Which lens should I get to cover an area of 76 in × 80 in (193 cm × 203 cm) from 3mt?

[nicemate1]

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]

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]

What format is your camera? That will be a major key to the lens necessary.

[Tin Can]

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]

... 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]

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]

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]

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]

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.

[Jerry Bodine]

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.