# Distance to subject

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<p>Hi and Happy New Year to all,<br>

Here is my question.<br>

Is there a way to know what distance one needs to be away from a subject? Here is what I mean. I am needing to take some photos of a business store front that is 45 feet wide. I realize that I can just move in and out with the lens that I have on the camera but I am curious if there is a calculation for this.<br>

If I used a 30 mm lens with DX system D90, being in the center how far back would I need to be to capture the entire store front. Sometimes this is restrictive as there may be objects that stop me from being at the right distance and I would like to know this info ahead of time. It would help in the lens selection.<br>

Thank You,<br>

phil b<br>

benton, ky</p>

<p> </p>

<p>Joe Smith</p>

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<p>You will need to figure out how much top-bottom areas of your image will be *lost* as well. If you had the use of a Nikon D700 body, the larger sensor size would result in a little more of the 45-feet wide area have more detail. You, with the DX sensor D90, will be squeezing in more area with a 16mm, 18mm, or 30mm lens.</p>

<p>The 30mm lens will crop like a 45mm lens (on the old 35mm film measurement system) on your D90.</p>

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<p>Hi. Yes there is! I have a magic spreadsheet which allows all matter of calculations to be performed. A 30mm lens on a D90 would require being about 17 meters back from your subject. An 18mm lens would allow you to be 10m away. PM me with more measurements if you need a more precise calculation.</p>
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<p>Use trigonometry - works every time. You need the angle of view of the lens and the width/height of the subject to do the calculation.</p>
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<p>About 60 ft - here is a calculator that allows you to get a feel for the distance: http://www.bobatkins.com/photography/technical/field_of_view.html <br>

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<p>A quick check reveals the Bob Atkins site to give more accurate results compared with the cambrigeincolour site. My spreadsheet also allows you to calculate the required focal length from subject size and working distance.</p>
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<p>It's fairly simple trigonometry using similar triangles.</p>

<p>Step (1) Multiply the width (or height) of the subject by the focal length of the lens.<br /> Step (2) Divide the number arrived at in (1) by the width (or height) of the camera's frame size.</p>

<p>You can mix your measurement units (e.g. feet and millimetres) as long as you stick to the same units for both the focal length and the frame size.<br /> To work through your example above:<br /> (1) 45feet x 30mm =1350. (2) 1350 / 24mm = 56.25. So a DX format camera would need to be just over 56ft from a 45ft storefront with a 30mm lens to get it all in.</p>

<p>To put this as a formula so that it can be transposed: D = Sf/W; where D is the subject distance, S is the subject width, f is the lens focal length and W is the width of the frame.<br /> If you want to work out the focal length needed, the formula becomes f = DW/S</p>

<p> </p>

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<p>WOW, Thanks to all of you for these great answers. I will be going over each of the suggestions and I can be assured that it will be worthwhile and fun.<br>

It always amazes me on just how smart a lot of people are on here.</p>

<p>phil b<br>

benton, ky</p>

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<p>double post</p>
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<p>Rodeo Joe has got it right. No need to calculate the angle of view (which is already calculated from the focal length).</p>

<p>With a 45 ft subject and a 30mm lens you get the distance 30mm (focal length)/24 mm (sensor width) x 45 ft (subject size) =56.25 ft for a horizontal shot. If you want 10% space around it you need to increase the distance 10%. If you want to do a vertical shot you need to increase the distance 50%.</p>

<p>If you use your 30mm a lot you just need to learn that the <strong>distance required is 1.25 times the subject size</strong>.</p>

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<p>Addendum.<br /> Phil, the formula I gave will only work reasonably accurately for fairly long subject distances - it's no good for close-ups. The reason being that lenses either extend or change focal length when focused. With a lens that simply extends to focus, the formula could be modified to make it accurate for any distance, but lenses that use internal focusing are becoming much more common these days. Such lenses are entirely an unknown quantity, and predicting their exact field of view at close distances is almost impossible.<br>

Anyhow, the formula given should work reasonably well for subject distances above 15ft or so.</p>

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<blockquote>

<p>Anyhow, the formula given should work reasonably well for subject distances above 15ft or so.</p>

</blockquote>

<p>No, it will work well much closer than that, especially with a prime lens. The change in focal length when focusing closer is usually very small but some lenses often zooms change a lot - the latest 70-200 f/2.8 for instance is known for this. Have a look at this for example: http://bythom.com/nikkor-70-200-VR-II-lens.htm</p>

<p>One should also keep in mind that some lenses have a different focal length than what's on the label. And it also makes a difference if you have a 100% viewfinder or not. But it's just an estimate so no need to calculate it with high precision.</p>

<p>Anyway, it pretty easy to check the focal length change. Just take your camera, put it in manual focus and rack the focus from infinity to minimum focusing distance forward and back. See how much the lens breaths (changes focal length) just by observing how much less of the initial scene you can view when you focus close. It might help to stop down and press the DOF preview button.</p>

<p> </p>

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