Welcome back Steve (if that is indeed you and not Bob Rosen pretending). - - - Received: from ruprect ([10.60.10.190] ident=Steve) - - - by Fruitloop with esmtp (Exim 3.35 #1 (Debian)) But where has all the useful and accurate factual detail about photography, focussing, lenses etc etc gone? The new "un(informed)Bob Rosen (Afterswift@xxxxxxxx) is all opinion and little substance in comparison. Have you by any chance gone digital? Is that why you no longer feel the need to worry about the technical aspects of photography that were always only a distraction from real creativity? I must say, I do miss the real Bob Rosen ... I also miss the old Steve Hodges ...the one who could post ... Steve Hodges wrote to PF: on Sunday, September 08, 2002 12:51 PM > OK, for simplicity we'll make the following asumptions: > > 1) film records objects in 2D, so all we need to consider is the > displacement as seen on the image, not absolute differences. In other > words we will be considering the angle that a certain distance subtends > with the lens. > > 2) we'll assume ideal lens (no aberrations), film (no grain), and > development (no adjacency effects). > > 3) we'll assume that the distances are measured from the nodal point of > the lens, and that both nodes are coincident. > > 4) we'll also assume that complaints that these assumptions somehow > invalidate the mathematics, or unduelt influence the result will be > proved rigourously ;-) > > 5) we'll consider only those rays of ligh which remain undeviated by the > lens (ignoring the effect of the nodal space) > > Lets assume that the first object (A) is 1000 metres away, the second > (B) 1100 metres, and the third (C) 1200 metres away A straight line > could be drawn connecting the tops of these objects. Also assume that > these objects are 2 metres in height and the top of object B is aligned > directly along the axis of the lens. Furthermore, the angle subtended > by the chord connecting the tops of objects A and B as measured from the > lens' nodal point is 0.01 radian. The objects are appropximatly > cylindrical in shape and are parallel. > > (If that doesn't make sense, think of three people standing on a field > in a line, such that you're 1 km from the first one, 1100m form the > second, etc.) > > Film is assumed to be 35mm (not that it matters) and lens focal lengths > are 28mm and 500mm > > Call the lens nodal point N > > The distance from the nodal point to the film plane (along the lens > axix) is approx: > > 28mm 500mm > > 1/f = 1/u + 1/v 1/f = 1/u + 1/v > > v = 1/(1/f - 1/u) v = 1/(1/f - 1/u) > > = 1/(1/28 - 1/1 100 000) = 1/(1/500 - 1/1 100 000) > > = 28.0007127 mm = 500.2273761 mm > > The distance between the objects A and B: > > > AB^2 = 1000^2 + 1100^2 - 2.1000 x 1100 x Cos(0.01) (cosine rule) > = 10110.00 > > AB = 100.55 m > > The angle at object A between the camera and B: > > Sin A = (Sin(0.01) x 1100) / 100.55 (sin rule) > = 0.109396 > > A = 3.0319768 > > Therefore angle at B between camera and C is: > > B = 0.01 + 3.0319768 (NBC supplimentary to ABN) > = 3.0419768 > > The angle at object C between the camera and B: > > Sin C = (1100 x Sin 3.0419768) / 1200 (Sin rule) > = 0.091163600 > > C = 0.09129 > > Thus the angle at the lens between onjects B and C is: > > BNC = pi - (3.0419768 + 0.09129) > = 0.00833 > > And the distance BC is: > > BC^2 = 1100^2 + 1200^2 - 2 x 1100 x 1200 x Cos 0.0833 > = 19154.04 > > BC = 138.40 m > > Now, knowing certain facts about nodal points, we know that the angle > subtended by a chord between 2 points on the object plane will equal the > angle subtended by those same 2 points on the image plane. > > Note that this does not hold for fisheye lenses, or other lenses where > barrel or pincushion distortion are significant. > > the distance between pts A and B on the film plane are: > > 28mm 500mm > > AB' = 28.0007127 Sin 0.01 AB' = 500.2273761 sin 0.01 > = 0.280002 mm = 5.002190 mm > > and between B and C are: > > BC' = 28.0007127 Sin 0.00833 BC' = 500.2273761 sin 0.00833 > = 0.233243 mm = 4.166846 mm > > This is actually an indication of perspective. Remember that the > distance AB is actually 100.55m and the distance BC is actually > 138.40m. But on the film, the distance BC' is actually shorter than > AB'. This is because A is closer to us than C. > > Look at the ratio between AB' values, BC' values for both lenses: > > 28mm / 500mm > > AB'/AB' > > = 0.280002 / 5.002190 > = 0.055976 > > BC'/BC' > > = 0.233243 / 4.166846 > = 0.055976 > > Aha! It appears that the ratio of the sizes is the same (0.055976:1), > even though the actual distances and angles subtended are different! > > Now this should be proof enough, but I know that many of you are far > more stubborn than I, and will want to see more proof. > > So I shall continue. > > The next step will be to look at the relative heights of the objects. > > We need to know the angle subtended by the objectds A, B, and C > > for A: sin a = 2/1000 > = 0.002 > a = 0.00200000133 > > for B: sin b = 2/1100 > = 0.00181818181818 > b = 0.00181818282 > > for C: sin c = 2/1200 > = 0.001666666666 > c = 0.00166666744 > > The height of B is: > > 28mm 500mm > > B = 28.0007127 x 0.00181818181818 B = 500.2273761 x 0.00181818181818 > = 0.0509104 mm = 0.909504 > > (note the ratio is once again 0.055976:1) > > The height of A is: > > A = sqrt(AB'^2 + 28.0007127^2) x 0.002 > = 0.0560042 A = sqrt(AB'^2 + 500.2273761^2) x > 0.002 > = 1.0005048 > > (And the ratio is again 0.055976:1) > > And finally, the height of C is: > > C = sqrt(BC'^2 + 28.0007127^2) x 0.001666666666 > = 0.0466695 C = sqrt(BC'^2 + 500.2273761^2) x > 0.001666666666 > = .833741 > > (And you would not be surprised to learn that the ratio is again > 0.055976:1) > > So what have we shown? > > 1) that the *only* difference between the image produced from such a > system with lenses of 2 different focal lengths is one of scale. > > 2) that the scale remains constant whether the objects are the same > height, but different distances, or indeed different distances from the > same point. > > 3) that greater displacement in the object space may show up as smaller > distances on the image but that the ratio again remains the same. > > Steve >
