How to convert light meter readings at a surface to Joules/cm2

Anyone know the math for converting visible light out of a strobe from the readings of one's light meter to energy units of mJ/cm2?

Although strobes are labelled with power settings, these represent the power calculated to be "stored" and not that delivered.

Of course, the light has a distribution of wavelengths and each has different energy.

So how does one approach these measurements and is a special instrument needed or is, for example a Minolta IV sufficient.

Thanks,

Asher

Try this..
http://www.translatorscafe.com/cafe/units-converter/luminous-intensity/c/
of course you will have to know what kind of thing your meter is reading...

Hi, Asher,

Asher Kelman said:

Anyone know the math for converting visible light out of a strobe from the readings of one's light meter to energy units of mJ/cm2?

Click to expand...

I do, subject to the various requirements and conditions I have already mentioned.

So how does one approach these measurements and is a special instrument needed or is, for example a Minolta IV sufficient.

Click to expand...

I have the manuals for the Minolta Autometer IV F and the Minolta Flash Meter IV. Is what you are speaking of by any chance one of those?

Neither directly read in lux, or lux•seconds for the flash mode. But we can get around that.

On both meters, the manual even gives a way to convert continuous luminance readings in so-called Ev to lux. I haven't figgered yet how to adapt that to the flash mode, although I should be able to reckon that. (I have only looked at each manual for maybe 35 seconds so far).

If I can't get back to sleep after I get up at 2:00 am to pee, I'll look into it a bit more.

As for the J/m^2 equivalence, as you know, a precise determination is a bit difficult because of spectral distribution issues.

But yes I know the math.

Best regards,

Doug

A quick look at wikipedia gives the relations between lux and Watts.

1 lx = 1 lm/m2

The lumen is linked to the Watt by the luminosity function. Some typical values of this function are given under luminous efficiency, for a 5800K black body radiator, 80 lm/W and for the same filtered to the visible 251 lm/W (perfect white light).

Joules are, of course, Watts per second.

This table will convert ev to lux.

So, taking ev 18, which is 656000 lux or 656000 lm/m2, we have 2613 W/m2 for a perfect white light. (Flashes are filtered and therefore much closer to perfect white light than to black body radiation, but if we want to assume unfiltered black body, we will just have to multiply the final result by about 3.)

Now for the link between joules x seconds and watts in photography. For a typical flash, the duration of light is about 1/1000s. You original question was about f/22 with a film equivalent of iso 3. At ev 18 (which is a sunny day) we would use f/22 and 1/3s with this kind of light. Therefore the flash will output as much energy in 1/1000s as the film would get in a sunny day in 1/3s, i.e. the flux is 333 times bigger. The flash will output 870310 W/m2 in a 1/1000s, which is 870 J/m2 or 0,87 J/cm2 (1 cm2 is more or less the size of the eye). For an "unfiltered black body flash", 3 times as much.

The thermic effects of such a small amount of energy will be negligible.

But, and this is the flaw in your reasoning, there is more to flash light than energy alone. The damage to the eye, if any, are likely to come from other effects than energy, in particular residual UV content.

This is further to converting the exposure recommendation of a flash exposure meter (incident mode) to the luminous energy density, denominated in lux•seconds. A later stage will be to convert that to radiant energy density in J/m^2.

The following derivation has been done "on the fly", and has not been audited. I will of course check it out as soon as I can. Please consider the intent of this note to show the train of thought and the procedure needed, not to report a reliable result. All work is done in SI units (I do not always take the time to mention all the units for each equation).

***********
We begin by assuming that the flash meter used practices this "exposure strategy":

When the exposure index setting of the meter (the "ISO setting") is set to the ISO speed of the presumed camera (not the ISO SOS rating), then with the f-number of the camera set to the value recommended by the meter after observing a test flash burst, an actual exposure of a test card with 100% reflectance, illuminated by an identical flash burst, will result in a photometric exposure, H, on the camera sensor of 1/2 stop below the "saturation" photometric exposure, Hsat. (Photometric exposure is the illuminance-time product on the sensor.)

Let's do our work assuming that the exposure index on the meter is set to ISO 100.

