Close up with a view camera

<p>Dear all,<br>

I am going to try some close up photography with my Ebony RW45 view camera and either of my 150mm or 210mm lenses. So far as I am aware they are both just normal lenses.<br>

I also understand that the 150mm lens must be twice its focal length away from the film plane, ie about 300mm out.<br>

Am I right saying that if I use the 210mm lens then it has to be 420mm away from the film plane?<br>

If this is correct do I also have to place the camera and tripod at a correct distance from the subject for the subject to be in focus?<br>

If this is so is there an easy way of doing this, for example using a sliding tripod head with knobs for moving the whole camera back and forth?<br>

Any help would be greatly appreciated just to get me started.<br>

Thanks<br>

Gary</p>

<p> </p>

<p>Sounds like you've got a good understanding of the issues, all your comments are correct. Yes, a macro focusing rail is a nice thing when working that close but they can be spendy.</p>

<p>Depth of field will be EXTREMELY shallow, you'll need to stop down quite a bit. Don't forget to compensate for bellows extension, additional exposure will be necessary based on how much extension and how much magnification you have. A quick google search on "bellows extension factor" should get you the info you need to make the necessary adjustments. </p>

<p> </p>

<p>If you want to do 1:1 then you should do as you ask and have the lens at double its focal length from the film. The distance from the lens to the subject will be the same distance as the lens to film distance. You should give two stops more exposure to compensate for the extension. Simple! Except that it always seems more complicated when you actually do it. Good luck!</p>

<p>Hello there,<br>

Thanks for your response.<br>

Is there a way of working out how far my subject needs to be from the lens to get the image in focus, on both the 150mm lens and the 210mm lens?<br>

I know the subject of macro and close ups is lengthy, I would jest like to get out and start experimenting.<br>

Also does anyone know of a macro focusing rail that's worth having a look at?<br>

Thanks<br>

Gary</p>

<p>At 1:1 the working distance will be twice the focal length of the lens from the lens itself (2x focal length from lens to film plane and 2x from lens to subject).</p>

<p>This would be a nice focusing rail, assuming that you're already using Arca Swiss style plates/clamps. </p>

<p>http://reallyrightstuff.com/ProductDesc.aspx?code=B150-B&type=0&eq=&desc=B150-B%3a-Macro-focusing-rail</p>

<p> </p>

Working with a large, heavy camera, it's most of the time easier to put the subject on a stage and move that, instead of the entire camera.<br><br>Often, a (spare, obviously) rear or front standard of the type of camera you are using, with a no hole 'lens board', put on the same base rail as the camera will do the trick nicely. Especially if the standard has gear driven movements. But it will limit you to more or less vertical shooting (unless you tilt the 'stage' by 90 degrees.)<br>But that doesn't work with folders like the ebony. Still, putting the subject on a moveable stage in front of the camera will be much easier than having to move the entire camera.

<p>Theres something here about exposure compensation when doing close ups:<br>

http://www.southbristolviews.com/pics/Graphic/CloseUpCalc.html<br>

It includes a little calculator you can print out and make.</p>

<p>You do know that you can't focus accurately by measuring distances? You have to achieve focus by examining the ground glass screen with a loupe - I apologise if this seems like a silly question. Good luck.</p>

One note about DoF and stopping down: there is very little DoF at 1:1. Stopping down will increase its size, but 2x or 4x very little is still very little.<br>So don't rely on stopping down. Compose the shot so that the plane of focus is on the main subject. Use tilt and Scheimpflug to move the plane of focus to the best position. and do that in a way that creates the best looking image even if you did not stop down at all in search of a little bit more DoF.<br>Then avoid stopping down too much, for though it will not produce significant amounts of extra DoF, it wil degrade image quality thanks to increased diffraction.

