The Weight Estimator program was develop by JMC McAlester to assist in characterizing munitions. It calculates volumes, surface areas and their associated weights based on material densities.

There are over 50 different shapes that can be selected to estimate the weight. Each shape has unique dimensions that can be entered. However, care must be taken to enter realistic dimensions since the system does not currently do any input validation. For example, it is possible to enter an inside radius larger than an outside radius for spheres or cylinders. However, the 3D model will help indicate impossible shapes by trying to generate unrealistic shapes. If the shape does not look correct, then the input values are most likely incorrect.

The 3D model for each shape is updated when you click on the "Update 3D Model". This will give an exact scale representation of the model. You can use the mouse to rotate and move the model. The scroll wheel will zoom the model in and out. Each time you update the model, it will reset to the initial view size and direction.

Each shape page also has a Help Link that accesses this help page. There is basic information for each shape, particularly if there are special conditions that should be noted.

The Anchor Pin object has two tapering conical sections, one at the top and one at the bottom. Both conical sections are assumed to taper inward towards the top. The middle section is a simple cylinder.

All sections are considered to be solid.

Technical Calculations

The equation for a tapered cylinder section is derived in the "Solid Ogive" (not a real ogive shape). The derivation will not be duplicated here. The the tapered cylinder volumes are

   V_{Top Tapered Cyl} = π D (E² + F² + EF) /12

   V_{Bottom Tapered Cyl} = π B (F² + A² + FA) /12

The outside surface area is just the average perimeter times the slope length (see "Solid Ogive" for derivation). This equation is only for the side and does not include the circular area at the top or bottom.

 S_{Top Tapered Side} = [π(F + E)/2] [(F - E)²/4 + D²]^0.5

 S_{Bottom Tapered Side} = [π(A + F)/2] [(A - F)²/4 + B²]^0.5

Outer Surface Area, A_outer, consists of two tapered cylinders, top and and bottom circular area, and one solid cylinder side,

  A_outer = [π(F + E)/2] [(F - E)²/4 + D²]^0.5 + π F C
              + [π(A + F)/2] [(A - F)²/4 + B²]^0.5
                 + π E²/4 + π A²/4

Inner Surface Area, A_inner, no inside surface,

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of two solid tapered cylinders and on solid cylinder

   V = π D (E² + F² + EF) /12 + π F²C/4
            + π B (F² + A² + FA) /12

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Block object is a multiple indented shape that can be used for various rectangular configurations. Each of the protrusions on the top can be a different height and thickness. One should be careful not to have one protrusion go negatively into another. The program does allow unrealistic shapes to be constructed with the 3D modeler.

Technical Calculations

Outer Surface Area, A_outer, consists of both top and bottom surface, plus the perimeter of the block. The top and bottom is divided into rectangular parts and added. The perimeter is just added using the dimensions in the diagram. Notice, B is the same as the bottom and top,

   A_outer = A_top-bottom + A_edge

       = 2[E(D - H) + DF + G(C - I) + C(B - E - F - G)]
            + A[B + C + B + I + (D + I - C) + H + (D - H)]

      = 2[ED - EH + DF - GI + CB - CE - CF]
             + A[2B + 2D + 2I]

Inner Surface Area, A_inner, no inside surface,

   A_inner =  0

   A_Total = A_outer + A_inner

Volume, V, consists of the block surface times the depth. Use the top area given above in the outer surface derivation,

   V = A[ED - EH + DF - GI + CB - CE - CF]

*Use diagram to count squares for area, perimeter
and volume

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

17 Dec 2020 - Kurt Gramoll
1) Surface equations incorrectly included inner surface with outer surface. New equation separated outer and inner surfaces.

The Solid Bolt object has an hexagon shape top and cylindrical shape shaft. The dimension A is from one side of the hexagon to the opposite side. The dimension A is not a single side width. It is assumed that the hexagonal head and cylindrical shaft are in line vertically.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = 2 [(3^0.5/2) A²] - πC²/4 + πCD + πC²/4
                       + 6 AB / 3^0.5
           = 3^0.5A² + πCD + 3^0.52AB

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of a single sphere.

   V = (3^0.5/2)A²B + πC²D/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

17 Dec 2020 - Kurt Gramoll
1) Previous equations included an extra circular area equal to the bolt shaft end. The area equations were corrected. Both volume and area equations were simplified.

The Bolt with Through Holes object has an hexagon shape top and cylindrical shape shaft. Both the top and shaft can have a single hole completely through the object. Each of the two holes can have different diameters. There is only one hole for each location.

The dimension A is from one side of the hexagon to the opposite side. The dimension A is not a single side width. It is assumed that the hexagonal head and cylindrical shaft are in line vertically.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = 2 [(3^0.5/2) A²] - πC²/4 + πCD + πC²/4
                       + 6 AB / 3^0.5 - 2E²π/4 - 2FE²π/4
           = 3^0.5A² + πCD + 3^0.52AB - π(E² + F²)/2

Inner Surface Area, A_inner,

   A_inner = πEA + πFC

   A_Total = A_outer + A_inner

Volume, V, consists of a single sphere.

   V = (3^0.5/2)A²B + πC²D/4 - πE²A/4 - πF²C/4
      = (3^0.5/2)A²B + π(C²D - E²A - F²C)/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

17 Dec 2020 - Kurt Gramoll
1) Previous equations included an extra circular area equal to the bolt shaft end. The area equations were corrected. Both volume and area equations were simplified.
2) Previous equations flipped E and F when calculating inside surface Equation was corrected.

The Cap object is a single cylindrical shape with a hollow cylindrical shape. The bottom is open, but the top is closed with a circular hole. The top closed end thickness is E. The wall thickness is the difference between the diameters A and B, divided by 2.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = πAC + π(A² - D²)/4 + π(A² - B²)/4

Inner Surface Area, A_inner,

   A_inner = πDE + πB(C - E) + π(B² - D²)/4

   A_Total = A_outer + A_inner

Volume, V, consists of a single sphere.

   V = πA²C/4 - πB²(C - E)/4 - πD²E/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Cartridge Case object has a cylindrical shaped hollow shell with an open end. The bottom has three different disks, each with a different radius. All three bottom circular plates have a center hole of radius E.

The upper cylindrical shell can up to 20 holes of similar type. The holes are assumed to go through the shell thickness. The location and spacing is not needed for the volume and surface area calculations.

Technical Calculations

Outer Surface Area, A_outer, all shapes are based on cylinders, ,

  A_outer = π(A² - E²) /4 + πA I + π(A² - C²) /4
                + πC (B - I) - πJ² K / 4 + π(C² - D²) /4

Inner Surface Area, A_inner, consists of just the inner surface of the hollow cone,

   A_inner = πEG + π(D² - E²) /4 + πD (B - H)
       - πJ² K /4 + πJ [(C - D)/2] K + πF (G - H)

   A_Total = A_outer + A_inner

Volume, V, consists several cylindrical hollow shapes. The diagram shows each of the cylindrical shapes that are used to derive the volume equation.

