Here is something that might be interesting to you - or maybe not.
I recently purchased a new reflector and lots of ads say that they fold to 1/3 the original size. This seemed strange to me - that somehow it should be a lot smaller, so I started thinking mathematically. Try to follow along:
Suppose we have a reflector that is 3ft in diameter when popped out. A 3 ft reflector will have pi*d circumference, which is about 9.42 ft. The area is pi*r^2, which is about 7.065 sq. ft.
When you fold the 3 ft reflector, the rim will go around 3 times making the new circle haev a circumference that is 1/3 the unfolded size, so that makes about 3.14 ft. Working the formula for circumference backwards, means that the diameter will be 1 ft. This is interesting, because the diameter is 1/3 the original size. However, the area of this circle is .785 sq ft, which is exactly 1/9 the area of the folded out size.
So, the next time someone tries to tell you that your reflector takes up 1/3 the amount of area when folded, you can confidently tell them that it's really 1/9, which is pretty neat when you think about it - folding reflectors is pretty efficient.
Ok, I know this is probably totally useless, but there it is!
Yes, I do know why the aperture numbers are a sqrt(2). It has to do with cameras being two dimensional and light falloff with the square of the distance. If yoiu double the distance from a light, the light drops by 1/4. If you move the sqrt(2) away, then the light halves.
dna86 wrote:
Yes, I do know why the aperture numbers are a sqrt(2). It has to do with cameras being two dimensional and light falloff with the square of the distance. If yoiu double the distance from a light, the light drops by 1/4. If you move the sqrt(2) away, then the light halves.
Nope. It has nothing to do with distance to the subject. It wasn't supposed to be a trick question.
A full-stop change in aperture, shutter speed or ISO causes either twice the amount of light to illuminate the sensor/film, or half the amount of light. This is equal to +1 Exposure Value (EV) or -1 EV. You get a +1 EV increase at the sensor/film plane by either: (i) doubling the area of the aperture opening; (ii) doubling the amount of time the shutter is open; or, (iii) doubling the ISO or sensitivity of the sensor/film. Since aperture f-stop is defined as a ratio of lengths*, and doubling the aperture area corresponds to increasing the aperture diameter by a factor of sqrt(2), the ratio of adjacent f-stops (or 1 full-stop changes) is sqrt(2) = approx. 1.4.
* f-stop = lens focal length divided by apparent diameter of aperture opening, as seen through the front element.
jcolwell wrote:
* f-stop = diameter of lens front element divided by apparent diameter of aperture opening, as seen through the front element.
The f-stop has nothing to do with the diameter of the lens front element... It's the focal length of the lens divided by the effective diameter of the aperture.
The effective diameter of the aperture is, as you say, the apparent diameter of the aperture as viewed through the front element, also termed "Entrance Pupil".
BigStuart wrote:
The f-stop has nothing to do with the diameter of the lens front element... It's the focal length of the lens divided by the effective diameter of the aperture.
Thanks Stuart, you're exactly right. I must have had a brain fart (original corrected).
