FStop Calculations
- Thread starter pmark23
- Start date
I don't think a light meter would be very useful, simply because an adapter GG doesn't diffuse light linearly. If you watch the light loss demonstration on Cinevate's website, you'll notice that when you stop down the taking lens on the Brevis one stop, you do not lose a stop of light, but if you do the same on the camera's stock lens, you do.
pmark23
Veteran
I want to make a chart of lumens vs. fstop (specifically for an HVX w. Brevis) to calculate how much (more) light is needed when stopping-down a lens.
Most of the required calculations are in the ASC manual, but there's nothing anywhere on how to combine two irises to get the stop. Is it averaged? A multiple? Summed?
Most of the required calculations are in the ASC manual, but there's nothing anywhere on how to combine two irises to get the stop. Is it averaged? A multiple? Summed?
Leo Versola
Well-known member
Perhaps using a waveform monitor and a grey card to record the IRE values for various combinations between the camera and taking lens while keeping the lighting constant, then plotting the values could create such a chart more easily...
Lenilenapi
Carbonite Member
Huy Vu and Leo are right. Try what Leo said and that will give a good idea of the Terrain, but be aware that metering as you do in film has to be radically adjusted - no mathematical rules will apply.
Shooter
Veteran
I am now a little lost as to what you want exactly (but intrigued to continue the thread).
Light physics maybe a little different on the planet you come from...
(your sig)
Why do you want "lumens" in the table? (Lumens is a kind of "efficiency" measure).
I maintain... double/halve the light (wattage )and you will be close to adding/subtracting one stop. The real world situation will exhibit some variation but what precision do you need a table for.
How will you measure your lumens?
Light physics maybe a little different on the planet you come from...
(your sig)Why do you want "lumens" in the table? (Lumens is a kind of "efficiency" measure).
I maintain... double/halve the light (wattage )and you will be close to adding/subtracting one stop. The real world situation will exhibit some variation but what precision do you need a table for.
How will you measure your lumens?
Lenilenapi
Carbonite Member
Nobody is going to give you an answer to this question. It does not exist. You need to experiment and learn for yourself with your gear.
von Mutius
Member
f-stop are adding up
f-stop are adding up
Hi, f-stop 1.0 means no light is lost passing through the lens, f 1.4 means half the light is passing, 2.8 is a quarter etc...
So if we have 2.8 on the camera we are at 25% of the original light , now we put the adapter in front which loses, just an example, another stop, we are at f 4.0 or at 12.5% of the original light because one stop means loosing half of the light .
Now we add a lense on the adapter with another fstop of 1.4 : Now we have f stop 5.6 or 6.25 % of the original light.
With a 2.8 lens on the adapter we are at f 8.0 (3.125 % of light).
With a 4.0 we are at f 11 (1.5625% of original light).
With a 5.6 we get f 16 (0.78125 gf original light).
With an 8.0 we are at f 22 (0.390625 of the light ).
With a lens at 11.0 we get a sum of f32 or 0.1 953125 of the original light reaches the sensor of your camera.
Mystery busted ?
Matthias :huh:
f-stop are adding up
Hi, f-stop 1.0 means no light is lost passing through the lens, f 1.4 means half the light is passing, 2.8 is a quarter etc...
So if we have 2.8 on the camera we are at 25% of the original light , now we put the adapter in front which loses, just an example, another stop, we are at f 4.0 or at 12.5% of the original light because one stop means loosing half of the light .
Now we add a lense on the adapter with another fstop of 1.4 : Now we have f stop 5.6 or 6.25 % of the original light.
With a 2.8 lens on the adapter we are at f 8.0 (3.125 % of light).
With a 4.0 we are at f 11 (1.5625% of original light).
With a 5.6 we get f 16 (0.78125 gf original light).
With an 8.0 we are at f 22 (0.390625 of the light ).
With a lens at 11.0 we get a sum of f32 or 0.1 953125 of the original light reaches the sensor of your camera.
Mystery busted ?
Matthias :huh:
Leo Versola
Well-known member
Not entirely. As mentioned by Huy in his post, adapters don't necessarily diffuse the light linearly. Meaning every additional stop down on the taking lens does not tranlate the way you would normally think it would.
And guess what? That's a good thing because it means you can stop down a few stops to gain sharpness without losing those same stops of light. A few other benefits include deeper DoF (instead of ultra razor thin DoF like it usually is at f1.4) and depending on the lens, usually more pleasing bokeh...