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jurgenestanislao
7-Dec-2020, 07:26
Hello! Calling on to the knowledge of the group.

How do I determine the correct amount of additional pops per stop?

Is my math correct?

1 pop; initial exposure e.g. f/8
2 pops; +1 stop e.g. f/16
4 pops; +2 stops. e.g. f/32
8 pops; +3 stops. e.g. f/64
16 pops; +4 stops. Etc.
32 pops; +5 stops
Etc.

So this being the case, if I needed to add 2 stops for my bellows compensation that means, I would have done 5 pops or 8 pops?

Thanks!

Jurgen

Sent from my Pixel 4 XL using Tapatalk

ic-racer
7-Dec-2020, 07:31
What happened to f11 and f22?

Tin Can
7-Dec-2020, 07:49
I count stops and pops on my fingers

sometimes more than 10 fingers

sharktooth
7-Dec-2020, 14:48
Hello! Calling on to the knowledge of the group.

How do I determine the correct amount of additional pops per stop?

Is my math correct?

1 pop; initial exposure e.g. f/8
2 pops; +1 stop e.g. f/16
4 pops; +2 stops. e.g. f/32
8 pops; +3 stops. e.g. f/64
16 pops; +4 stops. Etc.
32 pops; +5 stops
Etc.

So this being the case, if I needed to add 2 stops for my bellows compensation that means, I would have done 5 pops or 8 pops?

Thanks!

Jurgen


Your math is partly right and partly wrong. The part that's right is 2 pops is + 1 stop, 4 pops is + 2 stops, 8 pops is +3 stops ..... The part that's wrong relates to the f stop increments. f 16 is two stops smaller than f 8, not 1 stop. Similarly, f 64 is 6 stops smaller than f 8, not 3 stops.

Let me make this waaaaay more complicated than it needs to be, just for fun.

The f stop is really just the simple ratio of the lens focal length over the lens diameter (or diaphragm diameter). f=L/D where f= f number, L= Lens focal Length, D= Lens or diaphragm Diameter

The amount of light coming through the lens is related to the diameter of the lens (or diaphragm). The smaller the diameter, the less the amount of light getting through. One stop is a doubling or halving of the amount of light. Adding one stop would be doubling the amount of light, and reducing by one stop would provide half the amount of light. The amount of light coming through the diaphragm is relative to the area of the diaphragm opening. The area of a diaphragm opening for a circular diaphragm is Pi * D^2 / 4 (where D= Lens or diaphragm diameter). Note that the diameter is squared, so it's not a simple doubling of diameter.

Now, if we want to cut the amount of light in half (reduce it by one stop), we need to reduce the area of the diaphragm opening in half. If we call the starting diameter Da, then the starting area Aa = Pi * Da^2/4. Half that area would be Ab = Pi * Da^2/8 which would be one stop less light. The diameter of this smaller diaphragm opening would be Db, and the area would be Ab, so Ab = Pi * Db^2/4. Now that we have two different equations for Ab we can combine them to solve for Db.

Ab = Pi * Da^2/8 = Pi * Db^2/4

The Pi on each side cancels out leaving,

Da^2=2*Db^2
Sqrt(Da^2) = Sqrt(2*Db^2) Note that Sqrt is the terminology for square root
Da = Sqrt (2) * Db

or Db = Da / Sqrt(2)

This means that for one stop less light (or half the light), the diameter of the diaphragm opening has to divided by the square root of 2.

Now, let's see how this affects the f number. Remember that f = L / D from above.

If our starting f number is fa, then fa = L / Da. This can be rearranged as Da = L / fa

Similarly fb = L / Db

and we already determined Db = Da / Sqrt(2) for one stop less light

Therefor fb = L / (Da / Sqrt(2) ) = (L * Sqrt(2)) / Da

and since Da = L / fa

fb = (L * Sqrt(2)) / (L / fa)

The L cancels out leaving fb = fa * Sqrt(2)

fb =fa * Sqrt(2) is our one stop smaller relationship

Let's say our starting f stop is f8, then one stop smaller is 8 * Sqrt(2) = 11.3 (or rounded to f 11)

One stop smaller than f 11.3 is 11.3 * Sqrt(2) = 16 (f 16)

One stop smaller than f 16 is 16 * Sqrt(2) = 22.6 (or rounded to f 22)

One stop smaller than f 22.6 is 22.6 * Sqrt(2) = 32 (f 32)

..... and so on

If you want to look at one stop larger increments, you divide the original f number by Sqrt(2).

fb =fa / Sqrt(2) is our one stop larger relationship

Let's say our starting f stop is f8, then one stop larger is 8 / Sqrt(2) = 5.65 (or rounded to f 5.6)

One stop larger than f 5.6 is 5.6 / Sqrt(2) = 4 (f 4)

.... and so on

Here are the traditional one stop increments

f 0.5, f 0.7, f 1.0, f 1.4, f 2.0, f 2.8, f 4.0, f 5.6, f 8.0, f 11, f 16, f 22, f 32, f 45, f 64, f 90


I'll admit, it's a little more painful than counting fingers, but a lot less painful than a root canal.

Larry H-L
7-Dec-2020, 16:55
To keep it simple, double your previous number of pops for each additional stop.

In my experience, reciprocity failure starts to set in after 8 pops or so.

