Wikipedia:Reference desk/Archives/Mathematics/2010 August 18

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August 18

Optics: 35mm focal length equivalent

Hobbyist filmer here. I'd like to make mind-blowing ultra wide-angle images like Terry Gilliam, but in the more economical format of Super 8 mm film. What I do know is that I'm not even remotely looking like Gilliam with any focal length above 18mm...but that's my desired upper length in 35mm only. The focal length to achieve a particular angle of view (which is the thing that makes for the mind-blowing images with wide-angle images) is different with any format and sensor size you use, hence there's articles such as 35 mm equivalent focal length and crop factor. In other words, if you change the format (i. e. size of your film or sensor) but wanna have the same angle of view, you need a different focal length.

Now, I have a chance of acquiring a lens (for a Super8 camera made by the Austrian Eumig brand) which is labeled as ultra wide-angle, according to trade press this lens is guaranteed to be entirely rectilinear (no barrel aka fish-eye distortion, as I don't want this), and its focal length in Super8 is 4mm.

So what I'd like to know is, what's the 35mm equivalent of these 4mm in Super8, according to crop factor? Or in other words: If my desired upper limit is an 18mm focal length in 35mm, what equivalent focal length would that be in Super8?

I guess what might help are the dimensions of the Super8 frame area: 5.97mm horizontal x 4.01mm vertical, compared to 22mm horizontal x 16mm vertical in 35mm.

My second choice would be a 3CCD miniDV with a 1" chip size. What's the equivalent to 18mm there? --79.193.41.61 (talk) 06:50, 18 August 2010 (UTC)

I don't think this is the right forum for the question, unless there is a mathie who happens to be photography hobbyist as well. You might try the science desk.--RDBury (talk) 21:11, 18 August 2010 (UTC)

I've responded at science. Thegreenj 22:42, 18 August 2010 (UTC)

Asymptotic order

I have always been absolutely terrible at this type of problem. Any hints for making it easier to solve would be appreciated. The basic problem is, given two functions, which is asymptotically greater than the other? In other words, given function f and g, to be asymptotically greater than g, it must be proven that f is O(g). Here is an example and how I solve it (which is surely the most difficult way to solve it):

Given $f=2^{\sqrt {\lg {n}}}$ and $g=n(\lg {n})^{3}$.

I solve this by estimating a value for ${2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}$. I will show my steps because I am sure I do not remember the log rules correctly:

${{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}} \over {\lg {(n(\lg {n})^{3})}}}={{\sqrt {\lg {n}}} \over {\lg {n}+\lg {((\lg {n})^{3})}}}$

Now, if I look at this as ${\sqrt {x}} \over {x+lg(x^{3})}$, it is obvious that result will be less than 1. Therefore, I claim that g = O(f) and g is asymptotically greater than f. Correct? -- kainaw™ 20:48, 18 August 2010 (UTC)

Looks OK to me.--RDBury (talk) 21:05, 18 August 2010 (UTC)

Be careful,

${\frac {x}{y}}$ and ${\frac {\lg {x}}{\lg {y}}}$ are not always equal. What you have proven is that $\,\!\lg {f}=O(\lg {g})$. This in general does not imply that $\,\!f=O(g)$ (consider for example, $\,\!f=x$ and $\,\!g=x^{1/2}$).

In any case, your problem is a bit clearer if you substitute $\,\!x=\lg {n}$. Invrnc (talk) 03:45, 19 August 2010 (UTC)

If by "asymptotically greater" you mean "greater for every value starting at some point", then this is not what big O notation is. $f=O(g)$ means that f is less than (not greater than, as you wrote) g up to a constant. In your example, the (absolute value of the) ratio between f and g is bounded (it doesn't have to be less than 1), so $f=O(g)$. This is equivalent to showing that its logarithm is bounded from above. The correct log rule is $\lg \left({\frac {x}{y}}\right)=\lg x-\lg y$, so you just need to show that $\lg {{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}}-{\lg {(n(\lg {n})^{3})}}}<M$ eventually (this follows from the fact that it goes to $-\infty$, which is easy to see). -- Meni Rosenfeld (talk) 07:27, 19 August 2010 (UTC)

