Abstract

By means of a grid and three centered and astigmatic optical systems, it is possible to get an arbitrary orientation of the image of the slits of the grid with respect to the measuring direction. Such a device improves the accuracy of the observation of boundary layers encountered in aerodynamical, thermal, and mass transport phenomena. This process can also easily be applied to differential interferometry.

© 1971 Optical Society of America

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References

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  1. J. Surget, Onera, La Rech. Aérosp. No. 97, 37 (Nov.–Dec. 1963).
  2. G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).
  3. M. Parodi, Application de l’algebre moderne à quelques problèmes de physique classique (Gauthiers Villars, Paris, 1961).

1966 (1)

G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).

1963 (1)

J. Surget, Onera, La Rech. Aérosp. No. 97, 37 (Nov.–Dec. 1963).

Carr, P.

G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).

Gontier, G.

G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).

Henon, G.

G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).

Parodi, M.

M. Parodi, Application de l’algebre moderne à quelques problèmes de physique classique (Gauthiers Villars, Paris, 1961).

Surget, J.

J. Surget, Onera, La Rech. Aérosp. No. 97, 37 (Nov.–Dec. 1963).

C. R. Acad. Sci. Paris Ser. B (1)

G. Gontier, P. Carr, G. Henon, C. R. Acad. Sci. Paris Ser. B 262, 674 (1966).

La Rech. Aérosp. (1)

J. Surget, Onera, La Rech. Aérosp. No. 97, 37 (Nov.–Dec. 1963).

Other (1)

M. Parodi, Application de l’algebre moderne à quelques problèmes de physique classique (Gauthiers Villars, Paris, 1961).

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Figures (4)

Fig. 1
Fig. 1

Schematic of the optical assembly. Cylinder axis of the second lens is perpendicular to those of the other two.

Fig. 2
Fig. 2

Breakdown of the assembly into two separate systems.

Fig. 3
Fig. 3

Free convection thermal boundary layer along a vertical plane.

Fig. 4
Fig. 4

Free convection thermal boundary layer over a horizontal plane.

Equations (32)

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[ y β ] = [ 1 d 4 0 1 ] [ 1 0 - ϕ 3 1 ] [ 1 ( d 2 + d 3 ) 0 1 ] [ 1 0 - ϕ 1 1 ] [ 1 d 1 0 1 ] [ y β ] , [ z γ ] = [ z γ ] ,
[ y β ] = [ { 1 - d 4 [ ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) ] - ϕ 1 ( d 2 + d 3 ) } { - d 1 d 4 [ ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) ] + d 1 + d 4 + ( d 2 + d 3 ) [ 1 - ϕ 1 d 1 - ϕ 3 d 4 ] } { - [ ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) ] } { 1 - d 1 [ ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) ] - ϕ 3 ( d 2 + d 3 ) } ] [ y β ] .
ϕ = ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) .
- d 1 d 4 ϕ + d 1 + d 4 + ( d 2 + d 3 ) ( 1 - ϕ 1 d 1 - ϕ 3 d 4 ) = 0.
[ y β ] = [ g y 0 - ϕ 1 g y ] [ y β ] ,
g y = 1 - d 4 ϕ - ϕ 1 ( d 2 + d 3 ) = 1 1 - d 1 ϕ - ϕ 3 ( d 2 + d 3 ) .
g y = - ( d 2 + d 3 ) ϕ 1 + d 4 ϕ ( d 2 + d 3 ) ϕ 3 + d 1 ϕ ,
1 ( d 2 + d 3 ) ϕ 1 + d 4 ϕ + 1 ( d 2 + d 3 ) ϕ 3 + d 1 ϕ - 1 = 0.
[ y β ] = [ y β ] , [ z γ ] = [ 1 ( d 3 + d 4 ) 0 1 ] [ 1 0 - ϕ 2 1 ] [ 1 ( d 2 + d 3 ) 0 1 ] [ z γ ] ,
[ z γ ] = [ g z 0 - ϕ 2 1 g z ] [ z γ ] ,
1 ( d 1 + d 2 ) + 1 ( d 3 + d 4 ) - ϕ 2 = 0 ,
g z = - d 3 + d 4 d 1 + d 2 .
ϕ 2 = ϕ = ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) .
[ y g r β g r ] = [ 1 - ( l + d 2 ) ϕ 1 d 1 [ 1 - ( l + d 2 ) ϕ 1 ] + l + d 2 - ϕ 1 1 - ϕ 1 d 1 ] [ y β ] , [ z g r γ g r ] = [ 1 - l ϕ 2 ( l + d 1 ) ( 1 - l ϕ 2 ) + l - ϕ 2 1 - ϕ 2 ( l + d 1 ) ] [ z γ ] .
{ g y g r = 1 - ( l + d 2 ) ϕ 1 , g z g r = 1 - l ϕ 2 .
{ y g r = [ 1 - ( l + d 2 ) ϕ 1 ] λ , z g r = ( 1 - l ϕ 2 ) ( a λ + b ) ,
z g r = ( 1 - l ϕ 2 ) a 1 - ( l + d 2 ) ϕ 1 y g r + ( 1 - l ϕ 2 ) b ,
tan θ g r = a ( 1 - l ϕ 2 ) 1 - ( l + d 2 ) ϕ 1 .
tan ( θ - θ g r ) = ± = tan θ - tan θ g r 1 + tan θ / tan θ g r ,
1 + a 2 1 - l ϕ 2 1 - ( l + d 2 ) ϕ 1 = 0.
l = 2 - d 2 ϕ 1 ϕ 1 + ϕ 2 .
δ y g r = { d 1 [ 1 - ( l + d 2 ) ϕ 1 ] + ( l + d 2 ) } β , δ z g r = [ ( l + d 1 ) ( 1 - l ϕ 2 ) + l ] γ .
δ y g r / β g y g r = - δ z g r / γ g z g r ,
d 2 = ( l + 2 d 1 ) ( 1 - l ϕ 2 ) .
1 - g = ( d 2 + d 3 ) ϕ 1 + d 4 [ ϕ 1 + ϕ 3 - ϕ 1 ϕ 3 ( d 2 + d 3 ) ] , 1 - g = ( d 3 + d 4 ) ϕ 2 , g = - d 3 + d 4 d 1 + d 2 = ± ( d 2 + d 3 ) ϕ 1 + d 4 ϕ ( d 2 + d 3 ) ϕ 3 + d 1 ϕ .
g = - 1 , ϕ 1 = ϕ 3 = 1 k d , ϕ 2 = ( 2 1 + k ) 1 d , d 1 = d 4 , d 2 = d 3 = k d 1 , l = k ( 1 + k ) 1 + 3 k d ,
g = - 1 , ϕ 1 = ϕ 3 = 1 d , ϕ 2 = ( 2 1 + k ) 1 d , d 1 = d 4 = d , d 2 = d 3 = k d , 1 = ( 2 - k ) ( 1 + k ) d 3 + k ,
d 1 = d 2 = d 3 = d 4 = 1 ϕ 1 = 1 ϕ 2 = 1 ϕ 3 = d ;
ϕ = 0 ,             g y = g z = - 1 ,
δ y g r g y g r = { d 1 [ 1 - ( l + d 2 ) ϕ 1 ] + ( l + d 2 ) } β 1 - ( l + d 2 ) ϕ 1 ,
δ z g r g z g r = [ ( 1 + d 1 ) ( 1 - l ϕ 2 ) + l ] γ 1 - l ϕ 2 .
l = ( 2 - d 2 ϕ 1 ) / ( ϕ 1 + ϕ 2 )

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