Two23
7-Dec-2020, 17:16
Are you looking at the f scale on a very old lens that doesn't show f11? If so, remember that is the old "U.S." scale, not the modern f scale. F16 is f16, but what says f8 is really f11 and so on.


Kent in SD

jurgenestanislao
7-Dec-2020, 21:24
My bad, thanks for all the inputs!

The table I should have shared:

Initial exposure; 1 pop e.g. f/8
+ 1 stop; +2 pops; e.g. f/11
+ 2 stops; +4 pops e.g. f/16
+ 3 stops; +8 pops e.g. f/22
+ 4 stops; +16 pops e.g. f/32
+ 5 stops; +32 pops e.g. f/45
+6 stops; +64 pops e.g. f/64
... and so on.

My next question is if I needed +2 stops for example for bellows comp., my total pops should be 8 right? And not 7?

Sent from my Pixel 4 XL using Tapatalk

sharktooth
7-Dec-2020, 23:08
From your own corrected table, +2 stops extra exposure would require 4 pops total. 2 stops is 4 times as much light (4 pops).

Bellows compensation is often expressed as a multiplier factor. A multiplier of 1 would be normal exposure with no compensation. A multiplier of 2 means you need twice as much light. That could be achieved by opening up the lens by one stop (for 2 times as much light), or multiplying the exposure time by 2. Similarly, a multiplier of 4 means you need 4 times as much light. That could be achieved by opening up the lens by two stops (for 4 times as much light), or multiplying the exposure time by 4.

To make this easier, maybe avoid drinking any wobble pops prior to analysis. :)

Jim Michael
8-Dec-2020, 09:27
You have + in front of your number of pops. You’re not adding to the prior value in your table, just showing the number required.

jurgenestanislao
8-Dec-2020, 20:43
You have + in front of your number of pops. You’re not adding to the prior value in your table, just showing the number required.

Yeah figured that out now, thanks!

metcaldharv
12-Jan-2021, 09:51
Thanks for advice. Not it is clear for me.

eli
2-Jan-2024, 01:19
I find it sometimes very useful to think in terms of volume, not stops and pops, just because that's what you're really working with; a doubling or halving of light by aperture, or speed or, measured light or a combination, plus filters.

Halving each volume of light the film is receiving is just as easy but demands a very controlled flash(s) units.

One more thing you might want to print up, depending on if your lenses have 1/2 or 1/3ed F-Stop divisions, because sitting in a studio in the dark trying to work out that math really sucks and will drain positive energies we all need to confidently do our photography.

It also looks bad if your client or model has to wait for you to do your sums.

IMO.

Mal Paso
2-Jan-2024, 19:08
More Power, less math.

Daniel Unkefer
2-Jan-2024, 21:06
More Power, less math.

+1

Tin Can
3-Jan-2024, 06:04
I usually test using X-Ray

I set my studio scene

and have trays ready for action

Almost as fast as Fujiroid

When I had no shutter for the lens, I turned on Red Safelights in my entire studio

and shot pops like a Print Test Strip

My entire 650 sq Blacked out

Mal Paso
3-Jan-2024, 18:57
I added up the damage:
For 14,800 WS of power, 12 flash heads, 4 soft boxes, 16, 10+, and 5+ reflectors, barn doors, filter holders, stands, snoot, grids, umbrellas, triggers, cords, including tax and shipping. $2,400 and the market is so slow I doubt I could get that back but that's only 2/3s the price of a single new 2400 WS power pack, no flash head. Didn't need that much power but it's like I bought one studio for the reflectors and another power pack came with and who doesn't like spares?

Mark Sawyer
4-Jan-2024, 12:39
In my experience, reciprocity failure starts to set in after 8 pops or so.

Can anyone speak further to working with reciprocity failure during multiple strobe pops?

Daniel Unkefer
4-Jan-2024, 14:41
Color shifts are kinda inevitable if you're multiple popping and shooting color. I've gone up to 20 pops with 375J and I got Very tired of doing that! I have been bracketing, cheap enough with XRAY, not so with Panchromatics. I think you start getting into diminishing returns after so many pops. Sometimes I don't want the light in close, more power! is the answer. Like Mal Paso, I lucked into an astounding value of Biggie strobes for little money. Also Modifiers can cost you a lot of light, I have packs up to 3200J which can get you to F22 or F32 in one pop. The bigger the Modifier, the more the light is spread out, requiring more pops or power, your choice. My choice is more strobe power, enough to do the job. Films and Plates I like go all the way down to about EI 1.50 or a bit less. Like shooting wet plate, you need a lot of strobe power if you want to be spontaneous.

Tin Can
4-Jan-2024, 15:38
Flashbulbs can be very interesting

Of course, 1 Pop

Mal Paso
4-Jan-2024, 16:46
Flashbulbs can be very interesting

Of course, 1 Pop

I started with 50Bs shooting 4x5 for advertising before I had enough money for strobes. Had a shutter stick and one went off in my hand. Quite painful. They are the size of a 100W bulb with an Edison base.

LabRat
4-Jan-2024, 19:00
Or light with continuous light and keep the shutter open as long as your meter + reciprocity state...

Many multiple pops with a strobe negate the advantages of a fast single pop, so get out the hot lights...

Steve K