Is there anyone that can work out a mathematical lighting puzzle for me please?......

fatphotographer

fatphotographer

Suspended / Banned
Messages
574
Name
Gary
Edit My Images
No
Hi,

I was shooting a project with a pyrotechnic engineer yesterday, when he asked me if I could get a light measurement of the flash that I was photographing. I took a “photographic” light reading, but need to convert it back into a more industry accepted measurement – ideally, candela.

So here is the situation, the explosion goes off, and we took a flash reading of f64 at ISO 100 at 4 meters, just as if it was a studio flash.

I’m crap at puzzles, and I was off school the day we did algebra, but I am sure that there will be mathematicians, engineers or avid puzzlers who use the forum that could backward engineer the light reading to give the required value.

I had a quick look at http://www.mediacollege.com/lighting/measurement/ but decided to post it on here instead!

Rather than the “O level” answer of “the answer is…..” is there any chance that I could get the "GSCE type" answer of “using the inverse square law……..” as there is a good chance that I will need the equation again for the next shoot.

Any help would be greatly appreciated,

Thanks

Gary
 
Well it's quite long winded but there is a way...

First of all we need to figure out how powerful your light source is in lumens (which is the power from the source) - and this can be used to calculate the candela.

So let's start with the power of that light source and figure out the rating at 1m. We're halving the distance (one stop increase) then halving again (another stop) - so at 1m we'll have a guide rating of f/128 at ISO 100.

My Alien Bee 800 rates at f/32 at ISO 100 at 1m - 4 stops below the light reading you have for the explosion. According to the Alien Bee website the AB800 gives out 14,000Ls, so to get the same reading for the explosion we push this up by 4 stops equivalent. This gives us a figure of 224,000Ls at the source (Ls is lumens seconds and shows the amount of lumens per second are emitted by the source. I think when you talk in lumens it is taken as a given that it's lumens per second in same way watts are, technically, watts per second despite us only saying watts).

So now we have a technical figure for the lightsource. Thankfully someone else has done the hardwork in the next part...

This webpage will allow you to enter your lumens value (224,000) and the view angle. The view angle is the angle of light emitted by your source. If it's an explosion this will be 360 degrees (even though a percentage of those degrees will be lost into the ground or whatever the material is supported on) so we enter that.

The result is 17,825,354 mcd, which is milicandela, so the final figure is 17,825 cd.

If you want an equation...

Candela = Lumens / (2 times pi) times (1 - cos 1/2 a)

(where a is the view angle). So that's the lumens amount divided by one minus the cosine of half of the angle times two pi.

Sorry about the mess, it's hard to write equations in a forum!
 
Not sure I agree with the first bit.

The inverse square law states for electromagnetic waves in free space, if you half the distance you quadruple the amount.

Therefore halving the distance from 4 mtrs to 2 would give you 4x the amount of light (2 stops) and halving it again from 2 mtrs to 1 will give you another 4x (another 2 stops) the amount of light ie a total of 16x the amount of light (a total of 4 stops of addtional light) so from f64 to f256.

How you equate this to 2 stops beats me, perhaps you could explain further?
 
Last edited:
You're more than likely right Ed, I have to admit I was reading through five or six pages of equations and technical figures for lighting so I've probably made a basic error!

I was calculating based on flash use, so when I move a flash from 1m from a subject to 2m the increase in power to compensate is a stop. I think a caveat to add is that it depends on the nature of the light source - a concentrated beam will read differently to a diffused bulb effect.

Assuming that I've calculated incorrectly, and that the initial power reading is actually f/256, that should give you a lumens value of 896,000Ls (two stops more light than I calculated) - so the revised total is 71,301 Cd.

I don't profess to any great understanding of the equations (I dropped A-Level maths after 6 months!) so please feel free to amend wherever you think necessary.
 
adamgasson said:
You're more than likely right Ed, I have to admit I was reading through five or six pages of equations and technical figures for lighting so I've probably made a basic error!

I was calculating based on flash use, so when I move a flash from 1m from a subject to 2m the increase in power to compensate is a stop. I think a caveat to add is that it depends on the nature of the light source - a concentrated beam will read differently to a diffused bulb effect.

Assuming that I've calculated incorrectly, and that the initial power reading is actually f/256, that should give you a lumens value of 896,000Ls (two stops more light than I calculated) - so the revised total is 71,301 Cd.

I don't profess to any great understanding of the equations (I dropped A-Level maths after 6 months!) so please feel free to amend wherever you think necessary.
Click to expand...
Actually 2 stops, following the ISL
I think a caveat to add is that it depends on the nature of the light source - a concentrated beam will read differently to a diffused bulb effect.
Click to expand...
Yes, it will read differently. There are a few caveats (qualifications may be a better word) in Newton's definition, 'point source' and 'free space' being the ones that people sometimes miss. So, although the ISL will always work in principle, it rarely works in absolute, linear terms when applied to lights used for photography.

Well done you for taking all that trouble to answer the question:)
 
Thank you very much for all the effort put into getting an answer.

The energy from the flash and the explosion was immense. It is designed to disable "naughty people" in seige situations.

I was surprised at the relatively low value to the answer. I have seen some cheap rechargable torches bragging about "xx million candle power" and thought this flash would be a higher value than quoted. But when the energy was propelled in a shape similer to that of the millenium dome (it hit the floor and gave a semi domed shape flame), it looks like the measurement is taken differently to that of a direct beam.

Thanks again
 
Last edited:
fatphotographer said:
Thank you very much for all the effort put into getting an answer.

The energy from the flash and the explosion was immense. It is designed to disable "naughty people" in seige situations.

I was surprised at the relatively low value to the answer. I have seen some cheap rechargable torches bragging about "xx million candle power" and thought this flash would be a higher value than quoted. But when the energy was propelled in a shape similer to that of the millenium dome (it hit the floor and gave a semi domed shape flame), it looks like the measurement is taken differently to that of a direct beam.

Thanks again
Click to expand...
Yes, with these torches they forget to include the word 'effective' when they talk about the output.

A similar thing really to hotshoe flashguns, where the guide number is calculated at the maximum zoom setting, even though it's not relevant to the usage of most of the customers.

The result? A 60j or thereabouts hotshoe flash can have a higher guide number than a 300j studio flash, and a new model of flashgun that actually outputs less power can have a higher guide number than the previous model, that had a shorter zoom range.

And hotshoe flashes tend to have highly efficient reflectors - that again boost the guide number but at the expense of even lighting.

I think the term for this is good marketing, personally I prefer 'deceptive':)
 
Agreed, the bigger the number, the bigger the sales, it is down to us users to work out what the numbers actually mean in real life.

We just needed to be able to give an intensity for the light, and I only know how (as most photographers) light works in relation to photographic exposure.
 
Last edited:
You must log in or register to reply here.
Back
Top