Compound Angles

Date: 01/14/2001 at 02:37:08
From: Aaron Mangwiro
Subject: Compound Angles

I am installing a Ranch Rail vinyl fence with my Dad, and we have 
posts (5 x 5) with rails (2 x 3 1/2). The posts have 2.020 X 3.520 
holes routed into them, and the rails fit into these holes perfectly.

But we have one odd one. We have a situation where we have a post but 
the rail does not go into the post horizontally, but has a rise angle 
of 30 degrees and a side angle of 39 degrees. The hole to be routed 
needs to be a parallelogram instead of a square. I have tried to 
calculate the size of this parallelogram now for three weeks and 
asked every teacher, but no one could help me. Please help me - I am 
desperate. 

Thanks. 
Aaron

Date: 01/15/2001 at 09:51:02
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Here's a formula that you can use: cosA*cosB = cosC. It applies to the 
three angles on planes that meet at a point, where two of the planes 
meet at right angles. 

Specifically: take a piece of cardboard, draw angle A, and cut it out.  
Take a second piece of cardboard and draw angle B. Now tape the pieces 
of cardboard together so the two vertices coincide, and one line of 
each angle is in common, and adjust the tape hinge so the cardboard 
planes are at right angles to one another. The third angle formed by 
the free edges of the cardboard is angle C, which obeys the formula 
above.

(You can can convince yourself that this is true with some heavy-duty 
visualization, or by building a physical model.  If you know vector 
algebra, the vector derivation is much easier.)

Now picture the rail where it joins the plane of the post. Take as 
your vertex the point where the longest edge of the rail meets the 
post. Take angles A and B to be on the sides of the rail; they are 
(90-39) and (90-30) degrees. The third angle (C) is one angle of the 
parallelogram at the mouth of your hole in the post. By the formula 
above, cosC = cos(90-39)*cos(90-30), so C = 71.66 degrees, the acute 
angle in the parallelogram.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/21/2001 at 13:12:22
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Ever since you wrote this question to me, I've been going over it in 
my spare time and working on it. I gave you the wrong answer last 
week, and I think I have the right answer now.  

In the answer I gave you before, I assumed that the angles 30 and 39 
degrees were line angles between the axis of the rail and the line 
orthogonal to the post. I don't think that's what you intended - I 
think the given angles are probably plane angles: first between the 
top face of the rail and the horizontal, second between the vertical 
face of the rail and the vertical plane perpendicular to the post. So 
I've re-done the analysis under that assumption.

Last time, I started with the formula I call the hinge formula. As 
foundation, I want to clarify this and generalize it.

Start with a hinge. On one arm of the hinge, draw an angle with the 
axis of the hinge as one side. Call this angle alpha. On the other arm 
of the hinge, draw angle beta, again with the hinge axis as one side.  
You now have two angles in different planes, sharing a common ray.  

Open the hinge to a right angle. Measure the angle gamma between the 
outer rays of angles alpha and beta. Then the three angles are related 
by cos(gamma) = cos(alpha)*cos(beta). This is the hinge formula for 
right angle opening.

In general: Suppose the hinge is open to an arbitrary angle zeta.  
Then the formula becomes

  cos(gamma) = cos(alpha)*cos(beta) + sin(alpha)*sin(beta)*cos(zeta)

This is the general hinge formula for arbitrary opening angle zeta.

Now imagine three planes that come together at a point. They form 
three plane angles and three line angles, and you can apply the hinge 
formula in three different ways. Just as you can write cos(gamma) in 
terms of alpha and beta, you can also write cos(alpha) in terms of 
beta and gamma, and you can write cos(beta) in terms of gamma and 
alpha. In fact, as the hinge formula is set up, it gives you one line 
angle in terms of the other two line angles and the opposite plane 
angle; but any three of the angles are sufficient to solve the system. 

In your case, I believe you have started with a rail aligned along a 
horizontal axis, with its top face horizontal and its side faces 
vertical and perpendicular to the rail. You've turned the rail 
horizontally through an angle of 39 degrees, and then tilted it down 
through an angle of 30 degrees. So consider the point where the 
longest edge of the rail meets the rail. There are three faces coming 
together at that point: two of them are faces of the rail, and the 
third is the vertical face of the post. The two rail faces, of course, 
meet each other at a right angle. The horizontal face of the rail 
meets the vertical face of the post at an angle 90-30 = 60 degrees, 
and the vertical face of the rail meets the vertical face of the post 
at an angle 90-39 = 51 degrees.

So in this case, you have the three plane angles, and you'd like to 
find the line angles - in particular, the angle between the lines in 
the parallelogram, which you need to rout in the vertical face of the 
post.

So we go back to these three hinge formulas, specialized where one 
face angle is a right angle. We can solve the three equations for the 
three line angles. Call the three line angles alpha, beta, and gamma; 
call the two plane angles theta and phi - the third is 90 degrees.  
The three hinge formulas are:

1) cos(alpha)=cos(gamma)*cos(beta) + sin(gamma)*sin(beta)*cos(theta)
2) cos(beta)=cos(gamma)*cos(alpha) + sin(gamma)*sin(alpha)*cos(phi)
3) cos(gamma)=cos(alpha)*cos(beta)

You can solve this system for alpha, beta, and gamma, given theta and 
phi. Here are some hints how to do this:

1) Write x for cos(alpha), y for cos(beta) and z for cos(gamma).  
   Then sin(alpha)=sqrt(1-x^2), etc.

2) Substitute equation (3) for z into equations (1) and (2) to give 
   two equations in x and y. Solve either one, and substitute into the 
   other.  

Some fortunate simplifications are available along the way, and the  
result is surprisingly simple:

   cos(alpha) = cos(theta)/sin(phi)
   cos(beta) = cos(phi)/sin(theta)

   gamma is the angle in the parallelogram that you're looking for.  

 cos(alpha) = cos(51)/sin(60)  ==>  alpha = 43.4 deg
 cos(beta) = cos(60)/sin(51)  ==>  beta = 50.0 deg

 cos(gamma) = cos(alpha)*cos(beta)=0.468   ==>  gamma = 62.1 deg

- Doctor Mitteldorf, the Math Forum

Date: 01/24/2001 at 13:04:10
From: Dennis Mangwiro
Subject: Re: Compound Angles

Thank you very much - that was amazing. 

I have one more question. Will the width of the parallelogram be the 
same, taking into consideration the wall thickness of 0.150 inch, as 
the width of the the widened rectangle due to the side angle only? 
Similarly, will the length of the parallelogram side be the same as 
the length of the side of the rectangle, enlarged, due to the rise 
angle only?

Thank you.
Aaron

Date: 01/25/2001 at 06:52:12
From: Doctor Mitteldorf
Subject: Re: Compound Angles

Dear Aaron,

Last time, I described a method for calculating the three line angles 
where three planes meet. The three planes we're looking at here are 
the top and side of the rail and the vertical face of the post.

When you look at the point where the longest edge of the rail meets 
the post, one of the line angles we calculated gives you the acute 
angle in the parallelogram traced on the face of the post. The other 
two line angles, which we calculated as alpha and beta, appear on the 
top and the vertical side of the rail. In fact, you can draw a picture 
for yourself and see that the length and width of the rectangular 
cross-section of the rail are elongated by factors of 1/cos(alpha) and 
1/cos(beta).  

Does alpha go with the length and beta with the width or vice versa?  
I'll leave that to you to figure out.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/