UnitMath Example: ( Pyramids With 4 sides )
Last Modified 12/8/1999
I have tried to make sure the following formulas are correct, but I do not guarantee them.
 Regular  Truncated Pyramids

 Rectangular  Irregular  Truncated Pyramids

Regular  Truncated Pyramids TOP
The following drawing shows 3 views of a truncated pyramid with 4 sides.
Starting with the information given above we are to find the information needed to cut each side of the pyramid. The next figures show the information we need. Due to symmetry all four parts have the same dimensions, so we only need to find the information about only one part.
One side of a Regular  Truncated Pyramid TOP
Following are several example calculations. The calculations assume that angle <= 90 degrees. Example 1, I verified with a CAD program, example 4 is obviously true. ( See angle notes )
Example1: angle = ( 53.13013 pm 0 ) deg TOP
( 53.13013 pm 0 ) deg is the angle in the above CAD drawing. I built this model and it looks good.
 "Define the givens"

angle: ( 53.13013 pm 0 ) deg;
base: 3 in + 7/8 in;
thickness: 3/4 inch;
height: 1 inches;
 "Define points T, A, & B"

Tz: height; Ty: Tz/tan(angle); Tx: Ty;
T: Tx `i + Ty `j + Tz `k;
Ax: ft; Ay: 0 ft; Az: 0 ft;
A: Ax `i + Ay `j + Az `k;
By: thickness /sin( angle); Bx: By/tan(45deg); Bz: 0 ft;
B: Bx `i + By `j  Bz `k;
 "Define the Normals"

NTOB: T cross B;
NAOT: A cross T;
 "Define the remaining"

angle_1: acos( T dot A / ( abs(T) abs(A) ) ) as deg;
angle_2: 180 deg  acos( NTOB dot NAOT / ( abs( NTOB ) abs( NAOT ) ) ) as deg;
width: Tx tan(angle_1) as in & (in/16);
top: base  2 Tx as ft & in & (in/16);
 "Find the results"

angle_1; ~ 59.0 deg
angle_2; ~ 55.6 deg
width as in & (in/8); ~ in + 2.0 (in/8)
top as in & (in/8) ~ 2 in + 3.0 (in/8)
Example2: angle = 45 deg TOP
 "Define the changes from example 1"

angle: 45 deg;
base: 4 in;
thickness: inch/2;
height: 3/2 inches;
 "Find the angles and dimensions"

angle_1 ~ 54.7 deg
angle_2; = 60 deg
width as in & (in/16) ~ 2 in + 1.9 (in/16)
top as in = in
Example3: angle = 60 deg TOP
 "Define the changes from example 2"

angle: 60 deg;
 "Find the angles and dimensions"

angle_1; ~ 63.4 deg
angle_2 ~ 52.2 deg
width; ~ in + 11.7 (in/16)
top; ~ 2 in + 4.3 (in/16)
Example4: angle = 90 deg TOP
 "Define the changes from example 3"

angle: 90 deg;
 "Find the angles and dimensions"

angle_1; = 90 deg
angle_2 = 45 deg
width; = ( 1 + 1/2 ) in
top; = 4 in
Formula Derivation for the "Regular  Truncated Pyramid" TOP
Motivation:
Given the angles angle1 & angle2 all the other calculations are fairly simple, so I will concentrate on those angles. The three points T, O, & A, define vectors OT & OA, and angleTOA ( angle1 ) can then be found using the formula: "OT dot OA =  OT   OA  cos( angleTOA )" or "acos( OT dot OA / (  OT   OA  ) )". Angle2 can be found from the normals to the two surfaces TOA & TOB. Therefore I need to find points B, T, O, & A
Axis:
Judicious choice of axis simplifies everything, so place O on the left side of Part with coordinants ( 0,0,0 ) . Define the x axis to be parallel to the base of the part, the z axis to be pointing up from the base, the y axis is then given by the right hand rule.
 The following points are used in this derivation:

