Reader Question: Why Do Rim Lights Blow Out?
Reader Till Hamburg asks the above question in the Strobist Flickr Pool, and posts a series of photos just to prove his point. First of all, a little explanation of what Till means by these numbers. Then, an explanation and how to make that angle-dependent strength thing work for you.The series of photos at left have two lights which concern us. (Looks like there is a backgrond light going on there, too. But let's ignore that for the purposes of this discussion.)
In the top photo, the rim light is set one stop hotter than the light at camera right (which is illuminating the subject's face.) In the middle, they are equal. And in the bottom, the rim light is one stop lower than the main light.
__________
PLEASE NOTE: I think he has his f/stops reversed in the label on the bottom photo. As you can see, the main light is remaining constant, while the rim light drops a stop each frame, heading toward the bottom.
__________
This lighting quirk used to screw me up all of the time when I would shoot a rim-lit situation, until I figured out what was going on.
Why it happens, and a way to make this work for you, after the jump.
____________________________
Rimshot, Please...
The first component of the higher efficiency of rim light is the physical property of the way anything reflects when it hits something at a shallow angle. I always use the analogy of light behaving like a pool ball when it comes to reflections, as in this post on how to light people wearing glasses.
So let's stretch the light-as-pool-ball analogy a little more and think in terms of retained energy. A glancing reflection off of something will use up less energy than a direct hit. So to will light glancing off of a 3-d object. It is simply a more efficient way to reflect.
Second, rim lights are more likely (than main lights) to be hard light sources. As we have learned, hard light sources create stronger specular reflections than do larger light sources. I know the specular reflection strength is a component to rim lighting being more efficient because I see the effect lessened when my rim light is soft.
So, how can you use this info?
Well, you certainly want to use your weakest light as a rim light to take advantage of the efficient rim light thing. In fact , when I am rim lighting in close, I usually start my SB-800s at 1/128 power and adjust my working aperture to where that looks good as a starting point.
As you can see above, even one stop down is a little hot for a rim light. I tend to start out at two stops down. (If you work without a flash meter, as I do, use your guide number chart to get you close on the first pop.
Also, if a room is ambient-lit in a poor, muddy-ish, ASA-800 kinda way and you only have one light, consider using that light as a separation/rim light (instead of in front) to add depth to your scene. Because of the efficiency of the lighting angle, you can back that speedlight way back (adding depth and internal separation and slow fall-off to the whole scene) even though your speedlight is not very powerful.
___________________
Related posts:
:: L101: Lighting People With Glasses ::
:: L102: Specular Highlights ::
:: Guide Numbers: Your Free Flash Meter ::
Get the Full Monte: Follow Strobist on Twitter.
Now Shipping from USA and UK: Strobist Lighting Seminar DVDs
Feed your brain: My Favorite Lighting/Photo Books
by David, at 5:45 PM.
16 Comments:
Clever boy!
Could someone please explain to the novice how f-stop numbers are assigned to flash intensities. I understand f-stops on the lense, and fractions of total flash output (e.g. 1/16th), but I don't understand how f-stops are determined for the flashes. Many thanks in advance
The chief reason that rim light is so strong is because of the angle of reflectance.
Whatever blows out tends to be a part of your object that is towards the light source.
You can see this by using reflector cards on the backside of a billards ball. The outside edges capture reflections and light from sources behind the ball. The front side lights are fill, but have no effect on the edge of the ball themselves.
About to post the exact same question as 'anonymous' above... define the f/stop of the flash.
thanks for the explanation.
btw "1 stop hotter" is the sexiest term i've ever seen used in a lighting context :) def gonna use that from now on.
Assigning an f stop value to a light is the value taken from an incident light meter at the subject pointing towards the light source. This issue was very important when using film so that you could get good separation from the background and keep detail in the hair so that you could get good prints back from the Lab. Kodak sent master portraitist Frank Cricchio around the world giving seminars when VPS III film was released. Then it was considered that the incident value of the hairlight was to be the same as the main. The effect on the hair was then also a function of the relative size of the light and the surface efficiency of the hair e.g glossy oiled black hair v. frizzy blond. Look up Dean Collins on this for a full explanation and examples -
What about actual area where light is hitting the object, and apparent area? Rays of light are scattered when they hit the subject in all directions. The rays reflected to the camera from the side of the head come from an bigger area (the whole side of the head), but from the lens's point of view, this same number of rays are concentrated within a smaller apparent area.
Sorry to doubt you, you are the expert in photo lighting, but your theory seems a bit flawed. Is this your hypothesis, or do you have other sources to cite (like physics stuff)? I'm not a physicist, so please bear with me a little. I do have to say that this physics explanation of mine is speculative, and my main point is the one in the first and 3 last paragraphs. The reason for mentioning photons and stuff is to make the point that I don't think you can compare light to pool balls or macroscopic objects.
I think we can agree the reason the cue balls (or any macroscopic solid object) will lose less energy when bouncing at an angle than directly is because that energy gets passed on to either the other ball or whatever it hits. Depending on the mass of the other object, it will move faster or slower when hit (or not at all). And depending on the strike angle, the cue ball will bounce faster or slower too. Either way, energy is conserved as always.
With light, you have to account for the energy lost as well, which will be mostly in the form of heat. In this case, rays of light are coming with the same energy (for our practical purposes) to every point of the side of the head. You'd expect then, from your theory, for the ones that bounce straight to lose a lot of energy, and the ones that bounce at an angle to not lose that much energy, and hence appear brighter.
