spikes in other areas would need to be very large (15 arcsec or 0.5mm). Even with this rather improbable model, the fits as described above were not much better. I decided I had probably already crossed the line between processing the data to find what was there and massaging the data to get what I wanted - so I abandoned this line of investigation for now.

Conclusions

It is possible to explain most of the form of the LY and RY datasets in terms of the profile given by radial arm inclinometry, although there are residuals up to about 4 arcseconds. On the left hand side, the back wheel is much more important than the front in determining the overall slope of the beam. There is also a significant contribution from the right-hand front wheel. On the right hand side, the front and back wheels contribute much more evenly, and there is a contribution from both left-hand wheels. There is a puzzling symmetry between the residuals for the RY and LY data.
 
 
 
 

Figures follow: