Camera Expsosure - mostly @ Jeff but feel free to chime in

ernstbk said:

Steven Brooks said:

Great Bustard said:

GeorgianBay1939 said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

ernstbk said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

ernstbk said:

ernstbk said:

ernstbk said:

ernstbk said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

Table 1, (below Table 2) at https://en.wikipedia.org/wiki/Exposure_(photography)#Luminous_exposure is a bit of a bear, eh?

Pity that "Luminous Exposure" [ Hv , measured in lux - seconds or equivalently, lumen - sec per M^2 ] isn't equivalently described as "Luminous Energy Surface Density" [ the same Hv, measured in lumen - sec per M^2]

Click to expand...

I agree it would be helpful to note that Luminous Exposure is synonymous with Luminous Energy Surface Density, but they probably take it as a given that people reading such a page would readily make that connection.

ernstbk said:

The notion of exposure being the surface density of an accumulation of photons during a shutter cycle seems intuitive when exposing an emulsion and is critical when discussing the effects of Total Light on a sensor of effective surface area A, giving Total Light = Hv * A = Luminous Energy [measured in lumen - seconds] captured by the sensor during a shutter cycle.

Sorry for the alternate use of equivalently above.

Click to expand...

For a given spectral distribution, the units of exposure can be expressed as photons/mm². However, an interesting question is if a billion "red" photons has the same exposure as a billion "blue" photons since the "blue" photons have twice the energy as the "red" photons (keeping in mind that this example represents two different spectral distributions). Even more to the point is given that a Bayer CFA is RGGB, what if we are comparing a billion "green" photons to a billion "red" or a billion "blue" photons, as twice as many "green" photons will be recorded.

Click to expand...

Yes, but, The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure. Why do you have so much difficulty in understanding such a basic photographic concept?

On the other hand, you obviously love your tautology and obfuscation! And I must admit it is great fun. Keep up the good work.

Best regards

Steven Brooks

Click to expand...

@Steven: This question is not intended for you.

Click to expand...

ernstbk said:

Can somebody explain to me if the following I found in a book is correct or false ?

"Exposure is the amount of light reaching the film.

Click to expand...

ernstbk said:

Each film requires a specific amount of light to produce a picture of the proper brightness.

Click to expand...

ernstbk said:

How much light a film requires depends on its speed"

Click to expand...

gollywop said:

Great Bustard said:

Steven Brooks said:

tony field said:

Steven Brooks said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

gollywop said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

gollywop said:

gollywop said:

gollywop said:

gollywop said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

Or if you prefer to avoid tautological obfuscation: the term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure

Now isn't that much more, accurate, concise and easy to understand? Especially as every photographer in the World since 1890 has used that definition.

Click to expand...

Out of curiosity Steven (in the case of being much more, accurate, concise and easy to understand), please explain this:

Assume we are sitting in a nice west coast town on a day with a lovely summer rain. My house with a 30x20 foot patio is adjacent to park with a playground that has a 90x60 foot playground for the kids. Both the patio and the playground receive the same exposure to the wet weather and collect the rain drops for an hour.

Would the total amount of water collected in the patio and the park be the same - bearing in mind they have in, general terms, the same exposure to the weather? I would have thought that the playground receives nine times the total rain as does the patio.

You stated the desire for "being much more, accurate, concise and easy to understand". Would it add to the accuracy and understanding if we stated:

"The amount of water collected per square foot in the patio and the park would be the same". ?

Are the two versions tautological? - I think not.

It is reasonable to conclude that the book statement of "total amount of light" is just wrong, or at plausible best, an obfuscation - the words are quite precise and, for the "definition" of exposure, simply erroneous. A reader who understands the words will draw a vastly different interpretation between "total amount" and "amount per square unit".

I hope that a parallel between rain drops and photons and also between the surface areas are appropriate when discussing "photographic exposure". You could even draw a parallel between sensel quantum efficiency and the amount of rain that would evaporate per square foot while the rain falls

Click to expand...

Very good! You obviously love your tautology and obfuscation very much - or very intensely - or at a considerable quota of words per column inch. And I must admit it is great fun. Keep up the good work.

Click to expand...

Whereas you love your willful ignorance.

gollywop said:

Best regards

Steven Brooks

Click to expand...

a.k.a Andy Capp, Wee Eck, and David Bailey Wales.

gollywop said:

P.S. The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure.

Click to expand...

Getting something wrong multiple times doesn't make it any less wrong. But posting the same error multiple times after being corrected does portray a person as willfully ignorant.

Click to expand...

Steven Brooks has now been shown by three people to be wrong. His response is simply to repeat the OP's erroneous quote over and over. He would appear, therefore, indeed to have graduated from being simply ignorant to being willful ignorant.

However, his misuse of the word "tautology" and misapplication of the word "obfuscation" has a familiar smell to it. Combining this with his clear willingness to embarrass himself with a strange confidence from a person who just joined a few days ago, I'd have to think we have the reincarnation of a troll, and possibly a doppelgänger.

Ignoring the ignorant seems a reasonable policy, as does "don't feed the troll."

Click to expand...

Jeff said:

ernstbk said:

Steven Brooks said:

Yes, but, The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure. Why do you have so much difficulty in understanding such a basic photographic concept?

On the other hand, you obviously love your tautology and obfuscation! And I must admit it is great fun. Keep up the good work.

Best regards

Steven Brooks

Click to expand...

@Steven: This question is not intended for you.

Can somebody explain to me if the following I found in a book is correct or false ?

"Exposure is the amount of light reaching the film. Each film requires a specific amount of light to produce a picture of the proper brightness. How much light a film requires depends on its speed"

Click to expand...

False due to technicalities.

a. The 'how much light' should read 'how much exposure'.

b. 'reaching the film' should be modified to 'reaching the film per unit area of film'.

c. Speed should refer be called the exposure index. If calibrated to ISO standards, then it could be called ISO.

--
Jeff
http://www.flickr.com/photos/jck_photos/sets/

Click to expand...

ernstbk said:

Can somebody explain to me if the following I found in a book is correct or false ?

"Exposure is the amount of light reaching the film. Each film requires a specific amount of light to produce a picture of the proper brightness. How much light a film requires depends on its speed"

Click to expand...

tony field said:

ernstbk said:

Can somebody explain to me if the following I found in a book is correct or false ?

"Exposure is the amount of light reaching the film. Each film requires a specific amount of light to produce a picture of the proper brightness. How much light a film requires depends on its speed"

Click to expand...

I think this is a reasonably decent generalized description of exposure and how it relates to image capture by a film or digital camera.

Click to expand...

Great Bustard said:

GeorgianBay1939 said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

Great Bustard said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

Table 1, (below Table 2) at https://en.wikipedia.org/wiki/Exposure_(photography)#Luminous_exposure is a bit of a bear, eh?

Pity that "Luminous Exposure" [ Hv , measured in lux - seconds or equivalently, lumen - sec per M^2 ] isn't equivalently described as "Luminous Energy Surface Density" [ the same Hv, measured in lumen - sec per M^2]

Click to expand...

I agree it would be helpful to note that Luminous Exposure is synonymous with Luminous Energy Surface Density, but they probably take it as a given that people reading such a page would readily make that connection.

Click to expand...

Great Bustard said:

Great Bustard said:

The notion of exposure being the surface density of an accumulation of photons during a shutter cycle seems intuitive when exposing an emulsion and is critical when discussing the effects of Total Light on a sensor of effective surface area A, giving Total Light = Hv * A = Luminous Energy [measured in lumen - seconds] captured by the sensor during a shutter cycle.

Sorry for the alternate use of equivalently above.

Click to expand...

For a given spectral distribution, the units of exposure can be expressed as photons/mm². However, an interesting question is if a billion "red" photons has the same exposure as a billion "blue" photons since the "blue" photons have twice the energy as the "red" photons (keeping in mind that this example represents two different spectral distributions). Even more to the point is given that a Bayer CFA is RGGB, what if we are comparing a billion "green" photons to a billion "red" or a billion "blue" photons, as twice as many "green" photons will be recorded.

Click to expand...

Great Bustard said:

tony field said:

ernstbk said:

Can somebody explain to me if the following I found in a book is correct or false ?

"Exposure is the amount of light reaching the film. Each film requires a specific amount of light to produce a picture of the proper brightness. How much light a film requires depends on its speed"

Click to expand...

I think this is a reasonably decent generalized description of exposure and how it relates to image capture by a film or digital camera.

Click to expand...

For a given camera. But 4x as much light is projected on a FF sensor as is an mFT sensor for the same exposure, and, yes, that does make a difference. So, while distinguishing between the density of light and total amount of light is not particularly necessary within a format (just as distinguishing between mass and weight is not particularly necessary within a given gravitational field), it is *central* when comparing between formats.

Click to expand...

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Very good! You obviously love your tautology and obfuscation very much - or very intensely - or at a considerable quota of words per column inch. And I must admit it is great fun. Keep up the good work.

Click to expand...

Whereas you love your willful ignorance.

Great Bustard said:

Best regards

Steven Brooks

Click to expand...

a.k.a Andy Capp, Wee Eck, and David Bailey Wales.

Great Bustard said:

P.S. The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure.

Click to expand...

Getting something wrong multiple times doesn't make it any less wrong. But posting the same error multiple times after being corrected does portray a person as willfully ignorant.

Click to expand...

Steven Brooks has now been shown by three people to be wrong. His response is simply to repeat the OP's erroneous quote over and over. He would appear, therefore, indeed to have graduated from being simply ignorant to being willful ignorant.

However, his misuse of the word "tautology" and misapplication of the word "obfuscation" has a familiar smell to it. Combining this with his clear willingness to embarrass himself with a strange confidence from a person who just joined a few days ago, I'd have to think we have the reincarnation of a troll, and possibly a doppelgänger.