Point O is the origin located on the intersection of front Part & left Part on the outside of the pyramid.
Point T is above point A on the intersection of front Part & left Part on the outside of the pyramid.
Point A is any point on the x axis to the right of point A.
Point B is the intersection of front Part & left Part ( on the inside of the pyramid ) & where z = 0.
With O at the origin vector OA has the same coordinates as A, vector OT as T, and vector OB as B. With this selection of points the angles found are for the sides of the Part. The following figures show the locations of the various points.
The next figure shows how Tz, Ty, & By can be found. Given Ty and By, Tx and Bx can be found using the formula tan ( 45 deg ) = y/x. Bz = 0 by choice of axis.
The next figure gives more information on how angle2 is found.
Your Own Calculations for a Truncated Pyramid with 4 sides TOP
 Open UnitMath on your computer. If needed see downloads for Macintosh or Windows.
 Copy the following text into UnitMath.
" Define the Givens CHANGE THESE "
angle: ( 53.13013 pm 0 ) deg;
base: 3 in + 7/8 in;
thickness: 3/4 inch;
height: 1 inches;
" IN GENERAL DON'T CHANGE THESE  "
Tz: height; Ty: Tz/tan(angle); Tx: Ty;
T: Tx `i + Ty `j + Tz `k;
Ax: ft; Ay: 0 ft; Az: 0 ft;
A: Ax `i + Ay `j + Az `k;
By: thickness /sin( angle); Bx: By/tan(45deg); Bz: 0 ft;
B: Bx `i + By `j  Bz `k;
NTOB: T cross B;
NAOT: A cross T;
angle_1: acos( T dot A / ( abs(T) abs(A) ) ) as deg;
angle_2: 180 deg  acos( NTOB dot NAOT / ( abs( NTOB ) abs( NAOT ) ) ) as deg;
angle_1Gauge: 90 deg  angle_1 as deg;
angle_2Gauge: 90 deg  angle_2 as deg;
width: Tx tan(angle_1) as in & (in/16);
top: base  2 Tx as ft & in & (in/16);
" The Results will be here after the calculation "
angle_1;
angle_1Gauge;
angle_2;
angle_1Gauge;
width;
top;
" Stop Copying Here "
 Change the Givens
 For example:

angle: 50 deg;
base: 2 feet;
thickness: 1/2 inch;
height: 9 inches;
 Select all the text that you pasted in Step 2 and evaluate it.
Your results will be at the end of the text. Before building see
angle notes.
Rectangular  Irregular  Truncated Pyramids TOP
The following drawing shows 3 views of a rectangular truncated pyramid with 4 sides.
Starting with the information given above we are to find the information needed to cut each side of the pyramid. The next figures show the information we need.
Front side of the rectangular truncated pyramid TOP
Left side of the rectangular truncated pyramid TOP
Following are several example calculations. ( See angle notes )
Example1: angleF = 90 deg, angleL = 90 deg TOP
 "Define the givens"

angleL: 90 deg;
angleF: 90 deg;
thicknessL: inch;
thicknessF: inch;
height: 4 inch;
 "Define points T, A, B, & C"

Tz: height; Ty: Tz / tan( angleF ); Tx: Tz / tan( angleL );
T: Tx `i + Ty `j + Tz `k;
A: ft `i + 0 m `j + 0 m `k;
Bz: 0 inch; By: thicknessF / sin( angleF ); Bx: thicknessL / sin( angleL );
B: Bx `i + By `j + Bz `k;
C: 0 m `i + ft `j + 0 m `k;
 "Define the Normals"

