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- 1. Contactless Calliper Reporter –Giga Khizanishvili
- 2. CONTACTLESS CALLIPER Invent andconstruct an optical device that uses a laser pointer and allows contactless determination of thickness, refractive index, and other properties of a glass sheet. 25.02.2017 2Giga Khizanishvili
- 3. Problem Theoreticalmodel Experiments The margin of error Compare of theory and experiments Conclusion 25.02.2017 3Giga Khizanishvili
- 4. 𝛽 𝛽 𝛼 𝛼 d 𝑠 25.02.20174Giga Khizanishvili n2 n1 𝑥s = 𝑥 ∗ 𝑠𝑖𝑛𝑎 𝑠
- 5. Refractive index 1,0002926 𝒔𝒊𝒏𝜶 𝒔𝒊𝒏𝜷 = 𝒏𝟐 𝒏 𝟏 n1 (Air) ≈ 1 25.02.2017Giga Khizanishvili 5 n1 n2 𝒔𝒊𝒏𝜶 = 𝒔𝒊𝒏𝜷 ∗ 𝒏
- 6. 25.02.2017Giga Khizanishvili 6 cos𝛼 = 𝑠 𝑙 𝑐𝑡𝑔𝛽 = 2𝑑 𝑙 𝑙 = 2𝑑 𝑐𝑡𝑔𝛽 𝑐𝑜𝑠𝛼 = 𝑠 ∗ 𝑐𝑡𝑔𝛽 2𝑑 𝑑 = 𝑠 ∗ 𝑐𝑡𝑔𝛽 2 ∗ 𝑐𝑜𝑠𝑎 = 𝑠 ∗ 𝑐𝑜𝑠𝛽 2 ∗ 𝑐𝑜𝑠𝛼 ∗ 𝑠𝑖𝑛𝛽 = 𝑠 ∗ 1 − 𝑠𝑖𝑛2 𝛽 2 ∗ 𝑐𝑜𝑠𝛼 ∗ 𝑠𝑖𝑛𝛽 𝑠𝑖𝑛𝛽 ∗ n = 𝑠𝑖𝑛𝑎 ⇒ 𝑠𝑖𝑛𝛽 = 𝑠𝑖𝑛𝑎 𝑛 𝑑 = 𝑠 ∗ 𝑛 ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑛2 𝑠𝑖𝑛2𝑎
- 7. 25.02.2017Giga Khizanishvili 7 d n 𝑠1 𝑎1𝑎1 𝑠1 𝑎1 𝐹𝑖𝑟𝑠𝑡 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡
- 8. 25.02.2017Giga Khizanishvili 8 d n 𝑠2 𝑎2𝑎2𝑠2 𝑎1 𝑎2 𝑆𝑒𝑐𝑜𝑛𝑑 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡
- 9. 25.02.2017Giga Khizanishvili 9 𝑑= s1 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎1 𝑛2 𝑠𝑖𝑛2𝑎1 𝑑 = s2 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎2 𝑛2 𝑠𝑖𝑛2𝑎2 s1 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎1 𝑛2 𝑠𝑖𝑛2𝑎1 = s2 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎2 𝑛2 𝑠𝑖𝑛2𝑎2 𝒏 = 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏
- 10. 25.02.2017Giga Khizanishvili 10 𝑎 𝑎 𝑥 𝑦 𝑦= 𝑥 ∗ 𝑠𝑖𝑛𝑎 𝑠 𝐹 𝑑 cos(90 − 𝑎) − 𝑥 ∗ 𝑐𝑜𝑠𝑎 2 ∗ 𝐹 ∗ cos(90 − 𝑎) 𝑑 − 𝑥 ∗ 𝑠𝑖𝑛2𝑎 = 𝑠 𝑥 ∗ 𝑐𝑜𝑠𝑎 𝑠 = 2 ∗ F ∗ cos 90 − 𝑎 ∗ 𝑥 ∗ 𝑐𝑜𝑠𝑎 𝑑 − 𝑥 ∗ 𝑠𝑖𝑛2𝑎 𝑑 𝑠 = 𝐹 ∗ 𝑥 ∗ 𝑠𝑖𝑛2𝑎 𝑑 − 𝑥 ∗ 𝑠𝑖𝑛2𝑎 = 𝑠 𝑦
