Bulb exposure question.

[ryan_warner]

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<p >So, I've been thinking about doing some bulb exposures, once the weather warms up a bit, and I'm wondering how you determine the appropriate exposure time. I'm assuming it's a little bit of guess work , a little bit of luck and a little bit of knowledge. For the knowledge part, is there a mathematic equation to get your exposure time at least in the ballpark? Just for example, if the scene is properly exposed at 30sec f/2.8, but I want to expose at f/22, can I just multiply the shutter speed by the number of stops and get close to the time I'll need? Or is there another trick that I'm not thinking about?</p>

 

[mark_sirota1]

<p>Yep, that's it exactly.</p>

<p>It's six stops from f/2.8 to f/22, so you need 6 times as long an exposure.</p>

[lex_jenkins]

<p>Take a look at Fred Parker's <a href="http://www.fredparker.com/ultexp1.htm#Light%20Intensity%20Chart">Ultimate Exposure Computer</a> . It will help clarify the relationships between light (expressed in EV or Exposure Value), ISO, shutter speed and aperture. It can also be used to evaluate exposures in conditions where light metering is not practical, such as nighttime long exposure photography.</p>

[ryan_warner]

<p>Thank you!</p>

[rainer_t]

<p>-- "It's six stops from f/2.8 to f/22, so you need 6 times as long an exposure."</p>

<p>Half way correct only.</p>

<p>One stop is 2x ... or 2^1<br>

Two stops are 4x ... or 2^2<br>

Three stops are 8x ... or 2^3<br>

Four stops are 16x ... or 2^4<br>

five stops are 32x ... or 2^5<br>

and six stops are 64x ... or 2^6<br>

more or less light.<br>

So the exposure needed is not 6 times as long, it is 64 times as long.<br>

Nevertheless, the correcture to the exposure time equals 6 stops.</p>

<p>So, 30sec at f/2.8 is the same amount of light as 32minutes at f/22. (Not including any reciprocity effects).</p>

 

[frank_skomial]

<p>"It's six stops from f/2.8 to f/22, so you need 6 times as long an exposure." - not sure if this is correct ?</p>

<p>From 2.8 to 4 you need 2X the exposure time, then from 4 to 5.6 you need again 2X longer exposure time than the previous, then from 5.6 to 8 you need again double the exposure time. Thus going only from F=2.8 to F=8 you need to double the exposure time 2X then 2X then 2X that that would be already 2 * 2 * 2 = 8 , or EIGTH times longer exposure time needed at F=8 than at 2.8. </p>

<p>Since you are talking F=22, so certainly you will need much longer exposure than 6 times. </p>

<p>Use the recommended exposure computer...I like at the end it says: "the real Ultimate Exposure Computer will be YOU!"</p>

[frank_skomial]

<p>Yes, while I was typing, T. Rainer explained already...the same.</p>

[ryan_warner]

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<p >Oddly enough I knew what Mark was saying. I knew it wasn't 30x6, it's 30x2 6 times.</p>

</p>

[rainer_t]

<p>And pretty sure, this is what Mark really meant ... nevertheless, such things should be corrected just for the case someone else with the same question also reads the thread (or finds it later on with the search function).</p>

[Alan Marcus]

<p>

<p > </p>

<p >Metered normal exposure f/2.8 at 1/30 sec.</p>

<p > </p>

<p >This method has served me well for manipulation of shutter speeds – apertures or filter factors etc.</p>

<p > </p>

<p >The number sets used for f/numbers and shutter speeds are typically in 2x increments as to exposure differences.</p>

<p > </p>

<p >The number for f/numbers based on 2x change is:</p>

<p > </p>

<p >1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 – 22 -32 – 45- 64</p>

<p > </p>

<p >Note each number going right is 1.4 times its neighbor.</p>

<p >Each number going left is it neighbor to right divided by 1.4.</p>

<p > </p>

<p >The f/number set is based on changing the surface area of a circle in 2x increments. Draw any circle, measure its diameter. To construct a revised circle two times larger (2x) in surface area, multiply the diameter by 1.4.</p>

<p > </p>

<p >The number set for shutter speed (seconds) based on 2x chance is:</p>

<p > </p>

<p >1 – ½ - ¼ - ⅛ - 1/15 – 1/30 – 1/60 – 1/125 – 1/250 – 1/500 – 1/1000</p>

<p > </p>

<p >If the aperture setting is f/2.8 and you choose to use f/22, count the difference i.e. 6 stops difference using your fingers. 6 fingers difference.</p>

<p >Each finger is 2 x increment.</p>

<p > </p>

<p >2 – 4 – 8 -16 – 32 – 64 (6 fingers = 64x difference)</p>

<p > </p>

<p >Multiply exposure time by 64:</p>

<p > </p>

<p >1/30sec. = 0.3 (decimal form)</p>

<p > </p>

<p >0.3x 64 = 2.1 seconds (revised shutter speed for f/22.</p>

<p > </p>

<p >This works but the result will be underexposed. If the shutter speed change is extreme, exact mathematical equivalent will be under exposed due to reciprocity failure law. The exposing energy is governed by inertia, a tendency to remain sluggish as to change. Add more time. Likely ½ stop to 1 stop will be required.</p>