   V  = V₁ + V₂ + V₃ + V₄ + V_holes
   = π(C² - D²) (B - I) /4 + π(F² - E²) G /4
      + π(D² - F²) (H - I) /4 + π(A² - F²) I /4
       + πJ² [(C - D)/2] * K /4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

21 Dec 2020 - Kurt Gramoll
1) Previous equations was missing the inside surface of the small top washer shape, S = πF(G-H) The inside surface area equation was corrected.

The Hollow Cone object is a basic conical section with a hollow internal cone shape. The wall thickness is not constant, and is dependant on the relative height of the inside hollow cone, C. The horizontal thickness at the base is the difference between radius B and D, divided by 2.

Outer Surface Area, A_outer, consists of an outside of the outer cone, and the bottom edge,

  A_outer = A_cone + A_{bottom edge}
            = π (B/2) [ A² + (B/2)² ]^0.5
                              + π [ (B/2)² - (D/2)² ]

Inner Surface Area, A_inner, consists of just the inner surface of the hollow cone,

   A_inner = π (D/2) [ C² + (D/2)² ]^0.5

   A_Total = A_outer + A_inner

Volume, V, consists of a single cone minus the hollow inner cone.

   V = (π/3) [ A (B/2)² - C (D/2)² ]

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

7 Dec 2020 - Kurt Gramoll
1) Inner and outer surface equations were flipped in the program. Changed.
2) Inner equation multiplied the squared radius and squared height terms instead of adding them. Was also incorrect in original code.

The Cone object is a basic conical shape. The shape is solid and has only two dimensions, height A and base diameter B.

Technical Calculations

Equation details are documented at https://en.wikipedia.org/wiki/Cone

Outer Surface Area, A_outer, consists of a single cone including the bottom,

  A_outer= A_cone + A_bottom
          = π (B/2) [ (B/2)² + A²) ]^0.5 + π (B/2)²

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of a single cone.

   V = (π/3) (B/2)² A

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The "Hollow Cylinder A and C Closed" object is cylindrical shape that has cylindrical hollow section. Both ends are capped. The wall thickness is the difference between the diameters A and C, divided by 2. Both end cap thickness are the same, but can be different than the wall thickness.

The shape can have one type of hole, but there can be multiple instances of that hole. Each hole is assumed to goes through the wall (no partial indents).

The location of the holes is not needed for volume and surface area calculations. But it is assumed that the holes diameters are not large, and they will be contained inside the cylinder diameter, A and are in the hollow section. Also, it is assumed that a given hole depth does not exceed the cylinder wall thickness.

Outer Surface Area, A_outer, consists of the outer main cylinder and the two outside circular ends. Subtract the hole areas.

   A_outer = A_cylinder + A_ends - A_{hole ends}
            = (πA) B + 2 (πA²/4) - F (πE²/4)  

Inner Surface Area, A_inner, consists cylindrical inside surface and circular ends. Plus the inside cylindrical surface of the holes minus the end of the hole.

   A_inner = A_{inside cylinder} + A_{inside ends}
                + A_{hole cylinder} - A_{hole end}
           = (πC) D + 2 (πC²/4)
                 + F (πE) (A-C)/2 - F (πE²/4)

   A_Total = A_outer + A_inner

Volume, V, consists of the difference of the outside cylinder minus the hollow inside cylinder. Then subs tract all instances of the holes.

   V = (πA²/4) B - (πC²/4) D - F (πE²/4) (A - C)/2

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The "Hollow Cylinder A and C Open " object is cylindrical shape that has cylindrical hollow section. Both ends are open. This shape could also be described a simple tube. The wall thickness is the difference between the diameters A and C, divided by 2.

The shape can have one type of hole, but there can be multiple instances of that hole. Each hole is assumed to goes through the wall (no partial indents).

The location of the holes is not needed for volume and surface area calculations. But it is assumed that the holes diameters are not large, and are contained inside the cylinder diameter, A. Also, it is assumed that a given hole depth does not exceed the cylinder wall thickness.

Outer Surface Area, A_outer, consists of the outer main outside cylinder, the two end edges, and the subtraction of the hole areas.

   A_outer = A_cylinder + A_{edge ends} - A_{hole ends}
            = (πA) B + 2 π(A²-C²)/4 - F (πE²/4)  

Inner Surface Area, A_inner, consists cylindrical inside surface and circular ends. Plus the inside cylindrical surface of the holes minus the end of the hole.

   A_inner = A_{inside cylinder} + A_{hole cylinder} - A_{hole end}
           = (πC) B  + F (πE) (A-C)/2 - F (πE²/4)

   A_Total = A_outer + A_inner

Volume, V, consists of the difference of the outside cylinder minus the hollow inside cylinder. Then subs tract all instances of the holes.

   V = B π(A² - C²)/4 - F (πE²/4) (A - C)/2

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The "Hollow Cylinder A or C Open " object is cylindrical shape that has cylindrical hollow section. Only one end is capped. The wall thickness is the difference between the diameters A and C, divided by 2. The closed end cap thickness can be different than the wall thickness.

The shape can have one type of hole, but there can be multiple instances of that hole. Each hole is assumed to goes through the wall (no partial indents).

The location of the holes is not needed for volume and surface area calculations. But it is assumed that the holes diameters are not large, and are contained inside the cylinder diameter, A and will be in the hollow section. Also, it is assumed that a given hole depth does not exceed the cylinder wall thickness.

Outer Surface Area, A_outer, consists of the outer main cylinder, the top outside circular end, and the bottom edge. Subs tract the hole areas.

   A_outer = A_cylinder + A_end + A_{edge end} - A_{hole ends}
            = (πA) B + (πA²/4) + π(A²-C²)/4 - F (πE²/4)  

Inner Surface Area, A_inner, consists cylindrical inside surface and top circular end. Plus the inside cylindrical surface of the holes minus the end of the hole.

   A_inner = A_{inside cylinder} + A_{inside end}
                + A_{hole cylinder} - A_{hole end}
           = (πC) D + (πC²/4)
                 + F (πE) (A-C)/2 - F (πE²/4)

   A_Total = A_outer + A_inner

Volume, V, consists of the difference of the outside cylinder minus the hollow inside cylinder. Then subs tract all instances of the holes.

   V = (πA²/4) B - (πC²/4) D - F (πE²/4) (A - C)/2

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Cylinder with Holes object is a right constant cylindrical shape with four possible hole types.

The shape can have up to four different hole types. Not all four types must be used (just leave those input fields blank). Each hole type can have a different number of holes (N), depth (H), and diameter (D). Each hole type is labeled with its number after N, H and D, such as N₂, H₂ and D₂. Number may also be a subscript.

The location of the holes is not needed for volume and surface area calculations. But it is assumed that the holes diameters are not large, and they will be contained inside the cylinder diameter, A. Also, it is assumed that a given hole depth does not exceed the cylinder diameter, A.

