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← Revision 9 as of 2012-02-21 14:26:37 ⇥
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Comment: more tables
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* 85mm = 28 degrees | |
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<<TableOfContents>> = Working in landscape orientation = |
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'''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 = ||<rowbgcolor="lightblue"> || 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` ||<rowbgcolor="lightblue"> || 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 || |
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).
Contents
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
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.
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 |