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Anne_Raker
06-20-2008, 05:25 AM
I know the law - that moving the light twice the distance makes the light intensity 1/4 as strong as it was originally - but am I doing the math right here?

So, if my flash is 5' from my subject, and my subject is properly exposed at ISO 100, f/8 at 1/125, if I move my flash to 10' from the subject, I would need to adjust my exposure by two stops, so it would be come ISO 100, f/4 at 1/125?

Is that right? And conversely, if I move the light from 5' away to 2.5' away, my exposure would become ISO 100, f/16 at 1/125?

Or am I totally messing this up? :p Oh, and by the way... please tell me there aren't 100 "calculate this" questions on the CPP exam! Please??? :p

Erin_L._Clark
06-20-2008, 06:50 AM
oh gosh, please tell me too because that's math I don't like to figure out!

Hassel_Weems
06-20-2008, 06:57 AM
I don't know about the exam question but you are correct about how it works in practice.

There is an easier way to do it - put your light at distances you already know (2', 2.8', 4', 5.6', 8', 11', 16', etc.) and you will be working in 1 stop increments instead of 2 stop increments.

If your light is 8 feet from your subject, move it to 5.6 feet from your subject and gain 1 stop (instead of 2 stops by halving the distance.) If you light is 4 feet from your subject move it to 5.6 feet from the subject and lose 1 stop of light.

Heather_L._Smith
06-20-2008, 12:03 PM
Ladies - there ARE math questions on the test... not 100 of them, but a handful. Knowing how your settings change when things move around will be important.

Mark_Levesque
06-20-2008, 12:46 PM
Right on the money, Anne.

KirkDarling
06-20-2008, 02:11 PM
You need to be intensely intimate and comfortable with the relationships between ISOs, apertures, and shutter speeds. If they give you questions like "which of the following exposures is not equivalent to the others?" it should not worry you even if they throw an f/45 option at you.

Mark_Levesque
06-20-2008, 05:15 PM
Love the new avatar, Kirk. Love it even better bigger. link?

Anne_Raker
06-20-2008, 05:34 PM
Thanks guys! I am glad I'm calculating that right. :D I am expecting "math" questions about the exposure triangle, about filter factors and how they affect exposure, about lighting ratios and adjusting them with fill light, and about the inverse square law. Is there more "math" beyond that, and did anyone's test deal with any of that in half or third stops, or was it all in whole stops? If it's going to be all in whole stops, I think I'm going to be fine.

My bigger problem is that I still have four chapters to go in the book - The cyanotype/platinum printing chapter (which I fell asleep reading last night), the view camera chapter, the composition chapter and the history chapter.
So, if you see me online (anywhere, not just here ;) ), feel free to tell me to get off the computer and get back to the books!

Howard_Kier
06-20-2008, 05:40 PM
Anne,

Get back to the books!

Anne_Raker
06-20-2008, 05:41 PM
Thanks Howard! :D I am off right now to do just that!

Mark_Levesque
06-20-2008, 06:42 PM
If that's all you have left, Anne, your time would be better spent making sure you can construct a table of standard aperture values and a color wheel, IMO.

KirkDarling
06-20-2008, 08:06 PM
Mark is right. Draw yourself a color wheel and shutter speed and aperture tables. We old timers had a bit of help with that because the old-school cameras and lenses had the "tables" engraved right where we saw them constantly. But get them in your head, and when you go into the test, write them out on the scratch paper for reference.

Apertures: As a mnemonic help, notice that the aperture roughly doubles every other number: 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45. And be sure you understand that the aperture number is the denominator of a fraction, so like 1/4 of a pie is less pie than 1/2 a pie, f/4 is less exposure than f/2.

You do need to understand how ratios work--ask if you're not sure.

You also need to know how filter factors work--how you change both aperture or shutter speed--or both--depending on filter factor.

You could get a question like, "The correct aperture and shutter speed at ISO100 is f/8 at 1/250. Which of the following will provide a correct exposure when a filter with a factor of 4x is mounted?"

...And all of the options might vary all of the factors, so you will have to first apply the filter factor to the given exposure, then figure which of the options actually provide that corrected exposure.

Anne_Raker
06-20-2008, 08:39 PM
If that's all you have left, Anne, your time would be better spent making sure you can construct a table of standard aperture values and a color wheel, IMO.

I can do that. I have been practicing that, actually, and plan to grab some scratch paper and write it down as soon as I get to the test, so I have it to refer to. (Assuming they allow scratch paper... if not, I fully expect to be drawing imaginary charts in the air and looking just as crazy as I am!)

