How to calculate subject distance to fill image?

[PastInk]

Hi all.

After some searching I am still confused as to how to calculate lens plane to subject if I want to fill a 5 inch wide (4x5 Sinar) image with a flat subject whose maximum width is 48 inches. I am using a 150mm Symmar lens. Due to the peculiar camera set up I am using I can't actually focus through the ground glass so I am trying to put the system into the ball park orientation (lens to subject, lens to film plane) and then final focus using the Betterlight scanning back that I have. If someone can point me toward the correct set of formulae with a working example I would appreciate it greatly.

Thank you

 

[cramej]

Angle of view at f/16 for that lens is 70 degrees. A little trigonometry and paper and pencil will get you pretty close. You can look up solving for sides of a right triangle to understand the process but it ends up being just over 34". That is, of course, if I'm thinking straight today.

 

[Prof_Pixel]

Center of lens to subject distance of about 57 1/2 inches. It's just a simple matter of drawing 2 triangles


--------------
Edit
Oh good. Two different answers. Are there any more?

 

[photog_ed]

Here's another answer:

Recall the thin lens formula: 1/f = 1/d1 + 1/d2

where

f = focal length
d1 = lens to film distance
d2 = lens to object distance

Also recall magnification m = d1/d2

After a little algebra,

d2 = f * (1 + 1/m)

In your case, m = 5/48, f = 6", so I get d2 = 63.6"

HTH,

Ed

 

[wiltw]

So-called 5" Sinar image is actually 120mm if you measure a film holder like Lisco...the object being photographed is 1219mm wide, so you need to reproduce the object at 1/10.2 scale.
There are complicating factors to the computation, though

• Assuming the lens' actual FL is accurately reflected by its engraved value (which is NOT a good assumption to make), a 150mm lens results in Angle of View horizontally of 45.89 degrees -- not 70 degrees since the lens can cover 8x10 format (the 4x5 diagonal AOV is 56.93 degrees).
• the effective focal length of the SuperSymmar 150mm is 147.7mm, not 150mm
• I assume the FOV calculator accurately reflects actual 120mm long frame, not so-called 5" (which is 127mm), but I don't know with certainty what assumptions there! The FOV calculator says long dimenstion of 4x5 frame sees 46.53 degrees.

I solved with various techniques:

A) Using your formula, photog_ed, but plugging in Schneider's stated effective FL, and plugging in actual frame dimension (not 5') to determine true value of m, I get 64.88"

B) Assuming 147.7mm and an imprecise definition of frame size for determination of AOV, we have AOV horizontally (long dimension) of 46.53 degrees using the calculator program. Dividing frame size and object size in half so we can solve for right triangle, using trig functions I compute 'center of lens' to object as 55.6", and 'center of lens' to focal plane as 5.5", for a total of about 61" from object to film plane.

C) The calculator program I have says 61" subject distance using 147.7mm FL results in 47.45" FOV in long dimension.​

 

[cramej]

Here's the solution that I came up with - the 70* angle of view comes from Schneider's website. Edit -and I suppose this would be to the lens and not to the film. Who knows what I'm calculating now....

 

[wiltw]

Multiple solutions, as imperfect as they are, tell the OP to position the lens about 34 - 64" away from the target. The OP can tell us what works, but I would say the 61" solution is the target.

 

[photog_ed]

B) Assuming 147.7mm and an imprecise definition of frame size for determination of AOV, we have AOV horizontally (long dimension) of 46.53 degrees. Dividing frame size and object size in half so we can solve for right triangle, using trig functions I compute 'center of lens' to object as 55.6", and 'center of lens' to focal plane as 5.5", for a total of about 61" from object to film plane.

If you account for the fact that the AOV changes with magnification, you should get the same answer as with the thin lens formula.

 

[RobC]

5x4 image area isn't 5 inches wide. I'll assume its 12cm wide or 4.72 inches which gives us a magnification of 4.72/48 = -0.0983X
then using focal length of 147.7mm we get an object to film plane distance of 1812.01mm(71.34in)
lens extension will be 162.22mm(6.39in) and lens to object will be 1649.79mm(64.95in)
so there's another possibility using the software I have and my guesstimates of actual image size/magnification on neg.

using photog_ed formula then you need to know d1 which will be

d1 = (1/f - 1/d2)-1

[edit] corrected values above

 

[wiltw]

After some searching I am still confused as to how to calculate lens plane to subject if I want to fill a 5 inch wide (4x5 Sinar) image with a flat subject whose maximum width is 48 inches.