Outer Surface Area, A_outer, consists of the outer main cylinder, two circular ends, and subtraction of the holes (four possible types). Note, the equation ignores holes that pass all the way through. Only one end is subtracted.

   A_outer = A_cylinder + A_ends - A_{hole ends}
            = (πA) B + 2 (πA²/4)
                - (πD₁²/4) N₁ - (πD₂²/4) N₂
                - (πD₃²/4) N₃ - (πD₄²/4) N₄

Inner Surface Area, A_inner, consists of the hole cylindrical surface and one end. Note, the equation ignores holes that pass all the way through. One end is always added regardless.

   A_inner = A_{hole cylinders} + A_{hole ends}
            = (πD₁) H₁ N₁ + (πD₂) H₂ N₂
                        + (πD₃) H₃ N₃ + (πD₄) H₄ N₄
                 + (πD₁²/4) N₁ + (πD₂²/4) N₂ 
                        + (πD₃²/4) N₃  + (πD₄²/4) N₄ 

   A_Total = A_outer + A_inner

Volume, V, consists of the cylinder minus all four types of holes.

   V = (πA²/4) B - (πD₁²/4) H₁ N₁ - (πD₂²/4) H₂ N₂
                        - (πD₃²/4) H₃ N₃ - (πD₄²/4) H₄ N₄

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

24 Nov 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, holes 3 and 4 doubled the volume and surface area due to an error using 0.5 constant instead of 0.25. Hole types 1 and 2 were correct.
2) Inner surface was modified to include one end surface for each hole. This was previously not included.

The Cylinder object is one of the more basic shapes. It is a constant cylindrical tube or rod. The height, B and diameter, A can be changed.

Technical Calculations

Outer Surface Area, A_outer, consists of both the curved cylindrical section and both circular ends.

  A_outer = A_cylinder + A_ends
            = (πA) B + 2(πA²/4)

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of a single cylinder.

   V = (πA²/4) B

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Escapement Lever is a nonsymmetrical shape with various protrusions. The two long arms sticking out both have a single hole that goes through the arm thickness, F. There are two smaller arms near the center that has two parts each. Each of the two parts of the smaller inner arms are rectangular with width, height, and depth dimensions.

Technical Calculations

Outer Surface Area, A_outer, consists upper and lower trapezoidal arms with holes, center circular shape with hole, and projections on each side,

  A_outer = A_levers + A_center + A_projections
        = 2(C + A)B + 2F(B + A) + 2F[(A - C)² + B²]^0.5
           - πI² + 2π(G² - H²) + πGF - 2CF
             + 4(JD + 2FD + EF + JF)

Inner Surface Area, A_inner, includes the inside surface of the three holes,

   A_inner = 2IπF + HπF = πF(2I + H)

   A_Total = A_outer + A_inner

Volume, V, consists upper and lower trapezoidal arms with holes, center circular shape with hole, and projections on each side,

   V = V_levers + V_center + V_projections
      = 2[(C + A)/2]BF + π(G² - H²)F/4 - 2π I²F/4
           + 2JDF + 2EDF
      = (C + A)BF + πF(G² - H² - 2I²)/4 + 2(J + E)DF

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

26 Dec 2020 - Kurt Gramoll
Volume equation was correct but surface equations incorrect.
1) Equation for inside surface (two holes) incorrectly included with outside surface
2) Upper and lower arms incorrectly modeled as rectangular shape, they are trapezoidal shape
3) Incorrectly added I hole surface instead of subtract

The Fin object is a solid inclined rectangular shape with one side higher than the other. The shape has four dimensions to describe its volume and surface.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = 2A (C + B)/2 + BD + CD + AD
                  + D [(C - B)² + A ²]^0.5
      = A(C + B) + C(A + B + C) + D [(C - B)² + A ²]^0.5

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of a single cylinder.

   V = AD (C + B)/2

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Firing Pin is a solid object consisting of two cylindrical sections and a small semispherical ball on top. The two cylindrical sections can have different heights and diameters. The top semispherical ball is mounted at the center of the top cylinder section. The ball diameter must be less than the top cylinder diameter.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = πA²/4 + πAB + π(A² - C²)/4
                + πDC + π(D² - E²)/4 + (4/2) πE²/4
         = π(A²/2 + AB + DC + D²/4 + E²/4)

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of a single cylinder.

   V = πBA²/4 + πCD²/4 + (2/3) πE³/8
      = π(BA² + CD² + E³/3)/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

20 Dec 2020 - Kurt Gramoll
1) Previous surface equation excluded the top surface, D. Both volume and surface equations were simplified.

The Front Body shape has an ogive (a partial circular arc curve) top section with a flatten tip. The rate of curvature of the top section can be changed by modifying the radius dimension, B. The location of arc cure ('+' symbol) is determined by the program to match the other dimensions and D. The ogive curve does not have to be a vertical tangent at the bottom point (where A and C join).

It is possible to have a non-vertical tangent at the bottom edge (point at F and A).

The thickness is defined by the difference of A and E, and is constant in both curve sides and top flat tip.

The top vertical line at the joint does not technically exist. It is simply an artifact from the 3D modeling program.

The location of the arc center, the '+' symbol on diagram, is determined from the lower point of C and A dimension and the upper point of C and D dimension. Both these points are known.

As illustrated in the circle diagram, two points and the radius can define an arc curve without needing other dimensions. There is a second possible circular arc (dashed line), but it is not realistic so is not considered for this shape.

Technical Calculations
The Front Body shape is similar to the top section of the Projectile shape but has the top point removed and replaced with a solid circular disk. Also, the Front Body is hollow like the Hollow Projectile body.

First, the ogive curve needs to be modeled as an equation. Then it can be revolved around a given line to create the solid. Recall, an ogive curve is a section of a circular arc as mentioned in projectile shape section. The ogive needs to be described as an equation in an x-y cartesian coordinate system so it can be integrated about the x-axis to generate a solid using the infinitesimal slice dx.

The diagram at the right shows a general ogive curve (red line) that describes the edge of the front body. The curve starts at the x-y coordinate origin even though the shape will start a distance 'g' from the origin. That curve is rotated about the x-axis giving the outside Front Body shape. The final curve equation is (see Solid Projectile for derivation details - note dimension letters may be different),

   y = [B² - (x - B cosα)²]^0.5 - B sinα

    where α = cos^-1[ (R² + L²)^0.5 / (2B) ] - tan^-1(R/L)

This form is important so that the curve can be rotated about the x-axis to give a solid. The parameters A, B, and C are the input variables in the Weight Estimator program. Parameter g will be defined later, but can be considered as an given value. Parameter D will be used when integrating the ogive curve.

Before integration can be done to find the volume and surface area of a truncated ogive curve about an axis, the distance, 'g' needs to be defined. The distance g can be determined using the other dimensions A, B, C,and D.

