# Distance for Wide Shot

8 replies to this topic

### #1 David Schmüdde

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Posted 04 June 2014 - 09:56 AM

I'm building stage setup and I need a specific number to estimate the amount of space I need.

What I need to know is the distance necessary between a 6' tall subject and the camera for a wide shot. For some reason, I'm blanking on how to figure this. These are the constants:

• 35mm sensor size
• 50mm lens length
• 6' Tall Subject
• Wide Shot (includes top of the head to bottom of the feet)
• 16x9 frame dimension

What I need to solve for:

• Distance between camera and still subject

Could someone help me here?

Thanks,

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### #2 Mark Dunn

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Posted 04 June 2014 - 10:10 AM

It's a straight bit of similar triangle geometry.

Frame height for 16:9 is about 10mm. Focal length 50mm. 50/10=5 so distance is 5x height- about 10m.

Edited by Mark Dunn, 04 June 2014 - 10:14 AM.

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### #3 David Schmüdde

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Posted 04 June 2014 - 01:40 PM

35mm full frame is 36mm×24 mm - 1.5 aspect ratio
35mm HD TV is 36mmx20.25mm - 1.77 aspect ratio (16 : 9 = 36mm : X)

My formula
Hc (film height) / Lc (distance from lens to film) = Hh (human height) / Lh (distance from lens to human)

Hc = 20.25mm
Lc = 50mm (focal length of lens)
Hh = 1.8m (6' - a constant given above)

Hc / Lc = Hh / Lh
.020m / .050m = 1.8m / Lh

Lh = 4.5m or 15'.

Did I do something wrong here?

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### #4 Jaron Berman

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Posted 04 June 2014 - 09:29 PM

What do you mean by 35mm frame?  35mm still?  s35?

For super 35 frame (most d-cinema cameras, 7d, etc), 16:9 HDTV, its about 24' on a 50mm

For 35mm still frame (like 5d), it's about 16'

Edited by Jaron Berman, 04 June 2014 - 09:32 PM.

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### #5 Jon Rosenbloom

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Posted 05 June 2014 - 11:58 PM

I'm not so great at math, but, isn't 10 meters more than 30 feet? For a head to toe on a 50mm? It seems a little much. If you want to wait till tomorrow, I'll stand up my tape measure and check it through my 50mm on my old Nikon and then pace it off.

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### #6 John E Clark

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Posted 06 June 2014 - 04:47 PM

For those who don't want to 'do the math', which is tedious and if there are changes, needs to be recalculated...

For the iPhone:

http://www.davideuba...Calculator.html

Or

https://itunes.apple...d377419210?mt=8

Using the Field of View calculators you can easily get what lens length for what FoV/AoV, etc. is required for what sized object, etc. for a variety of cameras.

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### #7 Mark Dunn

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Posted 08 June 2014 - 05:40 AM

Did I do something wrong here?

Yes, you didn't specify what '35mm. sensor size' meant, so I assumed Academy. Turns out you meant Vistavision.

I calculated the AR from frame height, not width, so I was a bit off too. t would actually be about 5m.

The other thing you did wrong was asking the question when you'd already worked it out yourself.

This forum isn't an exam.

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### #8 Diego Treves

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Posted 29 October 2015 - 09:36 AM

It's a straight bit of similar triangle geometry.
Frame height for 16:9 is about 10mm. Focal length 50mm. 50/10=5 so distance is 5x height- about 10m.

Edited by Diego Treves, 29 October 2015 - 09:42 AM.

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### #9 Peter Hodgins

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Posted 23 January 2016 - 03:51 PM

I'm building stage setup and I need a specific number to estimate the amount of space I need.

What I need to know is the distance necessary between a 6' tall subject and the camera for a wide shot. For some reason, I'm blanking on how to figure this. These are the constants:

• 35mm sensor size
• 50mm lens length
• 6' Tall Subject
• Wide Shot (includes top of the head to bottom of the feet)
• 16x9 frame dimension

What I need to solve for:

• Distance between camera and still subject

Could someone help me here?

Thanks,

Can you just slap a 50mm lens on your dslr and frame a 6 foot subject or perhaps a doorway etc. in your 16x9 frame, and note your distance?

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