Stats for a full-frame 35mm camera, taken from lens manufacturers' spec sheets. * 50mm = 46 degrees * 70mm = 34 degrees * 85mm = 28 degrees * 135mm = 18 degrees * 200mm = 12 degrees Now, those are presumably across a circular diameter (ie. the diagonal of a frame). <> = Working in landscape orientation = For a 3m wide backdrop, your hypotenuse in 3:2 aspect ratio is sqrt(13), or 3.6m. Now use some trig to convert those angles into 3.6m. '''Cheating:''' divide your focal length (mm) by '''11.76''' to get the distance in metres == For a 50mm lens == * theta = 23 degrees * Opposite = 180cm {{attachment:50mm_triangle.png}} {{{ tan(23 degrees) = 180 / working distance 0.42 = 180 / working distance 1/0.42 = working distance / 180 180/0.42 = working distance working distance = 428cm? }}} '''4.3 metres''' My mathematical gut feeling says this is wrong... but it seems about right when I hold the 50mm up to my eye. == 70mm lens == {{{ theta = 17 degrees working distance = 180 / tan(17) = 180 / 0.306 = 588cm }}} '''5.9 metres''' == 85mm lens == Working distance = 180 / 0.249 = '''7.23 metres''' == 135mm lens == {{{ theta = 9 degrees working distance = 180 / tan(9) = 180 / 0.158 = 1136cm }}} '''11.3 metres''' == 200mm lens == {{{ theta = 6 degrees working distance = 180 / tan(6) = 180 / 0.105 = 1712cm }}} '''17.1 metres''' = Working in portrait orientation = For a 3m wide backdrop, your hypotenuse in 2:3 aspect ratio is sqrt(29.25), or 5.4m. Now use some trig to convert those angles into 5.4m. '''Cheating:''' divide your focal length (mm) by '''7.87''' to get the distance in metres == For a 50mm lens == {{{ theta = 23 degrees working distance = 270 / tan(23) = 270 / 0.424 = 636cm }}} '''6.3 metres''' == 70mm lens == {{{ theta = 17 degrees working distance = 270 / tan(17) = 270 / 0.306 = 883cm }}} '''8.8 metres''' == 85mm lens == Working distance = 270 / 0.249 = '''10.8 metres''' == 135mm lens == {{{ theta = 9 degrees working distance = 270 / tan(9) = 270 / 0.158 = 1704cm }}} '''17 metres''' == 200mm lens == {{{ theta = 6 degrees working distance = 270 / tan(6) = 270 / 0.105 = 2568cm }}} '''25.6 metres''' = Portrait mode with 2m fixed height subject = * 2m high * 1.33m wide * '''2.4m diagonal''' * 1.2m for right-angle triangle (120cm) '''Cheating:''' divide your focal length (mm) by '''17.71''' to get the distance in metres == 50mm lens == * theta = 23 degrees * working distance = 120 / tan(23) * '''2.82 metres''' == 70mm lens == Working distance = 120 / 0.306 = '''3.92 metres''' == 85mm lens == Working distance = 120 / 0.249 = '''4.82 metres''' == 135mm lens == Working distance = 120 / 0.158 = '''7.59 metres''' == 200mm lens == Working distance = 120 / 0.105 = '''11.4 metres''' = In a table = || || 3m landscape max. || 2m portrait max. || 2m fixed-height subject || || 50mm || 4.3 || 6.3 || 2.8 || || 70mm || 5.9 || 8.8 || 3.9 || || 85mm || 7.2 || 10.8 || 4.8 || || 135mm || 11.3 || 17.0 || 7.6 || || 200mm || 17.1 || 25.6 || 11.4 || = Diagrams = == Working room for subject == Assuming a 2m high subject, and a 3m x 3m backdrop, we have a certain amount of room to play with. {{attachment:clipping_and_maximum_working_distances.jpg}} Using the figures from the table, and assuming portrait mode: * Photographer can move back as far away as the second column ("2m portrait max.") * Subject must be further from photographer than the distance in the third column (aka. "clipping distance") * Subject could be right up against the backdrop, in theory * `$col3 - $col2 = playspace` || || 2m portrait max. || Clipping distance || Playspace in metres || || 50mm || 6.3 || 2.8 || 3.5 || || 70mm || 8.8 || 3.9 || 4.9 || || 85mm || 10.8 || 4.8 || 6.0 || || 135mm || 17.0 || 7.6 || 9.4 || || 200mm || 25.6 || 11.4 || 14.2 ||