You are correct. I stated it backwards. I do have a related question... Given that ${\frac {x}{y}}$ and ${\frac {\lg {x}}{\lg {y}}}$ are not always equal, is it valid to use that reduction here? If x is asymptotically greater than y, won't lg x be asymptotically greater than lg y? -- kainaw™ 15:43, 19 August 2010 (UTC)

Yes. Assuming everything is positive, ${\frac {x}{y}}<1\iff x<y\iff \lg x<\lg y\iff {\frac {\lg x}{\lg y}}<1$. However, as Invrnc explains, this will not work if you are talking about big O. -- Meni Rosenfeld (talk) 18:08, 19 August 2010 (UTC)

I believe that I found an example where it doesn't work. Proving that lg(f/g) is bounded from above did work. Thanks. -- kainaw™ 19:06, 19 August 2010 (UTC)

Space Curves

Consider a smooth space curve, parametrised by arc length. The Frenet frame {T,N,B} defines a rigid body at each point of the curve. This rigid body is the cube with sides T, N and B. The infinitesimal axis of symmetry of this rigid body is generated by τT + κB where κ is the curvature, and τ is the torsion, of the space curve. This means that infinitesimally the cube is rotating about the line spanned by τT + κB. Can someone show me how to calculate the infinitesimal angular frequency of the cube about this infinitesimal axis of symmetry? — Fly by Night (talk) 22:55, 18 August 2010 (UTC)

I'm not sure if the question is well posed as infinitesimally the box will be moving along the line as well as spinning. You also need to specify how fast the box is moving along the curve, you could say its a unit speed curve.

Those caverts taken into account you can consider rotation given by

${\frac {d\mathbf {v} }{dt}}={\begin{pmatrix}0&\kappa &0\\-\kappa &0&\tau \\0&-\tau &0\end{pmatrix}}\mathbf {v}$

The axis of this rotation is $(\tau ,0,-\kappa )\,$. To find the angular frequency take a vector perpendicular to the axis, say $(0,1,0)$ its tangental velocity is just $(\kappa ,0,-\tau )\,$, so the angular frequency is just $\omega =|(\kappa ,0,-\tau )|={\sqrt {\kappa ^{2}+\tau ^{2}}}$. --Salix (talk): 09:30, 19 August 2010 (UTC)

Thanks Salix. I think I mentioned that the curve was parametrised by arc length. Wouldn't that make it unit speed? You said that the axis of rotation is (τ,0,–κ) = τT – κB. Isn't there a sign error? I read that the axis of rotation is τT + κB. If you take a vector v = aT + bN + cB and solve dv/ds = 0 then you get, assuming v ≠ 0, that a = τ, b = 0 and c = κ; meaning that v = τT + κB. I've also seen it calculated by working out a vector perpendicular to dT/ds, dN/ds and dB/ds, i.e. perpendicular to κN, τB – κT and –τN. If κ and τ aren't both zero then you get τT + κB. Also, what did you mean by the "tangential velocity" of (0,1,0)? Do you mean the tangential velocity of N? And if so, what is it's tangential velocity? (It seems to be –dN/ds from your answer). I would be very pleased if you could fill in the details and answer some of the whys as well as the hows. Thanks in advance. — Fly by Night (talk) 12:35, 19 August 2010 (UTC)

Looks like brain not fully working this morning. Yes being parametrised by arc length does make it unit speed. Yes it is a sign error axis should be (τ,0,κ) = τT + κB. Tangential velocity comes from the angular frequency article, I'm meaning work out component of the velocity of the point (0,1,0) = N which is tangental a circle lying in a plane normal to the axis. The velocity is dN/ds = (–κ,0,τ) (another sign error on my part). Note that this has no component in the N direction and no component along the axis, so it is already the tangental velocity. The radius is 1 so the angular frequency is simply |dN/ds|. --Salix (talk): 18:57, 19 August 2010 (UTC)

Thanks again Salix! I appreciate your help. Let me check that I understand what I need to do. If the axis of symmetry is spanned by the vector v then I need to choose a unit vector, say w, at a right angle to v and then work out || dw / ds || to give the angular frequency? (Like you said, this all depends on the parametrisation of the curve, but once it's unit speed we can relax.) — Fly by Night (talk) 19:44, 19 August 2010 (UTC)

Yep. You may find the Angular velocity article more helpful as it has a few more details than angular frequency.--Salix (talk): 20:32, 19 August 2010 (UTC)