NF: A cross T;
NC: T cross B;
NL: T cross C;
NC2: B cross T;
 "Define the remaining"

angleF1: acos( T dot A / ( abs( T ) abs( A ) ) ) as deg;
angleL1: acos( T dot C / ( abs( T ) abs( C ) ) ) as deg;
angleF2: 180 deg  acos( NF dot NC / ( abs( NF ) abs( NC ) ) ) as deg;
angleL2: 180 deg  acos( NL dot NC2 / ( abs( NL ) abs( NC2 ) ) ) as deg;
widthF: Tx tan( angleF ) as in & in/16;
widthL: Ty tan( angleL ) as in & in/16;
 "Find the results"

angleF1; = 90 deg
angleF2; = 45 deg
widthF; = 4 in
angleL1; = 90 deg
angleL2; = 45 deg
widthL; = 4 in
Example2: angleF = 45 deg, angleL = 45 deg TOP
 "Define the changes from example 1"

angleL: 45 deg;
angleF: 45 deg;
 "Find the angles and dimensions"

angleF1; ~ 54.7 deg
angleF2; = 60 deg
widthF; ~ 4 in
angleL1; ~ 54.7 deg
angleL2; = 60 deg
widthL; ~ 4 in
Example3: angleF = 60 deg, angleL = 30 deg TOP
 "Define the changes from example 2"

angleL: 30 deg;
angleF: 60 deg;
thicknessL: inch/4;
 "Find the angles and dimensions"

angleF1; ~ 33.7 deg
angleF2; ~ 101.5 deg
widthF; = 12 in
angleL1; ~ 73.9 deg
angleL2; ~ 14.2 deg
widthL; = in + ( 5 + 1/3 ) in/16
Formula Derivation for the "Irregular Rectangular Pyramid" TOP
This derivation is very similar to the derivation for the "Regular  Truncated Pyramid". With the addition of a point C on the Y axis used to find the normal to the left part.
The next figures show how Tz, Ty, Tx, By, & Bx can be found. Bz = 0 by choice of axis.
Your Own Calculations for a Iregular Pyramid with 4 sides TOP
 Open UnitMath on your computer. If needed see downloads for Macintosh or Windows.
 Copy the following text into UnitMath.
" Define the Givens CHANGE THESE "
angleL: 90 deg;
angleF: 90 deg;
thicknessL: inch;
thicknessF: inch;
height: 4 inch;
" IN GENERAL DON'T CHANGE THESE  "
Tz: height; Ty: Tz / tan( angleF ); Tx: Tz / tan( angleL );
T: Tx `i + Ty `j + Tz `k;
A: ft `i + 0 m `j + 0 m `k;
Bz: 0 inch; By: thicknessF / sin( angleF ); Bx: thicknessL / sin( angleL );
B: Bx `i + By `j + Bz `k;
C: 0 m `i + ft `j + 0 m `k;
NF: A cross T;
NC: T cross B;
NL: T cross C;
NC2: B cross T;
angleF1: acos( T dot A / ( abs( T ) abs( A ) ) ) as deg;
angleF1Gauge: 90 deg  angleF1 as deg;
angleL1: acos( T dot C / ( abs( T ) abs( C ) ) ) as deg;
angleL1Gauge: 90 deg  angleL1 as deg;
angleF2: 180 deg  acos( NF dot NC / ( abs( NF ) abs( NC ) ) ) as deg;
angleF2Gauge: 90 deg  angleF2 as deg;
angleL2: 180 deg  acos( NL dot NC2 / ( abs( NL ) abs( NC2 ) ) ) as deg;
angleL2Gauge: 90 deg  angleL2 as deg;
widthF: Tx tan( angleF ) as in & in/16;
widthL: Ty tan( angleL ) as in & in/16;
" The Results will be here after the calculation "
angleF1;
angleF1Gauge;
angleF2;
angleF2Gauge;
widthF;
angleL1;
angleL1Gauge;
angleL2;
angleL2Gauge;
widthL;
" Stop Copying Here "
 Change the Givens
 For example:

angleL: 45 deg;
angleF: 30 deg;
thicknessL: inch;
thicknessF: inch;
height: 3 inch;
 Select all the text that you pasted in Step 2 and evaluate it.
Your results will be at the end of the text. Before building see
angle notes.
TOP