Based on interpretation of the ISO standard for the ISO speed of a digital camera, we may presume that the Hsat of the camera is given by:

Hsat = 78/S******(1)

where S is the ISO speed setting and Hsat is in lux•seconds.

In this next part (for conciseness) I will work with luminance and illuminance, not their time products. We can easily switch to the latter domain when we are ready. (This work is for the moment generic, not pertaining to flash exposure.)

Let E1 be the illuminance upon a Lambertian test object of reflectance R (R running from 0-1). Then the luminance, L1, exhibited by the object is given by:

L1=(E1*R)/pi (2)

When that test object is regarded by a camera with f number N, the resulting illuminance on the focal plane, E2, is given approximately by:

E2=(pi/4)*(1/N^2)*L1 (3)

Combining equations 2 and 3 we get:

E2=(1/4*N^2)*E1*R1 (4)

Now, for the assumption of an ISO speed of ISO 100 for the camera, Hsat is given by:

Hsat = 78/100 (5)

If we assume some arbitrary exposure time (remember, we are not yet actually working with flash bursts), then the saturation illuminance on the sensor, Esat, is given by:

Esat=Hsat/t (6)

Thus:

Esat=0.78/t (7)

Working back through equation 4, and now assuming a target with R=1, we find the illuminance, E3, needed on the target to produce Esat on the sensor (assuming ISO 100) is given by:

E3 = (0.78/t)*4*N^2 (8)

Now to convert illuminance to illuminance-time product (in which we are interested for a flash burst), we just multiply both sides by t, thus (I will use the symbol "H" for the illuminance-time product, H3 in this case):

H3=0.78*4*N^2 (9)

or,

H3=3.12*N^2 (10)

That is:

With the incident light exposure meter set to an exposure index of ISO 100, the illuminance-time product* of the flash burst (in lux•seconds) at the meter location is given by 3.12*N^2, where N is the f-number recommended by the meter for exposure under that flash burst.


(I don't know at the moment if that is even credible - I'll do a credibility test as soon as possible).

Well, I'll actually wake up now, have our nourishing Sunday breakfast (three kinds of fresh fruit, bacon, scrambled eggs, and hashed-brown potatoes, along with 12 IU of fast-acting insulin to support its processing), accompanied by a digest of world news from the New York Times and two local dailies (expecting a pic of mine in one of them today), and then carefully audit my result above.

Best regards,

Doug

Well, I erred with regard to the breakfast schedule. (It is influenced by one of Carla's medication protocol sequences.) So I'm back on the case. (But haven't yet verified my previous result.)

In my previous note, I suggested that the luminous energy density of the flash burst at the point where it is measured with an incident flash meter, would be 3.12 N^2, where N is the f number suggested by the meter with its exposure index set to ISO 100.

Our interest (maybe) is in the radiant energy density at this point. The luminous energy density is obtained from that by weighting over the band of visible wavelength. We need to work that backward.

But we do not know the spectral distribution of the emission of interest, so we cannot do that rigorously.

Some data I have seen suggests that we might reasonably make this equivalence:

1.00 lumen•second = 0.12 J/m^2 of radiant output

Putting that all together suggests that the radiant energy density at the meter site is approximately 0.37 N^2 J/m^2, where again N is the f-number recommended by the meter (with its exposure index set to ISO 100) after regarding a sample burst.

Expressed in other units that may be of interest, the estimate would be 0.037 N^2 mJ/cm^2 , or 37 N^2 µJ/cm^2.

Best regards,

Doug

Doug Kerr said:

1.00 lumen•second = 0.12 J of radiant output

Click to expand...

That would be an efficiency of 8.3 lm/W, about 10 times as small that what I have cited.

Doug Kerr said:

As for the J/m^2 equivalence, as you know, a precise determination is a bit difficult because of spectral distribution issues.

Click to expand...

Since we're trying to translate to human vision impact, it seems logical to convert to Lumen.

Cheers,
Bart


P.S. Never mind, I see it's already noted in a later post than the one I was just reading. I'll catch up ...

Hi, Jerome,

Jerome Marot said:

That would be an efficiency of 8.3 lm/W, about 10 times as small that what I have cited.

Click to expand...

My notes on this are a nightmare (I did this part mostly "in midair", while mostly thinking about something else), but it seems as if I was trying to work from an 40 lm/W, which would of course lead to 0.025 W/lm. I can't really tell where I got the 0.12 figure!