Ignoring the ignorant seems a reasonable policy, as does "don't feed the troll."

Click to expand...

This is at least his fourth nick in as many weeks. The previous three were deleted along with all the posts, which is unfortunate 'cause not only did I have a lot of good replies bookmarked, but because it's easier to have the same willfully ignorant "personality" consistently posting with the same nick rather than having to rediscover anew each time that it's just the same boring troll again.

Click to expand...

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Putting aside if most rendition of ET is pretty bad, isn't ET trying to describe the process of controlling the light i.e. the camera exposure?

What do you think?

If you don't care to discuss, don't feel obliged. No hard feelings.

However, be warned. Even though I really do not mean to, I frustrate golly to no end ( I thought I was having a pretty exciting (even scintillating) chat when we got into some German, for example ;-) , he didn't see it that way ) whenever he tries to engage me.

Be warned2: this conversation may lead to Read Amplifier Gain/PGA and what is "camera sensitivity" ;-)

BTW, I personally do not see much value in triangle graphics or list of f-stops and SS. This is not like a trig table. Nor it is level of nomograph.

Click to expand...

GeorgianBay1939 said:

<<Big Snip>>

GeorgianBay1939 said:

GeorgianBay1939 said:

GeorgianBay1939 said:

GeorgianBay1939 said:

Very good! You obviously love your tautology and obfuscation very much - or very intensely - or at a considerable quota of words per column inch. And I must admit it is great fun. Keep up the good work.

Click to expand...

Whereas you love your willful ignorance.

GeorgianBay1939 said:

Best regards

Steven Brooks

Click to expand...

a.k.a Andy Capp, Wee Eck, and David Bailey Wales.

GeorgianBay1939 said:

P.S. The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure.

Click to expand...

Getting something wrong multiple times doesn't make it any less wrong. But posting the same error multiple times after being corrected does portray a person as willfully ignorant.

Click to expand...

Steven Brooks has now been shown by three people to be wrong. His response is simply to repeat the OP's erroneous quote over and over. He would appear, therefore, indeed to have graduated from being simply ignorant to being willful ignorant.

However, his misuse of the word "tautology" and misapplication of the word "obfuscation" has a familiar smell to it. Combining this with his clear willingness to embarrass himself with a strange confidence from a person who just joined a few days ago, I'd have to think we have the reincarnation of a troll, and possibly a doppelgänger.

Ignoring the ignorant seems a reasonable policy, as does "don't feed the troll."

Click to expand...

This is at least his fourth nick in as many weeks. The previous three were deleted along with all the posts, which is unfortunate 'cause not only did I have a lot of good replies bookmarked, but because it's easier to have the same willfully ignorant "personality" consistently posting with the same nick rather than having to rediscover anew each time that it's just the same boring troll again.

Click to expand...

Thanks, you two, for uncloaking this vandal. ( I became suspicious when I saw the above bolded sentences also used in a reply to me.) Obviously a robotic, un-creative, mischievous dumb vandal wasting our time and words and this DPR space.

Googling Wee Eck ( "Lives in United States Baltimore, United States. Works as a Doctor. Joined on Jan 27, 2016") now returns a 404 error. The other two nicks, David Bailey Wales and Steven Brooks, appear to be legitimate photographers. Aren't there rules against trolls coming on here and impersonating real people?

Click to expand...

Steven Brooks said:

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Putting aside if most rendition of ET is pretty bad, isn't ET trying to describe the process of controlling the light i.e. the camera exposure?

What do you think?

If you don't care to discuss, don't feel obliged. No hard feelings.

However, be warned. Even though I really do not mean to, I frustrate golly to no end ( I thought I was having a pretty exciting (even scintillating) chat when we got into some German, for example ;-) , he didn't see it that way ) whenever he tries to engage me.

Be warned2: this conversation may lead to Read Amplifier Gain/PGA and what is "camera sensitivity" ;-)

BTW, I personally do not see much value in triangle graphics or list of f-stops and SS. This is not like a trig table. Nor it is level of nomograph.

Click to expand...

It is quite simple really. "Correct" exposure may be defined as an exposure that achieves the effect the photographer intended.

Click to expand...

Steven Brooks said:

A more technical approach recognises that a photographic film or sensor has a physically limited useful exposure range, sometimes called its dynamic range. If, for any part of the photograph, the actual exposure is outside this range, the film cannot record it accurately. In a very simple model, for example, out-of-range values would be recorded as "black" (underexposed) or "white" (overexposed) rather than the precisely graduated shades of colour and tone required to describe "detail". Therefore, the purpose of exposure adjustment (and/or lighting adjustment) is to control the physical amount of light from the subject that is allowed to fall on the film, so that 'significant' areas of shadow and highlight detail do not exceed the film's useful exposure range. This ensures that no 'significant' information is lost during capture.

