# A reminder for some memebers as to what the F stop is all about.

Started Jan 26, 2014 | Discussions
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Re: A reminder for some memebers as to what the F stop is all about.
1

bobn2 wrote:

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two'

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm.

Um, no Bob.

The f/ number is the focal length divided by the diameter of the entrance pupil.

Which means in your example above, if the element entrance diameter of the 50mm lens results in an f/2 maximum aperture and the element entrance diameter of the 25mm lens results in a f/2 maximum aperture, both lenses are in fact f/2. It's Elementary.

Sorry.

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8

gsergei wrote:

I am positive you have never finished your high school.

If you mean that I have never stopped learning, you are right.

Overall - a superb technical illiteracy. Can't argue with you, sorry.

I had realised that you can't argue with me, and yes you do display exactly the illiteracy you talk about.

BTW, the term "charge" is used for capacitance and is measured in Culons or Farades as in capacitors, not light sensitive diodes output , which is voltage/current.

'Coulombs' (for that is what it is) is the unit of charge. 'Farads' (for that is what they are) are the unit of capacitance. They are not the same thing, but related by the formula Q=CV - i.e. the charge is equal to the capacitance times the voltage. Maybe best to work those out before trying to explain how a sensor works.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
5

mapgraphs wrote:

bobn2 wrote:

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two'

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm.

Um, no Bob.

The f/ number is the focal length divided by the diameter of the entrance pupil.

The aperture diameter is the diameter of the entrance pupil. Let's call it A and the f-number N.

You tell me that N = f/A

I say that A = f/N. Can you not see these are the same thing? Lets do it one step at a time.

N = f/A

multiply both sides by A gives A*N = f

divide both sides by N gives A = f/N

Which means in your example above, if the element entrance diameter of the 50mm lens results in an f/2 maximum aperture and the element entrance diameter of the 25mm lens results in a f/2 maximum aperture, both lenses are in fact f/2.

I'm not sure what you mean by 'element entrance diameter' - let's stick to 'entrance pupil'. The entrance pupil (aperture) diameter of a f/2 50mm lens is 50/2 = 25mm. The entrance pupil (aperture) diameter of a f/2 25mm lens is 25/2 - 12.5mm. The two lenses have different entrance pupil sizes (as the must, since they have different focal lengths)

It's Elementary.

I'd have thought so, but you got it wrong nonetheless.

Sorry.

You don't have to be, we all make mistakes sometimes.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
1

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras. In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

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Re: A reminder for some memebers as to what the F stop is all about.
1

tray48 wrote:

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras.

Indeed it is, that si where we started.

In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

Yes, and in terms of the noise in the image, and therefore how small and exposure you can stand, and therefore how big an f-number you can tolerate for a given shutter speed, what matters is the total light.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
1

bobn2 wrote:

tray48 wrote:

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras.

Indeed it is, that si where we started.

In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

Yes, and in terms of the noise in the image, and therefore how small and exposure you can stand, and therefore how big an f-number you can tolerate for a given shutter speed, what matters is the total light.

You forgot to involve the width of the camera strap, the type of batteries used oh and the kitchen sink.

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Re: A reminder for some memebers as to what the F stop is all about.
7

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

Bob, the words have no "sound" until read out loud, in which case it sounds like "eff too". Who would read it out loud as "eff divided by too". Maybe in a formal setting a speaker giving a presentation would read it out as "an f ratio of two", or maybe even "a focal ratio of two". But most would read it out loud as "eff too"

As for "f2" sounding like the measure is in "effs", the only time we see the unit coming before the quantity is with currency \$2

So really you should have written "if you write", not "if you say", and "reads like", not "sounds like".
Yes it's a silly little thing, but quite important, especially when being critical of others.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
1

bobn2 wrote:

tray48 wrote:

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras.

Indeed it is, that si where we started.

In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

Yes, and in terms of the noise in the image, and therefore how small and exposure you can stand, and therefore how big an f-number you can tolerate for a given shutter speed, what matters is the total light.

So what would be the outcome if instead of a digital cameras with electronic sensors, you had two cameras with film, one say 35 mm and the other a medium format. Both cameras would have exactly the same film with the exception of the size of the frame. Now if you placed lenses on both camera such that you had equivalent angles of view, set them to same f-stop, photographed the same object from the same position and same light, would you expect a better image from the larger format camera?

