Документ взят из кэша поисковой машины. Адрес оригинального документа : http://site2010.sai.msu.ru/static/doc/ATokovinin_poster_site2010.pdf Дата изменения: Wed Oct 27 17:05:00 2010 Дата индексирования: Mon Oct 1 20:10:43 2012 Кодировка:

Atm
2

ospheric time constant with MASS and FADE
0

A. Tokovinin CTIO, Chile

Adaptive Optics time constant
0 = 0.314 r0/V0 = 0.057 6/5 [ Cn2(h) V5/3(h) dh MASS time constant:
0, MASS,

Numerical simulation
1. Set turbulent layer strength (seeing) & sp. response

Definition (1 rad phase difference between wavefronts at t= 0)

]

3/5

= 0.175 (2

DESI

)

3/5

; 2

DESI



= < [ IA(1ms)/ IA(3ms) ] >2
0, MASS,

This is an approximation! So scale it, 0 C

What is the best value of the corrective coefficient C? Previous simulations: C=1.27 while Travouillon et al. [4] use C=1.7

2. Generate intensity at the ground

3. Apply spatial filter, compute DESI for different wind speeds

h=5km

h=15km C=1.27

MASS overestimates 0 at slow wind speeds and under strong scintillation. The coefficient C=1.27 is good for typical situations (turbulence at 200mb, wind ~30m/s). However, C varies from 0.6 to 1.6. The MASS method is approximate!

Results

Corrective coefficient C vs. wind speed for different seeing

Fast Defocus = FADE
The temporal structure function (SF) of the Zernike defocus coefficient a4 (in rad2) has initial quadratic part D4(t) 0.036 (D/r0)1/3 (t/0)2 . This gives a clear method to measure 0 [2].

Comparison between FADE and MASS
SOAR Adaptive Module tests at Cerro Pachon, Chile Night 1 (Aug 31, 2009): clear, with strong and fast turbulence Night 2 (Oct. 2, 2009): better seeing, but only for 4 hours.

Method of Fusco et al. [3]: halftime
The halfwidth of the SF (or covariance) of Zernike aberrations is related to the wind speed V. For defocus and single layer, t0.5 = 0.30 (D/V), hence 0 = 1.05 t0.5 (r0/D) In the multilayer case, t0.5 measures the strongest layer, while 0 depends mostly on the fastest layer. Test case: two layers with 1" and 0.4" seeing, V=[10, 60]m/s. For each layer, 0 =[3.17, 1.32]ms, combined 0 =1.16ms The halftime method gives t0.5=0.105s 0 =2.39ms The true time constant is 0 =1.16ms. The halftime method works well for one layer, but is biased by 2x in this test case. Example of data (night 1). Points ­ measured SF, curve ­ twolayer model fitted to the first 0.1s segment

FADE vs. MASS with C=1.27

sum



Structure function of a4 for the 2layer test case

fast layer



[1 [2 [3 [4

] Tokovinin A. 2002, Appl. Opt., 41, 957 ] Tokovinin A. et al. 2008, A&A, 477, 671 ] Fusco T. et al. 2004, J. Opt., 6, 585 ] Travouillon et al. 2009, PASP, 121, 787

References

http://www.ctio.noao.edu/~atokovin/profiler/archive.html