It is worth noting that the photographer may carefully overexpose or underexpose the photograph to eliminate "insignificant" or "unwanted" detail; to make, for example, a white altar cloth appear immaculately clean, or to emulate the heavy, pitiless shadows of film noir. However, it is technically much easier to discard recorded information during post processing than to try to 're-create' unrecorded information.

In a scene with strong or harsh lighting, the ratio between highlight and shadow luminance values may well be larger than the ratio between the film's maximum and minimum useful exposure values. In this case, adjusting the camera's exposure settings (which only applies changes to the whole image, not selectively to parts of the image) only allows the photographer to choose between underexposed shadows or overexposed highlights; it cannot bring both into the useful exposure range at the same time. Methods for dealing with this situation include: using some kind of fill lighting to gently increase the illumination in shadow areas; using a graduated ND filter or gobo to reduce the amount of light coming from the highlight areas; or varying the exposure between multiple, otherwise identical, photographs (exposure bracketing) and then combining them afterwards in some kind of HDRI process.

Click to expand...

Jeff said:

Steven Brooks said:

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Putting aside if most rendition of ET is pretty bad, isn't ET trying to describe the process of controlling the light i.e. the camera exposure?

What do you think?

If you don't care to discuss, don't feel obliged. No hard feelings.

However, be warned. Even though I really do not mean to, I frustrate golly to no end ( I thought I was having a pretty exciting (even scintillating) chat when we got into some German, for example ;-) , he didn't see it that way ) whenever he tries to engage me.

Be warned2: this conversation may lead to Read Amplifier Gain/PGA and what is "camera sensitivity" ;-)

BTW, I personally do not see much value in triangle graphics or list of f-stops and SS. This is not like a trig table. Nor it is level of nomograph.

Click to expand...

It is quite simple really. "Correct" exposure may be defined as an exposure that achieves the effect the photographer intended.

A more technical approach recognises that a photographic film or sensor has a physically limited useful exposure range, sometimes called its dynamic range. If, for any part of the photograph, the actual exposure is outside this range, the film cannot record it accurately. In a very simple model, for example, out-of-range values would be recorded as "black" (underexposed) or "white" (overexposed) rather than the precisely graduated shades of colour and tone required to describe "detail". Therefore, the purpose of exposure adjustment (and/or lighting adjustment) is to control the physical amount of light from the subject that is allowed to fall on the film, so that 'significant' areas of shadow and highlight detail do not exceed the film's useful exposure range. This ensures that no 'significant' information is lost during capture.

It is worth noting that the photographer may carefully overexpose or underexpose the photograph to eliminate "insignificant" or "unwanted" detail; to make, for example, a white altar cloth appear immaculately clean, or to emulate the heavy, pitiless shadows of film noir. However, it is technically much easier to discard recorded information during post processing than to try to 're-create' unrecorded information.

In a scene with strong or harsh lighting, the ratio between highlight and shadow luminance values may well be larger than the ratio between the film's maximum and minimum useful exposure values. In this case, adjusting the camera's exposure settings (which only applies changes to the whole image, not selectively to parts of the image) only allows the photographer to choose between underexposed shadows or overexposed highlights; it cannot bring both into the useful exposure range at the same time. Methods for dealing with this situation include: using some kind of fill lighting to gently increase the illumination in shadow areas; using a graduated ND filter or gobo to reduce the amount of light coming from the highlight areas; or varying the exposure between multiple, otherwise identical, photographs (exposure bracketing) and then combining them afterwards in some kind of HDRI process.

Click to expand...

The trouble with your approach as described here is that provides no guidance on how to determine camera settings that will yield a useful and predictable image. That's what the ISO standards do ... and why its so useful to understand those definitions and methods.

--
Jeff
http://www.flickr.com/photos/jck_photos/sets/

Click to expand...

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

Click to expand...

"total quantity of light energy incident on a sensitive material" means, then, the luminous energy at the sensor plane. Yes?

Click to expand...

Yes, but that is not the exposure (the book is wrong). The exposure is the amount of luminous energy per area.

Click to expand...

The (photographic) exposure is total (exposure period) luminous energy incident at the sensor.

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

Click to expand...

I think saying "100 lumen second distributed over 20 m^2 means the luminous energy incident at the sensor plane regardless of the sensor size, is 5 lumen/m^2 second = 5 lux second...

Click to expand...

Correct.

Great Bustard said:

...and if the shutter was open for 2 second then the total luminous energy incident on the sensor plane is 10 lumen/m^2*second which is of course 10 lux second.

Click to expand...

That is wrong, and should be more than a little obvious since both the measure of "lumen second" and "lux second" already account for the exposure time, the big clue being the unit of "second" in each.

Click to expand...

You really should reread your reply.

Click to expand...

The first part of your paragraph is correct. The second part was not. Tell me why you feel you were not mistaken.

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

What has been hard for me (not sure why I have this problem :-( ) and as usual thanks for your keeping up, is that I now have reconciled lux as distance from candela and lux as lumen second distributed over an area...

