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March 25

relations between symmetries

I've noticed that in 2 dimensions the order of rotational symmetry for a shape is always greater than or equal to the order of reflectional symmetry. Firstly is this true, and secondly is this applicable to higher dimensions? Thanks. — Preceding unsigned comment added by 86.178.248.67 (talk) 13:42, 25 March 2012 (UTC)[reply]

An isosceles triangle has reflectional symmetry but not rotational. 84.197.178.75 (talk) 14:29, 25 March 2012 (UTC)[reply]

If a shape ~~has finite symmetries and~~ has any reflection symmetry, then the number of rotations and reflections must be equal if you count the identity (not moving the shape at all) as a rotation. So in the example given by 84.197.178.75, you have 1 reflection, and 1 rotation which is the identity but maybe you don't want to count that. The rotations form a normal subgroup of the symmetry group with index 2, so half the elements of the symmetry group correspond to orientation preserving moves and half to orientation reversing moves. This is true in any dimension. The alternative is the case with no reflection symmetry, where there's obviously more rotations than reflections. Rckrone (talk) 19:17, 25 March 2012 (UTC)[reply]

But why do you count not moving the shape at all as a rotation, but not a reflection ? StuRat (talk) 20:33, 25 March 2012 (UTC)[reply]

Because it's orientation preserving, rather than orientation reversing. Rckrone (talk) 20:45, 25 March 2012 (UTC)[reply]

It's convention mainly. See rotational symmetry. Everything in the world has rotational symmetry >= 1 - Cucumber Mike (talk) 20:48, 25 March 2012 (UTC)[reply]

Coordinate transformation

Consider the coordinate transformation $x_{1}=ar\cos \theta ,x_{2}=br\sin \theta$ where $x_{1}$ and $x_{2}$ are rectangular Cartesian coordinates. Calculate the scale factors $h_{r}$ and $h_{\theta }$ along the $r$ and $\theta$ coordinate curves. What is the general element of length $dl$ ? Is the transformation orthogonal? Widener (talk) 13:44, 25 March 2012 (UTC)[reply]

Where are you getting stuck? — Quondum ^☏✎ 15:32, 25 March 2012 (UTC)[reply]

How do you calculate the scale factors? Widener (talk) 15:54, 25 March 2012 (UTC)[reply]

I am guessing what is meant by the term scale factor; previous work in the exercise will probably have defined the symbols. At every point (i.e. at (r,θ) or equivalently at (x₁,x₂), there will be some transformation matrix of an infinitesimal distance vector dl expressed in terms of dr, dθ to get its expression in terms of dx₁, dx₂). Because this is somewhat complicated by the fact that there are two multipliers for each, I'm guessing that the scaling factor may relate to an initial transformation into (non-rotated) coordinates (α,β)=(f(r),g(θ)) for which the length of dl is √dα²+dβ². — Quondum ^☏✎ 16:45, 25 March 2012 (UTC)[reply]

My guess is that the term scale factor has the same meaning as it is used in Elliptic coordinate system, for example. The must be some general algorithm for calculating them. Widener (talk) 17:53, 25 March 2012 (UTC)[reply]

This is what I have got: Define

v=(ar\cos \theta ,br\sin \theta )

${\frac {\partial v}{\partial r}}=(a\cos \theta ,b\sin \theta )$
${\frac {\partial v}{\partial \theta }}=(-ar\sin \theta ,br\cos \theta )$
${\frac {\partial v}{\partial \theta }}\cdot {\frac {\partial v}{\partial r}}=-ar\sin \theta a\cos \theta +br\cos \theta b\sin \theta =(b^{2}-a^{2})\sin \theta \cos \theta$ . The transformation is orthogonal therefore if and only if $b=\pm a$
. Scale factors: $h_{r}=\left|{\frac {\partial v}{\partial r}}\right|={\sqrt {a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta }}$
$h_{\theta }=\left|{\frac {\partial v}{\partial \theta }}\right|={\sqrt {a^{2}r^{2}\sin ^{2}\theta +b^{2}r^{2}\cos ^{2}\theta }}$
General element of length $dl={\sqrt {\left|{\frac {\partial v}{\partial r}}\right|^{2}+\left|{\frac {\partial v}{\partial \theta }}\right|^{2}}}={\sqrt {(a^{2}+b^{2}r^{2})\cos ^{2}\theta +(a^{2}r^{2}+b^{2})\sin ^{2}\theta }}$
. Does this look right? Widener (talk) 18:37, 25 March 2012 (UTC)[reply]