 

<p > </p>

 

<p > </p>

</p>

<p > </p>

[ryan_warner]

<p>I completly agree Rainer T, they should be.</p>

[mark_sirota1]

<p>Right, I apologize for the confusion. I knew exactly what I <em>meant</em> -- if only I'd actually said that...</p>

[mark_sirota1]

<p>Alan wrote: <em>0.3x 64 = 2.1 seconds (revised shutter speed for f/22.</em></p>

<p>That's not right either. The time marked as 1/30th is actually 1/32nd; early photographers had this lazy rounding technique. And, of course, neither is precisely 0.3, so preserving that level of precision is inappropriate.</p>

<p>The actual calculated time would be exactly 2 seconds.</p>

[dbcooper]

<p>To maintain equivalent exposure, <em>each</em> full <em>f</em> -stop change = doubling (or halving) the shutter speed <em>once</em> -or- doubling (or halving) the ISO speed <em>once. </em></p>

<p>So 30 seconds @<em>f</em> 2.8 @ISO 100 = 32 minutes (1920 seconds) @<em>f</em> 22 @ISO 100 (exposure doubled 6 times), or the ISO could be raised to 6400 (doubled 6 times). You could also split changing the ISO and exposure times, like 4 minutes (240 sec.) <em>@f</em> 22 @ISO 800 (exposure doubled 3 times, ISO doubled 3 times to cover the 6 <em>f</em> -stop change). If film is used, the appropriate time must be added to the exposure time to compensate for the film's reciprocity failure per the film manufacturer's data.</p>

<p>Shooting with an SLR/DSLR at <em>f</em> 22 may result in loss of sharpness due to diffraction. If you need longer exposure times, you might consider shooting at <em>f</em> 8 or <em>f</em> 11 and using an ND filter.</p>

<p>If the camera is electronically controlled, long exposure times may exhaust the battery/batteries before the exposure is complete, never mind any time for in-camera long exposure noise reduction on a digital camera (often equal to the exposure time - check your camera's manual). You'll want to verify the maximum time your camera will expose regarding battery life before you go 'bulb' shooting, and plan accordingly (like bringing along spare batteries).</p>

<p> </p>

[James G. Dainis]

You should check the reciprocity characteristics of the film you are using. Some have little reciprocity failure at long exposures, some have a lot. At shutter speeds of less than one second, the law of reciprocity holds true. But for slower shutter speeds, as I assume you will be using with bulb exposures, the law of reciprocity fails. With some films if the call is for one second, you have to use two seconds, if the call is for two seconds you have to use ten seconds, if the call is for ten seconds you have to use one minute, etc.

[Alan Marcus]

<p >Mark, you’re a dreamer! Early mechanical shutter were never accurate as to the engraved speed declared on the bezel. Both manufactures and repair facilities adjusted and calibrated targeting the rounded values. We were lucky if the accuracy obtained was within 1/3 f/stop. Nowadays, the convention followed by young digital camera engineers is to shoot for the rounded values. </p>

[mark_sirota1]

<p>Alan, I timed my D200 with a nominal 30-second exposure, and the shutter stayed open for 32 seconds.</p>

[Alan Marcus]

<p >Mark is correct as to the geometric progression of the shutter speed number set.</p>

<p >The correct set is:</p>

<p > </p>

<p >1 – ½ - ¼ - ⅛ - ¹/₁₆– ¹/₃₂– ¹/₆₄ – ¹/₁₂₈ – 1/₂₅₆ – ¹/₅₁₂ – ¹/₁₀₂₄</p>

<p ></p>

<p >The traditional number set is:</p>

<p > </p>

<p >1 – ½ - ¼ - ⅛ - ¹/₁₅– ¹/₃₀– ¹/₆₀ – ¹/₁₂₅ – 1/₂₅₀ – ¹/₅₀₀ – ¹/₁₀₀₀</p>

<p ></p>

<p >Intentional adjustments are made to “smooth” the sequence by substituting 1/15 for 1/16 and 1/60 for 1/64 etc. Inserting these adjustments yields infinitesimal – undetectable – exposure changes.</p>

<p > </p>

<p >Likely the finest increment whereby measurable change is induced in the blackening of film is 1/6 of an f/stop. These tribulations are far tinier than that. </p>

<p > </p>

<p >A good percentage of my 50+ years in the business were dedicated to instrumentation and measurement of film and paper blackening i.e. densitometry and sensitometery.</p>

[craig_shearman1]

In addition to reciprocity failure as mentioned by James, if you're shooting with a digital camera you get into issues of digital noise with extremely long exposures. Check your manual to see if you camera has special noise reduction settings for long exposures.