<p>The RW45 is a light camera. Just move the tripod back and forth to get in the right focusing range. If you have no lighting behind the camera this can even be done without a darkcloth. If you need more bellows extension just order an extension tube, or back, or both. You can get 1:1 on that camera with a 120 macro lens, and probably with a 150. But if it is not a macro lens don't expect great results.</p>

<p>I'm a little late to the party, have a little to add to the good advice already given.</p>

<p>As has been pointed out, film plane-to-subject distance at 1:1 is approximately 4 * focal length and lens (measure from the diaphragm)-to-subject distance is approximately 2 * focal length. This for lenses of normal construction (neither retrofocus nor telephoto), which includes most LF lenses. Approximately because internodal distance is neglected in the 4* rule and because the lens' front node is usually a little in front of the diaphragm in the 2* rule.</p>

<p>Best practice is to set extension to get the desired magnification and then move the camera/lens assembly to focus. Coarse focus by moving the tripod, finer focus by using a focusing rail. Focusing by changing extension (move the rear standard, not the front) will work as long as film plane-to-subject distance is at least 4* focal length to start. This because the minimum film plane-to-subject distance comes at 1:1, where it is 4 * focal length + internodal distance.</p>

<p>The focusing rail suggested is a beauty, but pricey. I just took a look at "focusing rail" "macro slider" on eBay. I have one of this type (item 330936410415) that is at its limit with a 2x3 Graphic. I also have a Velbon slider (item 290928021963) that is more robust. I've had a Novoflex Castel, don't recommend it, poor ergonomics.</p>

<p>The RW45's maximum extension is 325 mm, so it won't go to 1:1 with a 210 mm lens.</p>

<p>The highest magnification available with a 210 on your camera is ~ 0.55, i.e., slightly larger than half life size. At that magnification extension will be 325 mm and film plane-to-subject distance will be ~ 924 mm. Approximate because I'm ignoring internodal distance and because your lens' focal length probably isn't exactly 210 mm.</p>

<p>For 1:1 on 5x4 you can get away with using an 80mm or 105mm enlarging lens. Remember that coverage increases with lens extension, so an 80mm lens set 160mm away from the film will have the same image circle as a 160mm lens focused at infinity. If you try to use normal focal length lenses for macro work you'll probably find you run out of bellows extension.</p>

<p>WRT exposure compensation, if you work in focal length units you simply need to multiply the scale F-number by the number of focal lengths in the bellows extension to work out the exposure. For example; at half scale the lens extension will be 1.5 focal lengths, so the effective F stop becomes 1.5 x the marked number (f/8 becomes f/12). At 1:1 the lens is 2 focal lengths away from the film and the F-number is doubled (f/8 becomes f/16). And so on.</p>

<p>In maths terms this is: Effective aperture = (bellows length / focal length) * f-number</p>

<p>Personally I wouldn't even contemplate doing macro work on LF. Smaller formats have too many advantages in this area.</p>

<p>Gary, about the focusing rail, yes it would be nice, and I wanted one when I decided to do some macro stuff on the studio some years ago, but too pricey. Better to move the tripod back and forth, it's a bit awkward at first but you will get used. I worked with a 4x5 as well. Had a 210 lens but not enough bellows extension. So I worked with my two other lenses: a 150mm and a (believe it or not) 90mm! This last one really worked for tiny subjects, though you have to bring the subject REALLY close to the lens. Bellows extension for this lens was sometimes 27cm, which meant that I had to correct exposure three stops. Even with this lens I could work moving the tripod back and forth, looking through the groundglass, under the cape. Of course, extremely tiny movements (awkward because of the rubber ends of the tripod legs) meant big differences in magnification, but when you start doing it you will get used. And for a 150mm or 210mm lens it will be easier.</p>

<p>Remember that 1:1 gives the same size image on the film regardless of format. So what fills up a 35mm frame will be a small part of a 5x4 frame.</p>

<p>You might need to magnify more than 1:1 depending on what your subject is.</p>

<i>"Remember that 1:1 gives the same size image on the film regardless of format. So what fills up a 35mm frame will be a small part of a 5x4 frame."</i><br><br>Indeed.<br>And it works the other way round too: what you can't get to fit inside a 35 mm frame at 1:1 will sit comfortably within a larger frame.<br><br>There are two approaches to macro.<br>The first is 'driven' by the need for a certain magnification. Then there is no reason to use larger format than that in which your subject will fit.<br>The other is driven by composition. You can fit any subject in any format frame, but the larger the frame, the higher the magnification needed to fit the same composition in it, and the more detail you pick up. Then it is a matter of how much detail you want (with the understanding that you wouldn't want to move into macro territory to begin with were it not that you want to reveal the detail you don't get to see at lower manifiactions. And that you need to print larger too: it's of little use to use a large format with high magnification and then print small, showing not much - if anything - more than you would see if you had used a small format and printed to the same size.)