Click to expand...

Both of those are wrong. What's disappointing is that it was spelled out to you in this very subthread.

Great Bustard said:

...both of which can be said to be the total light energy incident at the sensor plane during 1 second "exposure" ;-) for the photographic purposes.

I think I've got it

Click to expand...

Didn't you say once that you had a background in engineering? How can you get this so wrong so many times even though it's been spelled out *in detail* to you?

Great Bustard said:

And so the book is not wrong.

Click to expand...

You are wrong, the book is wrong, and it's been spelled out and explained, in detail, in this very subthread. In all seriousness, it is not much different from explaining that 5² means 5 x 5 to you over and over and over yet you continue to think that 5² means 5 x 2.

Click to expand...

Please reread my reply above. Somewhere you seem to have misread some part wrong.

Click to expand...

OK -- reread. What do you think I misread?

Click to expand...

roadtrpr said:

jrtrent said:

roadtrpr said:

jrtrent said:

Steven Brooks said:

Great Bustard said:

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Click to expand...

Exposure is the amount of light per area that falls on the recording medium, not the total quantity of light. . .

Click to expand...

So the definition of exposure in The Manual of Photography, first published in 1890 and now in its tenth edition, is wrong, is it? I am surprised that none of the millions of photographers, both amateur and professional, who have used it as the standard text on photographic principles have not noticed this important error before.

Click to expand...

Is it possible that the 10th edition is not the same as the previous editions? I don't have a 10th edition to read, but I do find this aspect of the 9th edition's explanation of photographic exposure useful:

"As the luminance of the subject varies from area to area, it follows that the illuminance on the emulsion varies similarly, so that the film receives not one exposure over the entire surface but a varying amount of light energy, i.e. a range of exposures. As a general rule the exposure duration is constant for all areas of the film, variation in exposure over the film being due solely to variation in the illumination that it receives."

It has always been helpful in my thinking about exposure to not worry about the total quantity of light, if by that is meant all the light hitting all parts of the film or sensor (and if I had a 10th edition to look at, it's possible that they're referring to the total quantity of light energy hitting one particular spot on the film or sensor), but to use a spot meter to measure a particular object in the scene and adjust my exposure so that it comes out with the tonality I desire in the picture (e.g., I might meter off of a dark bush and set exposure so that it is at zone iv) and let all other values fall around it however they will (that is, green grass will then likely be at zone v, Caucasian skin at zone vi, and so on). Alternatively, I might not care to deal with reflected light values at all and just use an incident light meter to measure the light hitting the subject instead (or forget a meter entirely and use an exposure table for the light conditions I'm in), then whether it's a black cat or a white dog, the subject will turn out with the proper tonality in my picture, and so will everything else (within the dynamic range limits of the equipment I'm using).

Click to expand...

That same statement is in the 10th edition also in chapter 8.

"When a photograph is taken, light from the various areas of the subject falls on corresponding areas of the film. The photographic exposure, H (the effect produced on the emulsion), is, within limits, proportional to the product of the illuminance E and the exposure time t. We express this by the equation:

H = Et (8.1)

Before international standardization of symbols, the equation was written as E = It (E was exposure, I was illuminance) and this usage is still sometimes found.

The SI unit for illuminance is the lux (lx). Hence the exposure is measured in lux seconds (lx s). It should be noted that the lux is defined in terms of the human observer, who cannot see radiation in either ultraviolet or infrared regions of the spectrum. Inclusion of either of these bands in the imaging exposure may therefore yield erroneous results with some imaging systems.

As the subject luminance varies from area to area, it follows that the illuminance on the emulsion varies similarly, so that the film receives not one exposure over the entire surface but a varying amount of light energy, i.e. a range of exposures. As a general rule the exposure dura-tion is constant for all areas of the film, variation in exposure over the film being due solely to variation in the illumination that it receives."

Click to expand...

So chapter 8 of the 10th edition covers the same ground as chapter 15 of the 9th edition. It appears that chapter 12 of the 10th edition is similar to chapter 19 of the 9th edition. I only scanned through it quickly so may have missed it, but I did not find the 10th edition's statement, "The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure," in the 9th edition. Does chapter 12 of the 10th edition go on to explain the meaning of that sentence in any more detail?

Click to expand...

I was wondering if that sentence was added in the 10th edition. The curious thing is that the sentence is followed by this one:

"As described in Chapters 1, 8 and 11, two parameters are used to control photographic exposure: the amount of light incident on the imaging sensor and the duration of the sensor's exposure. This is described by Eqn 8.1, the reciprocity equation:

H = Et (12.1)

where H is exposure, E is the subject illuminance and t is exposure duration."

I say curious because illuminance, of course, is a measure of light per unit of sq area.

I think they may have said total light instead of light per unit of area simply because what follows is a discussion of how the total amount of light passed through to the sensitive material is controlled by the lens iris diaphragm and shutter speed and that various combinations of aperture and shutter speed pass the same total amount of light.

