F number equivalence ?

[YT2095]

just a quick question, can anyone corroborate that 512 seconds at f512 is the equiv of 1/125 Sec at f8 (assuming that ISO 100 is used for both)?

[Sayonara]

What do you mean by equivalent? Do you mean total light let through?

[YT2095]

yes sir.

[Sayonara]

Not sure to be honest. By f8 do you actually mean f/8?

[YT2095]

yup, f/8 (often in italics too).

my calculations So far, based on a 0.5mm aperture and a focal length of 230mm, gives an f-number of 460.

so what I`v done is look at the pattern in the number progression of f numbers and continued it onwards to f 2048.

I did the same with a variation on the "sunny 16" rule" using ISO 100 at 1/125 sec but using f/8 rather than 16 (it was a bit overcast that day).

so, cutting it short, in the columns as the f-number got smaller I doubled the time to keep the total light the same.

 

I noticed that f/512 was the closest to my f/460 And that the time in seconds was also 512.

 

so in effect I just need someone (better at maths than me) to verify that my working out is correct.

[Klaynos]

You need to know the focal length to work it out (as the f/# is a ratio of focal length to the diameter of the lens)

[YT2095]

the f Number is 460.

as I mentioned Focal length times Aperture (230mm x 0.5mm in my case).

 

what I`m trying to work out is the exposure time, and according to my calcs it should be ~16 mins using ISO 100.

oh yeah, and there is No lens, it`s a pinhole camera.

[Klaynos]

If there's no lens then you can't use the f/# because that has the focal length of the lens in it, surely?

 

But as long as the light is evenly distributed and you're hitting the same detector size (which will require optics) then just making the aperture bigger will increase teh light level at the same rate as the area increase, NOT the radius/diameter increase, so you will need to do a bit of maths to move between the two.

[YT2095]

well, according to the formula that I looked up, the maths that I derived f460 from is accurate.

and an online calculator that I used supports my number progression, but not as high up as I need to go

 

if it`s of any use to you in assisting me, the number progression I assumed is:

8,11,16,22,32,45,64,90,128,180,256,360,512,720,1024,1440,2048.

I have verification that it`s accurate up until 90, I had to assume the rest myself.

 

and I assure you, it requires no lens or focusing mechanism, a pinhole camera is quite efficient for it`s simplicity.

[Klaynos]

well, according to the formula that I looked up, the maths that I derived f460 from is accurate.

and an online calculator that I used supports my number progression, but not as high up as I need to go

 

if it`s of any use to you in assisting me, the number progression I assumed is:

8,11,16,22,32,45,64,90,128,180,256,360,512,720,1024,1440,2048.

I have verification that it`s accurate up until 90, I had to assume the rest myself.

 

I'm not really aware of where the progression comes from, but I assume it's related to the internal optics, so I can't answer your question, I'd have to know the whole optical train and work it out from first principles

[YT2095]

there is no "internal optics" at all, anywhere, I made this camera myself.

 

and although I`m quite willing to explain how (later), it`s not required to work out the correct progression for the shutter speed.

 

if we take the sunny 16 rule of perfect exposure, we have f16 @ 1/125s for ISO 100.

and as you know, for each f-stop you move, you either have to Double or half the shutter speed depending on which way you move the stop.

 

so f16 @ 1/125 sec is equal in light to f22 @ .25 sec and so on.

 

now using my f-number sequence I derived, and starting at f8 @ 1/125sec, when you get down to f512 will the time be 512 secs?

And is my f-number sequence correct?

===============================================================================================

 

 

as promised,

and although I`m quite willing to explain how (later), it`s not required to work out the correct progression for the shutter speed.

 

the details of what I did are here:

http://www.scienceforums.net/forum/showthread.php?p=451785#post451785

Edited December 13, 2008 by YT2095
multiple post merged

[Sayonara]

But the stops are worked out for lenses of a given focal length.

 

Will it work for a pinhole camera?

[YT2095]

yup perfectly...

 

"The f-number of the camera may be calculated by dividing the distance from the pinhole to the imaging plane (the focal length) by the diameter of the pinhole. For example, a camera with a 0.02 inch (0.5 mm) diameter pinhole, and a 2 inch (50 mm) focal length would have an f-number of 2/0.02 (50/0.5), or 100 (f/100 in conventional notation)."

 

taken from: http://en.wikipedia.org/wiki/Pinhole_camera#Calculating_the_f-number_.26_required_exposure

[Sayonara]

Aces!

Looking forward to seeing some shots from this.

[John Cuthber]

You need to know about reciprocity failure too.

http://en.wikipedia.org/wiki/Reciprocity_(photography)

[YT2095]

indeed, Happily that info comes with each pack of Photopaper that you buy and also some films, or at least it does with Ilford paper.

it helps when I`m doing long exposure Night time shots that require exposures of 10 seconds or more (sometimes Minutes), however this would be a whole lot easier to calculate if my working out (as posted in this thread) can be verified as correct or not.

 

Aces!

Looking forward to seeing some shots from this.

 

Done

Edited December 13, 2008 by YT2095
multiple post merged

[John Cuthber]

I think each doubling of the f number quadruples the time so

1/125 @f/8

1/31.25 @f/16

1/7.8@f/32

1/1.95@f/64

i.e 0.512 sec @f/64

2.05@f/128

8 sec @ f/256

32.768 @ f/512

131.072@ f/1024

 

I think you need to check your arithmetic, but I'm not sure. I have a cold and my head's not working properly.