<p>All good stuff here.</p>

<p>Also consider using auxillery close-uplenses. These are +diopters that screw into the front filter ring . . .like a filter. There is no exposure adjustment for them. They allow you to uas a closer subject to lense distance for the same bellows extension ( and so, more magnification).</p>

<p>Drawbacks are: Additional air-glass surfaces.<br>

Possible distortion at edges.</p>

<p>I have a 77mm set that are much larger than the 67mm front threads of my lenses, so edge distortion is minimised.<br>

Subjects with a central area of interest (flowers etc) wwill also minimise any concern in that regard.</p>

<p>They are light, compact and easy to use, and can be screwed onto nay lens with the right step-up rings.</p>

<p>Just another option in your quest for Macro.</p>

 

<p ><a name="00bj9Z"></a><a href="/photodb/user?user_id=5151169">Gary Ritson</a>, Jun 10, 2013; 01:43 p.m.</p>

 

<p>"Is there a way of working out how far my subject needs to be from the lens to get the image in focus, on both the 150mm lens and the 210mm lens?"</p>

<p>Here's where you use the general lens equation. For s = distance from lens (node) to subject, i = distance from (rear) lens (node) to film, and F = focal length, 1/F = 1/s + 1/i. So the answer to "how far my subject needs to be from the lens to get the image in focus" depends on how far the lens is from the film plane.</p>

<p>We can make it even easier. Taking implicit derivatives, delta s / delta i = -(s/i)^2. If you've got it set up for magnification of 1, then s = i which reduces the derivative equation to delta s / delta i = -1. That means if you move the lens backwards by 1 cm or move the film plane forwards by 1 cm, delta i = -1 cm so delta s = + 1 cm. Thus, for every change in lens to film distance, you must change the lens to subject distance by the same amount, in the opposite direction.</p>

<p>HTTH</p>

 

 

<p>Daniel, I'm all for solidarity among Daniels but this</p>

 

<blockquote>

<p>Thus, for every change in lens to film distance, you must change the lens to subject distance by the same amount, in the opposite direction.</p>

</blockquote>

<p>is a bit much. If I understand you correctly, when I move the lens backwards towards the film, I have to move the lens away from the subject. If I move the lens forwards away from the film then I have to move the lens closer to the subject. And in both cases, the absolute values of the two distances moved are identical. Is that what you meant?</p>

<p>Now consider situations where magnification is, respectively, 1 and 0. Lifesize image, subject at infinity, respectively. In the first case lens' rear node to film distance = lens' front node to subject distance = 2 * f. In the second, lens' rear node to distance = f and lens front node to subject distance is considerably larger. Are you sure that's what you meant to say? In general, f isn't infinite.</p>

<p>The symmetry you asserted doesn't exist.</p>

<p>Cheers,</p>

<p>Dan</p>

<p>Gary, if you want to learn as much as you need to know about working closeup (= more than you think you need to know) buy a copy of Lester Lefkowitz' book The Manual of Closeup Photography and read it. Daniel, you might want to read the book too.</p>

<p>Not quite. While it's easy to focus by moving the lens forward or back (to and fro'), its very hard to do it correctly. In moving the lens forward or back, you are changing both the lens to film distance and the lens to subject distance!</p>

<p>The OP first asks, "If this is correct do I also have to place the camera and tripod at a correct distance from the subject for the subject to be in focus?" The answer is, yes, you can do it that way.</p>

<p>He then seeks a way of determining the subject to lens distance so that the image will be in focus. You can do this by moving the subject or, as in the former inquiry, moving the camera / tripod (but NOT just the lens). You can also focus the image by moving the film plane. The last one is the preferred method.</p>

<p>I showed how you could use the standard lens equation 1/F = 1/s + 1/i to solve for s, where s = distance from lens (node) to subject, i = distance from (rear) lens (node) to film, and F = focal length. Solving for s, all other quantities fixed, gives 1/s = 1/F - 1/i and so s = 1/(1/F - 1/i).</p>

<p>Now we come to the part which I claim simplifies things and which Dan Fromm disputes. Let's say we have it set up for magnification 1 and the image is in focus or close to being in focus. Now we change i, the distance from the lens to the film. The change is di. If that change is small, say less than 10%, we restore focus by moving the subject a small distance ds in the opposite direction. ds = -di. These distances are the same. </p>