It is unfortunate that they say photographic exposure is total light and then in the next sentence define exposure as light per unit area.

Click to expand...

Great Bustard said:

Steven Brooks said:

Great Bustard said:

GeorgianBay1939 said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

Great Bustard said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

Great Bustard said:

Great Bustard said:

Great Bustard said:

Great Bustard said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

Table 1, (below Table 2) at https://en.wikipedia.org/wiki/Exposure_(photography)#Luminous_exposure is a bit of a bear, eh?

Pity that "Luminous Exposure" [ Hv , measured in lux - seconds or equivalently, lumen - sec per M^2 ] isn't equivalently described as "Luminous Energy Surface Density" [ the same Hv, measured in lumen - sec per M^2]

Click to expand...

I agree it would be helpful to note that Luminous Exposure is synonymous with Luminous Energy Surface Density, but they probably take it as a given that people reading such a page would readily make that connection.

Great Bustard said:

The notion of exposure being the surface density of an accumulation of photons during a shutter cycle seems intuitive when exposing an emulsion and is critical when discussing the effects of Total Light on a sensor of effective surface area A, giving Total Light = Hv * A = Luminous Energy [measured in lumen - seconds] captured by the sensor during a shutter cycle.

Sorry for the alternate use of equivalently above.

Click to expand...

For a given spectral distribution, the units of exposure can be expressed as photons/mm². However, an interesting question is if a billion "red" photons has the same exposure as a billion "blue" photons since the "blue" photons have twice the energy as the "red" photons (keeping in mind that this example represents two different spectral distributions). Even more to the point is given that a Bayer CFA is RGGB, what if we are comparing a billion "green" photons to a billion "red" or a billion "blue" photons, as twice as many "green" photons will be recorded.

Click to expand...

Yes, but, The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure.

Click to expand...

The total quantity of light energy per area. I believe you've been corrected on this point well more than once.

Click to expand...

Great Bustard said:

Great Bustard said:

It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure. Why do you have so much difficulty in understanding such a basic photographic concept?

On the other hand, you obviously love your tautology and obfuscation! And I must admit it is great fun. Keep up the good work.

Click to expand...

Of course a willfully ignorant troll would consider self-harming posts of gross ignorance to be "great fun".

Great Bustard said:

Best regards

Steven Brooks

Click to expand...

a.k.a. Andy Capp, Wee Eck, and David Bailey Wales.

Click to expand...

mostlyboringphotog said:

Great Bustard said:

Steven Brooks said:

Great Bustard said:

GeorgianBay1939 said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

mostlyboringphotog said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

Table 1, (below Table 2) at https://en.wikipedia.org/wiki/Exposure_(photography)#Luminous_exposure is a bit of a bear, eh?

Pity that "Luminous Exposure" [ Hv , measured in lux - seconds or equivalently, lumen - sec per M^2 ] isn't equivalently described as "Luminous Energy Surface Density" [ the same Hv, measured in lumen - sec per M^2]

Click to expand...

I agree it would be helpful to note that Luminous Exposure is synonymous with Luminous Energy Surface Density, but they probably take it as a given that people reading such a page would readily make that connection.

mostlyboringphotog said:

The notion of exposure being the surface density of an accumulation of photons during a shutter cycle seems intuitive when exposing an emulsion and is critical when discussing the effects of Total Light on a sensor of effective surface area A, giving Total Light = Hv * A = Luminous Energy [measured in lumen - seconds] captured by the sensor during a shutter cycle.

Sorry for the alternate use of equivalently above.

Click to expand...

For a given spectral distribution, the units of exposure can be expressed as photons/mm². However, an interesting question is if a billion "red" photons has the same exposure as a billion "blue" photons since the "blue" photons have twice the energy as the "red" photons (keeping in mind that this example represents two different spectral distributions). Even more to the point is given that a Bayer CFA is RGGB, what if we are comparing a billion "green" photons to a billion "red" or a billion "blue" photons, as twice as many "green" photons will be recorded.

Click to expand...

Yes, but, The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure.

Click to expand...

The total quantity of light energy per area. I believe you've been corrected on this point well more than once.

Click to expand...

Luminance of 1 lux is 1 lux regardless of luminance area is 1 m^2 or 10 m^2.

Click to expand...

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure. Why do you have so much difficulty in understanding such a basic photographic concept?

On the other hand, you obviously love your tautology and obfuscation! And I must admit it is great fun. Keep up the good work.

Click to expand...

Of course a willfully ignorant troll would consider self-harming posts of gross ignorance to be "great fun".

mostlyboringphotog said:

Best regards

Steven Brooks

Click to expand...

a.k.a. Andy Capp, Wee Eck, and David Bailey Wales.

Click to expand...

Click to expand...

Jeff said:

Steven Brooks said:

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Putting aside if most rendition of ET is pretty bad, isn't ET trying to describe the process of controlling the light i.e. the camera exposure?

