# Multiple exposure calculation

Discussion in 'Exposure Discussion' started by naeroscatu, Jul 7, 2009.

1. ### naeroscatuSubscriber

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My math skills are fairly poor so help me out. If I need to do a multiple exposure (where I cannot make a very long exposure) how do I calculate?
For example my meter shows 1/30 sec as the correct exposure and I want to make 5 exposures; ho do you figure out in the field 1/30 divided by 5? would that be 5 times 1/ 150 sec? (guessing). thanks

2. ### Christopher WalrathSubscriber

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Your five exposures need to equal 1/30 so your closest bet would five exposures at 1/125th of a second. Actually four of these would give you the correct exposure. The fifth would increase your exposure by 0.5 stops over the required exposure. Now, if you are doing multiple exposures over Zone 0 areas on your negative from previous exposures and there is no overlap of exposure then expose as recommended each time there is no overlap of negative densities.

Hope that doesn't confuse too much.

3. ### Q.G.Inactive

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5 x 1/125 is only 0.3 stops more than 4 x 1/125.

4. ### Christopher WalrathSubscriber

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Wirelessly posted (BlackBerry9000/4.6.0.167 Profile/MIDP-2.0 Configuration/CLDC-1.1 VendorID/102 UP.Link/6.3.0.0.0)

I stand corrected. Geez! ;p

5. ### Ian GrantSubscriber

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You usually have to give extra as the short underexposes cause reciprocity, usually around half a stop.

It's more normal to be using much longer exposures of a second and then make that up with multple exposures., not all of the same length

Ian

6. ### naeroscatuSubscriber

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Very useful info, thank you. Never occured to me to consider reciprocity on short underexposures, thanks Ian.