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Spiral Inside a Hexagonal Room

Date: 09/03/2003 at 11:18:47
From: Bill 
Subject: Trigonometry

I want to construct a spiral up in the walls of a hexagon, and I need 
to know the relationship of the angle at which it moves to the angle 
that two of the spiral sides make to one another.

You can imagine this as two walls meeting at 120 degrees, and you have
a piece of cardboard with an angle of 137.5 degrees that you want to
tilt until its sides are snug against the wall.  What is the rule for 
figuring the angle of tilt?  I know is has something to do with the 
cosine, but I can't figure it because more than on plane is involved.

I'm looking for a math solution so I don't have to cut out and fit a 
bunch of angles and work on them empirically.  Thanks!


Date: 09/03/2003 at 16:44:37
From: Doctor Rick
Subject: Re: Trigonometry

Hi, Bill.

Let me carefully define what I think you are asking for. Let's say we 
put up two vertical walls meeting at that 120 degree angle. Then we 
draw a line on each wall, meeting at the intersection of the walls. 
Each line makes an angle theta with the horizontal; one goes up from 
the intersection and the other goes down. You are asking what angle 
theta should be so that the two lines form a given angle, say, 137.5 
degrees. Correct?

I find it easier to visualize the reverse problem: knowing theta, 
what is the angle between the lines? I solve this problem using 
vectors. Suppose that the intersection of the two lines is (0,0,0); 
one wall is in the x-z plane and the other wall is rotated phi 
degrees from the first; and the lines are each one unit long at an 
angle of theta degrees to the horizontal. Then the coordinates of the 
endpoints of the lines are

  (cos(theta), 0, -sin(theta))

  (cos(theta)cos(phi), cos(theta)sin(phi), sin(theta))

The dot product (scalar product) of the two vectors is the cosine of 
the angle between them times the product of their lengths. Since the 
lengths are unity, the dot product is just the cosine of the angle. 
Thus

  cos(alpha) = cos^2(theta)cos(phi) - sin^2(theta) 

       alpha = arccos(cos^2(theta)cos(phi)-sin^2(theta))

That's the formula to find the angle between the lines if we know the 
angle of inclination and the angle of the walls. In your problem, we 
know phi = 120 degrees and alpha = 137.5 degrees; can we solve the 
equation for theta? Yes:

  cos^2(theta)cos(phi) - sin^2(theta) = cos(alpha)
  
  (1-sin^2(theta))cos(phi) - sin^2(theta) = cos(alpha)
 
  cos(phi) - sin^2(theta)(cos(phi)+1) = cos(alpha)
 
  sin^2(theta) = (cos(phi)-cos(alpha))/(cos(phi)+1)
 
  theta = arcsin(sqrt((cos(phi)-cos(alpha))/(cos(phi)+1)))

Let's try it out. If phi = 120 degrees and alpha = 120 degrees, then

  theta = arcsin(sqrt((-0.5 - -0.5)/(-0.5 + 1)))
 
       = arcsin(sqrt(0))

        = 0 degrees

as you noted. When we change alpha to 137.5 degrees, we get

  theta = arcsin(sqrt((cos(phi)-cos(alpha))/(cos(phi)+1)))

        = arcsin(sqrt((cos(120)-cos(137.5))/(cos(120)+1)))

        = arcsin(sqrt((-0.5 - -0.737277)/(-0.5+1)))

        = arcsin(sqrt(0.474555))

        = arcsin(0.688879)

        = 43.541 degrees

Check that out with your cardboard to see if I did it right!

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Polyhedra
College Trigonometry
High School Polyhedra
High School Trigonometry

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