tdtaylor Posted March 2, 2011 Share #1 Posted March 2, 2011 Advertisement (gone after registration) My questioning mind wants to know, and I know others in the Forum possess the knowledge... I have a APO 90 Cron (in process of selling) and a 90 Elmarit. When focusing, the Cron extends considerably less than the Elmarit does by a factor of two to three times. I am suspicious that it has to do with the elements/groups- the Elmarit is 4/4 and the Cron 5/5- but just can't put it together in my mind. Thanks for humoring me on this subject. Link to post Share on other sites More sharing options...
Advertisement Posted March 2, 2011 Posted March 2, 2011 Hi tdtaylor, Take a look here Lens extension question. I'm sure you'll find what you were looking for!
01af Posted March 2, 2011 Share #2 Posted March 2, 2011 I have a APO 90 Cron (in process of selling) and a 90 Elmarit. When focusing, the Summicron extends considerably less than the Elmarit does by a factor of two to three times. Impossible. Link to post Share on other sites More sharing options...
Xmas Posted March 2, 2011 Share #3 Posted March 2, 2011 Impossible. Could be internal focusing? Noel Link to post Share on other sites More sharing options...
jaapv Posted March 2, 2011 Share #4 Posted March 2, 2011 Could be internal focusing? Noel Which these lenses don't have. It may be that the difference in bulk of the lenses gives the impression of different extensions, but it is an optical illusion in that case. Link to post Share on other sites More sharing options...
lars_bergquist Posted March 2, 2011 Share #5 Posted March 2, 2011 Both have a close focus limit of 1m. None has internal focusing or a floating element. So the change of extension from infinity to 1m should be the same in both cases. With my 90mm Emarit, last version, it is c. 9.5mm. The same with my only other 90mm lens, a v.2 Tele-Elmarit. Go measure. The old man with the folding ruler Link to post Share on other sites More sharing options...
tdtaylor Posted March 2, 2011 Author Share #6 Posted March 2, 2011 Both have a close focus limit of 1m. None has internal focusing or a floating element. So the change of extension from infinity to 1m should be the same in both cases. With my 90mm Emarit, last version, it is c. 9.5mm. The same with my only other 90mm lens, a v.2 Tele-Elmarit. Go measure. The old man with the folding ruler Definitely have a task for tonight when I get home. And I have many folding rulers- I am Architect and Woodworker. I also have the TE, so I'll measure all three. None internally focus. Thanks, Link to post Share on other sites More sharing options...
Guest #12 Posted March 2, 2011 Share #7 Posted March 2, 2011 Advertisement (gone after registration) Should be 1.0cm for a 90.0mm lens (f constant) with 1.0m closest focus (measured to the film plane). Link to post Share on other sites More sharing options...
tdtaylor Posted March 2, 2011 Author Share #8 Posted March 2, 2011 Tonight I'll check it out and post- all three. Just sold the 90 APO so tonight's the last chance to check. Link to post Share on other sites More sharing options...
tdtaylor Posted March 3, 2011 Author Share #9 Posted March 3, 2011 I stand corrected. All three move 1.0 cm. near to far focus. On a quick look, due to the different spacing of the aperture ring to the focus ring, it does definitely appear the Elmarit moves much farther - that, and as Jaap mentioned, the barrel diameter changes the perspective. And Lars, I didn't use a folding scale- mine are all Imperial, not metric. Always happy to be corrected by a knowledgable group. As a side note, the focus ring moves smoothly on all three, with just about the same force. So even though the very different diameters and weights, they all feel very similar. Never had them lined up to try together before- and tonight is the last time. Thanks to all- although I still do not understand why it is 1 cm. Link to post Share on other sites More sharing options...
Guest #12 Posted March 3, 2011 Share #10 Posted March 3, 2011 ...still do not understand why it is 1 cm. oh they got an equation for that x^2 + (2f - s)x + f^2 = 0 "f" is the focal length and "s" is the closest focus distance Link to post Share on other sites More sharing options...
tdtaylor Posted March 3, 2011 Author Share #11 Posted March 3, 2011 Much appreciated- Link to post Share on other sites More sharing options...