- 11. 25.02.2017Giga Khizanishvili 11 1 𝑛𝑎𝑖𝑟 = 1 1,0002926 = 0,9997 ⇒ 𝐸𝑟𝑟𝑜𝑟 = 1 − 0,9997 =0,03% The margin of error of the refractive index of the air
- 12. 25.02.2017Giga Khizanishvili 12 The margin of the error of the angle ( 𝛼1) 𝑙 ℎ = 132 ± 0,5 cm 𝑥 𝑎 𝑥 = 126 ± 0.5 cm ℎ 𝑡𝑔𝑎 𝑚𝑎𝑥 = 126 + 0.5 132 − 0.5 = 0,961965 𝑡𝑔𝑎 𝑚𝑖𝑛 = 126 − 0.5 132 + 0.5 = 0.94749
- 13. 25.02.2017Giga Khizanishvili 13 The margin of the error of the angle ( 𝛼1) 𝑙 𝑥 𝑎 ℎ 𝑎𝑟𝑐𝑡𝑔 0,961965 = 43.88° 𝑡𝑔𝑎 𝑚𝑖𝑛 = 126 − 0.5 132 + 0.5 = 0.94749 𝑎𝑟𝑐𝑡𝑔 0.94749 = 43.45° 𝑎 = 𝑎 𝑚𝑎𝑥 + 𝑎 𝑚𝑖𝑛 2 = 43.88 + 43.45 2 = 43.665° ± 0.215° ∆𝑎 𝑎 = 0.215° 43.665° ≈ 0.005 ≈ 0.5% 𝑡𝑔𝑎 𝑚𝑎𝑥 = 126 + 0.5 132 − 0.5 = 0,961965
- 14. 25.02.2017Giga Khizanishvili 14 The margin of the error of the angle ( 𝛼2) 𝑙 𝑥 𝑎 ℎ ℎ = 130 ± 0,5 𝑐𝑚 𝑥 = 81 ± 0,5 𝑐𝑚 𝑎 = 𝑎 𝑚𝑎𝑥 + 𝑎 𝑚𝑖𝑛 2 = 32.44° + 31,4° 2 = 31.92° ± 0.26° ∆𝑎 𝑎 = 0.26° 31.95° ≈ 0.008 ≈ 0.8% 𝑎 𝑚𝑎𝑥 = 32.18° 𝑎 𝑚𝑖𝑛 = 31,66°
- 15. 25.02.2017Giga Khizanishvili 15 𝑑= s ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑛2 𝑠𝑖𝑛2𝑎 ∆𝑓 𝑎, 𝑏, 𝑐, = 𝜕𝑓 𝜕𝑎 ∆𝑎 + 𝜕𝑓 𝜕𝑏 ∆b + 𝜕𝑓 𝜕𝑐 ∆c ∆𝑑 𝑑 = 4.8 %⇒ 𝑠 𝑛 𝑎 𝑠𝑖𝑛𝑎 𝑠𝑖𝑛2𝑎 ∆𝑠 𝑠 = 0.5 36 = 0.0138 ≈ 1.4% ∆𝑎1 𝑎1 = 0.5% ∆𝑎2 𝑎2 = 0.8%
- 16. 25.02.2017Giga Khizanishvili 16 𝑐= 1 𝜇0 ∗ 𝜀0 𝑣 = 1 𝜇𝜇0 ∗ 𝜀0 𝜀 𝑐 = 𝑣 ∗ 𝑛 𝑐 𝑛 = 𝑣⇒ 𝑛 ∗ 𝜇0 ∗ 𝜀0 = 𝜇𝜇0 ∗ 𝜀0 𝜀 𝑛 = 𝜀 𝑛2 = 𝜀⇒ 𝜀 𝑜 − 𝑑𝑖 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝜇 𝑜 − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑝𝑒𝑟𝑚𝑒𝑎𝑖𝑙𝑖𝑡𝑦 𝜇 − 𝑖𝑛𝑑𝑒𝑥 𝑜𝑓𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑝𝑒𝑟𝑚𝑒𝑎𝑖𝑙𝑖𝑡𝑦 𝜀 − 𝑖𝑛𝑑𝑒𝑥 𝑜𝑓 𝑑𝑖 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑝𝑒𝑟𝑚𝑒𝑎𝑏𝑖𝑙𝑖𝑡𝑦
- 17. 2.37 2.28 2.19 2.28 2.2 0.0001 0 0.5 11.5 2 2.5 I Exp II Exp III Exp IV exp V Exp Vi asfas I Exp II Exp III Exp IV exp V Exp Vi asfas Di electrical permeability 2.37 2.28 2.19 2.28 2.2 0.0001 Di electrical permeability 25.02.2017Giga Khizanishvili 17
- 18.