I don't think non-uniform light loss is occurring at all. Whatever light is lost, it is pretty much uniform, and all bounced light rays carry pretty much the same energy for all practical purposes (i.e. are just as bright) whether bounced straight or not. Photons don't behave like macroscopic objects, I don't think they just "bounce" off a surface. An individual photon would have the same effect on a molecule whatever its angle of incidence, from what I can tell from the uncertainty principle (the photon doesn't have a place in space until it has been disturbed). Any physicist can jump in if I'm wrong, which I could totally be.
Picture a laser and a mirror, the simplest form of this experiment. The bounced laser won't register less energy when bounced straight than when bounced at an angle.
So I think most if not all of this effect comes from the apparent area that the lens sees, because it sees about the same amount of light rays concentrated in a smaller area than it would if if wasn't at an angle.
However, I am not an expert in photo lighting, so I will happily accept any corrections on my thoughts.
"F-stop numbers assigned to flash intensities" simply refers to the f-stop (aperture) required to get a normal exposure for that flash. Thats it, pretty simple.
If you are using multiple flashes you can have multiple intensities but you must pick one aperture to use for your exposure (duh). So if your main light (key) is at f4 and your edge light (or hair, fill, whatever) is at f5.6, then your edge is +1. Translation: "normal" flash exposure at f5.6 is one stop more intense than "normal" at f4.
Hope that clears it up.
With regards to Andyo, the apparent area isn't the issue here. If it was, then the camera to subject distance would be as important as flash to subject distance. I haven't done the math, but i would guess that this is because the 1/r^2 reduction in intensity from subject to camera is exactly compensated for by the 1/r^2 reduction in area as viewed from the camera. Thus, if you set off a strobe in the forest without anyone to see it, were the trees still illuminated and at what ev?
I think the real issue is that the skin and hair cannot be treated as lambertian (ie, reflecting light equally in all directions). For normal non-made-up skin, there is some diffuse reflection and some specular from oil. This would be why you get hotspots on the face in pictures with on camera flash, at the point where the incident angle and reflected angle are the same, the oil on the skin reflects like a mirror. Now imagine the same effect over the entire side of the head where the rimlight angle matches the camera angle, which it will along a vertical band. I would bet that if the side of the head were made-up, the effect would be less.
David....at the risk of making this sound too simple, are you saying then that since the rim light tends to be more efficient due to it's typical hardness then you should one set the power at 2 stops less rather than one and play with backing it away from your subject if it still seems too intense?
With the rim light, you're experiencing a direct reflection.
Direct reflections on non-metalic surfaces are always the color of the light source. They're also quite a bit more intense than diffuse reflections, because they cover less area. So you get white highlights for two reasons.
I'll refer you to the "family of angles" discussion in Lighting: Science and Magic.
Larger light sources allow the diffuse reflections to compete with the direct reflections, so things look less "blown out" when correctly exposed. Same deal with a weak rim light: it gives diffuse reflection a chance to compete.
You can also decrease the reflectivity of the subject to control the highlight, using makeup.
Unless I am missing something....
It looks like the problem with the rim light is more to do with the angle rather than the power. Sure it's a little hot but surely moving it more above & further behind the model would be more helpful?
It doesn't look like the OP's question was fully answered, although bit and pieces are found here and there. It was a very valid question indeed: "I understand f-stops on the lense, and fractions of total flash output (e.g. 1/16th), but I don't understand how f-stops are determined for the flashes."
The answer is: Flashes don't give a damn about f-stops. The only thing they know is "I can generate X amount of light. Do you want me to give you everything I have or only a portion of it?" So we tell it "Give me 1/2 of your power." or " Give me 1/64 of your power." That is all a flash cares about.
Now the f-stops settings shown in the examples in this article are only useful to illustrate the relative power of each light source compared to each other. In other words, in the top example, the rim light is set to give 1 stop more light than the key light. What this means is twice as much light. Whether those numbers are 8 and 5.6 or 22 and 16 or whatever doesn't make any difference. The only thing that counts is that it there is a one stop difference.
So it would probably have been clearer to say that, in the top photo, the light ratio was 2 for the rim versus 1 for the key. In the second photo, it was 1 for rim and 1 for key, etc... As for flash setting, this would mean that if the rim was set to full power (top photo), the key would be set to 1/2 power, etc...
The "f-stop on the flash" business is simply another way to determine power output but this time by taking into consideration two additional variables which are set on the camera: the ISO setting and the lens aperture. So there is no such thing as f-stops on a flash. It is simply a way of saying for example, "Since my camera is set at 200 ISO and my lens is set to f8, I want my flash to give me 1/4 of its power in order to have my subject properly exposed."
Hope this helps.
Bo
The rim light is not necessarily brighter because of the shallow angle (a mirror, for example, reflects at a shallow angle just as well as at a steep one). As Mithrandir points out, the difference is one of direct reflection vs. diffuse reflection.
Diffuse reflection (the kind of reflection that allows us to see things that don't generate their own light -- most everything) is omnidirectional, so it can be less efficient than the direct reflection off the hair of the rim light, at least when the angle of the direct reflection lands the light in the lens. The angle is important in this case not because one angle reflects somehow more efficiently than another, but because the angle happens to allow reflection into the lens.
Check out the dramitic lighting in this movie clip from the game Unreal Tournament 3 (UT3):
http://www.gametrailers.com/player/usermovies/146406.html
-Eric
I am confused. When you say one stop hotter for the rim light do you mean the f8 aperture or the f4 aperture? The f8 rim light seems brighter to me but is a stop down from f5.6. Therefore a smaller aperture of f8 would mean less light would it not compared to a rim light of f4?
Post a Comment
<< Home