Exposure Compensation (Stops)

TheVDM

TheVDM

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Hello everybody.

This may seem a bit of a stupid question, but its something I just want to clarify that I am getting it right.

When people mention over or under exposing by x ammount of stops would it go like this.

if the metered exposure was at 1/100 seconds then

-4 Stop = 1/1600 second
-3 Stop = 1/800 second
-2 Stop = 1/400 second
-1 Stop = 1/200 second

Metered Exposure = 1/100 second

+1 Stop = 1/50 second
+2 Stop = 1/25 second
+3 Stop = 1/12.5 second (or 1/12)
+4 Stop = 1/6.25 second (or 1/6)

And if the metered exposure was 1 second then

-4 Stop = 1/16 second
-3 Stop = 1/8 second
-2 Stop = 1/4 second
-1 Stop = 1/2 second

Metered Exposure = 1 second

+1 Stop = 2 seconds
+2 Stop = 4 seconds
+3 Stop = 8 seconds
+4 Stop = 16 seconds

and so on

Please let me know if this is the correct way of working this out!

All the best
Jym
 
Assuming your apperture is constant then yes, that's perfectly correct.

If you want to increase or decrease appertures they alter by a factor of 1.4 (square root of 2)

so 1->1.4->2->2.8->4->5.6->8
 
Uneducated_Rick said:
Assuming your apperture is constant then yes, that's perfectly correct.

If you want to increase or decrease appertures they alter by a factor of 1.4 (square root of 2)

so 1->1.4->2->2.8->4->5.6->8
Click to expand...

Hmm, thats lost me a bit :gag:

I usually shoot in aperture priority to control the depth of field so hopefully that should be enough to know for now!!

Many thanks for confirming this for me

Jym
 
TheVDM said:
Hmm, thats lost me a bit :gag:

I usually shoot in aperture priority to control the depth of field so hopefully that should be enough to know for now!!

Many thanks for confirming this for me

Jym
Click to expand...

Ok, think of it this way - do increase exposure by 1 stop you double the light getting to the sensor.

The way you are currently doing this is by halving the shutter speed or doubling the time the shutter is open.

The alternative to this is to let twice as much light through in exactly the same amount of time. The area of a circle is PI*r^2, or PI multiplied by the radius squared, so increasing the apperture by 1.4 would double the area letting through twice as much light in the same time
 
So the difference between an aperture of 4.5 & 5.6 would be about 1.2 stops...

You may have to bare with me as my maths is no where near as good as it used to be.

Jym
 
TheVDM said:
So the difference between an aperture of 4.5 & 5.6 would be about 1.2 stops...

You may have to bare with me as my maths is no where near as good as it used to be.

Jym
Click to expand...

4.5 to 5.6 is 2/3 of a stop- unfortunately to make things fun most cameras work in 1/3 of a stop increments for apperture.

Depending on your camera make and model you might be able to switch to 1/2 stop increments

Have a look at this

http://en.wikipedia.org/wiki/F-number#Standard_full-stop_f-number_scale

makes it easier when you can see it all written down and colour coded
 
TheVDM said:
So the difference between an aperture of 4.5 & 5.6 would be about 1.2 stops...

You may have to bare with me as my maths is no where near as good as it used to be.

Jym
Click to expand...
not quite the differance would be a full stop4.5-5.6=1.1 oops no the differance would be half doh!!!!
 
I think I have mastered it now... f3.5 - f5 is +1 stop, and so is f5 to f7.1

It makes it nice and easy when you have your camera in your lap on aperture priority
 
Uneducated_Rick said:
Assuming your apperture is constant then yes, that's perfectly correct.

If you want to increase or decrease appertures they alter by a factor of 1.4 (square root of 2)

so 1->1.4->2->2.8->4->5.6->8
Click to expand...



Uneducated_Rick said:
Ok, think of it this way - do increase exposure by 1 stop you double the light getting to the sensor.

The way you are currently doing this is by halving the shutter speed or doubling the time the shutter is open.

The alternative to this is to let twice as much light through in exactly the same amount of time. The area of a circle is PI*r^2, or PI multiplied by the radius squared, so increasing the apperture by 1.4 would double the area letting through twice as much light in the same time
Click to expand...


... and you call yourself "Uneducated_Rick" :shrug: :lol: ...

.. so, then, the educated Rick would give us the trajectory to circumnavigate my D700 around the moon :p

Seriously, thank you for this, it has been very educational and explained it very well too :thumbs:
 
The term 'stop' dates from the dawn of time, when a piece of metal with a hole in it was used as a lens aperture to 'stop' a certain amount of light reaching the film. That term has now become photographers' shorthand for 'One Exposure Value' (1EV) which refers to any halving or doubling of exposure. eg 1/125sec to 1/250sec is one stop, and f/4 to f/5.6 is one stop (relationship sq root of 2=1.4) and ISO100 to ISO200 is one stop.

For reference, the formal arrangement of 'full stops' is 1sec, 1/2sec, 1/4sec, 1/8sec, 1/15sec etc for shutter speeds, and f/1, f/1.4, f/2, f/2.8, f/4 etc for lens apertures. All the inbetween numbers that you see on modern cameras and lenses are correctly known as either 'half stops' or 'third stops'.
 
It is better to think of exposure values (EV). Then you will know what aperture is needed with a timed shot. I do wish posters proof read their postings before sending posts.
 
For anybody struggling to figure out the stops you dont really need to know the whole maths behind it. On canon camera's set to 1/3 stops all you have to do is move either dial 3 clicks to adjust it 1 stop. I just counted them that way as I found it quicker than trying to calculate in your head.

Now I just know the stops and can work it out quickly in my head anyway but at the start it's probably just as easy to work it out on the camera.
 
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