First, the location of x_o, y_o (center of circular arc for outside surface) needs to be determined. This point will also be used in other calculations. Note, the coordinate system is different in this diagram used to find x_o, y_o.

Knowing two points on a circular arc (points 1 and 2 in diagram), and the radius of the arc, B, the center point location can be determined (derivation not given here, basic geometry exercise). Points 1 and 2 are,

    (x₁, y₁) = (A/2, 0)
    (x₂, y₂) = (D/2, C)

then

   x_o = (D + A)/4 - (C/2) (b / a)
   y_o = (C/2) + [(D - A)/4)] (b / a)
     where a = [(D - A)²/16 + C²/4]^0.5
              b = [B² - (D - A)²/16 - C²/4]^0.5

This gives 'g' distance as,

   g = B cosα - C + y_o

Note, y_o can be ether positive or negative, depending on the location of point o.

With x_o known, angle α can be defined as

   α = sin^-1(x_o / B)

Note, α can be negative, depends on the sign of x_o.

Using distance 'g', the ogive curve can now be rotated about its center axis as shown in the first diagram (diagram with red ogive curve). Using the ogive curve, the volume can be calculated by considering a series of disks, and integrating them the from g to L (= C + g),

For simplification, let a₁ = B cosα and b₁ = B sinα. Completing the integration using math tables (steps not shown) gives,

V = h(x = C+g) - h(x= g)
   where h(x) = π [(B² + b₁² - a₁²) x + a₁ x² - x³/3]
                        + π b₁ { (x - a₁) [B² - (x - a₁)²]^0.5
                             + B² sin^-1[ (x - a₁)/B] }

This equation is similar to the volume equation for projectile shapes except this starts at g instead of 0.

Similarly, the surface can be determined by integrating the circumference of the element dx. The slope distance is included (square root term), giving

where a₁ = B cosα and b₁ = B sinα. While this looks messy, using basic algebra, it actually simplifies to

Completing the integration using math tables (steps not shown) gives,

   S = 2πB [ (C+g) + b₁ sin^-1[(a₁ - (C+g)) / B]
            - 2πB [ g + b₁ sin^-1[(a₁ - g) / B]

      = 2πB [C + b₁ sin^-1[(a₁ - C - g) / B]
                        - b₁ sin^-1[(a₁ - g) / B]

For the inside ogive, two points on the curve are needed, just like the outside curve. These two point will be referred to as the top and bottom points as noted in the diagram. To determine the top point, the shell thickness, t, and the top vertical thickness, t_p, are needed. From the diagram

   sinβ = y_o/B
and
   cosβ = t / F

Square both equations and add,

   t = [1 + (y_o/B)²]^0.5

The t and t_p thickness are related, (see Oval shape for derivation details),

   t_p = (2B/A) t

So the two points are,

   (x_top, y_top) = (D/2, C - t_p)

   (x_bottom, y_bottom) = (D/2, C - t_p)

Finally, the integration equations for the inside volume and surface (see above) need A_in, B_in, C_in and g_in for the inside ogive curve.

   A_in = A - 2F

   B_in = B - t
        = B - F [1 + (y_o/B)²]^0.5

   C_in = C - t_p

   g_in = B_in cosα_in - C_in + y_o

       where α_in = sin^-1(x_o / B_in)

and a_1-in = B_in cosα_in and b_1-in = B_in sinα_in.

For the final volume and surface equations, it is best to split the shape into three regions, 1, 2 and 3 as shown in the diagram. These regions are also referred to as out, in and tapered cylinder regions, respectively. Region 1 includes region 2 and 3 (2 and 3 overlap region 1).

Volume, V, total for the full Front Body is

   V = V_out - V_in - V_{tapered cylinder}

if E <= t_p. The last term is positive if E > t_p

Due to the large number of terms in this equation, they are not shown. In Weight Estimator code functions are used for each volume. See code for details.

Outer Surface Area, A_outer, includes the outer ogive surface plus the top and bottom edges.

  A_outer = 2πB [C + b₁ sin^-1[(a₁ - C - g) / B]
                        - b₁ sin^-1[(a₁ - g) / B]
                 + π(D/2)² + π[(A/2)² - (A/2 - F)²]

Inner Surface Area, A_inner, includes the inner ogive surface plus an approximate tamper cylinder above the inside ogive curve,

   A_inner = A_{inner ogive} - V_{tapered cylinder}

   A_inner= 2πB_in [C_in + b₁ sin^-1[(a₁ - C - g_in) / B_in]
                            - b₁ sin^-1[(a₁ - g_in) / B_in]
               + π (D/2 - t_p + E)²
                    + π(D - t_p + E) 2^0.5(t_p - E)

   A_Total = A_outer + A_inner

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

29 Dec 2020 - Kurt Gramoll
1) Previous equations were convoluted and could not be verified. New equations were derived. New equations match the old equations within 5%.

The Fuze Adapter object consists of a truncated conical top section and a solid cylindrical bottom section. The top section has an internal hollow cylindrical hole through its total depth. This hole does not go through the bottom cylinder section.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = π [(E+A)/2] [(A/2 - E/2)² +D² ]
      + π(E² - F²) /4 + π(A² - B²) /4 + πBC + π B²/4

Inner Surface Area, A_inner,

   A_inner = π FD + π F²/4

   A_Total = A_outer + A_inner

Volume, V, consists of a single cylinder.

   V = πD(E² + A² + AE) /12 + πB²C /4 - πF²D /4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Gear object is a basic gear shape. The gear teeth size can be changed, both depth and width. Also, the number of teeth can be modified. The program does not check if the teeth intercept each other if too many teeth or if the teeth are too big. The user should visually check and confirm that the shape is realistic and acceptable. The internal hole diameter can also be changed.

Technical Calculations

To simplify the volume and surface calculations, the gear teeth are modeled as trapezoid even though the top and bottom edges are slightly curved. .

Outer Surface Area, A_outer, includes two main surfaces, top/bottom and outer edge. The top and bottom are identical,

 A_top/bottom = 2 [π(A/2 + B)² - πA²/4 + (F/2 + E/2)CG]

The edge has three surfaces, end of the teeth, sides of the teeth, and gap between teeth. The diagram shows the dimensions of the teeth.

 A_{outer edge}= A_end + A_side + A_gap

     = FDG + [(E/2 - F/2)²+ C²]^0.5 D2G
                                     + [(A + 2B)π - GE] D

Note, the gap was calculated by taking the full circle perimeter at the gap, and subtracted the tooth root length, E.