Michael Geschlecht Posted March 16, 2011 Share #12 Posted March 16, 2011 Hello Everybody, To put it another way: For a lens's image size to equal the size of the object it is focussed on the lens's extension must be equal to its focal length. This paradigm is always shown in the following relationship: Image Size : Object Size. Sometimes also written: Image Size / Object Size. Never the other way around. 1 : 1 Sometimes also written 1 / 1 For the lens's image size to be 1 / 2 the object's size the extension must be 1 / 2 of the focal length. 1 / 5 the image size requires 1 / 5 the focal length & so on. A 90mm lens focussing @ 1 meter produces an image 1 / 9 life size. 1 / 9 = 0.111111...... 90mm X 0.111111...... = 10mm = 1 cm extension. A 135mm lens focussed @ 1.5 meters also produces an image 1 / 9 life size. 1 / 9 = 0.111111...... 135mm X 0.111111...... = 15 mm = 1.5cm extension. Best Regards, Michael Link to post Share on other sites More sharing options...
sparkie Posted March 16, 2011 Share #13 Posted March 16, 2011 Go measure. LOL Link to post Share on other sites More sharing options...
Guest #12 Posted March 16, 2011 Share #14 Posted March 16, 2011 ...A 90mm lens focussing @ 1 meter produces an image 1 / 9 life size. ... from the brochure, right? Link to post Share on other sites More sharing options...
Michael Geschlecht Posted March 17, 2011 Share #15 Posted March 17, 2011 Hello # 12, From the math. If you place a metric rule on a wall & photograph it w/ a 90mm lens set @ 1 meter & then again w/ a 135mm lens set @ 1.5 meters you will find that if you measure a horizontal portion of the processed negative/slide counting 270mm on the rule it will measure 30mm on the negative. 270 / 30 = 9 The image on the negative is 1 / 9th of the size of the object photographed. Please see my other Post above. The format is 24mm X 36mm. 24 X 9 = 216 just as 36 X 9 = 324 The coverage @ 1 : 9 is in actuality 216mm X 324mm The brochure says 220mm X 330mm. Accurate in that you can measure that slightly larger space but not the limit @ the actual focus point of 1 & 1.5 meters respectively. Best Regards, Michael BTW: We are talking about what is recorded on film. Not what is within the frames in the viewfinder. There was a Thread about what frames in viewfinders actually cover not that long ago. There is a range of variation in what frames cover when & how. It is dependent on camera model, focal length of lens, which side of the frame line you are on & so on. What we are talking about here is actual coverage on film. I don't know what the parameters are for measuring images created on digital sensors. Someone else may be able to help w/ that information. The 1 : 9 portion of the relationship will remain 1 : 9 digital or film because that is determined by the lens & its extension. Not by the image capture medium/mechanism. Link to post Share on other sites More sharing options...
Guest #12 Posted March 17, 2011 Share #16 Posted March 17, 2011 ... Thank you Michael. Here's how I read your post: since the product of magnification and focal length is the extension, take the magnification at closest focus from the brochure, and 1/9 X 9cm = 1cm. Very easy, if you have that additional information. Link to post Share on other sites More sharing options...
Michael Geschlecht Posted March 19, 2011 Share #17 Posted March 19, 2011 Hello #12, Yes. If they give you a reproduction ratio such as 1 : 9 but not if they give you a size like 220mm X 330mm which is an approximation in an easy to remember form. We have already seen 1 : 9 = 216 X 324 for a format of 24 X 36. For example if we round off reasonably closely the reproduction ratio for a full frame camera using a 16464 on a Visoflex II, IIa or III with a 65mm Elmar = 1 : 2.4 If we compute 65/2.4 = 27mm, for a 90mm Elmarit = 1 : 3.3 If we compute 90/3.3 = 27mm or for a 135 Tele-Elmar = 1 : 5 If we compute 135/5 = 27mm. Pretty much the extension of the 16464. Best Regards, Michael Link to post Share on other sites More sharing options...
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