So are WE! Your use of our estimates, along with your report back about theory vs. reality will go a long way toward a number of us abandoning certain computation methods (or using them only with a great amount of caution in the future)!

 

[Dahod]

I know you were looking for help finding the answers based on first principles and you got a lot of good responses.

These days there are a lot of apps kicking around and I just wanted to recommend one that I use a lot and seems to get me pretty close. It's the Mark II Artists Viewfinder and lets me set up my film size (4x5) and my various lenses. The downside is that it's expensive at $25 but it saves me a lot of time and also let's me get a quick snapshot of the scene for a record.

 

[RobC]

for most lens calcs there is a free android app written by one of Rodenstocks lens designers. It will give you all the numbers you want for your lens extension and lens to subject distance. It can never be 100% accurate for what you are doing but you won't find anything more accurate unless you know the refractive indexes of all your lens elements and its true focal length.

https://play.google.com/store/apps/details?id=net.opticalsoftware.calclensthin

and for those that don't know, rodenstock were taken over by Linos a few years ago and now they come under the Qioptiq company umbrella. Qioptiq are Rodenstock.

I note you are using a scanning back so you may infact have a full 4x5 inch scan area which hadn't registered with me when I suggested 12cm instead of 5 inches. You can plug your values into the software.

[edit]
if you don't have an android device you can get the windows desktop version from here

http://www.winlens.de/index.php?id=70

[edit]
I use the windows desktop version of this software but also have the android version. I would recommend the windows desktop version over the android version as its easier to get paramaters right with the desktop version and know they're right. The android version is too finicky on a small handset and doesn't give an easily seeable clue if your paramaters are wrong. And the android version has a tendancy to second guess your parameters if you use the wrong combinations which the desktop version doesn't and is better for that.

 

[ic-racer]

Just to emphasize what has already been mentioned, the 'thin lens equation' can be used to calculate both subject/object distances or subject/object sizes. You, of course, don't need to know the angle of view, but you can also calculate that by solving the simple lens equation for both subject size (opposite) and distance (adjacent) and then use trigonometry. 'Opposite over adjacent' = Tangent.

 

[RobC]

object angle is 75.818.

Since I've been using this qioptiq software which is quite a lot of years, I've forgotten all the lens formula I used to know. Which only goes to show that automation and access to software makes you lazy.

 

[wiltw]

I don't see why the Object Angle even needs to be considered directly in the computation.

An equation which I just found from Computar (CCTV lenses) tells the derivation of FL value needed to cover a certain size object to fill a certain size image:

FL = image size * ( Object distance / Object size )​

If we plug in FL = 147.7mm, image size = 120mm, Object size = 1219.2mm,

147.7 = (120 * (D/1219.2)
D = (147.7 * 1219.2) / 120
​

the value of D computed is 1500.6mm, or 59.08"

Once again, it is important for OP to come back and report what was the final solution distance!

BTW, ( Object distance / Object size ) = 0.4067 = Tan (22 degrees 8 minutes) or Tan (22.13 degrees)
and the calculator program says 156mm lens at 59" distance captures 44.3 degrees, so now I know the program wrongly uses 5" or 127mm, not the correct frame size determine by the film holder of 120mm.

 

[gone]

All the fancy smancy math in the universe isn't going to be of any help to you, as each subject has a different requirement for filling the frame (I don't understand why people are talking about center distance when you specifically said filling the frame), and your different distances to the subject are close to infinity. How do you figure that? Simple, you figure out a way to sight through the lens somehow and ck the actual GG. Otherwise, it's impossible.

Any lens can be used for any subject if you can get close enough to it or far enough away, but you don't want to shoot a 4x5 portrait w/ a 90mm lens due to distortion and other bad things. That's just one example of what can go wrong when you use the wrong lens for a subject. There's a lot more. Make your life simple and figure out what subject you want to photograph, buy an appropriate lens, and get a camera that allows the sort of focus that is required for this.

 

[tedr1]

Here is a different approach. Do trial set up before you get to the location. Set up the camera in front of something 48 inches across, focus the camera, measure the lens to subject distance, measure the lens to film plane distance, use these distances when you get to the location.

 

[photog_ed]

I don't see why the Object Angle even needs to be considered in the computation.