Combining surfaces, gives

   A_outer = A_top/bottom + A_{outer edge}

            = 2 [π(A/2 + B)² - πA²/4 + (F/2 + E/2)CG]
                 + FDG + [(E/2 - F/2)²+ C²]^0.5 D2G
                     + [(A + 2B)π - GE] D

Inner Surface Area, A_inner, is just the inside ring,

   A_inner = π FD + π F²/4

   A_Total = A_outer + A_inner

Volume, V, is the top or bottom surface, multiplied by the gear thickness,

   V = = D [π(A/2 + B)² - πA²/4 + (F/2 + E/2)CG]

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

22 Dec 2020 - Kurt Gramoll
1) Previous equations were convoluted and could not be verified. New equations were derived. New equations match the old equations within 0.1%.

The Head object is basically a truncated conical section for the top, a middle cylinder and a cylindrical section for the bottom. The bottom cylindrical section has two internal hollow shapes, a cylinder and a cone.

Outer Surface Area, A_outer, is based on the top tampered cylinder, middle cylinder and bottom cylinder.

   A_outer = A _{tapered cyl} + A _{middle cyl} + A _{bottom cyl}

            = π (F/2 + G/2) [ (G/2 - F/2)² + E² ]^0.5
                + πF²/4 + π(G² - H²)/4 + πHD
                + π(A² - H²)/4 + π(B + C) + π(A² - I²)/4

Inner Surface Area, A_inner, is the inside hollow cone and cylinder,

   A_inner = π (I/2) [ (I/2)² + C²) ]^0.5 + π I B

   A_Total = A_outer + A_inner

Volume, V, involves the three main shapes and subtracting the two inside hollow shapes,

   V = V_{tapered cyl} + V_{2 cyl} - V_{hollow cone and cyl}

     = π E (F² + G² + FG) /12 + π H²D /4
          + π A²(C + B)/4 - π I²B /4 - π I²C /12

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

22 Dec 2020 - Kurt Gramoll
1) Previous equations did not separate inside and outside surfaces. The new equations have both the inside and outside surface areas. All equations were rearranged to match the equation format given in the help page.

The Housing object is one of the more complex shapes. Its basic shape consists of a top and bottom hollow cylinder and a conical transition section. The top cylindrical and conical section have the same internal radius, F. The lower cylindrical section has a separate internal radius, G.

The top section has two types of holes at different depths. The J holes (total of 3) has a depth equal to D plus C. The I holes (total of 3) has a depth that can be set by the user. Both hole types can have different diameters. But the 3 holes in either hole type will have the same diameter. I and J are the diameters of their respective holes.

Outer Surface Area, A_outer, is based on the top cylinder, middle tampered cylinder and bottom cylinder. To and bottom surface are included. Hole areas need to be subtracted The I hole ends needs to be subtracted from A_outer using an if statement in code.

   A_outer = A_{top cyl} + A_{tapered cyl} + A_{bot cyl} - A_{hole ends}

            = π(E² - F²)/4 + πED
                 + π (E/2 + A/2) [ (E/2 - A/2)² + C² ]^0.5
                + πAB + π(A² - G²)/4 - 3π(J² + I²)/4

Inner Surface Area, A_inner, is the inside is two hollow cylinders, and surface between the two inside cylinders, The I hole end needs to be subtracted from A_inner using an if statement in code.

   A_inner =  πF(D + C) + π(G² - F²)/4 + πGB
                    + 3 πJ(D+C) + 3π IH - 3πJ²/4

   A_Total = A_outer + A_inner

Volume, V, is based on the top cylinder, middle tampered cylinder and bottom cylinder. Inside hollow cylinders are subtracted.

   V = V_{tapered cyl} + V_{2 cyl} - V_{hollow cone and cyl}

      = πE²D /4 + πA²B /4 + πC(A² + E² + AE) /12
           - πG²B /4 - πF²(D + C) /4
              - 3πJ²(D + C) /4 - 3π I²H /4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

23 Dec 2020 - Kurt Gramoll
Various problems with original equations,
  1) Volume is missing top center hole going through section C, corrected
  1) Missing J hole end for inside
  1) Missing inside surface
  1) Result page has wrong J value (used I => _HoursingResult.cshtml page corrected)

The Lower Body represents a circular rotated shape similar to a circular sink bowl with a drain. The curvature of the side is defined by 2 points, G and A, and D and B, plus the radius length E. But the location of the radius center (the "+" symbol in the diagram) is not known, and must be determined by the program.

It is assumed that the thickness F is constant at all location.

The location of the arc center, the '+' symbol on diagram, is determined from the lower point of D and B dimension and the upper point of A and D dimension. Both these points are known.

As illustrated in the circle diagram, two points and the radius can define an arc curve without needing other dimensions. There is a second possible circular arc (dashed line), but it is not realistic so is not considered for this shape.

Technical Calculations
The Lower Body shape is similar to the top section of the Projectile shape but has the top point removed and is flipped vertically. Also, the Lower Body is hollow like the Hollow Projectile body. There are also various cylinders and disks attached to top and bottom (all have thickness of F.

First, the ogive curve section needs to be modeled as an equation. Then it can be revolved around a given line to create the solid. Recall, an ogive curve is a section of a circular arc as mentioned in projectile shape section. The ogive needs to be described as an equation in an x-y cartesian coordinate system so it can be integrated about the x-axis to generate a solid using the infinitesimal slice dx.

The diagram at the right shows a general ogive curve (red line) that describes the edge of the front body. The curve starts at the x-y coordinate origin even though the shape will start a distance 'g' from the origin. That curve is rotated about the x-axis giving the outside and inside Lower Body curved shape. The final curve equation is (see Solid Projectile for derivation details - note dimension letters may be different),

   y = [E² - (x - E cosα)²]^0.5 - E sinα

    where α = cos^-1[ (R² + L²)^0.5 / (2E) ] - tan^-1(R/L)

This form is important so that the curve can be rotated about the x-axis to give a solid. The parameters A, D, and E are the input variables in the Weight Estimator program. Parameter g will be defined later, but can be considered as an given value. Parameter B will be used when integrating the ogive curve.

Before integration can be done to find the volume and surface area of a truncated ogive curve about an axis, the distance, 'g' needs to be defined. The distance g can be determined using the other dimensions A, B, C,and D.

First, the location of x_o, y_o (center of circular arc for outside surface) needs to be determined. This point will also be used in other calculations. Note, the coordinate system is different in this diagram used to find x_o, y_o.

Knowing two points on a circular arc (points 1 and 2 in diagram), and the radius of the arc, E, the center point location can be determined (derivation not given here, basic geometry exercise). Points 1 and 2 are,

    (x₁, y₁) = (B/2, -D)
    (x₂, y₂) = (A/2, 0)

then

   x_o = (A + B)/4 - (D/2) (b / a)
   y_o = -D/2 + [(A - B)/4)] (b / a)
     where a = [(A - B)²/16 + D²/4]^0.5
              b = [E² - (A - B)²/16 - D²/4]^0.5

This gives 'g_i' distance as,

   g_i = E cosα_i - D - y_o

Note, y_o can be ether positive or negative, depending on the location of point o.

With x_o known, angle α_i can be defined as

   α_i = sin^-1(x_o / E)

Note, α_i can be negative, depends on the sign of x_o.