But quickly, from memory:
ISOs in full stops: 50, 100, 200, 400, 800, 1600, 3200, 6400
Av in full stops: 1, 1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45, 64 (I know it can go higher in theory, and I think the next ones are 90 and 126, but I'm not 100% sure...)
Tv in full stops: 1 sec, 1/2 sec, 1/4, 1/8, 1/15, 1/30, 1/60, 1/125, 1/250, 1/500, 1/1000

And the color wheel, clockwise, starting with red at the top, would be: red, yellow, green, cyan, blue, magenta

Okay, off to check my work...

Anne_Raker
06-20-2008, 08:48 PM
Mark is right. Draw yourself a color wheel and shutter speed and aperture tables. We old timers had a bit of help with that because the old-school cameras and lenses had the "tables" engraved right where we saw them constantly. But get them in your head, and when you go into the test, write them out on the scratch paper for reference. Whew! I'm glad you said that... I was counting on being able to use scratch paper! :)


Apertures: As a mnemonic help, notice that the aperture roughly doubles every other number: 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, 45. And be sure you understand that the aperture number is the denominator of a fraction, so like 1/4 of a pie is less pie than 1/2 a pie, f/4 is less exposure than f/2. Ah! So my above calculation is wrong. The last one would be 128 (double of 64), not 126. That definitely helps - thanks Kirk!


You do need to understand how ratios work--ask if you're not sure.
I think I get it... a 1:1 ratio is even - no difference between lit and shadow side. a 2:1 ratio is one stop darker on the shadow side. A 4:1 ratio is two stops darker on the shadow side (a 3:1 ratio would be 1.5 stops darker). an 8:1 ratio is three stops darker on the shadow side (a 6:1 ratio would be 2.5 stops darker, a 5:1 ratio would be 2 1/3 stops darker and a 7:1 ratio would be 2 2/3 darker, I think... right? I am pretty confident with the full stops, but the 1/2 and 1/3s throw me.


You also need to know how filter factors work--how you change both aperture or shutter speed--or both--depending on filter factor.

You could get a question like, "The correct aperture and shutter speed at ISO100 is f/8 at 1/250. Which of the following will provide a correct exposure when a filter with a factor of 4x is mounted?"
This would require a 2-stop increase in exposure, right? So, the answer would be either ISO 100 at f/4 at 1/250, or an equivalent exposure, like ISO 200 at f/2.8 at 1/250, or ISO 100 at f/5.6 at 1/125... along those lines?

Michael_Gan
06-20-2008, 09:54 PM
I think I get it... a 1:1 ratio is even - no difference between lit and shadow side. a 2:1 ratio is one stop darker on the shadow side. A 4:1 ratio is two stops darker on the shadow side (a 3:1 ratio would be 1.5 stops darker).

Heh, here we go again...:) http://ourppa.com/forums/showthread.php?t=10405

Anne_Raker
06-20-2008, 10:20 PM
Michael, Do I have the whole stops right? I think I do, and I'm just getting the half stops wrong... is that right? The half stops totally throw me.

So if the light has a power of 8, and you cut the light in half (reduce by one stop), it would have a power of 4. Relative to the original power, it would be 8:4 or 2:1. If you take that new power of 4 and cut it in half again (reducing by a second stop), it would have a power to 2. Relative to the original power, it would be 8:2 or 4:1.

But what is the math to reduce by 1.5 stops? You'd cut it in half first, right? So reduce by one stop, bringing power in the example above to 4, and the ratio to 8:4 or 2:1. But instead of reducing by half again, you'd reduce by half of a half (or one quarter power). So instead of bringing the power from 4 to 2, you'd bring it from 4 to three. So you'd have a ratio of 8:3, which does NOT equate to 3:1...

Am I getting this at all? Thanks for the link - I have read and reread that thread, and I "think" I'm getting it, but I may need a ton of bricks dropped on my head! LOL!

KirkDarling
06-20-2008, 10:26 PM
Don't sweat half stops for the test, not even for ratios. If the right answer seems to be a half stop, that's the wrong answer.

For us old guys, a half stop was just the umarked click between two full stops: "f/8 and a half."

Zack_Davis
06-21-2008, 08:55 AM
Right on the money, Anne.

GEEK! :p :D

KirkDarling
06-21-2008, 02:20 PM
So if the light has a power of 8, and you cut the light in half (reduce by one stop), it would have a power of 4. Relative to the original power, it would be 8:4 or 2:1. If you take that new power of 4 and cut it in half again (reducing by a second stop), it would have a power to 2. Relative to the original power, it would be 8:2 or 4:1.