- 19. Angle oflaser’s declination - 𝛼1 = 43,7° Angle of laser’s declination - 𝛼2= 32° 25.02.2017Giga Khizanishvili 19
- 20. 25.02.2017Giga Khizanishvili 20 𝑠1= 2.8 𝑚𝑚 𝒏 = 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝑠2 = 2.4 𝑚𝑚 𝑎1 = 43.7° 𝑎2 = 32° ⇒ 𝑛 = 1.53
- 21. 25.02.2017Giga Khizanishvili 21 𝑑= s ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑛2 𝑠𝑖𝑛2𝑎 𝑛 = 1.53 𝑠𝑖𝑛𝑎 = 0.69 𝑠𝑖𝑛2𝑎 = 0.99 𝑠 = 𝐹 ∗ 𝑥 ∗ 𝑠𝑖𝑛2𝑎 𝑑 − 𝑥 ∗ 𝑠𝑖𝑛2𝑎
- 22. 25.02.2017Giga Khizanishvili 22 𝑎 𝑎 𝑥 𝑦 𝑠 𝑑 𝑑1 𝑑2 𝑑1= 953 cm 𝑑2 = 𝐹 ∗ 𝑠𝑖𝑛𝑎 𝑑 = 𝑑1 + 𝑑2 𝑑 = 953 + 15 ∗ 0.69 𝑑 = 953 + 10.35 𝑑 = 963 cm
- 23. 25.02.2017Giga Khizanishvili 23 𝑛= 1.55 𝑠𝑖𝑛𝑎 = 0.69 𝑠𝑖𝑛2𝑎 = 0.9989 𝑠 = 𝐹 ∗ 𝑥 ∗ 𝑠𝑖𝑛2𝑎 𝑑 − 𝑥 ∗ 𝑠𝑖𝑛2𝑎 𝐹 = 15 𝑐𝑚 𝑑 = 963 𝑐𝑚 𝑑 = s ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑛2 𝑠𝑖𝑛2𝑎 𝑠𝑖𝑛2𝑎 = 0.99 𝑥 = 𝑠 = 15 ∗ 0.999 ∗ 6.5 172.35 ∗ 2 = 2.8 𝑚𝑚⇒ 𝑥 = 36 𝑐𝑚 𝑥 = 36 𝑐𝑚
- 24. 25.02.2017Giga Khizanishvili 24 𝑛= 1.53 𝑠𝑖𝑛𝑎 = 0.69 𝑠𝑖𝑛2𝑎 = 0.9989 𝑑 = s ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑛2 𝑠𝑖𝑛2𝑎 𝑠 = 2.8 𝑚𝑚 𝑛 = 1.53 𝑠𝑖𝑛2𝑎 = 0.999 𝑠𝑖𝑛𝑎 = 0.69 𝑑 = 2.8 ∗ 1.55 ∗ 1 − 0.47679 2.4 0.999 = 4.39 ∗ 0.8 = 3.83 𝑚𝑚
- 25. 𝑑 = 3.83𝑚𝑚𝑛 = 1.53 𝐶𝑎𝑙𝑙𝑖𝑝𝑒𝑟 25.02.2017Giga Khizanishvili 25
- 26. 1.5 4.96 1.54 5.09 0 1 23 4 5 6 n d n d Experiment 1.54 5.09 Theory 1.5 4.96 d = 4.96mm Experiment Theory 25.02.2017Giga Khizanishvili 26
- 27. 1.5 2.39 1.51 2.44 0 0.5 11.5 2 2.5 3 n d n d Experiment 1.51 2.44 Theory 1.5 2.39 d = 2.39mm Experiment Theory 25.02.2017Giga Khizanishvili 27
- 28. 1.5 6.21 1.48 6.14 0 1 23 4 5 6 7 n d n d Experiment 1.48 6.14 Theory 1.5 6.21 d = 6.21mm Experiment Theory 25.02.2017Giga Khizanishvili 28