I think normally they start from

dA=dx_{1}^{2}+dx_{2}^{2}with:dx_{1}=a\cos \theta dr-ar\sin \theta d\theta and:dx_{2}=b\sin \theta dr+br\cos \theta d\theta

; with a=b, you get your result, the scale factors are ar and 1 and other terms cancel each other; not sure if you're supposed to do it starting with dx*dy; also Orthogonal matrix is unitary, det = 1 or -1 84.197.178.75 (talk) 20:00, 25 March 2012 (UTC)[reply]

So can you guide me exactly through what you have done? Widener (talk) 20:32, 25 March 2012 (UTC)[reply]

Sorry, didn't do the whole thing, I knew the result had to be that with a=b... 84.197.178.75 (talk) 16:45, 26 March 2012 (UTC)[reply]

So is this the answer simply:

dl={\sqrt {(a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta )dr^{2}+2r(b^{2}-a^{2})\cos \theta \sin \theta drd\theta +r^{2}(a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta )d\theta ^{2}}}

? Widener (talk) 21:34, 25 March 2012 (UTC)[reply]

It seems from Orthogonal coordinates that the scale factors are only defined when the coordinates are orthogonal (when the off-diagonal metric components are zero). So when they are not orthogonal, one cannot really use the concept. Your answer for the general element $dl$ is correct. Going further, one can see the multiplier $2r(b^{2}-a^{2})\cos \theta \sin \theta$ of the cross term, which must be zero, i.e. $a^{2}=b^{2}$ , for the transform and (in the case of (x₁,x₂) being orthogonal) for the (r, θ) coordinates to be orthogonal. When this is the case, one can then calculate the required scale factors as the square roots of the other two multipliers as done by 84.197.178.75, strictly 1 and |ar| (=|br|). — Quondum ^☏✎ 13:46, 26 March 2012 (UTC)[reply]

There's a bit of confusion about orthogonal and orthonormal; orthogonal transformation "is a transformation which preserves a symmetric inner product. In particular, an orthogonal transformation (technically, an orthonormal transformation) preserves lengths of vectors and angles between vectors" (from Wolfram Mathworld). So by that definition it's only when a and b are 1 or -1. 84.197.178.75 (talk) 16:45, 26 March 2012 (UTC)[reply]

Oops, quite right. Thanks for pointing that out. I sometimes misuse terms like that when the same word gets used in quick succession but has different meanings due to context. :) — Quondum ^☏✎ 17:25, 26 March 2012 (UTC)[reply]

(Except, even then it surely isn't globally an orthogonal transformation, because the multipliers of dr² and dθ² must be 1, and we'd have 1 and r²?) — Quondum ^☏✎ 17:31, 26 March 2012 (UTC)[reply]

The online articles shouldn't confuse the meaning of orthogonal and orthonormal too much: orthogonal essentially means "doesn't change angles" and orthonormal essentially means "doesn't change angles or distances". Angles and distances are determined by whatever metric is in play, so any metric preserving transformation automatically preserves both distances and angles. Rschwieb (talk) 18:23, 26 March 2012 (UTC)[reply]

Orthogonal transformation seems to be stricter than that, namely:

Q^{\mathrm {T} }=Q^{-1}

. Any full-rank linear transformation would be orthogonal under your argument, since the metric would always transform so as to preserve angles and lengths. However, we wouldn't refer to the transformation e.g. given by the matrix

Q={\begin{pmatrix}1&0\\0&r^{2}\end{pmatrix}}

as being orthogonal. — Quondum ^☏✎ 09:26, 27 March 2012 (UTC)[reply]

As 84.197.178.75 pointed out before, the "orthogonal transformation" article really refers to orthonormal transformations. I was speaking in general about the adjectives. The transformation which doubles vector length would be a transformation which preserves angles but not lengths. You need to reconsider your claim about full-rank transformations. In R², let F be the transformation sending [1,0] to [1,0], and [0,1] to [1,1]. F is clearly full rank, but the angles between images and preimages of these two vectors are clearly different. In general, full rank transformations "skew" angles between vectors.

