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nicemate1
11-Oct-2015, 09:24
Hello I need some technical help by someone with more experience and knowledge here.
I need to do a reproduction of big paintings of a maximum size of 76x80 inches (193x203cm), but can only position my camera so far.
To be precise, I will not be able to position my camera more than 3 mt away (app. 120 inches).
I only have a 90mm, but I doubt I will be able to use it. How far would I need to go, with a 90mm anyway, is someone knows?
Many thanks for your help,
Pietro

Tin Can
11-Oct-2015, 09:54
I don't have the answer and I think it depends.

Why not draw the situation on paper and calculate angle of view needed and compare that number to your existing lens official statistics.

Or draw the proposed 'painting' dimensions on a wall, put your tripod at 3m and have a look?

Jim Noel
11-Oct-2015, 10:14
What format is your camera? That will be a major key to the lens necessary.

Tin Can
11-Oct-2015, 10:45
What format is your camera? That will be a major key to the lens necessary.

I forgot to add format size, Jim is of course correct.

It's all in the angle, both forward and rearward.

Taija71A
11-Oct-2015, 14:05
... How far would I need to go, with a 90mm anyway, is someone knows?

Quick Answer:

Based upon your above cited parameters...

M = 1:20
f = 90mm

... and 'assuming' that you are going to be using 4x5 Film:

You will need approximately 2 Meters (Lens to Subject Distance).
Best regards,

-Tim.

nicemate1
11-Oct-2015, 23:54
Hi Tim, Randy and Jim!
Thanks.
You are right, I forgot to mention that I am using a 4x5 camera, with film (not a digital back).
I would trace the image (or it's corners) on a wall, maybe with masking tape and see if it works with my 90mm, but as it stands now I don't even have that lens with me as I borrowed it to a friend, and will pick it up just the day before the shoot. However, if I need a wider angle, I need to know that beforehand because I need to plan a rental accordingly, hence the question on the LF Forum.
Tim you say 2mt would be enough: that is great - I thought it would have had to be many more meters away !

Lachlan 717
12-Oct-2015, 00:02
Do you have a CF for the 90?

It's not likely to be a flat-field design, so you might have to consider distortion correction in post.

Jerry Bodine
12-Oct-2015, 13:25
It'll depend a lot on which 90mm lens we're talking about. Some have significantly more covering power than others. The subject will require an image circle with 2.8m diameter at the chosen f-stop. A few examples below:

Angulon/6.8 has a 40.5* viewing half-angle (at f/16) and would need a lens-to-target distance of 1.64m.

Super Angulon/8 has a 50* half-angle (at f/22) and would need a lens-to-target distance of 1.17m.

Super Angulon/5.6 has a 52.5* half-angle (at f/22) and would need a lens-to-target distance of 1.07m.

All these distances increase if illumination level requires wider apertures.

Add the lens-to-film distance when focused closer than infinity + sufficient distance for working behind the camera to compose/focus.

ic-racer
12-Oct-2015, 14:10
The lens, of course only has to cover 150mm if the film taking format is 4x5. At 3 feet the lens probably only needs to be spec'd at 120mm infinity to cover your film, but this has nothing to do with your problem and don't know why it has been mentioned.
My guess based on experience is 65mm lens.
The math to figure the lens you need is not easy, as the angle of view on a 4x5 negative gets smaller as you focus closer.
140927

Jerry Bodine
12-Oct-2015, 22:11
The subject will require an image circle with 2.8m diameter at the chosen f-stop.

Perhaps I didn't say that very clearly. What I meant was that the subject painting has a diagonal of 2.8m (not the spec image circle of the lens, which of course would be outrageous). It's clear to me now where I went wrong re getting the entire painting contained on the groundglass. Please excuse my errors.

B.S.Kumar
13-Oct-2015, 07:09
According to this nifty calculator (http://www.scantips.com/lights/fieldofview.html) a 90mm lens used on 4x5 film at a distance of 1.85 meters from the painting will cover an area of 2.47m x 1.95m - just about its shorter size.

Kumar

Doremus Scudder
14-Oct-2015, 03:19
Pietro,

Your 90mm will probably "work" if all you want is to squeeze the image onto the film at a limited distance.

That said, you have to take into account that when doing wide-angle close-up work (which this turns out to be), you will get significant distortion in the corners of the image. Circles on the reproduced art work will end up being uneven ellipses in the corners. This may or may not be acceptable for your purposes.

When I was reproducing art works for museum catalogs, shooting farther back with a longer-focal-length lens was the norm in order to reproduce the shapes in the artwork accurately (I usually used a 210-240mm lens on 4x5 and filled the frame with the artwork by positioning the camera - a 150mm lens would work, but I liked the longer renderings).

I'm not sure how much you can doctor distortion from working close to the subject digitally, but it certainly won't be ideal.

Is it not possible to move the artwork to a larger space? That's what I often did (helped with lighting too).

Best,

Doremus

ic-racer
14-Oct-2015, 06:04
That said, you have to take into account that when doing wide-angle close-up work (which this turns out to be), you will get significant distortion in the corners of the image. Circles on the reproduced art work will end up being uneven ellipses in the corners.


...reproduction of big paintings...

The original poster is photographing paintings, not sculpture.

IanG
14-Oct-2015, 06:17
The original poster is photographing paintings, not sculpture.


You will still get the distortion.

Ian

nicemate1
15-Oct-2015, 09:43
Hello thanks to everyone! I don't really understand how to use the nifty calculator Kumar posted :confused:...
I also got a bit lost in the calculus, but, correct me if I am wrong, I will be able to cover the entire surface of the painting, that has a diagonal of 2.8 mt., by positioning myself app. 2 mt. away
from the subject, with a 90mm lens. Right ?

Taija71A
15-Oct-2015, 09:48
... by positioning myself app. 2 mt. away from the subject, with a 90mm lens. Right ?



You will need approximately 2 Meters (Lens to Subject Distance).

As per all previous message replies...

You will of course, still need to allow space:
For yourself (behind the Camera) and for your Tripod Legs etc...

Drew Wiley
15-Oct-2015, 11:48
Doremus already mentioned the Achilles heel. Few things in this focal length will be even close to rectilinear, though the pricey Zeiss Biogon 90 would probably be the best option.

Lachlan 717
15-Oct-2015, 13:06
Don't forget the CF that I mentioned earlier....

xkaes
18-Oct-2015, 09:22
Check out www.subclub.org/fujinon/index.htm

Dan Fromm
18-Oct-2015, 10:16
Doremus already mentioned the Achilles heel. Few things in this focal length will be even close to rectilinear, though the pricey Zeiss Biogon 90 would probably be the best option.

Pricey? Worse than that, non-existent.

Kirk Fry
18-Oct-2015, 17:53
Seems to me even lighting and square alignment are going to be killers...