- 29. 1.5 3.24 1.51 3.28 0 0.5 11.5 2 2.5 3 3.5 n d n d Experiment 1.51 3.28 Theory 1.5 3.24 d = 3.24mm Experiment Theory 25.02.2017Giga Khizanishvili 29
- 30. 1.5 3.91 1.485 3.85 0 0.5 11.5 2 2.5 3 3.5 4 4.5 n d n d Experiment 1.485 3.85 Theory 1.5 3.91 d = 3.91mm Experiment Theory 25.02.2017Giga Khizanishvili 30
- 31. We cameup with a mechanism that measures the thickness of glass sheet The thickness of a glass sheet depends upon 𝑎, 𝑛, 𝑥. We have derived a formula for the refractive index of glass Experimental Errors we are measured Experiments have been conducted Thickness of glass sheet and refractive index have been determined Theoretical model was compared to experiments We have solved the problem 25.02.2017 31Giga Khizanishvili
- 32. Baldwin, G.C.; Solem, J. C. (1997). "Recoilless gamma-ray lasers". Reviews of Modern Physics 69 (4): 1085–1117. Vogel, Werner. Physics of Glass. Wiley, 1985. Bundschuh, Bernhard; Himmel, Jörg: Optische Informationsübertragung. Oldenbourg Verlag, München, Wien 2003 Baldwin, G. C.; Solem, J. C. (1982). "Is the time ripe? Or must we wait so long for breakthroughs?". Laser Focus 18 25.02.2017 32Giga Khizanishvili
- 33. https://www.google.com/patents/US4902902 http://russianpatents.com/patent/242/2429447.html http://physics.aidio.net/index.php/18-teoria/73-geometriuli-optika-ziritadi-kanonebi http://www.google.com/patents/US7414740 http://vlab.amrita.edu/?sub=1&brch=189&sim=1519&cnt=1 https://www.google.com/patents/US5636027 25.02.2017 33Giga Khizanishvili
- 34.
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- 36.
- 37. სინათლის გაძლიერება იძულე- ბულიგამოსხივების მეშვეობით მოწყობილობა, რომელიც ენერგიას გარდაქმნის მიმარ- თულ სინათლის სხივად. 25.02.2017 4
- 38.
