Multiply for using 16mm lens on Super8 cameras

[David Odell]

Hi, I have a Beaulieu with an Ang 6-80 lens. Since I wonder how the effect would be to use a Kern Vario-Switar 12.5-100mm,

to achieve longer zoomdistance and hopefully good resolution images, is there a factor how to multiply using 16mm optics on Super8/SuperDuper8?

 

As when using a Nikon F optic with adapter on the Beaulieu I calculate (x)mm times 6,3. So do anybody know what the same is for 16mm?

 

I tried a search but didn't find much.

 

Thanks!

Edited July 10, 2009 by David Odell

[David Mullen ASC]

Hi, I have a Beaulieu with an Ang 6-80 lens. Since I wonder how the effect would be to use a Kern Vario-Switar 12.5-100mm,

to achieve longer zoomdistance and hopefully good resolution images, is there a factor how to multiply using 16mm optics on Super8/SuperDuper8?

 

As when using a Nikon F optic with adapter on the Beaulieu I calculate (x)mm times 6,3. So do anybody know what the same is for 16mm?

 

I tried a search but didn't find much.

 

Thanks!

 

Focal length is focal length. It's a 12.5 to 100mm zoom, just now on an 8mm camera instead of a 16mm camera. But the field of view would be the same if you found a 12.5-to-100mm lens made for an 8mm camera, so there's no conversion factor. So instead of a lens that only goes to 80mm, you've found one that goes to 100mm.

[David Odell]

I see, then it's not such a big advancement after all. Thanks for the reply!

 

David

Edited July 11, 2009 by David Odell

[Mark Dunn]

The difference is that 12.5mm is a fair wide angle on 16 but quite long on Super-8. David is quite correct- a given focal length will have about half the angle of view on Super-8 compared with 16mm. So the Super-8 equivalent, in angle of view, of the 12.5-100 would be about 6-50, or a little longer, because Super-8 is a bit bigger than half the size of 16mm.

I hope that's not too much information.