I think you're bringing "ratios" (2:1, 3:1, et cetera) into a place they don't belong, Anne. We don't use the term "ratio" to discuss changes made to a single light.

"Ratio" refers to the comparison of the powers of two or more lights; for test purposes, let's keep it to just two, the main light and the fill light.

Setting the ratio is how you determine the drama of the lighting contrast of the scene--how bright the highlight area is compared to the shadows. Do you want the shadows to be very dark (a high ratio) or very light (a low ratio)?

But wait. Why do we care about the numbers? Just set up the lights, shoot off a few frames, and check the LCD, right? Yes, we can do that in this digital age. However, we want to be professionals who know up front how to get the results we want quickly and efficiently, and how to duplicate an effect that we know works for a given result.

Also, there will be a good number of ratio problems on the test.

This the key thing about ratios to understand: Light on the subject is additive. If you put one light on the subject (say, a 200 watt second flash) that provides an exposure of f/8 and then put a second identical light on the subject lighting the same area, the exposure of the combined lighting would be f/11--twice the number of identical lights gives you twice as much light, hence, you can close the aperture down one stop more.

When you place your main light at an angle to the subject, from the camera's viewpoint it lights part of the subject and keeps another part of the subject in shadow.

When you set your fill light near the camera, then from the camera's viewpoint it lights the entire subject. It puts light onto the shadow area that the main light did not (which is why it's a "fill" light) as well as the areas the mainlight illuminated. So the fill light adds its light to the highlighted area already illuminated by the main light. It's that additive factor you have to keep in mind when calculating what ratio of highlight to shadow that the two lights are giving you.

Here is an illustration:
http://www.lord-and-darling.net/tech/images/ratios.jpg

The main light is 200 watt seconds, the fill is 100 watt seconds. The camera sees 100 watt seconds on the shadowed part of the subject and a combined total of 300 watt seconds on the highlighted part of the subject. That's a 3:1 ratio of highlight to shadow.

But we're using light meters that tell us f-stops, not watt-seconds, so we have to transfer this concept to f-stops.

But do you get it so far?

Anne_Raker
06-22-2008, 04:08 AM
GEEK! :p :D

Why, thank you kindly, sir! :D

Anne_Raker
06-24-2008, 01:50 AM
I think you're bringing "ratios" (2:1, 3:1, et cetera) into a place they don't belong, Anne. We don't use the term "ratio" to discuss changes made to a single light.

"Ratio" refers to the comparison of the powers of two or more lights; for test purposes, let's keep it to just two, the main light and the fill light.

Setting the ratio is how you determine the drama of the lighting contrast of the scene--how bright the highlight area is compared to the shadows. Do you want the shadows to be very dark (a high ratio) or very light (a low ratio)?

But wait. Why do we care about the numbers? Just set up the lights, shoot off a few frames, and check the LCD, right? Yes, we can do that in this digital age. However, we want to be professionals who know up front how to get the results we want quickly and efficiently, and how to duplicate an effect that we know works for a given result.

Also, there will be a good number of ratio problems on the test.

This the key thing about ratios to understand: Light on the subject is additive. If you put one light on the subject (say, a 200 watt second flash) that provides an exposure of f/8 and then put a second identical light on the subject lighting the same area, the exposure of the combined lighting would be f/11--twice the number of identical lights gives you twice as much light, hence, you can close the aperture down one stop more.

When you place your main light at an angle to the subject, from the camera's viewpoint it lights part of the subject and keeps another part of the subject in shadow.

When you set your fill light near the camera, then from the camera's viewpoint it lights the entire subject. It puts light onto the shadow area that the main light did not (which is why it's a "fill" light) as well as the areas the mainlight illuminated. So the fill light adds its light to the highlighted area already illuminated by the main light. It's that additive factor you have to keep in mind when calculating what ratio of highlight to shadow that the two lights are giving you.

Here is an illustration:
http://www.lord-and-darling.net/tech/images/ratios.jpg

The main light is 200 watt seconds, the fill is 100 watt seconds. The camera sees 100 watt seconds on the shadowed part of the subject and a combined total of 300 watt seconds on the highlighted part of the subject. That's a 3:1 ratio of highlight to shadow.

But we're using light meters that tell us f-stops, not watt-seconds, so we have to transfer this concept to f-stops.

But do you get it so far?

Got it, and the visual really helps! And the complicating factor when you add f/stops is that each f/stop is double the light of the one before it (or half of the one after it, depending on which direction you're going), right?

I was happy that the ratio questions seemed pretty straightforward on the test - I hope they actually were as straightforward as they seemed! ;)