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Re: A reminder for some memebers as to what the F stop is all about.
3

tray48 wrote:

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras. In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

I don't often agree with Bob, however one needs to consider that the image made with the larger sensor and the 2x longer lens shows the same scene, but has been generated from 4x as many photons. It's different if one uses the same fl lens on both systems and then crops the ff image down to show the same scene

Peter

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One more time
1

bobn2 wrote:

mapgraphs wrote:

bobn2 wrote:

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two'

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm.

Um, no Bob.

The f/ number is the focal length divided by the diameter of the entrance pupil.

The aperture diameter is the diameter of the entrance pupil. Let's call it A and the f-number N.

You tell me that N = f/A

I say that A = f/N. Can you not see these are the same thing? Lets do it one step at a time.

N = f/A

multiply both sides by A gives A*N = f

divide both sides by N gives A = f/N

Which means in your example above, if the element entrance diameter of the 50mm lens results in an f/2 maximum aperture and the element entrance diameter of the 25mm lens results in a f/2 maximum aperture, both lenses are in fact f/2.

I'm not sure what you mean by 'element entrance diameter' - let's stick to 'entrance pupil'. The entrance pupil (aperture) diameter of a f/2 50mm lens is 50/2 = 25mm. The entrance pupil (aperture) diameter of a f/2 25mm lens is 25/2 - 12.5mm. The two lenses have different entrance pupil sizes (as the must, since they have different focal lengths)

It's Elementary.

I'd have thought so, but you got it wrong nonetheless.

Sorry.

You don't have to be, we all make mistakes sometimes.

I'll try and make it more understandable,

The f/ number is the Focal Length (FL) divided by the Diameter (D) of the entrance pupil.

so the equation is f/ = FL/D

You're defining f as focal length one moment and then defining f as aperture the next. It just doesn't work that way. So no they aren't the same.

Sorry.

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Bingo nt

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Re: A reminder for some memebers as to what the F stop is all about.
2

Messier Object wrote:

I don't often agree with Bob, however one needs to consider that the image made with the larger sensor and the 2x longer lens shows the same scene, but has been generated from 4x as many photons.

Why don't you often agree with Bob, then?  The above, or some derivative of it, is basically all he ever says.

Julie

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Keep it clear and simple
1

mapgraphs wrote:

I'll try and make it more understandable,

The f/ number is the Focal Length (FL) divided by the Diameter (D) of the entrance pupil.

so the equation is f/ = FL/D

You're defining f as focal length one moment and then defining f as aperture the next. It just doesn't work that way. So no they aren't the same.

Sorry.

Bob is saying that relative aperture (note relative) is defined as f/D, where f is the focal length and D is the entrance pupil diameter.

The trick is to make it clear whether you are talking about relative aperture (e.g. 'f/2') or physical aperture (e.g. '25 mm').

To get f/2 at f = 50 mm requires an aperture of 25 mm: 50/2 = 25. So, as an example:

• FourThirds: the ZD 50 mm f/2 lens has a maximum aperture of 50/2 = 25 mm
• 135: a 100 mm f/2 lens requires a maximum aperture of 100/2 = 50 mm.

Both the above lenses would have the same EFL of 100 mm (EFL means 'Equivalent Focal Length' in 135 terms).  The 135 lens requires twice the physical aperture to achieve f/2 because the lens must cover a much larger sensor.

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Re: A reminder for some memebers as to what the F stop is all about.

Dimitri_P wrote:

bobn2 wrote:

tray48 wrote:

bobn2 wrote:

veroman wrote:

Sergey_Green wrote:

There is no f2, there is f/2 ...

Wow. Big difference. Surprised nobody caught this before. What a terrible "error."

It's a silly little thing, but quite important. The 'f2' notation doesn't tell you what's going one. For instance, if you say

'The aperture of the lens if f2' it sounds like the aperture is 2 and it's measured in 'effs'. That is quite wrong, the aperture isn't two.

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two' and is measured in units of distance, usually millimetres.

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm. If both lenses are collecting light from the same angle of view (for instance because the 50mm lens is working with a sensor of twice the linear dimensions of the 25mm lens, then it is collecting from the same solid angle over four times the area, so it collects four times the amount of light.

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras.

Indeed it is, that si where we started.

In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

Yes, and in terms of the noise in the image, and therefore how small and exposure you can stand, and therefore how big an f-number you can tolerate for a given shutter speed, what matters is the total light.

You forgot to involve the width of the camera strap, the type of batteries used oh and the kitchen sink.

can't say those affect my photography so much, but as you like.

