21mm Equivalent focal length in LF

[ndwgolf]

My New Chamonix 8x10 will arrive in a few days time. I have a 360mm lens for it already and plan to use that for the first few months. My next lens will be in the 21mm FL range.
Any recommendations for a nice wide angle landscape lens?

Neil

 

[adelorenzo]

This may be useful it has specs for all lenses for the format:

http://www.largeformatphotography.info/lenses/LF8x10in.html

Your three modern choices in the ~21 mm equivalent range would be:

Nikon Nikkor SW 150
Schneider Super-Symmar XL 150
Rodenstock Grandagon N 155

 

[photog_ed]

If you're looking for a lower-cost option, I suggest an early Fujinon W 180 (lettering on the front rather than on the outside of the barrel), which is what I recently bought. The early ones barely cover 8x10, while the later ones do not. I got mine for under $200. I think it's closer to the 21mm equivalent than simply following the multiply-by-six rule. Here's my rationale:

If you're enlarging a 35mm negative to 8x10 and masking 1mm on each side of the short dimension, you're using about 22x27.7mm of negative area. Contact printing an 8x10 negative and masking off 5mm on each side, you're using 254-10 = 244mm length of negative. Then the equivalent focal length for 8x10 is 21mm * 244/27.5 = 186mm.

Ed

 

[Steve Goldstein]

The lenses Anthony mentioned are all big, heavy, relatively uncommon, and expensive. An older lens you might consider is the 159mm Wollensak f/12.5 Extreme Wide Angle.

 

[wiltw]

The different aspect ratios of the different formats makes the Diagonal a poor way to calculate equivalence! It is better to simply use the SHORT DIMENSION of very format so that the vertical AOV equivalence is met, and then you let the horizontal (and diagonal) 'fall where it may'. After all, what is stupid is to shoot an overly wide format (like 6x9 or 135) and then crop it to 16x20" aspect ratio, having calculated AOV equivalence based on a longer diagonal of 3:2!

So if 21mm LF : 24mm (135 format vertical), equivalence is 21/24 = n/(8" format height)
Assuming (wrongly, as 4x5" format is truly 93mm and not the assumed 4" nor 90mm!) 8" format is 200mm, n = 175mm

 

[ic-racer]

On a budget look at the Fuji superwide 125mm or 120mm. It might cut the corners a little, but once you crop, it has about the same angle of view as those very expensive 150mm 8x10 lenses.
I got one for a couple hundred dollars a while back.

 

[EdSawyer]

Nikkor SW 120mm.

 

[Alan Gales]

I've got an old single coated Schneider SA 121mm f/8 lens which I paid around $180 for including shipping. I use it for 4x5 but it will just cover 8x10 stopped down and straight on with no movements. Not a great option but a dirt cheap option.

 

[Old-N-Feeble]

If you're cropping 135 format to 4:5 ratio (1:1.25) then a normal lens on 135 format is 38mm. Normal focal length for 8x10 inch film is 315mm. If you want a similar look on 8x10 vs. 135 format with 21mm lens.....

21/38=.5526
.5526x315=174mm

So a 174mm lens will have a similar look on 8x10 inch film as a 21mm lens does on a frame of 135 format film... if you crop 135 format to 8x10 prints.

Perhaps a 165mm Super Angulon, 155mm Grandagon, 150mm Super Angulon XL, 150mm Nikkor SW??

FWIW, I partially agree with wiltw. To determine 'normal' focal length lens on any format, I prefer to simply multiply the longest film edge by 1.25. So, normal FL for 135 format cropped to 24x30 would be 38mm (rounded up). Normal for 8x10 inch (long edge 245mm) is 306mm.