What do you think?

If you don't care to discuss, don't feel obliged. No hard feelings.

However, be warned. Even though I really do not mean to, I frustrate golly to no end ( I thought I was having a pretty exciting (even scintillating) chat when we got into some German, for example ;-) , he didn't see it that way ) whenever he tries to engage me.

Be warned2: this conversation may lead to Read Amplifier Gain/PGA and what is "camera sensitivity" ;-)

BTW, I personally do not see much value in triangle graphics or list of f-stops and SS. This is not like a trig table. Nor it is level of nomograph.

Click to expand...

It is quite simple really. "Correct" exposure may be defined as an exposure that achieves the effect the photographer intended.

A more technical approach recognises that a photographic film or sensor has a physically limited useful exposure range, sometimes called its dynamic range. If, for any part of the photograph, the actual exposure is outside this range, the film cannot record it accurately. In a very simple model, for example, out-of-range values would be recorded as "black" (underexposed) or "white" (overexposed) rather than the precisely graduated shades of colour and tone required to describe "detail". Therefore, the purpose of exposure adjustment (and/or lighting adjustment) is to control the physical amount of light from the subject that is allowed to fall on the film, so that 'significant' areas of shadow and highlight detail do not exceed the film's useful exposure range. This ensures that no 'significant' information is lost during capture.

It is worth noting that the photographer may carefully overexpose or underexpose the photograph to eliminate "insignificant" or "unwanted" detail; to make, for example, a white altar cloth appear immaculately clean, or to emulate the heavy, pitiless shadows of film noir. However, it is technically much easier to discard recorded information during post processing than to try to 're-create' unrecorded information.

In a scene with strong or harsh lighting, the ratio between highlight and shadow luminance values may well be larger than the ratio between the film's maximum and minimum useful exposure values. In this case, adjusting the camera's exposure settings (which only applies changes to the whole image, not selectively to parts of the image) only allows the photographer to choose between underexposed shadows or overexposed highlights; it cannot bring both into the useful exposure range at the same time. Methods for dealing with this situation include: using some kind of fill lighting to gently increase the illumination in shadow areas; using a graduated ND filter or gobo to reduce the amount of light coming from the highlight areas; or varying the exposure between multiple, otherwise identical, photographs (exposure bracketing) and then combining them afterwards in some kind of HDRI process.

Click to expand...

The trouble with your approach as described here is that provides no guidance on how to determine camera settings that will yield a useful and predictable image. That's what the ISO standards do ... and why its so useful to understand those definitions and methods.

Click to expand...

mostlyboringphotog said:

Great Bustard said:

Steven Brooks said:

Yes, but, The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure.

Click to expand...

The total quantity of light energy per area. I believe you've been corrected on this point well more than once.

Click to expand...

[Il]Luminance of 1 lux is 1 lux regardless of luminance area is 1 m^2 or 10 m^2.

Click to expand...

Steven Brooks said:

mostlyboringphotog said:

So, another maxed out exposure triangle thread...

If you care to continue the scintillating (today's word for golly ;-) ) conversation, I'd like to start from the following quote from The Manual of Photography Tenth Edition - Chapter 12

"The term exposure in photography describes the total quantity of light energy incident on a sensitive material, which in general terms is the photographic exposure. It may alternatively describe the process of controlling the light energy reaching a sensitive material in a camera, which is more specifically the camera exposure." - bold added.

Putting aside if most rendition of ET is pretty bad, isn't ET trying to describe the process of controlling the light i.e. the camera exposure?

What do you think?

If you don't care to discuss, don't feel obliged. No hard feelings.

However, be warned. Even though I really do not mean to, I frustrate golly to no end ( I thought I was having a pretty exciting (even scintillating) chat when we got into some German, for example ;-) , he didn't see it that way ) whenever he tries to engage me.

Be warned2: this conversation may lead to Read Amplifier Gain/PGA and what is "camera sensitivity" ;-)

BTW, I personally do not see much value in triangle graphics or list of f-stops and SS. This is not like a trig table. Nor it is level of nomograph.

Click to expand...

It is quite simple really. "Correct" exposure may be defined as an exposure that achieves the effect the photographer intended.

A more technical approach recognises that a photographic film or sensor has a physically limited useful exposure range, sometimes called its dynamic range. If, for any part of the photograph, the actual exposure is outside this range, the film cannot record it accurately. In a very simple model, for example, out-of-range values would be recorded as "black" (underexposed) or "white" (overexposed) rather than the precisely graduated shades of colour and tone required to describe "detail". Therefore, the purpose of exposure adjustment (and/or lighting adjustment) is to control the physical amount of light from the subject that is allowed to fall on the film, so that 'significant' areas of shadow and highlight detail do not exceed the film's useful exposure range. This ensures that no 'significant' information is lost during capture.