The 40 lm/W value was inferred from a report on the impact of streetlights and so forth on wildlife which gave the results of a certain survey of an urban area both in terms of luminous flux density and radiant power flux density.

I had not earlier noticed the table in Wikipedia that you cited. That seems quite valuable.

It may well be that something in the area of 80 lm/W is reasonable for this inquiry. I'll rework my end-to-end result based on that.

Thanks for your catch on my number, and for your other inputs.

Best regards,

Doug

Hi, Bart,

Bart_van_der_Wolf said:

Since we're trying to translate to human vision impact, it seems logical to convert to Lumen.

Click to expand...

I don't think we are concerned with human vision, but rather physiological impact on the eye.

Best regards,

Doug

Jerome has called attention to a table in the Wikipedia article on luminous efficacy that shows for a 7000 K black body radiator (perhaps as good a model as any), the overall efficacy is 80 lm/W, about 10 times that suggested by my value of 0.12 W/lm.

I had started with a value of 40 lm/W from another source (very tenuous), and then apparently bungled its conversion to W/lm, getting the doubly-inappropriate value 0.12 W/lm.

Based on the 80 lm/W value, we would have 0.0125 W/lm.

Plugging that into my overall chain (still not validated!) would give a radiant power density of 0.039 N^2 W/m^2.

That would be 3.9 N^2 µJ/cm^2.

Best regards,

Doug

Doug Kerr said:

That would be 3.9 N^2 µJ/cm^2.

Click to expand...

And then we find approximatively the same value. I computed that for a ISO 3 film at f/22, we have 870 J/m2. For an ISO 100 film, we would have 33 times less: 26 J/m2.

Your formula: 0.039 N^2, gives (N=22): 19. Close enough, considering the approximations.

Hi, Jerome,

Jerome Marot said:

And then we find approximatively the same value. I computed that for a ISO 3 film at f/22, we have 870 J/m2. For an ISO 100 film, we would have 33 times less: 26 J/m2.

Your formula: 0.039 N^2, gives (N=22): 19. Close enough, considering the approximations.

Click to expand...

Nice to see that agreement.

Best regards,

Doug

Hi, Duke,

Duke Beattie said:

Try this..
http://www.translatorscafe.com/cafe/units-converter/luminous-intensity/c/
of course you will have to know what kind of thing your meter is reading...

Click to expand...

This specific converter of course pertains to the quantity luminous intensity which, while we could consider it in overall chain of events, is not one in which we are really interested.

In fact our concern is not really with changing the units in which a quantity is denominated, but rather in finding the relationship between one quantity and a different quantity. For example, our problem is not that the same quantity can be denominated in either J/M^2 or lm•s and we just need to know the "conversion" .

But those unit converters can be very handy. Thanks for the link.

Best regards,

Doug

Jerome Marot said:

And then we find approximatively the same value. I computed that for a ISO 3 film at f/22, we have 870 J/m2. For an ISO 100 film, we would have 33 times less: 26 J/m2.

Your formula: 0.039 N^2, gives (N=22): 19. Close enough, considering the approximations.

Click to expand...

Jerome,

For ISO 3 at f=22 then energy received at the skin would be 870/10^4 j/cm^2

=0.87 j/m™ or 870 mJ/cm^2

Is that correct?

Asher

Doug Kerr said:

Jerome has called attention to a table in the Wikipedia article on luminous efficacy that shows for a 7000 K black body radiator (perhaps as good a model as any), the overall efficacy is 80 lm/W, about 10 times that suggested by my value of 0.12 W/lm.

I had started with a value of 40 lm/W from another source (very tenuous), and then apparently bungled its conversion to W/lm, getting the doubly-inappropriate value 0.12 W/lm.

Based on the 80 lm/W value, we would have 0.0125 W/lm.

Plugging that into my overall chain (still not validated!) would give a radiant power density of 0.039 N^2 W/m^2.

That would be 3.9 N^2 µJ/cm^2.

Best regards,

Doug

Click to expand...

Doug,

Do we know what the cited references mean when they talk about lumens coming from watts? Are these measures of efficiency of the output of the device or the simple translation of the stored power being converted at 100% efficiency.