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Bob

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Re: One more time
3

mapgraphs wrote:

bobn2 wrote:

mapgraphs wrote:

bobn2 wrote:

If you say 'the aperture of the lens is f/2' what you are saying is 'the aperture of the lens is f divided by two', so then you ask, what is 'f' - of course it's the focal length, so 'the aperture' is 'the focal length divided by two'

So, if someone tells you that the aperture of a 50mm f/2 lens and a 25mm f/2 lens is the same, they are wrong. in the case of the 50mm lens it is 50/2 mm, = 25mm and in the case of the 25mm lens it is 25/2 mm = 12.5mm.

Um, no Bob.

The f/ number is the focal length divided by the diameter of the entrance pupil.

The aperture diameter is the diameter of the entrance pupil. Let's call it A and the f-number N.

You tell me that N = f/A

I say that A = f/N. Can you not see these are the same thing? Lets do it one step at a time.

N = f/A

multiply both sides by A gives A*N = f

divide both sides by N gives A = f/N

Which means in your example above, if the element entrance diameter of the 50mm lens results in an f/2 maximum aperture and the element entrance diameter of the 25mm lens results in a f/2 maximum aperture, both lenses are in fact f/2.

I'm not sure what you mean by 'element entrance diameter' - let's stick to 'entrance pupil'. The entrance pupil (aperture) diameter of a f/2 50mm lens is 50/2 = 25mm. The entrance pupil (aperture) diameter of a f/2 25mm lens is 25/2 - 12.5mm. The two lenses have different entrance pupil sizes (as the must, since they have different focal lengths)

It's Elementary.

I'd have thought so, but you got it wrong nonetheless.

Sorry.

You don't have to be, we all make mistakes sometimes.

I'll try and make it more understandable,

I understand it already, thanks - it's you that's having the difficulty, I think.

The f/ number is the Focal Length (FL) divided by the Diameter (D) of the entrance pupil.

so the equation is f/ = FL/D

You're defining f as focal length one moment and then defining f as aperture the next. It just doesn't work that way. So no they aren't the same.

Sorry.

That's a not helpful notation. You're using '/' in two different senses. Allow me to replace it with '#' which is a notation some use:

f# = FL/D

D*f# = FL

D = FL/f#

same result, just different symbols. Whatever, the aperture diameter is given by the focal length divided by the f-number.

You're defining f as focal length one moment and then defining f as aperture the next.

No, I'm not, my usage of 'f' for the focal length is consistent.

It just doesn't work that way. So no they aren't the same.

'f' always stands for 'focal length', yielding a formula which gives the 'aperture'. Aperture in optics is a distance, the diameter of the entrance pupil, consistently for telescopes, binoculars and so on. In photography we use the 'relative aperture' or 'f-number', and the derivation is as the formula I give.

Sorry.

No need to be sorry, unless you're intentionally being obtuse for effect.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
1

windsprite wrote:

Messier Object wrote:

I don't often agree with Bob, however one needs to consider that the image made with the larger sensor and the 2x longer lens shows the same scene, but has been generated from 4x as many photons.

Why don't you often agree with Bob, then? The above, or some derivative of it, is basically all he ever says.

I don't know, sometimes I've been known to say 'who's buying the next round?'.

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Bob

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Re: Keep it clear and simple

boggis the cat wrote:

mapgraphs wrote:

I'll try and make it more understandable,

The f/ number is the Focal Length (FL) divided by the Diameter (D) of the entrance pupil.

so the equation is f/ = FL/D

You're defining f as focal length one moment and then defining f as aperture the next. It just doesn't work that way. So no they aren't the same.

Sorry.

Bob is saying that relative aperture (note relative) is defined as f/D, where f is the focal length and D is the entrance pupil diameter.

The trick is to make it clear whether you are talking about relative aperture (e.g. 'f/2') or physical aperture (e.g. '25 mm').

To get f/2 at f = 50 mm requires an aperture of 25 mm: 50/2 = 25. So, as an example:

• FourThirds: the ZD 50 mm f/2 lens has a maximum aperture of 50/2 = 25 mm
• 135: a 100 mm f/2 lens requires a maximum aperture of 100/2 = 50 mm.

Both the above lenses would have the same EFL of 100 mm (EFL means 'Equivalent Focal Length' in 135 terms). The 135 lens requires twice the physical aperture to achieve f/2 because the lens must cover a much larger sensor.

and of course as the format gets smaller a desireable-for-marketing small f-number becomes more and more effortlessly achievable. I have a super-8 cine camera with an f/1.0 3x zoom lens.

