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First of all - I am sorry if it is the wrong forum or if this is a very trivial question. I am not a mathematician nor a trigonometry genius - and therefor I would ask a simple answer that someone like me could understand (and not just fancy formulas if possible)

Giving a circle with a known radius (r) and another circle with an offset of (t) I would like to fill the "gap" (t) with non-overlapping triangles with the closest possible angle to 45 degrees..

  • 1 - How can I know how many triangles will enter the space ?
  • 2 - How can I calculate their exact angles (a) (b) ?
  • 3 - assuming I want an EXACT angle of (b) = 90 (and not approximation) - how can I know the number of triangles and also calculate the "left-over" ??

enter image description here

UPDATE I: as per comment : a visual example of wanted result. enter image description here

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You seem to have drawn the triangles with all their corners on the two circles, such that the ones with their long edge on the inner circle (e.g. the yellow edge) extend beyond the gap (the mathematical term for which is "annulus" by the way) whereas the ones with their long edge on the outer circle are entirely inside the gap -- is that what you want? Also, angle $a$ seems to be approximately $360-45=315$ degrees, not $90$ -- isn't it $b$ that you want to be exactly $90$ degrees? –  joriki Oct 19 '12 at 13:06
    
yes, it is b that I want 90 (sorry) .. and about the yellow line - what I need is actually to fill the gap. but because it is a circle, I do not know how (if ever) it is possible to calculate that angle. hence I made the yellow line.. –  Obmerk Kronen Oct 19 '12 at 13:10
    
I don't understand how you're using "fill". Are you saying that you want the triangle with the yellow edge to be entirely inside the gap but you don't know how to do that, or do you just want to cover the gap and don't mind if parts of the triangles extend beyond the gap? If by "fill" you mean that you want to cover the entire gap with triangles that don't extend beyond the gap, then that's impossible since the gap is curved and the triangles aren't. (Or perhaps you're thinking of generalized triangles that are allowed to have curved edges?) –  joriki Oct 19 '12 at 13:12
    
@joriki - please see my update. as for your other comment about that being impossible - what I am doing is a real-world project in which I do not need the "base" of the triangles to be precise (it is done manually) - but I do need the angles in order to execute that project. this is the reason for the straight yellow line . I know that without it it would not be a triangle . but for me, it is also ok to know the calculation as if it were a triangle (even if it is going outside the bounds of the circle. so your definition of "generalized triangles that are allowed curved edges" is on the spot –  Obmerk Kronen Oct 19 '12 at 13:21
    
Please clarify the question itself; people shouldn't have to read through the comments to find that it's actually $b$ and not $a$ that's supposed to be $90$ degrees. –  joriki Oct 19 '12 at 13:21

1 Answer 1

On this figure :

triangulation of annulus

What you want is to have $c$ = $r_2 - d$. This leads you to the equation $\frac{r_2}{r_1} = \cos{\theta} + \sin{\theta}$. We assume here that $\theta > 0$.

Use the trigonometry formulas to convert the cosine and sine as functions of $t = \tan{\frac{\theta}{2}}$, and this gives you t as the root of a degree $2$ polynomial : $$ \begin{aligned} \frac{1+2t-t^2}{1+t^2} &= \frac{r_2}{r_1} \\ - \left(1+\frac{r_2}{r_1}\right)t^2 +2t +\left(1-\frac{r_2}{r_1}\right) &= 0\\ \end{aligned} $$

Solving this you get $$ \begin{aligned} \Delta &= 4\left(2-\frac{r_2^2}{r_1^2}\right)\\ t &= \frac{ 1\pm \sqrt{2-\frac{r_2^2}{r_1^2}} }{ \left(1+\frac{r_2}{r_1}\right) } \end {aligned} $$ and we assumed $\theta$ positive.

Using $\arctan$, and rounding to find the best integer $k$ (since the number of samples is integral) such that $\theta = \frac{\pi}{k}$ should give you the angle $\theta$ to use as half of the sampling frequency for both circles. Then both circles should be sampled using twice this angle frequency, but shifted with respect to one another with an angle $\theta$.

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ok - thanks for the answer - but I lost you somewhere around (K) . like I said before , I am not a mathematician. can you maybe give a series of calculations that a simple person like me , ignorant in trigonometry can follow ? also, like in OQ - is there a simple wy to know how many of these can enter ?? –  Obmerk Kronen Oct 19 '12 at 17:09
    
I edited my answer, I hope this is better. –  Vincent Nivoliers Oct 20 '12 at 9:43
    
Since the base edge rotates a bit when moving from one triangle to the next, the angle $b$ will be (slightly) exceeding $90$ degrees in this set-up; in situation 3 therefore, $k$ will increase. Note furthermore that in general the triangles won't fit perfectly (i.e. the last one won't match up with the first), so that an exact angle of $90$ degrees won't always be possible. –  Lord_Farin Oct 20 '12 at 13:30

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