## If $(N,\|.\|)$ is a non trivial vector space with norm, then the diameter of the unit sphere is $\ge2$

It is easy to see that $$\delta(B_1(0))\le 2$$ because $$\|x-y\|\le\|x\|+\|-y\|=\|x\|+|-1|\|y\|=\|x\|+\|y\|<1+1=2$$ so $$\sup\{\|x-y\|:x,y\in B_1(0)\}\le 2$$

But to show that $$\delta(B_1(0))\ge 2$$ we chose $$a\in B_1(0)\backslash\{0\}$$ and $$s\in[0,1)$$ and observe that $$\|s\frac{a}{\|a\|}\|=s$$ $$\implies s\frac{a}{\|a\|},-s\frac{a}{\|a\|}\in B_1(0)$$ and $$\|s\frac{a}{\|a\|}-(-s\frac{a}{\|a\|})\|=2s$$

But why does that imply that $$\delta(B_1(0))\ge2s$$?

This is how I think I understand it: the diameter of $$B_1(0)$$ is greater than that of $$B_s(0)$$ for all $$s\in[0,1)$$ because we it contains it. And the diameter of $$B_s(0)$$ is $$2s$$ for each $$s$$, which is true because there exists a sequence $$(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})$$ of pairs of points, with $$0\le\epsilon_n and $$\lim\limits_{n\rightarrow\infty}\epsilon_n=s$$ such that it is always contained in $$B_s(0)$$ and $$\lim\limits_{n\rightarrow\infty}d(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})=2s$$

I’m not sure if I got it right…

## What is the relationship between lens diameter and exposure?

All of the online calculators for exposure, I have seen, factor in time, film speed, and appature (which is a function of the real appature diameter and the focal length of the lens)

However I would have thought a lens with a larger diameter (with the same focal length and appature) would collect more light and therefore be faster!

Am I just looking at rubbish exposure calculators, or am I missing something?

## How is the diameter of a non-circular (effective) aperture actually measured?

An f-number is determined as the ratio of the focal length to the diameter of the entrance pupil. But the actual aperture diaphragm is usually not a perfect circle. Its shape is determined by the number of diaphragm blades and their edge (curved or straight). Depending on those the aperture might be a pentagon, hexagon, octagon… and with straight or curved edges. So how is the diameter actually determined? Is it for example twice the distance between the center and the widest point at an intersection of the blades? Twice the distance between the center and the nearest point on an edge?

## How to find the vole of a cube remaining after drilling a cylinder with a diameter larger than the cube’s side length?

So, for example, say there’s a cube with side length L, and you drill a cylinder with a diameter larger than L through the cube, what volume remains?