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Pyramid Construction


Date: 05/24/97 at 02:39:31
From: Don Mohr
Subject: Pyramid building

I'm not a student - I am a professor of history. My son is an advanced 
math tenth grader. Neither he nor his teacher could help me.

I need to construct a four-sided pyramid for a children's exhibit at 
the Anchorage Museum. I want to use plywood. I need to figure out the 
compound angles (angle of side, and mitre) that will join the sides. 

After a much-needed review of trigonometry (and with my son's help), 
I was able to figure out the mitre angle needed.  The angle of the 
sides of the pyramid are directly calculable by the Pythagorean 
theorem. I found the mitre angle (joints between sides) by laying out 
a right angle triangle on a side one foot from the point of the base. 
Knowing the side angle, one foot length of side, and ninety degree 
angle allows the calculation of the other sides. Then looking down at 
a corner of the base from above the pyramid, one has a triangle that 
is perpendicular to the joining edge of two sides. The length of all 
sides of this triangle are known. The angle of the joined sides may 
be calculated and the mitre revealed.  

My son has even programmed his TI-85 to calculate the compound mitre 
of any four-sided pyramid by entering only the base length and height.

Now my question is this:

It seems to me that there should be a more direct path to calculating 
the angle of the sides. It appears that the angle of the sides must 
change in a linear fashion determined by the base length and height.
Is there a formula that will do this?  Once this angle is known, the 
mitre calculation can be done in one's head. Thus, one would not have 
to carry around a TI-85.

Thanks for your help.


Date: 05/24/97 at 07:57:10
From: Doctor Mitteldorf
Subject: Re: Pyramid building

Dear Don,
    
A picture would really help here! I gather that the pyramid you are 
working with has 4 triangular faces and a square base (as opposed to a 
tetrahedron, which has 4 triangular faces, including the base).
    
A square pyramid like this can be constructed with four copies of any 
isoceles triangle, as long as the height of the triangle is more than 
half its base. Once you decide the dimensions of the isoceles 
triangle, you can use the fact that the apex is above the center of 
the square base to calculate the mitre angle of the base: the cosine 
of this angle is half the base of the triangle divided by the height 
of the isoceles triangle.
    
The only other angle left to compute is the mitre angle where any two 
adjacent triangular faces meet. There's a formula I carry around in
my head that is useful in this and many situations like it, involving
angles between lines and between planes. I'll describe it to you, with 
three special cases:
     
Open a notebook and draw a line from a point on the book's spine 
diagonally across one page. Draw another line starting at the same 
point on the spine, going diagonally across the facing page at a 
different angle. Let's say that the angle between the spine and the 
first line is a and the angle between the spine and the second line 
(measured on the facing page) is b. Let x be the angle between the two 
lines.
      
Now this angle x will depend on how wide open the book is.  If the
book is open flat, then the angle x becomes a + b. If the book is 
closed tight, the angle x becomes a - b. The interesting case, of 
course, is when the book is open to an intermediate angle. If x is a 
right angle so that the book is open to 90 degrees, then 
cos(x) = cos(a)*cos(b)
     
You can write the other two cases in terms of cosines as well:

cos(x) = cos(a)*cos(b) - sin(a)*sin(b)     Book open flat
cos(x) = cos(a)*cos(b) + sin(a)*sin(b)     Book closed tight

These come from the formulas for the cosine of the sum and difference 
of two angles.
     
Now here comes the general case: For any angle c that the book may be 
open:

cos(x) = cos(a)*cos(b) + sin(a)*sin(b)*cos(c)     

Notice that this formula subsumes the three special cases we did 
first. The general formula can be derived easily using vector dot 
products. It can also be derived straight from trigonometry, if you 
have a lot of imagination and are good at visualizing 3-dimensional 
objects.

I leave it to you to calculate your pyramid angles using this formula.

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Practical Geometry

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