Gaffer & Gear 69 - Moving a Light, Distance for Exposure

[@YeahWhiplash]

This is so helpful! Thanks Andrew for providing the best G&E knowledge on the internet!

[@floripilsl2671]

An older colleague showed me his trick, a simple long rope with knots at each distance, he even colored them, so he could tell a new kid to move to the green knot Thanks on the above! Well done

[@chrisgee2610]

Sir, i learned so much from you. The fact you care so much is greatly appreciated!

[@vincentzavier8109]

Thank you for sharing your knowledge. Your videos are helpful and informative. I got interest in learning gaffer techniques after seeing your very first video on HMI leads (How to lock them properly) You are the only channel I regularly watch for hours. God bless you

[@cubul32]

Wow this is such a useful piece of info Andrew - especially if the subject moves in the frame. What you say implies that the f stop ratios are constant: 8/5.6 = 5.6/4= 2.8/2 = 1.4 /1 (and yes the math checks out). So multiples of 1.4. Which means that if the subject moves away 40% (40% of 1 unit is 1.4) or goes in towards the light 28.57% (0.4/1.4) of the initial distance, a stop is lost or gained on the subject. This certainly puts things in perspective. Say you want to have exactly 1/3 stop light increase in a frame, as the subject moves towards the light, and the distance of moving on set is d . Then d is 0.1 / 1.4 = 7.14 % of the light-subject distance, call it D ( 0.1 because that's the difference between 1.3 and 1.4 on the f stop to get 1/3, you can use any 1/3 ratios in the scale like 7.1 to 8 etc, doesn't matter, it will give the same % because they are all 1.4 ratios from stop to stop). The light has to be 19.6 x d away from the subject; 19.6 = 1.4 / (0.1/1.4) = 1.4 / 0.0714 (

[@flickcine]

Brilliant explanation. Thanks, Andrew! I find it hard to get my head around inverse square but this explanation and the last video have been very helpful.

[@gaffergear]

I can't do complex math

[@DonaldDork]

Thanks for such an invaluable lesson. Cheers.

[@4CardsMan]

To figure the change when moving closer, multiply by square root of 2 divided by 2, or .7. When moving farther away, multiply by the square root of 2, 1.4.

[@4CardsMan]

Sorry, I had that backwards. To get 1 stop more light by moving closer, divide the distance by the square root of 2, which is the same as multiplying by half the square root of 2, which is .7 So, if the light is 10 feet away and you want to get 1 stop more brightness, move the light to 7 feet away )10 * .7). To get 1 stop less light by moving the light away, multiply by the square root of 2, 1.4, and you get 14 feet. I used the same trick years ago to determine the exposure time when making traditional photographic prints. If the exposure time under the enlarger is 10 seconds, 1 stop more would be 10 * 1.4, 14 seconds. 1 Stop less would be 10/1.4 or 10 * .7, 7 seconds.

This video is priceless. Thanks load

[@cameraworks2623]

park cars as fstop/distance reference - brillant ;-)

[@EldenEleSinema]

The Logic behind this: 1) One stop means you should double the light. 2) Inverse square law says: intensity1/intensity2=distance1^2/distance2^2 3) so intensity1 being x, intensity2 being 2x (as we want to double the light) 1/2 is our intensity ratio. 4) so distance1/distance2 = 1/√2 5 so to double the light you should √2 change the distance. Which is approximately 1.4. Where you get the ratio between distances (also f-stops) 1 meter, 1,4 meter, 2 meter, 2,8 and so on… 1.4 is the ratio when you double the intensity or half it…

[@philpritchard5173]

Great rule of thumb.

[@sebastianestrada4699]

Great channel!

[@OppaiDaisukiful]

I didn't quite get in when you explain the different unit measurement, is it the same equivalent when moving 4 feet to 5,6 feet to 4 park cars to 5,6 park cars ? i mean the distance is obviously different between feet and park cars right ?

[@gaffergear]

Sounds to me like you got it! You can use ANY unit of distance measurement. Half the distance, or double the distance gives you 2-stops difference. It makes no difference what the initial distance is, same rule applies. To help you relate that to F-stop numbers, every second f-stop number is double..... 1,2,4,8,16.

[@OppaiDaisukiful]

@@gaffergear oh yeah i got it now, i was tricked by thinking about the intensity of the light with different unit measurement. but now i understand it's about light fall off. Thank you sir now it is easier for me to use this law of physics when i'm on set.

[@dennismenace4188]

Great idea using car spaces!

[@gaffergear]

One gaffer I spoke to imagines football fields when he's lighting epic battle scenes.

[@dennismenace4188]

@@gaffergear Jesus that's some serious light output.

[@rmatiasbravo]

dude just great! thanks a lot!

[@michaelnoronha7386]

Sir, please explain what wattage the light source to be. The information will help us beginners. Thank you and God Bless.

[@gaffergear]

Ok, Good topic for a video. 1st thing, wattage has nothing to do with brightness (anymore). Wattage is a measurement of how much power draw the light uses. In the old days wattage made sense as a measurement of brightness, when we only had 3 different types of lights (tungsten, HMI, flourescent), but now we have LED lights and different types of LED lights, and not all LED lights are as efficient, so wattage is very confusing as a gauge of brightness. So wattage is useless as a way of thinking about brightness. It's just how much power you need to run the light of power or batteries.

[@michaelnoronha7386]

Thank you for the reply. My inherent doubt about how much light be addressed in the new video please. Confused for the sake of digital sensor and bleach concepts. God bless you.

[@Pk_patra]

Thanks sir thank u so much Plz make more info videos Love you

[@jean-patrickpeuvion5373]

Thanks, very interesting 🙂

[@shaukat863]

thank you sir 👍😊

[@kellyeaton7252]

Your association of distance with F-stop numbers is a very clunky way to think of this phenomenon. It does demonstrate what an F-stop is doing, but presents the fact as if those distances are important, when in fact it is the ratios that are important. And the ratios are even easier to memorize. When looking at the distance between your subject and the light, divide it in half. That mark is two stops of difference. If you divide the distance between that new point and the light in half again, you'll find the location of a single stop of difference. In the opposite direction consider the light half way away. If you double the distance, you'll get two stops of light. Half the double is one stop. 2 = 2×1, 4 = 2×2, 8 = 2×4, 16 = 2×8 2.8 = 2×1.4, 5.6 = 2×2.8, ~11 = 2×5.6, 22 = 2×11 F-stops are doublings/halvings/ doublings/quarterings/ and 1.5Xs of distance.

[@kellyeaton7252]

And for clarities sake, the ratio being measured in an F-stop is the diameter of the aperture to the focal length of the lens. If you are at a 2 F-stop, the aperture diameter is 1/2 of the focal length. So if you have a 100mm lens at an F2.0, that aperture is 50mm in diameter. If you have that same 100mm lens at a 5.6 F-stop, the aperture 17.857mm, or 100/5.6.

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