- 39. 25.02.2017Giga Khizanishvili 39 s1 2∗ (𝑛2 −𝑠𝑖𝑛2 𝑎1) 𝑠𝑖𝑛22𝑎1 = s2 2 ∗ (𝑛2 − 𝑠𝑖𝑛2 𝑎2) 𝑠𝑖𝑛22𝑎2 s1 2 ∗ 𝑛2 −s1 2 ∗ 𝑠𝑖𝑛2 𝑎1 𝑠𝑖𝑛22𝑎1 = s2 2 ∗ 𝑛2 − s2 2 ∗ 𝑠𝑖𝑛2 𝑎2 𝑠𝑖𝑛22𝑎2 s1 2 ∗ 𝑛2 ∗ 𝑠𝑖𝑛22𝑎2 − s1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 = s2 2 ∗ 𝑛2 ∗ 𝑠𝑖𝑛22𝑎1 − s2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 s1 2 ∗ 𝑛2 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑛2 ∗ 𝑠𝑖𝑛22𝑎1 = s1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 𝑛2 ∗ (s1 2 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑠𝑖𝑛22𝑎1) = s1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 𝑛2 = s1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 s1 2 ∗ 𝑠𝑖𝑛22𝑎2 − s2 2 ∗ 𝑠𝑖𝑛22𝑎1 𝒏 = 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏
- 40. 25.02.2017Giga Khizanishvili 40 𝑑= s1 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎1 𝑛2 𝑠𝑖𝑛2𝑎1 𝑑 = s2 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎2 𝑛2 𝑠𝑖𝑛2𝑎2 s1 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎1 𝑛2 𝑠𝑖𝑛2𝑎1 = s2 ∗ n ∗ 1 − 𝑠𝑖𝑛2 𝑎2 𝑛2 𝑠𝑖𝑛2𝑎2 s1 2 ∗ n2 ∗ 𝑛2 −𝑠𝑖𝑛2 𝑎1 𝑛2 𝑠𝑖𝑛22𝑎1 = s2 2 ∗ n2 ∗ 𝑛2 − 𝑠𝑖𝑛2 𝑎2 𝑛2 𝑠𝑖𝑛22𝑎2 s1 2 ∗ (𝑛2 −𝑠𝑖𝑛2 𝑎1) 𝑠𝑖𝑛22𝑎1 = s2 2 ∗ (𝑛2 − 𝑠𝑖𝑛2 𝑎2) 𝑠𝑖𝑛22𝑎2
- 41. 25.02.2017Giga Khizanishvili 41 𝒅= 𝐬 ∗ 𝐧 ∗ 𝟏 − 𝒔𝒊𝒏 𝟐 𝒂 𝒏 𝟐 𝒔𝒊𝒏𝟐𝒂 𝒏 = 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝐬 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟐 − 𝐬 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝟐𝒂 𝟏 𝑑 = s ∗ 𝑠1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 − 𝑠2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 𝑠1 2 ∗ 𝑠𝑖𝑛22𝑎2 − 𝑠2 2 ∗ 𝑠𝑖𝑛22𝑎1 ∗ 1 − 𝑠𝑖𝑛2 𝑎 𝑠1 2 ∗ 𝑠𝑖𝑛2 𝑎1 ∗ 𝑠𝑖𝑛22𝑎2 − 𝑠2 2 ∗ 𝑠𝑖𝑛2 𝑎2 ∗ 𝑠𝑖𝑛22𝑎1 𝑠1 2 ∗ 𝑠𝑖𝑛22𝑎2 − 𝑠2 2 ∗ 𝑠𝑖𝑛22𝑎1 𝑠𝑖𝑛2𝑎
- 42. 25.02.2017Giga Khizanishvili 42 𝒏= 𝒙 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟏 𝒙 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 𝒏 = 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐(𝒙 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟏) 𝟒 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐(𝒙 𝟏 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟏) 𝒏 = 𝒙 𝟏 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟏 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒔𝒊𝒏 𝟐 𝒂 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟏 𝒙 𝟏 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟐 − 𝒙 𝟐 𝟐 ∗ 𝒄𝒐𝒔 𝟐 𝒂 𝟏 s = 𝑥 ∗ 𝑠𝑖𝑛𝑎⇒
- 43. 25.02.2017Giga Khizanishvili 43 დახრილი სიბრტყის ცდომილება 𝛼 𝑜 𝑠 𝑙 𝑙 = 𝑠 𝑐𝑜𝑠𝑎 ∆𝑠 𝑠 = 𝑠 𝑠 ∗ 𝑐𝑜𝑠𝑎 − 1 ∆𝑠 𝑠 = 𝑙 𝑠 − 1 ∆𝑠 𝑠 = 1 𝑐𝑜𝑠𝑎 − 1 𝑎 = 2.5° 𝑐𝑜𝑠𝑎 = 0.999048221582 ∆𝑠 𝑠 = 1 0.999048221582 − 1 ∆𝑠 𝑠 = 1.0009526851631775 − 1 ∆𝑠 𝑠 = 0.0009526851631775 ∆𝑠 𝑠 ≈ 0.01%