Asher

Hi, Asher,

Asher Kelman said:

Do we know what the cited references mean when they talk about lumens coming from watts? Are these measures of efficiency of the output of the device or the simple translation of the stored power being converted at 100% efficiency.

Click to expand...

We run into two rather differ things:

• If we know that a light source is emitting x watts of electromagnetic radiation (a measure that has nothing to do with human perception), how many lumens of luminous flux does that constitute (a measure that is predicated on human perception)?

The reckoning of that is explicit if we know the spectrum of the radiation, since the amount of luminous flux is the amount of radiate power weighted by wavelength (using an agreed weighting curve) and given a certain scaling factor.

The ratio of the amount of luminous flux to the amount of radiant power is properly said to be the luminous efficacy of the radiation (a function only of its spectral distribution).

• If a light source device draws y watts of electrical power, how many lumens of luminous flux does it emit?

The ratio of the amount of luminous flux emitted to the amount of electrical power consumed is properly said to be the luminous efficiency of the device.

It can be thought of as being built up in two steps:


In some literature it can be tricky to know which of these is meant.

Best regard,

Doug

Asher Kelman said:

Jerome,

For ISO 3 at f=22 then energy received at the skin would be 870/10^4 j/cm^2

=0.87 j/m™ or 870 mJ/cm^2

Is that correct?

Asher

Click to expand...

870/10^4 j/cm^2, 0.087 J/cm2 or 87mJ/cm2. I wrote 0.87j/cm2 in error above.

Jerome Marot said:

A quick look at wikipedia gives the relations between lux and Watts.

1 lx = 1 lm/m2

The lumen is linked to the Watt by the luminosity function. Some typical values of this function are given under luminous efficiency, for a 5800K black body radiator, 80 lm/W and for the same filtered to the visible 251 lm/W (perfect white light).

Joules are, of course, Watts per second.

This table will convert ev to lux.

So, taking ev 18, which is 656000 lux or 656000 lm/m2, we have 2613 W/m2 for a perfect white light. (Flashes are filtered and therefore much closer to perfect white light than to black body radiation, but if we want to assume unfiltered black body, we will just have to multiply the final result by about 3.)

Now for the link between joules x seconds and watts in photography. For a typical flash, the duration of light is about 1/1000s. You original question was about f/22 with a film equivalent of iso 3. At ev 18 (which is a sunny day) we would use f/22 and 1/3s with this kind of light. Therefore the flash will output as much energy in 1/1000s as the film would get in a sunny day in 1/3s, i.e. the flux is 333 times bigger. The flash will output 870310 W/m2 in a 1/1000s, which is 870 J/m2 or 0,87 J/cm2 (1 cm2 is more or less the size of the eye). For an "unfiltered black body flash", 3 times as much.

The thermic effects of such a small amount of energy will be negligible.

But, and this is the flaw in your reasoning, there is more to flash light than energy alone. The damage to the eye, if any, are likely to come from other effects than energy, in particular residual UV content.

Click to expand...

870/10^4 j/cm^2, 0.087 J/cm2 or 87mJ/cm2. I wrote 0.87j/cm2 in error above.


Jerome,

All the UV, is of course, is removed with filters, so that does not damage anything, luckily!

87 mJ/cm^2 is then you best estimate and would be multiplied by a factor to account for removal of the UV component. I can't think of other corrections at this moment. Can you find any orders of magnitude errors?

Asher

Asher Kelman said:

All the UV, is of course, is removed with filters, so that does not damage anything, luckily!

Click to expand...

Of course not. Filters have residual transmission. A fraction of percent of residual transmission of the kind of blasting strobes you consider will quickly add up. I am afraid that this is the main flaw in your reasoning.

Jerome Marot said:

Of course not. Filters have residual transmission. A fraction of percent of residual transmission of the kind of blasting strobes you consider will quickly add up. I am afraid that this is the main flaw in your reasoning.

Click to expand...

Jerome,

Well, of course you're right there. I mean that's one gets the UV down to a level that's safe. However, in order to proceed with the calculation, we need to put that aside, just for the moment. Certainly it's critical, but we're not forgetting it, 'promise!

Thanks,

Asher