Mathematics in Photography

I always find it interesting to to discover or find/work out the maths behind our hobby - one was how the F number sequence is devised.

This post is about determining what lens to use to cover a certain angle of view.

The scenario is, that I will be sat in the boot of a Range Rover (harnessed in for safety), looking back at a following convoy of cars, all doing between 30 and 40 mph around the oval of a race track.

I used AutoCAD to to come up with various scenarios:

IMAGE LINK: https://flic.kr/p/Tmex​mE

IMAGE LINK: https://flic.kr/p/Tmex​od

Math is fun. I don't completely follow your scenario, but that's besides the point. Don't the online field of view calculators calclate this for you. http://www.scantips.co​m/lights/fieldofview.h​tml

Was this scenario for fun, or are you actually planning to be strapped in the back of a Range Rover?

I wished maths was more fun when I was at school - perhaps now, in the real world, it makes more sense.

Yes, I will be strapped into the boot of a Range Rover for a day of moving vehicle photos.

The event is not until next month, but I'm putting together some graphics to work out the different shots I hope to get and also act as notes for the participants on the day. Here's another:

IMAGE LINK: https://flic.kr/p/SPT2​R1

Would you please post a link to your F-number explanation. I've often wondered why it's different to SS and ISO interval calculation.

My F stop interpretation...

Take a theoretical 50mm lens with an internal 50mm aperture - this is 1:1 (50:50), which we'll call f1:1 or f/1

We can work out the area of the aperture using Pi, which equates to 1963.195mm²

Each "stop" is a reduction of light by 1/2, so we need to reduce the area of the aperture by one half, which is 981.7477mm². All we need to do is then work out the appropriate diameter... which is 17.6758mm

If we now look at this new ratio of 17.6458:50, we see it is (near as dammit) 1:2.8

Again we halve the area, determine the diameter and work out the ratio etc

IMAGE LINK: https://flic.kr/p/S8se​nw

You left out f/1.4 and f/2

Diameter corresponding to area of 981.748 is 35.355 which gives the ratio 1:1.4
Diameter corresponding to area of 490.874 is 25 which gives the ratio 1:2

All the ratios are off since you computed the radius of the aperture instead of the diameter.

You guys are making the math way more complicated than it needs to be. No pi, no exponents needed. By the definition of aperture (f/stop = focal length / aperture) we have:

FL 50mm
f/1 = 50 / 1 = 50mm
f/1.4 = 50/1.4 = 35.7mm
f/2 = 50/2 = 25mm
f/2.8 = 50/2.8 17.9mm
Yadda-yadda...

None of which is really relevant to anything Canon makes because no lens with the diaphragm matching this formula was ever made. My 200mm f/2.8L II has a max aperture of 2.8 (obviously). According to our calculations, that yields an aperture of 71.4mm. Neat trick on a lens that is only 72mm in diamter, and the 72mm part isn't even glass. Ditto the 85L II. An f/1.2 aperture would need to be 70.8mm across. Good luck with that in a 72mm wide lens.

Yes, of course, this is just a purely theoretical exercise. But still useful to understand how Aperture Priority works.

AnnieMacD wrote in post #18315150
Would you please post a link to your F-number explanation. I've often wondered why it's different to SS and ISO interval calculation.

Wikipedia has as good a summary as any on f numbers. It can be quite easily comprehended by the layperson.

https://en.wikipedia.o​rg/wiki/F-number

Trugga wrote in post #18315170
My F stop interpretation...

Take a theoretical 50mm lens with an internal 50mm aperture - this is 1:1 (50:50), which we'll call f1:1 or f/1

We can work out the area of the aperture using Pi, which equates to 1963.195mm²

Each "stop" is a reduction of light by 1/2, so we need to reduce the area of the aperture by one half, which is 981.7477mm². All we need to do is then work out the appropriate diameter... which is 17.6758mm

If we now look at this new ratio of 17.6458:50, we see it is (near as dammit) 1:2.8

Again we halve the area, determine the diameter and work out the ratio etc

IMAGE LINK: https://flic.kr/p/S8se​nw
f calc by Lawrence Fowler, on Flickr

Does that make sense ?

Lawrence

Perfect. The key is that it's area rather than radius or diameter. I've never bothered to look up the explanation but now I get it.
Thanks.

Absolutely correct. Area is the key to understanding aperture. It is impossible to separate area, radius, and diameter, though
.

The f/stop progression on the lens is actually a rounded value to make them easier to remember the values. I created this chart assuming a 100mm FL to make it easier for folks to see the numeric progression of the aperture diameter and the aperture area for that diameter.

.
It reminds me one story about Pavel Lebeshev, who was cameraman for soviet comedy Kin-dza-dza.
He was taking pictures from helicopter. Open door, him sliding outside. After they landed, they find out what assistant never stripe him in. Lebeshev turned white skin and didn't talk to anyone for three days.
Math is good, but safety is first.

The general equation for the change in ƒ/ratio required for any sized change of exposure in stops is:

ƒ/N+n = 2N^(n/2)

(Or N times the square root of 2 times n, which is really hard to represent using HTML code.)

Where N is the current ƒ/ratio and n is the change in stops.

This equation works for any arbitrary series of ratios, not just the usual series that we see that is based around ƒ/1. It also works in either direction and for fractional stop values, and if you want to know what one stop wider than ƒ/1 is, i.e. when n=-1, its ƒ/0.707, followed by ƒ/0.5.

Alan

A quick correction, because I was not paying full attention. If you use the form of the equation that uses the square root of 2n, it will only work with positive values of n.