Re: Compound miter brainteaser

somehow after atepting to try and understand

i declare myself (STUPID)
"Josh" <jcaron2@xxxxxxxxx> wrote in message
news:1137431541.505102.131100@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>I agree with your algorithm, DJ. Several such algorithms can be found
> on the web, often for computing compound miters for crown molding,
> which is essentially the same problem. But many of them give slightly
> different answers. Why is yours right, and what's wrong with the other
> ones?
>
> If you don't want to see a bunch of crazy math, stop reading now.
>
> One of the other common algoriths for computing the miter (x in your
> notation) is x=1/2*arccos(cos^2(b)*cos(a)+sin^2(b)). This is a pretty
> simple equation to derive using simple vector algebra. Going back to
> first semester Calculus, recall that the dot product of two vectors is
> defined as a scalar value equal to the product of the vector magnitudes
> times the cosine of the angle between them.
>
> A.dot.B = |A| * |B| * cos(alpha),
>
> where alpha is the angle formed between them, and the | | notation
> means magnitude (i.e. length, independent of direction).
>
> A second way to calculate the dot product is to write the vectors as
> functions of the unit vectors i, j, and k which are simply vectors of
> length 1 along the x, y, and z axes. If the vectors are written as A =
> ax*i + ay*j + az*k and B = bx*i + by*j + bz*k then their dot product is
> simply
>
> A.dot.B = ax*bx + ay*by + az*bz
>
> Now if we simply find two vectors which form one wedge of the ten-sided
> roof, we can easily compute the angle between them by using the two
> definitions of dot product. It's an easy construction:
>
> If we take the peak of the roof to be the point (0,0,0), and we imagine
> ten rafters radiating outward, angled down with a pitch (slope) of
> 5/12, then it's easy to find their endpoints (which will define our
> vectors). If we assume for simplicity's sake that they have a length
> of 1 foot, then one of the rafters would stretch from (0,0,0) to
> (cos(atan(5/12)),0,-sin(atan(5/12))). Since the choice of 5/12 for a
> pitch gives us a 5-12-13 right triangle, we can simpify the second
> coordinate of the vector to (12/13,0,-5/13). A second rafter would
> start at (0,0,0) and go to (12/13*cos(36), 12/13*sin(36), -5/13). The
> 36 degree angle is the angle of one wedge of roof when viewed from
> directly above (i.e. there are ten sides so the the angle is a tenth of
> 360).
>
> Now that we have two vectors, we can compute their dot product both of
> the ways desribed above.
>
> A.dot.B = |A| * |B| * cos (angle) = 1 * 1 * cos (alpha) = cos(alpha)
> A.dot.B = 12/13*12/13*cos(36) + 0*12/13*sin(36) + 5/13*5/13
>
> Equating the two different definitions we get ((12/13)^2*cos(36) +
> (5/13)^2)= cos (alpha)
>
> Thus alpha = arccos(0.83727) = 33.147 degrees.
>
> This is the angle between the two vectors (i.e. the two roof rafters).
> The miter angle is simply going to be half of this angle, or
>
> x = 16.57 degrees.
>
> This is pretty close to what DJ's formula gives us, yet it's slightly
> different. Why?
>
> I'll post the reason next. I don't want this one post to get too long.
>
> Josh
>


.

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