An equation which I just found from Computar (CCTV lenses) tells the derivation of FL value needed to cover a certain size object to fill a certain size image:

FL = image size * ( Object distance / Object size )

If we plug in FL = 147.7mm, image size = 120mm, Object size = 1219.2mm,

147.7 = (120 * (D/1219.2)
D = (147.7 * 1219.2) / 120
​

the value of D computed is 1500.6mm, or 59.08"

Once again, it is important for OP to come back and report what was the final solution distance!

This results from setting d1 = f in the formula for magnification, i.e. m = d1/d2 becomes m = f/d2. It's close enough when d2 is much greater than f, but gives a result that is off by a few inches in this case.

 

[wiltw]

This results from setting d1 = f in the formula for magnification, i.e. m = d1/d2 becomes m = f/d2. It's close enough when d2 is much greater than f, but gives a result that is off by a few inches in this case.

But the equation which I cited, from the Computar web site (they sell lenses for various size CCTV sensor cameras and can compute directly the FL needed to accomplish a given goal) does not directly compute any angle, and while ( Object distance / Object size ) = Tangent of angle, one does not need to look up the Angle value itself ...at all!

Alter their equation to compute D, and assuming we have an accurate statement (for the right model of Schneider Symmar lens -- there are multiples!) we know FL = 147.mmm, without knowing explicitly any value for Angle

 

[wiltw]

This results from setting d1 = f in the formula for magnification, i.e. m = d1/d2 becomes m = f/d2. It's close enough when d2 is much greater than f, but gives a result that is off by a few inches in this case.

But the equation which I cited, from the Computar web site (they sell lenses for various size CCTV sensor cameras and can compute directly the FL needed to accomplish a given goal) does not directly compute any angle, and while ( Object distance / Object size ) = Tangent of angle, one does not need to look up the Angle value itself ...at all!

Alter their equation to compute D, and assuming we have an accurate statement (for the right model of Schneider Symmar lens -- there are multiples!) we know FL = 147.mmm, without knowing explicitly any value for Angle.

If we continue...

thin lens formula: 1/f = 1/d1 + 1/d2, where

f = focal length
d1 = lens to film distance
d2 = lens to object distance
​

we can compute, using your equation...

1/147.7 mm = 1/d1 + 1/1500.6 mm​

to derive the value of d1, and then we know d1 + d2 = object to film distance

...all without any angles in the computation.

​

 

[photog_ed]

But the equation which I cited, from the Computar web site (they sell lenses for various size CCTV sensor cameras and can compute directly the FL needed to accomplish a given goal) does not directly compute any angle, and while ( Object distance / Object size ) = Tangent of angle, one does not need to look up the Angle value itself ...at all!

Alter their equation to compute D, and assuming we have an accurate statement (for the right model of Schneider Symmar lens -- there are multiples!) we know FL = 147.mmm, without knowing explicitly any value for Angle

Agreed, no angles are needed. My formulas do not contain any angles.

 

[RobC]

All the fancy smancy math in the universe isn't going to be of any help to you, as each subject has a different requirement for filling the frame (I don't understand why people are talking about center distance when you specifically said filling the frame), and your different distances to the subject are close to infinity. How do you figure that? Simple, you figure out a way to sight through the lens somehow and ck the actual GG. Otherwise, it's impossible.

Any lens can be used for any subject if you can get close enough to it or far enough away, but you don't want to shoot a 4x5 portrait w/ a 90mm lens due to distortion and other bad things. That's just one example of what can go wrong when you use the wrong lens for a subject. There's a lot more. Make your life simple and figure out what subject you want to photograph, buy an appropriate lens, and get a camera that allows the sort of focus that is required for this.

he specifically said a 48in wide object size to fill a 4x5in image size. There is only one distance for his specific lens focal length hence all the calcs for it being given. And he said appoximate values are good enough as he can fine tune once he gets close to right focus.

 

[Mr Bill]

Photog_ed had it right from the beginning. I think Prof_Pixel disregarded the need to extend the lens while focusing, otherwise he would have had the same answer.

For practical purposes, Ed's formula is plenty good. The major problem is that the exact measuring point on the lens is unknown (unless the two nodal points are either measured or supplied by the lens maker). Given this, I think that calculated precision beyond a half inch or so doesn't have any practical value.

 

[Prof_Pixel]

Photog_ed had it right from the beginning. I think Prof_Pixel disregarded the need to extend the lens while focusing, otherwise he would have had the same answer.

That's correct - I did NOT include any lens extension from the infinity position.

 

[Drew Bedo]

Why not just look at the ground glass?