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Forgive me if this question is a little too strange or maybe even off. Mathematics has never been my strong point, but I definitely think it's the coolest...

Anyway, I was looking into tau, pi's up-and-coming sibling. I started rethinking why pi worked. The thing about tau is that it supposedly skips the step of doubling πr, since tau is twice pi. I tried this, and it works! (Of course it works. The fact that this surprises and amazes me shows how little I get out...)

Then I started wondering why we do 2πr instead of πd. It does give the same answer... I checked. Is there any reason why using the radius is preferred over using the diameter?

Here's my work:

r = 6 (ergo diameter must equal 12)

C = 2πr

C = 37.68

and...

C = Tr

C = 37.68

and...

C = πd

C = 37.68

General Nuisance
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    Only for historical reasons. If you know $\pi$ then you know $\tau$. If you know radius, then you know diameter. Read more at this related question – JMoravitz Jul 21 '16 at 01:54
  • The area of a circle is $A=\pi r^2=\frac{\tau}{2} r^2$ you don't get to skip squaring the radius. – n1000 Jul 21 '16 at 01:55
  • It is quite conventional to define $\pi$ as $\dfrac{\text{circumference}}{\text{diameter}}$. $\qquad$ – Michael Hardy Jul 21 '16 at 02:07
  • @n1000 I'm sure you realize we're talking about circumference instead of area (happens to the best of us...). And also, since tau is just twice pi, if you were to cut it in half the calculation would be the same and squaring would be required, but if you leave tau be it, in my extremely limited experience (I literally only tested it on the sample problem above in like ten minutes late last night...) it eliminates the need to square. And that makes sense, although I would be at a loss to explain it. – General Nuisance Jul 21 '16 at 02:20
  • @MichaelHardy Right, so why would we double r instead of just using the diameter? – General Nuisance Jul 21 '16 at 02:23
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    @GeneralNuisance There are a lot of formulas that are easier to work with in terms of $r$. Even in calculus there is a differential $dr$ when talking about polar coordinates. That is the same $r$ as when talking about concentric circles with the Origin as center and some point $P$ on it. Surface area formulas of revolutions (about x or y-axis) depend on a (variable) $r$ and not on some $d$. And so there are many reasons – imranfat Jul 21 '16 at 02:32
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    @imranfat I think this could be incorporated into a very good answer. – General Nuisance Jul 21 '16 at 02:45
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    "skips the step of squaring $\pi r$, since $\tau$ is twice $\pi$": Do you mean doubling instead of squaring? I think this may be why n1000 mentioned the area formula, since this involves squaring (regardless of whether $\pi$ or $\tau$ or $r$ or $d$ is used). – stewbasic Jul 21 '16 at 02:57

2 Answers2

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You can define a circle knowing the centre and the radius (distance $r$).   A circle is the set of all points, on a 2D-plane, at distance $r$ from the centre.

That's a concise and elegant definition; try doing so using the diameter (distance $d$).

Then, having defined circles using the radius, it becomes convenient to also define the radian measure of angles in terms of the radius of a circle.   One radian is the measure of an angle subtended at the centre of a circle by an arc length equal to the radius.

Then we asked: what is the radian measure of a straight angle (that formed by two rays of a line)?   Why it is that irrational number we have decided to call $\pi$, to honour Pythagorus.

What then is the angle subtended by the circumference of a circle?   Well, we could call it $\tau$, but it is $2\pi$, and we just happened to have named $\pi$ first.

Thus $$\begin{array}{cc}C&=&2\pi r &=& \pi d &=& \tfrac 12 \tau d &=& \tau r & \text{circumference of circle}\\[1ex]A&=& \pi r^2 &=& \tfrac 14 \pi d^2 &=& \tfrac 18 \tau d^2 &=& \tfrac 12\tau r^2 & \text{area of circle/disc} \\[2ex] S &=& 4\pi r^2 &=& \pi d^2 &=&\tfrac 12 \tau d^2 &=&2\tau r^2 & \text{surface area of sphere}\\[1ex] V &=& \tfrac 43 \pi r^3 &=& \tfrac 16 \pi d^3 &=&\tfrac 1{12}\tau d^3 &=& \tfrac 23 \tau r^3 & \text{volume of sphere/ball}\end{array}$$

Graham Kemp
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There is actually a good reason for writing the circumference as $\pi d$ rather than $2\pi r$: the first formula actually holds for the perimeter of any closed curve of constant diameter $d$, not just for circles. An example of such a curve is the Reuleaux triangle.

Especially Lime
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