Re: high school trigonometry?

In article <F-WdnasL3-yD4BvYRVnyjwA@xxxxxx>,
"Nick" <tulse04-news1@xxxxxxxxxxx> wrote:

"ram" <ram@xxxxxxxxxxxxxxxx> wrote in message
news:PJdhh.198$4Q6.32@xxxxxxxxxxx
hi!

I have been working on some formulas for compound miter cuts in carpentry.

The formulas and a diagram are presented at

http://www.woodcentral.com/bparticles/miter_formula.shtml

The assertion that the formulas could be derivied with high school math
was a challenge to me. I was able to derive a formula for the miter or
crosscut angle, but the bevel or blade tilt angle is eluding me.

The author actually says that his father who derived the formulae is a math
whiz. Being able to derive them from high school math principles hardly
means that it is simple.


I used unit vectors and 3D rotation matrices and the dot product operation
to calculate the angle between two planes starting perpendicular, then
both tilted back by the slope and finally one rotated to the angular bend
(as an excess to 90 degrees). This turns out to be twice the blade tilt as
one might expect, and the formula I got is

planesAngle = ArcCos[ Cos[Theta]^2 Sin[Phi] + Sin[Theta]^2 ]

Here my Theta is the supplement of the website's B or slope, and my Phi is
the supplement to the website's A or flat miter.

But the website formula is much nicer leading me to believe that something
could be done to my formulation to simplify it.

bladeTilt = ArcSin[Sin[B] Sin[A/2]]

I have looked at all the standard trig identities and fooled around a bit,
I can only make things worse. I would appreciate any help you can give on
this; I suppose I am admitting to giving up at this juncture.

See this Miter Saw Calculator http://www.csgnetwork.com/sawmitercalc.html

and http://ca.geocities.com/xpf51/SQUARE_CUT_FASCIA/crown_molding.html which
contains some trig.

The canonical way to approach this is by drawing a
spherical triangle, and identifying the angles of
interest. The spherical triangle can be solved using
vector calculus or the formulas of spherical
trigonometry. Take the center of the sphere at the
intersection of the base plane and the inside edge of
the miter. The ray perpendicular from the base plane at
O intersects the sphere at N. The other two rays are
the reference lines for angle X in the diagram that you
reference. The ray along the miter edge intersects the
sphere at E. The other reference line for angle X
intersects the sphere at D. Now we have

D E





N

angle NDE = 90 deg.
side DE = X
side ND = 90 - B
angle DNE = A/2
Notice that the plane NOE is the plane of the saw cut so that
angle NED = 90 - Y

The law of cosines gives
cos NED = -cos NDE * cos DNE + sin NDE * sin DNE * cos ND
sin Y = -cos 90 * cos A/2 + sin 90 * sin A/2 * sin B
sin Y = sin A/2 * sin B

For a right triangle
sin ND = tan DE * tan (90 - DNE)
cos B = tan X * cot A/2
tan X = cos B * tan A/2.

--
Michael Press
.

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