Using distance 'g' (g_i for inside curve and g_o for outside curve), the ogive curve can now be rotated about its center axis as shown in the first diagram (diagram with red ogive curve). Using the ogive curve, the volume can be calculated by considering a series of disks, and integrating them the from g to L (= D + g),

For simplification, let a₁ = E cosα and b₁ = E sinα. Completing the integration using math tables (steps not shown) gives,

V = h(x = D+g) - h(x= g)
   where h(x) = π [(E² + b₁² - a₁²) x + a₁ x² - x³/3]
                        + π b₁ { (x - a₁) [E² - (x - a₁)²]^0.5
                             + E² sin^-1[ (x - a₁)/E] }

This equation is similar to the volume equation for projectile shapes except this starts at g instead of 0. This equation can be used for either the inside or outside ogive curve.

Similarly, the surface can be determined by integrating the circumference of the element dx. The slope distance is included (square root term), giving

where a₁ = E cosα and b₁ = E sinα. While this looks messy, using basic algebra, it actually simplifies to

Completing the integration using math tables (steps not shown) gives,

   S = 2πE [ (D+g) + b₁ sin^-1[(a₁ - (D+g)) / E]
            - 2πE [ g + b₁ sin^-1[(a₁ - g) / E]

      = 2πE [D + b₁ sin^-1[(a₁ - D - g) / E]
                        - b₁ sin^-1[(a₁ - g) / E]

For the outside ogive, two points (3 and 4) on the curve are needed, just like the inside curve. These two point are noted as 3 and 4 in the diagram. With these points, E_o, g_o, and p can be determined, and used in the integration equation developed previously.

   E_o = E + F

   g_o = E_o [cos(α_o) - cos(β_o)]
        where α_o = sin^-1(x_o / E_o)
        and β_o = sin^-1[(-x_o + B/2 + F) / E_o]

   p = E_o cosα_o - F - y_o

The volume and surface integration equations are from g_o to p. Note, the integration equation constants are now a₁ = E_o cosα_o and b₁ = E_o sinα_o.

For the final volume and surface equations, it is best to split the shape into three regions, 1, 2 and 3 as shown in the diagram. These regions are also referred to as out, in and tapered cylinder regions, respectively. Region 1 includes region 2 and 3 (2 and 3 overlap region 1).

Volume, V, total for the full Lower Body is

   V = V₁ - V₂ + V₃ + V₄ - V₅

Due to the large number of terms in this equation, they are not shown. In Weight Estimator code functions are used for each volume. See code for details.

Outer Surface Area, A_outer, includes the outer ogive surface plus the flat outside surfaces. Note the top angle theta is used to locate the horizontal location of point 4,

  A_outer = π [(A/2 + G)² - (A/2 + G - F)²]
              + π [(B / 2 + F)² - (B / 2)²]
              + π (A + 2 G) (H + F)
              + π (B + 2 F) (C - F - p + g_o)
              + π [(A/2 + G)² - (E_o cosθ_o + x_o)²]
              + 2π E (p - g_o + b₁ sin^-1[(a₁ - p) /E_o]
                    - b₁ sin^-1[(a₁ - g_o) /E_o)]

Inner Surface Area, A_inner, includes the inner ogive surface plus inside flat surfaces,

   A_inner = π (A + 2 * G - 2 * F) H
               + π [(A/2 + G - F)² - (A/2)²]
               + π B (C - D)
               + 2π E (D + b₁ sin^-1[(a₁ - g_i - D) / E]
                     - b₁ sin^-1[(a₁ - g_i) / E]

   A_Total = A_outer + A_inner

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

4 Jan 2021 - Kurt Gramoll
1) Previous equations were convoluted and could not be verified. New equations were derived. New equations match the old equations within 5% using graphical methods.

The Nut shape is a basic hexagon shape that is extruded a distance B. There is a hollow hole at the center with radius C. The diameter, A, is the distance from one flat side to the opposite side. A is not the length of a single side.

Technical Calculations

Outer Surface Area, A_outer,

  A_outer = 3^0.5A² + 3^0.52AB - πC²/2

Inner Surface Area, A_inner,

   A_inner = πCB

   A_Total = A_outer + A_inner

Volume, V, consists of a single sphere.

   V = (3^0.5/2) A²B - π C²B/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

17 Dec 2020 - Kurt Gramoll
1) Previous equations excluded the bottom (or top) surface. Both volume and surface equations were simplified.

It should be noted, this shape is technically not based on an Ogive curve but has been given this name regardless. The shape is basically a tapered circular tube. The rate of taper and thickness can be set. The thickness is perpendicular to the side and is constant.

Technical Calculations

Volume, V

The volume is simply the outside tapered cylinder minus the inside tapered cylinder. The input value D is NOT the horizontal distance. The horizontal wall thickness, D_x, can be determined with trigonometry,

   tanθ = (A - C) / (2B)

   D_x = D / cosθ

Using the wall thickness, D_x, the volume is,

 V = (π/12) B (A² + C² + A C)
              - (π/12) B [(A - 2D_x)² + (C - 2D_x)²
                                      + (A - 2D_x) (C - 2D_x))]

Outer Surface Area, A_outer,

The outer surface consists of the outside side surface and the top and bottom edge. The outside surface equation was developed in the Ogive solid shape (next section below) and can be used here. The top and bottom edge surfaces are simple circular surfaces,

   A_outside = π (A/2 + C/2) [ (A/2 - C/2)² + B² ]^0.5

   A_edges = (π/4) [A² - (A - 2D_x)² + C² - (C - 2D_x)² ]

   A_outer = A_outside + A_edges

Inner Surface Area, A_inner,

The inner surface is just a smaller tapered cylinder. The same surface equation can be used for the sloped surface with the horizontal wall thickness, D_x, included,

  A_inner = (π/2) (A + C - 4 D_x)  [ (A/2 - C/2)² + B² ]^0.5

   A_Total = A_outer + A_inner

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

It should be noted, this shape is technically not based on an Ogive curve but has been given this name regardless. The shape is basically a solid tapered cylindrical object. The rate of taper can be set with the upper and lower radius.

Technical Calculations

Volume, V

The simplest way to calculate the volume of a tapered cylinder is to consider a cone with the top cut off. Simply subtract the top cone from the full cone, and what remains is a tapered cylinder. So, using the volume equation for a cone the volume will be,

   V = V_{full cone} - V_{top cone}
     = (π/3) (A/2)² h - (π/3) (A/2)² (h - B)
     = π [h(A² - C²) + C² B ] /12

The dimension h can be written in terms of A, B, and C giving h = B A / (A - C). Substitute and simplify,

   V = π [ A B (A² - C²) / (A - C) + B C²] /12

This can be rearranged, and simplified,

   V = π B [ A (A + C) (A - C) / (A - C) + C²] /12

   V = π B (A² + C² + A C) /12

Outer Surface Area, A_outer,

The side surface can be calculated as the slope length times the average circumference. It could also be integrated, but that would be more complex. The top and bottom circular surfaces also need to be included,

   L = [(A/2 - C/2)² + B²]

   s = (A + C) / 2

  A_outer = π (A/2 + C/2) [ (A/2 - C/2)² + B² ]^0.5
                + π [(A/2)² + (C/2)²]

Inner Surface Area, A_inner, does not exist.

