Formulae

I was wondering through some books and I found the following formula.

 

A 50mm lens focus an objet 3m away.

A 25mm lens focus the same object 1,5m away.

Same opening f/11

d = distance

A = opening

f = focal length

 

 

Pc(alpha) = (A*dˆ2) / fˆ2

 

Pc - Proportionality.

 

Is the formula valid?

Please elaborate on what is Pc means? The size of the subject?

Why do you want to make the vote Yes/No?

What a load of tosh. Really.

With what ever lens you have on the camera, fill your viewfinder with what your eye & brain want. Yes you need some knowledge of materials, but your eye balls / brain can not be put into a box. Aloha, Bill

What a load of tosh. Really.

Despite the fact that I found it on an old physics book, I can't seem to validate it, not in a physics sense of photography sense.

Please elaborate on what is Pc means? The size of the subject?

Why do you want to make the vote Yes/No?

It is a proportionality coefficient, a ratio of some kind that would enable you to compare different focal lengths ...

Please elaborate on what is Pc means? The size of the subject?

Why do you want to make the vote Yes/No?

The Yes or No, just to see if you know the equation or not.

It is a proportionality coefficient, a ratio of some kind that would enable you to compare different focal lengths ...

If you plug the numbers into the formula it gives you the result about 4 and 2 so what does that mean to you?

If you can scan at least the page of the book it would help.

Edited April 16, 2020 by BeBu Lamar

Moderator Note

 

The Yes or No, just to see if you know the equation or not.

 

That was not what you originally wrote. You wrote, "Is this formula OK?"

 

I have edited the question to reflect your intention.

 

William

If you can scan at least the page of the book it would help.

 

Correct: the formula in context would greatly assist repsondents.

 

WW

Think of a number; square it.

Think of another number.

 

Multiply the two numbers you dreamed up. Now divide them by 4 and the square of another random number.

 

And shouldn't the distance and focal length be measured in the same units? Metres or millimetres - choose one or the other.

Think of a number; square it.

Think of another number.

 

Multiply the two numbers you dreamed up. Now divide them by 4 and the square of another random number.

 

Hmm... I get E=MC2

http://bayouline.com/o2.gif

McTwo? I remember them. An ancient Scottish clan that lived fast and died hard. Or did they just dye the plaidy? (That's pronounced to rhyme with 'tidy' BTW.)

Another banana for the Deep Thinker.

I haven't heard of it before. However, I think I understand why it exists. You can fill the frame with the same subject, but with different focal lengths, and the amount of visible background will change in proportion to the subject. Am I correct?

You can fill the frame with the same subject, but with different focal lengths, and the amount of visible background will change in proportion to the subject.

Nope! At some point a simple ratio will break down, because there's a constant of 2 focal lengths to add into the formula for magnification versus distance.

 

And the subject-background ratio can never remain constant with a change in distance. It's called 'perspective'.

Edited April 17, 2020 by rodeo_joe|1

All I can get out of it is three fifths of five eighths of not much at all.

If you plug the numbers into the formula it gives you the result about 4 and 2 so what does that mean to you?

If you can scan at least the page of the book it would help.

It is in Portuguese but I see what I can do.

Relations between depth of field and Aperture, focal distance and distance from object. Depth of Field (profundidade de campo in portuguese).

PC1/PC2=a1/a2 ----- PC1/PC2=((d1)ˆ2/(d2)ˆ2) ----- PC1/PC2=((f2)ˆ2/(f1)ˆ2) -> PC(alpha)= (a.dˆ2)/fˆ2

Ah ha! So it's a depth of field formula? If I remember right, you need a blur circle diameter (circle of confusion) to do depth of field, so this doesn't look quite right, or I don't see how it's in a useful form. If you can describe what you want to calculate, and from what inputs, any number of us can provide the correct formulas.

Relations between depth of field and Aperture, focal distance and distance from object. Depth of Field (profundidade de campo in portuguese).

PC1/PC2=a1/a2 ----- PC1/PC2=((d1)ˆ2/(d2)ˆ2) ----- PC1/PC2=((f2)ˆ2/(f1)ˆ2) -> PC(alpha)= (a.dˆ2)/fˆ2

 

Now I get it. It related to which lens has more depth of field

Ah ha! So it's a depth of field formula? If I remember right, you need a blur circle diameter (circle of confusion) to do depth of field, so this doesn't look quite right, or I don't see how it's in a useful form. If you can describe what you want to calculate, and from what inputs, any number of us can provide the correct formulas.

It is I think. Not the one I am used to. I have never seen it and I have never found a proper way to apply it. Overthinking things, in my experience, always seem to ruin the picture.

Nope! At some point a simple ratio will break down, because there's a constant of 2 focal lengths to add into the formula for magnification versus distance.

 

And the subject-background ratio can never remain constant with a change in distance. It's called 'perspective'.

The relationship is certainly logarithmic.

No. It doesn't work for depth-of-field either.

My depth of field calculator gives a total D-o-F of 2.63 metres for the 50mm lens @ f/11 and 3 metres distance. While the 25mm lens @ f/11 and 1.5 metres distance has a D-o-F of 5.21 metres.

 

The subject magnification is the same, but the aperture used has absolutely no bearing on that, and doesn't have to be included in the formula for magnification.

 

So let's work out Pc (1) and Pc (2) with the focal length and subject distance in the same units - millimetres.

We get:

Case (1) Pc = (11*3000^2)/50^2

= 11* 60^2 = 39600

Case (2) Pc = (11*1500^2)/25^2

= 11* 60^2 = 39600

 

We get the same (meaningless?) figure in each case, which is only to be expected if the distance is kept proportional to the focal length. And squaring both d & f is totally pointless - if the formula given in the original post is followed and correctly transcribed.

 

Incidentally, the photographic measurement of subject distance is from focal-plane to subject. If that convention is followed, then even the proportionality of magnification falls apart.

Edited April 18, 2020 by rodeo_joe|1