1
$\begingroup$

Why aren't turns used more often in math? It seems much more fundamental of a unit compared to degrees and radians. The value of a turn is definite, and other units can be derived from it. The problem with radians is we'll never know exactly what a radian is, unless you're putting it in terms of pi or tau. And degrees are quite arbitrary (360? Where did we get that from?) Turns could be put in terms of transcendental numbers, using fractions. This doesn't need to be done as often, but it's not too inconvenient to do.

360 degrees = 1 turn

2π radians or 𝝉 radians = 1 turn

1/400 gradians = 1 turn

I just think it's more fundamental. The Planck unit of arc, if you will. Probably just my opinion.

My question is simply why radians are used far more than turns as a unit.

$\endgroup$
  • 5
    $\begingroup$ Tradition/history/inertia is a major reason your proposal will go nowhere. But for more mathematical reasons, look at the derivatives for the sine and cosine functions in radians: $\sin'(x) = \cos x$ and $\cos'(x) = -\sin x$. If you use other angle units then you don't get this property. Let t be the notation for turns, so $1 \, {\rm t} = 2\pi$ radians and thus the function $f(x) = \sin(x \, {\rm t})$ is $\sin(2\pi x)$ and $g(x) = \cos(x \, {\rm t})$ is $\cos(2\pi x)$. Thus $f'(x) = 2\pi g(x)$ and $g'(x) = -2\pi f(x)$: ugly scaling factors! The power series for $\sin x$ and $\cos x$ (contd.) $\endgroup$ – KCd Nov 11 '17 at 1:30
  • 3
    $\begingroup$ have coefficients that are all rational, while the power series of $f(x)$ and $g(x)$ will have coefficients with lots of factors of $\pi$ floating around. This is the reason why anyone who uses calculus writes trig functions in terms of radians. (The sine and cosine functions in radians are a basis of the clean ODE $y'' + y = 0$. What 2nd-order ODE do $f$ and $g$ fit?) Issues that arise when using turns or degrees with trig functions would arise if we used exponential functions with a base other than $e$, and accounts for why anyone who uses calculus writes exponentials with base $e$. $\endgroup$ – KCd Nov 11 '17 at 1:31
  • 1
    $\begingroup$ Radian measure is defined by the ratio of two lengths (arc & radius). This seems quite natural geometrically. And as KCd points out, it leads to simple equations and formulas related to the trigonometric functions. $\endgroup$ – Michael E2 Nov 11 '17 at 2:12
  • $\begingroup$ The word more often used in the literature is 'revolutions', not 'turns'. Revolutions have sometimes been used as a unit in astronomical computing (e.g. for mean elements in 'Improved Lunar Ephemeris', US Govt Printing Office 1954.) The convenience in that application included simpler accurate numerical range-reduction of angular measures to the range 0-360 degrees. For the development of symbolic calculations, the previous comments of course provide the ample reasons why revolutions are not used. $\endgroup$ – terry-s Nov 11 '17 at 10:51
  • 1
    $\begingroup$ From a certain perspective a "turn" is not a unit for angles: an angle can be thought of as an element of the group $S^1$, and a turn is the same thing as $0$, hence not a generator (unit) of that group. $\endgroup$ – Michael Bächtold Nov 12 '17 at 21:57
3
$\begingroup$

Turns are used in mathematics when convenient. For example, some authors write Fourier transform as $$\int e^{-2\pi i xz}f(x)dx,$$ while other authors as $$\int e^{-ixz}f(x)dz.$$ In the first formula $x$ is in "turns" (1 turn=$2\pi$ radians). This is a question of convenience. Using turns, the Taylor expansion look more complicated: $$e^{2\pi x}=\sum_{n=1}^\infty \frac{(2\pi x)^n}{n!}.$$ In many mathematical formulas $\pi$ is unavoidable. So people choose units to minimize the number of occurrences of $\pi$ in the formulas.

Same in physics. There is angular frequency $\omega$ (radians per second) and there is frequency $\omega/(2\pi)$ (oscillations/turns per second).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.