   A_inner = 0

   A_Total = A_outer + A_inner

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

10 Dec 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used formulation that gave incorrect values if C > A, or if C = A. Now there are one set of equations.

The Oval shape is a circular arc rotated about a center line. If the arc length B is equal to half of A, then a sphere would be generated. The thickness can be set, and it is assumed to be constant.

The top and bottom vertical lines at the joint do not technically exist. They are simply an artifact from the 3D modeling program.

The dimension B is for the arc length from the arc circular center ('+' location) to the outside surface.

Technical Calculations
The hollow Oval consists of an outer and inner oval shape. Both ovals are similar but with different A and B variables, shown in the diagram as A_in and B_in.

The new B_in variable is simply the B minus the shell thickness, C. Giving,

   B_in = B - C

However, the inside oval A_in variable is not simply A minus C. The change in length of the oval depends on the angle a, as shown in the diagram. The angle can be calculated as,

   cosα = (A/2) / B

Using trigonometry equations on the blue triangle, the change in the shape length for one end is

   cosα = C / S

This is not exact since the angle α is based on the outside oval, and should be based on the inside oval. But the inside oval is not yet know since S is unknown. Using the outside oval is extremely close, and is used here.

Combining the two equations gives,

   S = (2B/A) C

So the new A_in variable is,

   A_in = A - 2C

The volume and surface equation were derived in the solid oval section below, and will not be repeated here. The final equations are,

   V = π [ (B² A - A³ /12
                     - B² [4B² - A²]^0.5 sin^-1[A / (2B)]

and

   Area_out = 2πBA - 2π [4B² - A²]^0.5 sin^-1[A / (2B)]

   Area_in = 2πB_inA_in
                    - 2π [4B_in² - A_in²]^0.5 sin^-1[A_in / (2B_in)]

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

11 Dec 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used the incorrect value (A-C) for the inside oval length, A. On the examples, this cause errors of 5-20%

The Oval shape is a circular arc rotated about a center line. If the arc length B is equal to half of A, then a sphere would be generated. This shape is a solid.

The dimension B is for the arc length from the arc circular center ('+' location) to the outside surface.

Technical Calculations
The shape is a simplification of the solid projectile shape. A circular arc (also called an ogive curve) is rotated about an arbitrary axis as shown in the diagram. The axis generally will not go through the center of curvature. The resulting shape is similar to a football (US), but not exactly (footballs are actually a prolate spheroid).

Since the shape is symmetrical about both axis, only one half of the curve will be used to calculate the volume and surface. Then the result will be doubled.

The input variables are only A (length from tip to tip) and B (radius of curvature for arc). Since the radius of curvature pivot point is at the centerline, the width of the shape can be determined from A and B

First, the ogive curve needs to be established. Then it can be revolved to create the solid. Recall, an ogive curve is a section of a circular arc as mentioned in introduction section above. The ogive needs to be described as an equation in an x-y cartesian coordinate system so it can be used in math equations to generate a solid.

The diagram at the right shows the ogive curve (red line) that describes the edge of the upper section of the oval. That curve is rotated about the x-axis giving half shape. The lower half shape is identical, so only the upper half is derived.

The ogive curve can then be modeled as a section of a circle the x'-y' coordinate system,

   x'² + y'² = D²

Next, t he x'-y' coordinate system can be translated into the x-y coordinate system. Using Pythagoras' theorem on the blue triangle for translation distances,

   (x - A/2)² + (y + [B² - A²/4]^0.5 )² = B²

Rearranging gives,

   y = [B² - (x - A/2)²]^0.5 - [B² - A²/4]^0.5

This form is important so that the curve can be rotated about the x-axis to give a solid. The parameters A and B are the input variables in the Weight Estimator program.

Using the ogive curve, the volume can be calculated by considering a series of disks, and integrating them the length of projectile, C,

For simplification, let b₁ = [B² - A²/4]^0.5. Rearranging the integration terms,

Completing the integration using math tables (steps not shown) gives,

  V = π [ (B² A - A³ /12 - B² [4B² - A²]^0.5 sin^-1[A/(2B)]

The above equation was doubled to give volume for both halfs.

Similarly, the surface can be determined by integrating the circumference of the element dx. The slope distance is included (square root term), giving

where b₁ = [B² - A²/4]^0.5. While this looks messy, using basic algebra, it actually simplifies to

Completing the integration using math tables (steps not shown) gives,

   S = 2πBA - 2π [4B² - A²]^0.5 sin^-1[A / (2B)]

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Pallet Lock Slide object is a unique shape for a particular application. The dimensions can be different for each pallet configuration. The hole location is not needed to determine volumes and surface area, thus there is not input dimension for the hole center. The program always places the hole at the origin.

Technical Calculations

Outer Surface Area, A_outer, consists of both top and bottom surface minus the area of the hole plus the edge surface.

   A_outer = 2 [ B(G + E) + C(E + F+ G) +DI ] - πH²/2
                + 2 [(E+F+G) + (B+C+D) + B] A

Inner Surface Area, A_inner, consists of the inside surface of the one hole

   A_inner =  πHA 

   A_Total = A_outer + A_inner

Volume, V, consists of the block, minus the one hole,

   V = A [ B(G + E) + C(E + F+ G) +DI ] - πAH²/4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

17 Dec 2020 - Kurt Gramoll
1) Surface equations incorrectly included inner surface with outer surface. New equation separated outer and inner surfaces.

Similar to the Anchor Pin, the Plate No. 1 Pin object has two tapered cylinders at each end. The tapered cylinders are assumed to go inward, thus A and should be smaller than D. The middle section is a cylindrical shape with a diameter of D.

All sections are considered to be solid.

Technical Calculations

The equation for a tapered cylinder section is derived in the "Solid Ogive" (not a real ogive shape). The derivation will not be duplicated here. The the tapered cylinder volumes are

   V_{Tapered Cyl} = π B (A² + D² + AD) /12

The outside surface area is just the average perimeter times the slope length (see "Solid Ogive" for derivation). This equation is only for the side and does not include the circular area at the top or bottom.

   S_{Tapered Side} = [π(D + A)/2] [(D - A)²/4 + B²]^0.5

Outer Surface Area, A_outer, consists of two tapered cylinders, top and and bottom circular area, and on solid cylinder side,

  A_outer = π(D + A) [(D - A)²/4 + B²]^0.5
              + πDC + πA²/2

Inner Surface Area, A_inner, no inside surface,

   A_inner = 0

   A_Total = A_outer + A_inner

Volume, V, consists of two solid tapered cylinders and on solid cylinder

   V = π B (A² + D² + AD) /6 + πD²C /4

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

While this object is called a Plate, it is assumed to be circular. For rectangular shaped plate, the Rectangular Solid shape should be used.