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Bob

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Re: A reminder for some memebers as to what the F stop is all about.
5

tray48 wrote:

bobn2 wrote:

tray48 wrote:

Yes it is true that 4 times the amount of light will enter through a 50mm lens at f/2 than a 25 mm lens. But since area of the sensor of the of the camera with a 50mm lens is 4 times larger than the camera with the 25 mm lens, the amount of light falling on the sensor per unit area is the same for both cameras.

Indeed it is, that si where we started.

In other words the light intensity at the sensor is the same for both cameras even though the camera with the 50 mm lens is working with 4 times the amount of light as the camera with the 25 mm lens. I think the problem arises from confusion about the difference between light intensity and total light

Yes, and in terms of the noise in the image, and therefore how small and exposure you can stand, and therefore how big an f-number you can tolerate for a given shutter speed, what matters is the total light.

So what would be the outcome if instead of a digital cameras with electronic sensors, you had two cameras with film,

Exactly the same.

one say 35 mm and the other a medium format.

Of you could say 110, which is the same frame size as Four Thirds, and 135.

Both cameras would have exactly the same film with the exception of the size of the frame.

Let's go with that just for the present.

Now if you placed lenses on both camera such that you had equivalent angles of view, set them to same f-stop, photographed the same object from the same position and same light, would you expect a better image from the larger format camera?

Absolutely, that's why you bought a larger format camera in the first place. The four times as many photons reaching the 135 frame produce four times as many photoelectrons which reduce four times as many grains. So, enlarged to the same size, the 135 image contains four times as many grains as the 110 image so will look smoother and contain more information. If one really needed a certain fast shutter speed, and had a minimum quality threshold, you could load the 135 camera up with a four times faster film with four times the grain size and get the same result as the 110 and a two stop larger f-number.

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Bob

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bobn2 wrote:

gsergei wrote:

I am positive you have never finished your high school.

If you mean that I have never stopped learning, you are right.

Overall - a superb technical illiteracy. Can't argue with you, sorry.

I had realised that you can't argue with me, and yes you do display exactly the illiteracy you talk about.

BTW, the term "charge" is used for capacitance and is measured in Culons or Farades as in capacitors, not light sensitive diodes output , which is voltage/current.

'Coulombs' (for that is what it is) is the unit of charge. 'Farads' (for that is what they are) are the unit of capacitance. They are not the same thing, but related by the formula Q=CV - i.e. the charge is equal to the capacitance times the voltage. Maybe best to work those out before trying to explain how a sensor works.

Bob, this is hopeless.

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Renato.
http://www.flickr.com/photos/rhlpedrosa/
OnExposure member
http://www.onexposure.net/
Good shooting and good luck
(after Ed Murrow)

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rhlpetrus wrote:

bobn2 wrote:

gsergei wrote:

I am positive you have never finished your high school.

If you mean that I have never stopped learning, you are right.

Overall - a superb technical illiteracy. Can't argue with you, sorry.

I had realised that you can't argue with me, and yes you do display exactly the illiteracy you talk about.

BTW, the term "charge" is used for capacitance and is measured in Culons or Farades as in capacitors, not light sensitive diodes output , which is voltage/current.

'Coulombs' (for that is what it is) is the unit of charge. 'Farads' (for that is what they are) are the unit of capacitance. They are not the same thing, but related by the formula Q=CV - i.e. the charge is equal to the capacitance times the voltage. Maybe best to work those out before trying to explain how a sensor works.

Bob, this is hopeless.

Hopeless for what, providing factually correct information? OP tried to post yet another thread which will make things straight, yet made couple mistakes, so those should be pointed out.

BTW do you know what CCD stands for? "A charge-coupled device (CCD) is a device for the movement of electrical charge, usually from within the device to an area where the charge can be manipulated, for example conversion into a digital value." (courtesy of Wikipedia).

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Renato.
http://www.flickr.com/photos/rhlpedrosa/
OnExposure member
http://www.onexposure.net/
Good shooting and good luck
(after Ed Murrow)

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Fujifilm X-T3 Fujifilm XF 18-55mm F2.8-4 R LM OIS Sigma 18-35mm F1.8 DC HSM Art Sigma 50-100mm F1.8 DC HSM Art Canon PowerShot G7 X +13 more
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