# Flash Zoom - Difference in Stops?



## entlassen (Jul 14, 2014)

Using a 600EX-RT here. Does anybody know the difference in stops at the center of the light circle created by the flash for:
1) 24mm flash zoom vs 70mm flash zoom?
2) 70mm flash zoom vs 200mm flash zoom?

I've always assumed it was just a 1 stop difference for both (1) and (2), but I can't be sure. I don't have a light meter and am wondering if anybody has done any actual tests.

Thanks!


----------



## IMG_0001 (Jul 24, 2014)

From the manual of Canon 600EX-RT
24mm - GN28m
70mm - GN50m
200mm - GN60m

So, thinking it as I write...

This means that a subject at those respective distances should be lit to correct exposure, our reference exposure value. Now the light intensity is the amount of light (or power) divided by the area and the area in turns varies to the square of the subject distance. Therefore, for a given power the intensity would vary to the inverse square of the distance. Since each additional stop, by definition, doubles the light intensity of the former, the gain in stops also follows a square progression. This means that, if I'm not mistaken, for a subject at a fixed distance from the flash, the ratio of the guide numbers should equate to the difference of light intensity in stops.

Assuming that 24mm is the reference:
28m/28m - 1 
50m/28m - 1.8, so a 0.8 stop gain.
60m/50m - 1.2, so a further 0.2 stop gain.

Which is not as bad as I would have intuitively expected... may be someone can verify my line of thought. I'll try to have a second look tomorrow, with a fresh head, a pen and paper.

Edit: I was mistaken, stops are a base 2 exponential, not a square law so I definitely need to revise that tomorrow... the solution should not be hard to correct.

Nighty night!


----------



## privatebydesign (Jul 24, 2014)

24mm - GN28m = f2.8 @ 10m and 100 iso
70mm - GN50m = f5.0 @ 10m and 100 iso
200mm - GN60m = f6.0 @ 10m and 100 iso

Therefore:-

24mm - 70mm = 1 & 2/3 stops
70mm - 200mm = 1/2 stop


----------



## IMG_0001 (Jul 24, 2014)

privatebydesign said:


> 24mm - GN28m = f2.8 @ 10m and 100 iso
> 70mm - GN50m = f5.0 @ 10m and 100 iso
> 200mm - GN60m = f6.0 @ 10m and 100 iso
> 
> ...



Ok, usign the definition of guide number F=GN/d certainly was the simplest solution from the onset. Thamks for reminding me to stop overthinking stuff.


----------



## IMG_0001 (Jul 24, 2014)

IMG_0001 said:


> privatebydesign said:
> 
> 
> > 24mm - GN28m = f2.8 @ 10m and 100 iso
> ...



But to my defense, my straightened line of thought leads to the same results:

2^F = (GN2/GN1)^2

Then from 24mm to 70mm, F=1.67 stops (+1 2/3)
70mm to 200mm, F=0.52 stop (+1/2)
So of course 24mm to 200mm, F=2.19 stops.


----------



## entlassen (Jul 29, 2014)

privatebydesign said:


> 24mm - GN28m = f2.8 @ 10m and 100 iso
> 70mm - GN50m = f5.0 @ 10m and 100 iso
> 200mm - GN60m = f6.0 @ 10m and 100 iso
> 
> ...




I'm a little unfamiliar with the "GNXXm" nomenclature when it comes to guide numbers.

On p. 115 of the 600EX-RT's manual, I'm seeing, for 1/1 manual flash power:

24mm: "28/91.9"
70mm: "50/164"
200mm: "60/196.9"

I'm going to guess that means 28m (91.9 ft.), 50m (164 ft.), and 60m (196.9 ft.).
Are you saying that by the inverse square law, that directly translates to f/2.8, f/5.0, and f/6.0?
Also, I'm not sure where your statement of "@10m and 100 ISO" comes from.

Thanks for the clarification!


----------



## entlassen (Jul 29, 2014)

IMG_0001 said:


> 2^F = (GN2/GN1)^2
> 
> Then from 24mm to 70mm, F=1.67 stops (+1 2/3)
> 70mm to 200mm, F=0.52 stop (+1/2)
> So of course 24mm to 200mm, F=2.19 stops.



OK, so you're taking the log base 2 of both sides of that formula to solve for F.
But can you tell me where that formula comes from? All I know is the GN = f * d formula and the Inverse Square Law formula (Intensity = 1 / distance^2). Would like to know how you derived it.


----------



## privatebydesign (Jul 29, 2014)

The second number is the GN in feet, 28 meters equals 91.9 feet.

Take your GN in feet and divide it by your subject distance in feet to get your f stop. So if your subject is 10 feet away and you have the flash set to 24mm then you get an f stop of 91.9 feet/10 feet = f9.2. If your flash is set to 200mm then 196.9/10 = f19.7


----------



## entlassen (Jul 29, 2014)

privatebydesign said:


> The second number is the GN in feet, 28 meters equals 91.9 feet.
> 
> Take your GN in feet and divide it by your subject distance in feet to get your f stop. So if your subject is 10 feet away and you have the flash set to 24mm then you get an f stop of 91.9 feet/10 feet = f9.2. If your flash is set to 200mm then 196.9/10 = f19.7



OK, I get it thanks, so you're just plugging in F = GN/d. So with GN = 28 and d = 10m, then F = 28/10 = 2.8. Cool. 

It's not immediately clear in the manual's GN chart though that "flash coverage" actually equates to "flash zoom". Can we actually assume that the flash zoom head is set to the focal length (i.e. the horizontal row labeled "Flash Coverage" in the chart)?


----------



## privatebydesign (Jul 29, 2014)

entlassen said:


> Can we actually assume that the flash zoom head is set to the focal length (i.e. the horizontal row labeled "Flash Coverage" in the chart)?



Yes we can.


----------



## IMG_0001 (Jul 29, 2014)

entlassen said:


> IMG_0001 said:
> 
> 
> > 2^F = (GN2/GN1)^2
> ...



First, to clear things off, I advise to just stick to the basic GN = F*d which gives the right results and is much more tractable on the field.

As for my formula, I just went on from basic physics.

Intensity I is Power over Area --> P/A

A, the area lit, increases to the square of the distance from the source -->A proportional to d^2

Therefore, I is proportional to P/d^2 (not equal to)

Then, given that power remains the same --> P1 = P2 entails that I1*d1^2 = I2*d2^2 or, I1/I2 = (d2/d1)^2

Now, that is for the flash in a fixed zoom setting.

Now considering that the subject distance is fixed and posing that Case1 is for 24mm and GN28m and Case2 is for 70mm and GN50m. Zooming from 24mm to 70mm is like moving the subject from a distance of 50m to a distance of 28m. Then d2/d1 = GN2/GN1 and :

I1/I2 = (GN2/GN1)^2

Finally, in photography the aperture F goes up one stop every time the light intensity doubles so I is proportional to 2^F. It ensues that:

2^F = (GN2/GN1)^2

I hope this helps.


----------

