Hasselblad FOV

[BertL]

I was wondering about the field of view on the Apollo's Hasselblad cameras.

The "standard" seems to be a focal length of 60mm with 70mm square film. Using the equation found here, I come to the following numbers:

α = 2arctan( d / (2f) )

f = 60mm
d = 70mm (horizontal/vertical)
d = (70 * sqrt(2) ) mm (diagonal)

α = ~ 60.5° (horizontal/vertical)
α = ~ 79° (diagonal)

However, when I just do a Google search on the Hasselblad's FOV, websites give varying different numbers. For example, Clavius gives a diagonal FOV of 71° without sourcing or providing calculations. I've also seen other (even lower) numbers crop up on somewhat more shady websites.

What I was wondering is, what gives? Can there be such a big difference between lenses? Should I have considered focus? I understand that the actual FOV would be lower in scans of pictures due to the edges being cut off and factors like that, but surely that alone wouldn't account for such a big difference?

[Al Johnston]

There is a difference in field of view with focal length: that's pretty much the point of having a camera where you can change lenses to use different focal lengths.

IIRC the usual 'standard' lens for a Hasselblad is 80mm.

[ChrLz]

I've gotta ask - what are you going to do with any number you might come up with..? 

A couple of points - yes, the lens was a Zeiss Biogon 60mm f5.6, not the 'normal' 80mm.  So it was a 'semi-wide' lens - a bit like a 38mm if you're an oldster like me and mostly think in 35mm SLR equivalent terms.

Your formula is correct, but I think the problem you have is that 70mm isn't right for the size of the actual image captured and therefore the actual recorded field of view - which is quite different to the lens' theoretical fov.

Yes, it's called 70mm film, but that isn't the size of the area of film that is exposed..  I used to have the numbers for the Hasselblad's 'gate' size, but can't lay my eyes on them right now and it's late, I'm tired and going to bed..  I do have some of my old medium format images filed away - if someone else doesn't chime in, I'll dig them out tomorrow and actually measure them.  I don't think I've got any from a Hasselblad, just a Bronica/Yashica, but they'll be similar.

[Al Johnston]

According to here it's nominally 6cm x 6cm, actually 56mm x 56mm.

[cjameshuff]

The reseau plate crosses were 10 mm apart, in a 5x5 grid with 7 mm outside the outermost crosses (the reseau plate itself was 5.4x5.4 cm). The cross separation in angular terms was 10.3 degrees with the 60 mm lens, for a horizontal/vertical angle of 55.6 degrees and a diagonal of 78.7 degrees, but this is subject to projection distortion (it was a rectilinear lens, meaning it preserves straight lines and areas but not angles). Working it out from the reseau plate size and focal length, I get 48.4 degrees horizontal and 64.9 degrees diagonal.

As for the Clavius figure, if you round up to 6x6 cm and go by focal length, the diagonal works out to 71 degrees...perhaps this is where that figure came from.

http://www.hq.nasa.gov/alsj/alsj-reseau.html

edit: that 10.3 degrees figure seems suspiciously large. Due to projection distortion, the center ones subtend the largest angle. A 60 mm focal length gives a spacing of 9.46 degrees...so 10.3 can't even be the maximum. You have to drop focal length down to 55 mm to get a spacing of 10.3 degrees, which gives a horizontal/vertical field of view of 52.2 degrees and a diagonal of 69.5 degrees. If the crosses on the reseau plate were about 5 mm from the film, the figures make a bit more sense...does anyone know such details?