It is worth noting that the photographer may carefully overexpose or underexpose the photograph to eliminate "insignificant" or "unwanted" detail; to make, for example, a white altar cloth appear immaculately clean, or to emulate the heavy, pitiless shadows of film noir. However, it is technically much easier to discard recorded information during post processing than to try to 're-create' unrecorded information.

In a scene with strong or harsh lighting, the ratio between highlight and shadow luminance values may well be larger than the ratio between the film's maximum and minimum useful exposure values. In this case, adjusting the camera's exposure settings (which only applies changes to the whole image, not selectively to parts of the image) only allows the photographer to choose between underexposed shadows or overexposed highlights; it cannot bring both into the useful exposure range at the same time. Methods for dealing with this situation include: using some kind of fill lighting to gently increase the illumination in shadow areas; using a graduated ND filter or gobo to reduce the amount of light coming from the highlight areas; or varying the exposure between multiple, otherwise identical, photographs (exposure bracketing) and then combining them afterwards in some kind of HDRI process.

Click to expand...

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

Great Bustard said:

mostlyboringphotog said:

gollywop said:

Exposure is measured in Lux-seconds, i.e., Lux*s.

A Lux is a lumen/m^2.

Luminous energy is lumen-seconds, i.e., lumen*s.

So, Exposure = Lux*s = (lumen/m^2)*s = (lumen*s)/m*2 = (Luminous-energy)/m^2.

Click to expand...

I get the math.

Click to expand...

Do you get the concept, though? Specifically, do you understand that exposure is defined as the luminous energy per area that falls on the sensor?

Click to expand...

The classic definition of exposure: Hv = Ev*t. Meaning illuminance at sensor times duration of shutter opening. I think that is still the better definition for photography than "luminous energy per area that falls on the sensor".

Click to expand...

You understand those are the same thing, right? Illuminance is luminous energy per second per area, so the product of illuminance and time gives luminous energy per area.

Click to expand...

"total quantity of light energy incident on a sensitive material" means, then, the luminous energy at the sensor plane. Yes?

Click to expand...

Yes, but that is not the exposure (the book is wrong). The exposure is the amount of luminous energy per area.

Click to expand...

The (photographic) exposure is total (exposure period) luminous energy incident at the sensor.

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

In any case, I realize this may be a nit but I think "per unit area" reads better ;-)

Click to expand...

A luminous energy of 100 lumen seconds distributed over 20 square meters results in an exposure of 5 lux seconds.

Click to expand...

I think saying "100 lumen second distributed over 20 m^2 means the luminous energy incident at the sensor plane regardless of the sensor size, is 5 lumen/m^2 second = 5 lux second...

Click to expand...

Correct.

mostlyboringphotog said:

...and if the shutter was open for 2 second then the total luminous energy incident on the sensor plane is 10 lumen/m^2*second which is of course 10 lux second.

Click to expand...

That is wrong, and should be more than a little obvious since both the measure of "lumen second" and "lux second" already account for the exposure time, the big clue being the unit of "second" in each.

Click to expand...

You really should reread your reply.

Click to expand...

The first part of your paragraph is correct. The second part was not. Tell me why you feel you were not mistaken.

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

mostlyboringphotog said:

Luminous energy is not a density of light, but luminous-energy per unit area is, and it's called exposure.

Click to expand...

However, respectfully, what is the illuminance 1m away from a 1 candela light source?

I think it's 1 lux; if this is wrong, this is source of my confusion.

Click to expand...

A source with a luminous intensity of 1 candela will result in an illuminance of 1 lux at a distance of 1 meter and is subject to the inverse square relationship for other distances.

Click to expand...

Thanks - I wondered if I confused with lumen per m^2 ;-) joke

Click to expand...

Bottom line: the exposure is the total amount of visible light (luminous energy) per area that falls on the sensor during the time the sensor is exposed to the light. Just not that hard to understand, really.

Click to expand...

What has been hard for me (not sure why I have this problem :-( ) and as usual thanks for your keeping up, is that I now have reconciled lux as distance from candela and lux as lumen second distributed over an area...

Click to expand...

Both of those are wrong. What's disappointing is that it was spelled out to you in this very subthread.

mostlyboringphotog said:

...both of which can be said to be the total light energy incident at the sensor plane during 1 second "exposure" ;-) for the photographic purposes.

I think I've got it

Click to expand...

Didn't you say once that you had a background in engineering? How can you get this so wrong so many times even though it's been spelled out *in detail* to you?

mostlyboringphotog said:

And so the book is not wrong.

Click to expand...

You are wrong, the book is wrong, and it's been spelled out and explained, in detail, in this very subthread. In all seriousness, it is not much different from explaining that 5² means 5 x 5 to you over and over and over yet you continue to think that 5² means 5 x 2.

Click to expand...

Please reread my reply above. Somewhere you seem to have misread some part wrong.

Click to expand...

OK -- reread. What do you think I misread?

Click to expand...

Exposed for 2 seconds at 1 Talbot would be 2 Talbots, no?

Click to expand...