Sheet Metal Mathematics first created 02/12/05 - last modified 20/09/06 Author: Ty Harness
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I think it is important to explain how the software calculates the properties
required to draw a square to round not only for the engineer to asses the limitations of the software but also from an educational point of view to help
pupils and students apply mathematics to engineering problems.
The circumference of a circle:
$C = pi*D$
The area of the circle:
$R_A = pi*r^2 = pi*(D^2)/4 $
The perimeter of the base square or rectangle:
$R_P = 2*(S_x + S_y)$
The area of the base rectangle:
$S_A = S_x*S_y$
Traditionally (on the drawing board) you would divide your circle into 24 divisions (15 degree segments) using a 90-30-60 and 90-45-45 set squares
or you can do it all with dividers set to the circle radius. The greater the number of divisions the less truncation error is accumulated over the pattern making process.
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Figure 1 - How to make 15 and 75 degrees using your set squares or you can just use your dividers to segment the circle.
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Figure 2 - The square-to-round software allows you to increase the numbers circle divisions up to 96.
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The above graphic shows a silly example of a square-to-round with no height. It's just a plate with a hole in and you would never go to the trouble of
triangulating this pattern out on the drawing board, but it does highlight the
truncation error and the greater the number of divisions in the circle not only means a smoother curve but a greater pattern accuracy.
True Lengths: Again when setting out a pattern it was necessary to calculate the true lengths that
could not be measured directly from either the plan or the elevation. This can be done by setting plan length against the elevation height
as a right angle triangle and then the true length is the hypotenuse of the triangle easily measured with your dividers.
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Figure 3 - True length diagram
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Mathematically this is described by Pythagoreon theorem:
$TL = sqrt (PL^2 + H^2) $
The software works by storing the 3D vertices of the square-to-round so it is actually very simple to find the true length between
two 3D vertices.
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Figure 4 - The magnitude or length of a 3D vector
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From the above figure the hypotenuse of the triangle in the XY plane is $sqrt(x^2 + y^2)$ (Pythagorean Theorem) and the triangle
out of plane where L is the hypotenuse $L^2 = (sqrt(x^2 + y^2))^2 + z^2 = x^2 + y^2 + z^2$ and hence $L = sqrt(x^2+y^2+z^2)$
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Figure 5 - Changing the height of the front elevation then the true length increase forcing the pattern apart
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Triangulation
We found the true lengths from Pythagoreon theorem therefore we know the length a,b and c with reference to figure xx. So all we need
to do is find w and h to define the next point in the planar pattern.
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Figure XX - Calculating the pattern points with triangulation
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$a^2 = (w-c)^2 + h^2$
$a^2 = w^2 - 2cw + c^2 + h^2$
$h = b*sin(A) $ and $ w=b*cos(A)$
$a^2 = b^2*cos^2(A) - 2*c*b*cos(A) + c^2 + b^2*sin^2(A) $
$a^2 = b^2*(sin^2(A) + cos^2(A)) - 2*c*b*cos(A) + c^2 $
where $1 = sin^2(A) + cos^2(A) $
$a^2 = b^2 + c^2 - 2*c*b*cos(A) $ {well known as the cosine rule}
We can then simply transpose for A and then w and h can be found.
Incidently if we did need Angle B and C we could just use the sine rule
$a/sin(A) = b/sin(B) = c/sin(C)$
The square-to-round is nicely described by triangular faces alone so the surface area is the summation of all the triangle areas.
Heron's formula is easily applied because you can find the triangle edge lengths (a,b and c) from the vertices.
$A_(tri) = sqrt( s*(s-a)*(s-b)*(s-c) )$
where s is the semi perimeter:
$s = 1/2 *(a+b+c)$
It's useful to calculate the minimum bounding rectangle of the pattern. Some material can be saved by nesting but this can be time consuming by
trying different nesting solutions by hand (CAD CAM software is starting to apply nesting algorithms these days). You may only have a guillotine
(straight cuts) method so finding the bounding rectangle is still useful.
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Figure 6 - Minimum Bounding Box Area Calculation
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Finding the bounding rectangle is simply a case of examining all the vertices and storing the smallest and largest. If the bounding rectangle was
rotated then perhaps the area may be smaller. Mathematically this can be achieved by rotating the coordinate system where the angle
theta is the relative link.
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Figure 7 - Absolute to relative coordinates
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Once you've found the bounding rectangle in the relative coordinate system you may wish to transform the answer back into the absolute
coordinate system.
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Figure 8 - Relative to absolute coordinates
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The face normals and angles between adjacent face normals can be found with V1.032 and above of the Square - To - Round
software.
Face Normals
The internal volume:
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Figure 9 - Reducing the circle diameter to zero creates a pyrimid
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If we set the diameter of the circle to zero then we obtain a pyramid. A pyramid is a special case of the tetrahedron family. The volume
of a pyramid is well known (even from school text books):
$V = 1/3 A_(base) * height $
A more general formula for a Tetrahedron in terms of the 3D vertices:
$V = 1/(3!) * | (x_(1), y_(1), z_(1),1), (x_(2), y_(2), z_(2),1),(x_(3), y_(3), z_(3),1),(x_(4), y_(4), z_(4),1) | $
Placing the 4th vertex point (x4,y4,z4) at the 0,0,0 origin simplifies the solution of the determinant:
$V = 1/6 * | (x_(1), y_(1), z_(1),1), (x_(2), y_(2), z_(2),1),(x_(3), y_(3), z_(3),1),(0,0, 0,1) | $
$V = 1/6 * (x_1* y_2*z_3 - x_1*y_3*z_2 - x_2*y_1*z_3 + x_2*y_3*z_1 + x_3*y_1*z_2 - x_3*y_2*z_1)$
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Figure 10 - Use the faces to make up tetrahedrons by adding a fourth vertex inside the square-to-round
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Then it's just a matter of summing up all the tetrahedron volumes for each of the triangular faces. The choice of setting the rectangular base at z = 0 means I do not need to
triangulate the base square, but I've triangulated the circle area shown below with the blue tetrahedrons. I should of course of translated the whole structure down
leaving the circle on the z plane which would of resulted in n-2 calculations.
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Figure 11 - Sum up all the Tetrahedrons to determine the internal Volume
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Just one last case is when you set the area of the base to zero and you create a cone
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Figure 12 - 96 Sided polyhedron approximating a cone
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Again from school days:
$V = 1/3 A_(base) * height $
References and Further reading
mathworld.wolfram.com/HeronsFormula.html
mathworld.wolfram.com/Tetrahedron.html
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