<p>This analysis is based on differentials and thus requires that ds and di be small. Furthermore, I have limited the analysis to magnification = 1, or is close to 1.</p>

<p>This analysis is inapplicable to situations where m = 0 for the following reasons: 1. the analysis assumes m is close to 1; 2. s = infinity so i = F, ds/di = -(s/i)^2 = -infinity. The subject, an infinite distance from the lens, is moved by another infinite distance. That is meaningless. However, we CAN consider moving the lens since such movement changes the subject-to-lens distance a negligable amount.</p>

<p>This analysis would be useful for estimating how to bring the subject in focus. It is still necessary to refine the focus using a groundglass or accurate rangefinder.</p>

<p>I don't recall what this process is called in applied mathematics / calculus, but it is similar to perturbation analysis used in physics.</p>

<p>Under the conditions specified, yes, that is what I meant.</p>

<p>Am I redeemed, fellow Dan?</p>

<p>My oh my,it's amazing how in depth the simple act of stretching out the bellows and moving towards the subject (while looking at the ground glass until,voila,I actually see what's happening) can be!<br>

Regards,Peter</p>

<i>"While it's easy to focus by moving the lens forward or back (to and fro'), its very hard to do it correctly."</i><br><br>Depends on what "correctly" would mean.<br>If it means "such that the thing you focus on is in focus", it is very easy to focus moving the lens, i.e. change both lens to film and lens to subject distances.<br>If it would mean "such that the desired magnification is maintained" it's impossible. Lens to film distance has to remain set, and only lens to subject distance may vary.<br>If the thing is set up for 1:1, and you change lens to film distance, changing lens to subject distance will restore focus, but you will have lost 1:1.<br><br>I also don't see that symmetry. To keep focus, a change in lens to film distance of, say, 5 mm isn't balanced by an equal, 5 mm change of the lens to subject distance.<br>1/10 + 1/10 =/= 1/5 + 1/15.

<p>Brother Dan, you are not redeemed. The OP needs (a) a practical solution that can be applied with minimal calculation and (b) a general solution that works at all distances and magnifications.</p>

<p>The simple practical solution is to take advantage of the relationship rear node-to-film distance = focal length * (magnification +1). This is an easy calculation and since most lenses used on LF cameras have rear nodes near the diaphragm, allows magnification to be set, approximately but usually well enough, with a tape measure. As QG has already pointed out, after magnification is set focus is obtained by moving the camera/lens assembly, not by changing extension, which changes magnification.</p>

<p>By an odd coincidence that won't bear close examination, the magic formula given above works for all magnifications.</p>

<p>There's a parallel magic formula for the distance from front node to subject, viz., distance = f*(m + 1)/m.</p>

<p>If you play with the two formulas' sum, you'll see that at 1:1 film-to-subject distance is, ignoring the lens' internodal distance, 2f and that this is the minimum film-to-subject distance. At all other magnifications film-to-subject distance > 2f. Also, if magnification != 1 there are <em>two</em> magnifications with the same film-to-subject distance. This is another reason for setting magnification in advance and then focusing by moving the camera/lens assembly.</p>

<p>As I said, buy the book, read the book.</p>

<p>Cheers,</p>

<p>Dan</p>

<p>Mr. de Bakker:</p>

<p>Your inequality is based on i = 10 and s = 10 or vice-versa, right? That should yield 1/F = 1/10 1/10 = 1/5 = 0.20 and thus F = 1/(1/i + 1/s) = 5. You then show that for di = -5 and ds = +5, i = 5, s = 15, 1/i + 1/s = 0.27 which is not equal to 0.20. In fact, it differs by 35%.</p>

<p>As I wrote, the algorithm, which only gives a first-order approximation, applies when the magnitudes of ds and di are small relative to s and i. In Mr. de Bakker's example, ds and di are each 50% of s and i, respectively. That is not small.</p>

<p>Now let's look at this proposed counterexample with di = -0.5 and ds = .5, i = 9.5 and s = 10.5. di = -5% of i and ds = 5% of s. Now, 1/i + 1/s = 0.20050125. This differs from 0.20 by 0.00050125 which is 0.250627%. Much closer.</p>

<p>Okay, I ordered the book, sight unseen.</p>

<p>Congratulations! I don't think it will disappoint you.</p>