Add graphic image_5
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@ -245,7 +245,7 @@ Besides its to and fro movement the lens has lateral movements.
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These adjust the position of the original frame's image on the print frame.
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For example, if the lens is raised a bit...
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**![Graphic depicting a lens' central position between two frames demonstrating a rise adjusting framing](#)**
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**![Graphic depicting a lens' central position between two frames demonstrating a rise adjusting framing](img/image_5.jpg)**
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At 1:1 moving the lens up a distance d raises the viewed field by twice d.
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Likewise for down, right, and left.
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@ -242,7 +242,7 @@
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<p>An optical printer is a device for photographing the frames of one film so as to make another film.</p>
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<p><strong><img src="img/image_1.jpg" alt="Graphic depicting labelled components camera, bellows, lens, gate and lamp" /></strong></p>
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<p><strong><img src="../img/image_1.svg" alt="Graphic depicting labelled components camera, bellows, lens, gate and lamp" /></strong></p>
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<p>It consists essentially of a camera (C) connected by a bellows (B) to a lens (L) aimed at a film in a gate (G) illuminated from behind by a lamp (I).</p>
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<p>The camera and gate each have motorized intermittent film movements so that any frame of the “original” film can be conveniently photographed onto any frame of the “print” film.</p>
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<p>The camera can be an ordinary cine camera, less its lens, and the gate can be an ordinary cine projector, less its lens. Ideally they have identical systems of film registration, as if one were the lens’ image of the other. The lens can be any bellows mountable lens. Ideally it is specially corrected for the small and nearly equal sizes of this object and image.</p>
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@ -250,11 +250,11 @@
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<p><a name="magnification"></a></p>
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<h2 id="magnification">MAGNIFICATION</h2>
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<p>If the lens is (nominally) midway between the films when one is focused on the other, then the magnification is 1. At <code>M = 1</code> (also called 1:1) the whole of the original frame is photographed at a size which fills the whole of the print frame.</p>
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<p><strong><img src="img/image_2.jpg" alt="“M = 1” Graphic depicting two frames with a lens at their midpoint with a lightbulb illuminating from the right" /></strong></p>
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<p><strong><img src="../img/image_2.svg" alt="“M = 1” Graphic depicting two frames with a lens at their midpoint with a lightbulb illuminating from the right" /></strong></p>
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<p>If the lens is moved closer to the gate, then the camera must be moved back, farther from the gate, to keep the one film focused on the other. Then the magnification is greater than 1. At <code>M > 1</code> a part of the original frame is photographed at a size which fills the whole of the print frame.</p>
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<p><strong><img src="img/image_3.jpg" alt="“M = 3” Graphic depicting two frames with a lens closer to the right projection source image with the lamp demonstrating an enlargement" /></strong></p>
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<p><strong><img src="../img/image_3.svg" alt="“M = 3” Graphic depicting two frames with a lens closer to the right projection source image with the lamp demonstrating an enlargement" /></strong></p>
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<p>If, starting from the 1:1 setup, the lens is moved farther from the gate, then the camera must also be moved back, farther from the gate, to keep the one film focused on the other. Then the magnification is loser than 1. At <code>M < 1</code> the whole of the original frame is photographed at a size which does not fill the whole of the print frame. The remainder of the print frame is filled with a photograph of the gate as it surrounds the original frame (ideally perfectly black).</p>
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<p><strong><img src="img/image_4.jpg" alt="“M = 1/3” Graphic depicting two frames with a lens closer to the left camera image demonstrating a reduction" /></strong></p>
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<p><strong><img src="../img/image_4.svg" alt="“M = 1/3” Graphic depicting two frames with a lens closer to the left camera image demonstrating a reduction" /></strong></p>
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<p>For each position of the lens there is exactly one correct (focused) position for the camera. But for each position of the camera (except the 1:1 position) there are two correct positions for the lens. One gives <code>M > 1</code>, the other <code>M < 1</code>.</p>
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<p><a name="blowup-and-reduction"></a></p>
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<h2 id="blowup-reduction">BLOWUP & REDUCTION</h2>
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@ -305,7 +305,7 @@
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<p><a name="x-y-adjustment"></a></p>
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<h2 id="x-y-adjustment">X-Y ADJUSTMENT</h2>
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<p>Besides its to and fro movement the lens has lateral movements. These adjust the position of the original frame’s image on the print frame. For example, if the lens is raised a bit…</p>
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<p><strong><img src="#" alt="Graphic depicting a lens’ central position between two frames demonstrating a rise adjusting framing" /></strong></p>
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<p><strong><img src="../img/image_5.svg" alt="Graphic depicting a lens’ central position between two frames demonstrating a rise adjusting framing" /></strong></p>
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<p>At 1:1 moving the lens up a distance d raises the viewed field by twice d. Likewise for down, right, and left.</p>
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<p>At <code>M > 1</code> lateral adjustment effects a scan of the original frame. This is not geometrically equivalent to a pan, bad it been made in the original photography.</p>
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<p>On simple optical printers the only lateral adjustment is of the lens (rather than the heavier camera or gate). This is geometrically adequate.</p>
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