# Help with exposure factors.

Discussion in 'Exposure Discussion' started by jstraw, Dec 3, 2013.

1. ### jstrawMember

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I'm sure this is something I shouldn't be confused about.

Here's an example. If I'm using an extension tube with a factor of 2.15 and a filter with a factor of 2.5 and I have an aperture of f4, id this the calculation or do I have this all wrong:

4 x 2.15 x 2.5 = 21.5(f22) and I adjust my shutter speed as though my aperture was f22 and then compensate for reciprocity as needed?

2. ### MattKingSubscriber

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You use exposure factors this way with shutter speeds, not apertures.

If your shutter speed was 1 second, the new shutter speed would be 1 x 2.15 x 2.5= 5.375 seconds.

If you need to use them with apertures, you need to use the base 2 log of the factor instead.

3. ### jstrawMember

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Can you break that base 2 log thing down for a dummy?

4. ### Greg DavisMember

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I find it easier if I divide my film speed by the factor. For example, if I am using a 400 speed film and have a factor of 2, then divide 400 by 2 (400/2= 200), so I set my meter to ISO 200 and use the shutter and aperture from that.

5. ### jstrawMember

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When applying more than one factor do you add them or multiply them?

6. ### MattKingSubscriber

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If you are adjusting shutter speeds, you multiply.

If you are adjusting apertures, you add the base 2 logs of them to get the change in stops.

The base 2 logs are the number you "raise" 2 to to get your factor. And they are the stop equivalent of the factor.

Here is a calculator: http://www.miniwebtool.com/log-base-2-calculator/

so for your factor numbers 2.15 and 2.5 the result is 1.1 + 1.3 = 2.4 stops (after rounding off).

7. ### jstrawMember

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Thanks for helping me understand this.

So for 1 sec at f4:

Adjusting shutter speed it's 1 x 2.15 x 2.5 = 5.375 sec (adjusted for RF) at f4.

Adjusting aperture it's 1 sec at f4 + 1.1 + 1.3 = roughly f1.8 (2.4 stops).

Or my original exposure becomes effectively 1 sec at roughly f9.

Do I have that correct yet?

8. ### Greg DavisMember

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No. It would be 1 sec at f/1.8, if you change to f/9 you will underexpose your film. To answer your question to me: multiply the factors together first, then divide the film speed by that number. For example, if you have two filters, one with a factor of 2, and the other with a factor of 1.6 then:

2 x 1.6 = 3.2

ISO 400 / 3.2 = ISO 125

Since a factor of 2 is one stop, and 1.6 is 2/3 of a stop, this works out to adjusting your exposure 1 2/3 stop total.

9. ### jstrawMember

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So then yes, I had it right. A said 1 sec at f1.8.

I described this, perhaps clumsily, as "effectively" 1 sec at f9. f9 being simply a way of describing the effect of the amount of illumination the factors reduced. For 1 second, the amount of light reaching the film plane at f4 with the extension and filter in place is roughly equivalent to the amount of light reaching it at f9 without the extension and filter...if I understand things correctly.

10. ### jstrawMember

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Does anyone know how to translate a 2 stop loss from a teleconverter back into an exposure factor so that I can incorporate it into the math with other factors such as filter factors?

11. ### Christopher WalrathSubscriber

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Just treat it like a filter factor. KISS.

12. ### MattKingSubscriber

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Two stops is a filter/exposure factor of 4.

13. ### jstrawMember

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Thanks! 2^2...I'm getting this...slowly.