Rereading the example you gave makes me want to emphasize that I made no claims about orthogonality of columns of the matrix. The condition for orthogonality I have in mind is: "a.b=0 implies Fa.Fb=0 for all a, b." After scribbling a bit it seems clear this is the same as preserving angles. The stronger condition "a.b=C implies Fa.Fb=C for all a,b,C" is of course preservation of the inner product, and implies the weaker condition I mentioned. Rschwieb (talk) 14:36, 27 March 2012 (UTC)[reply]

(This is going off-topic. I accept what you are saying, except that I will stick with my statement about nonsingular transformations of the metric: the transformed metric (aka inner product) will compensate for any skew etc. in the transformed vectors, better considerd as the same vectors expressed in terms of a new basis.) I was trying to address the original question "Is the transformation orthogonal?", by which I assume the standard meaning of orthogonal transformation (i.e. orthonormal) is meant. Of course your statement applies if only angle-preserving is meant, in which case it is orthogonal when a²=b². — Quondum ^☏✎ 17:17, 27 March 2012 (UTC)[reply]

OK, all straightened out then. It's unfortunate that "orthogonal transformation" has its current usage, but things could be worse. Rschwieb (talk) 17:55, 27 March 2012 (UTC)[reply]

Free software for plotting data

Does any free application exist for Mac OS X that I can use to plot ordered pairs? It should be graphical and available as binaries. Also, it should be able to connect the points and have different sets of points so I can compare the plots. --Melab±1 ☎ 20:31, 25 March 2012 (UTC)[reply]

gnuplot is very powerful if you don't mind a little "programming". Works great in OS X. Staecker (talk) 13:48, 26 March 2012 (UTC)[reply]

Best angle for launching a projectile for the longest distance

Assuming that you have perfectly flat terrain, no wind, a projectile of regular shape, etc., what angle of launch would cause said projectile to fly for the longest distance? I always assumed that it would be 45º, but Point-blank range says that it's actually somewhat less. I can't find a specific number in the citation for this sentence, Unknown (1848). Instruction Upon the Art of Pointing Cannon. J. and G.S. Gideon.. Nyttend (talk) 21:54, 25 March 2012 (UTC)[reply]

I think 45 degrees is correct, if you add the additional constraint of no air resistance. The presence of air means you want to shorten the amount of air through which the projectile travels, so less than 45 degrees is then ideal. StuRat (talk) 22:19, 25 March 2012 (UTC)[reply]

You also need to assume the projectile is launched from ground level. If you're firing a pistol from several feet off the ground, then you would again want slightly less than 45 degrees.--121.74.125.218 (talk) 06:01, 26 March 2012 (UTC)[reply]

I can understand the first one, but why does the elevation of the launch location matter? Nyttend (talk) 15:27, 26 March 2012 (UTC)[reply]

Well if you're on top of a high cliff the best bet is to put it out near horizontally for instance. If you put iut at 45 degrees that will start a bit further out, but all the rest of the way down it only goes at one over root 2 of the speed horizontally. Dmcq (talk) 17:00, 26 March 2012 (UTC)[reply]

Is that also because of air resistance? Aside from that, I can't understand how anything would reduce total distance as you get closer to the 45° angle, as long as you don't launch so hard that you achieve escape velocity. Nyttend (talk) 22:08, 26 March 2012 (UTC)[reply]

No, it's true in a vacuum. I don't have a good intuition to give you, other than that's how the calculations work out.--121.74.125.218 (talk) 05:54, 27 March 2012 (UTC)[reply]

I take it back; here's a way to think about it. When you increase the angle, you're exchanging horizontal velocity for vertical velocity. When you fire from the top of a cliff, the trajectory can be divided in to two parts: the part where the projectile is above the starting height, and the part where it is below. If your cliff is high enough, the distance traveled in the first part is insignificant compared to the total distance. A shallower angle will give you a higher horizontal velocity during the second part, so the projectile will travel farther in total.--121.74.125.218 (talk) 06:02, 27 March 2012 (UTC)[reply]

There's a rule for the longest distance with air resistance: the sum of the launch angle and impact angle should be 90°. Don't know how to prove this... 84.197.178.75 (talk) 17:16, 26 March 2012 (UTC)[reply]

This works when there is no air resistance (but the initial height is not at ground level), as can be proven with a straightforward (if cumbersome) calculation. But based on my simulations I don't think it applies when there is air resistance, not exactly anyway. -- Meni Rosenfeld (talk) 12:21, 27 March 2012 (UTC)[reply]