The plate can have up to four different hole types. Not all four types must be used (just leave those input fields blank). Each hole type can have a different number of holes (N), depth (H), and diameter (D). Each hole type is labeled with its number after N, H and D, such as N₂, H₂ and D₂.

The location of the holes is not needed for volume and surface area calculations. But it is assumed that the holes diameters are not large, and they will be contained inside the plate diameter, A. Also, it is assumed that a given hole depth does not exceed the plate thickness, C.

Technical Calculations

Outer Surface Area, A_outer, consists of the cylinder and subtraction of the holes (four possible types). Note, if any of the hole lengths, H, are equal to the block thickness, C, then there is no hole end. The other hole end needs to be subtracted from A_outter using an if statement in code.

   A_outer = A_block - A_{hole ends}
            = πA²/4 + πAB
                - (πD₁²/4) N₁ - (πD₂²/4) N₂
                - (πD₃²/4) N₃ - (πD₄²/4) N₄

Inner Surface Area, A_inner, consists of the hole cylindrical surface and one end. Note, if any of the hole lengths, H, are equal to the block thickness, C, then there is no hole end. The hole end needs to be subtracted from A_inner using an if statement in code.

   A_inner = A_{hole cylinders} + A_{hole ends}
            = (πD₁) H₁ N₁ + (πD₂) H₂ N₂
                        + (πD₃) H₃ N₃ + (πD₄) H₄ N₄
                 + (πD₁²/4) N₁ + (πD₂²/4) N₂ 
                        + (πD₃²/4) N₃  + (πD₄²/4) N₄ 

   A_Total = A_outer + A_inner

Volume, V, consists of the block, minus all four types of holes.

   V = πA²B/4 - (πD₁²/4) H₁ N₁ - (πD₂²/4) H₂ N₂
                     - (πD₃²/4) H₃ N₃ - (πD₄²/4) H₄ N₄

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

14 Dec 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, inside surface of holes only included one hole end even when N > 1.

The basic Prism object is a three sided extruded shape. For the "Hollow Prism A and C Open" shape, there is a hollow three sided shape going the length, D, of the prism. Both ends are open. The thickness, A, is constant and equal on all sides.

Technical Calculations
Before actual surfaces and volumes are calculated, various dimensions and angles are required. The diagram shows new constants that are needed.

Using the Law of Cosines,
   θ_A = cos^-1[ (B² + C² - A²) / (2 B C) ]
   θ_B = cos^-1[ (A² + C² - B²) / (2 A C) ]
   θ_C = cos^-1[ (A² + B² - C²) / (2 A B) ]

Inside triangle edge dimensions,
   A_i = A - E [ 1/tan(θ_C/2)+ 1/tan(θ_B/2) ]
   B_i = B - E [ 1/tan(θ_C/2)+ 1/tan(θ_A/2) ]
   C_i = C - E [ 1/tan(θ_B/2)+ 1/tan(θ_A/2) ]

To find the area A₁ (triangle area of ABC) and A₂ (triangle area of cut out A_iB_iC_i), use Heron's Formula,
   A₁ = [ P₁ (P₁ - A) (P₁ - B) (P₁ - C) ]^0.5
         where P₁ = (A + B + C) / 2
   A₂ = [ P₂ (P₂ - A_i) (P₂ - B_i) (P₂ - C_i) ]^0.5
         where P₂ = (A_i + B_i + C_i) / 2

Using the new inside triangle dimensions with the calculated areas A₁ and A₂, the volume, inside surface area and outside surface area are,
   V = (A₁ - A₂) D
   A_outer = (A + B + C) D + 2 (A₁ - A₂)
   A_inner = (A_i + B_i + C_i) D

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

27 Nov 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used arcsin which gave incorrect angle when any angle was over 90^o. The correct method uses the arccos formulation for the triangle angles. Derivation details are given above.

The basic Prism object is a three sided extruded shape. For the "Hollow Prism A or C Open" shape, there is a hollow three sided shape going the length, D, of the prism with one end capped. Only one end is open. The thickness, A, is constant and equal on all sides, including the capped end.

Technical Calculations
Before actual surfaces and volumes are calculated, various dimensions and angles are required. The diagram shows new constants that are needed.

Using the Law of Cosines,
   θ_A = cos^-1[ (B² + C² - A²) / (2 B C) ]
   θ_B = cos^-1[ (A² + C² - B²) / (2 A C) ]
   θ_C = cos^-1[ (A² + B² - C²) / (2 A B) ]

Inside triangle edge dimensions,
   A_i = A - E [ 1/tan(θ_C/2)+ 1/tan(θ_B/2) ]
   B_i = B - E [ 1/tan(θ_C/2)+ 1/tan(θ_A/2) ]
   C_i = C - E [ 1/tan(θ_B/2)+ 1/tan(θ_A/2) ]

To find the area A₁ (triangle area of ABC) and A₂ (triangle area of cut out A_iB_iC_i), use Heron's Formula,
   A₁ = [ P₁ (P₁ - A) (P₁ - B) (P₁ - C) ]^0.5
         where P₁ = (A + B + C) / 2
   A₂ = [ P₂ (P₂ - A_i) (P₂ - B_i) (P₂ - C_i) ]^0.5
         where P₂ = (A_i + B_i + C_i) / 2

Using the new inside triangle dimensions with the calculated areas A₁ and A₂, the volume, inside surface area and outside surface area are,
   V = (A₁ - A₂) D + E A₂
   A_outer = (A + B + C) D + (2A₁ - A₂)
   A_inner = (A_i + B_i + C_i) (D - E) + A₂

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

28 Nov 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used arcsin which gave incorrect angle when any angle was over 90^o. The correct method uses the arccos formulation for the triangle angles. Derivation details are given above.

The basic Prism object is a three sided extruded shape. For the "Hollow Prism B Open" shape, there is a hollow cut out on one side (labeled B). The cutout shape is also a basic three sided prism scaled the same as the full prism. Thus, the wall thickness is the same for all sides, including the two ends.

Technical Calculations
Before actual surfaces and volumes are calculated, various dimensions and angles are required. The diagram shows new constants that are needed.

Using the Law of Cosines,
   θ_A = cos^-1[ (B² + C² - A²) / (2 B C) ]
   θ_B = cos^-1[ (A² + C² - B²) / (2 A C) ]
   θ_C = cos^-1[ (A² + B² - C²) / (2 A B) ]

Using trigonometry,
   C_A = E / sinθ_B
   C_B = E / sinθ_A

Inside triangle edge dimensions,
   A_i = A - E [ 1/tanθ_B + 1/tan(θ_C/2) ]
   B_i = B - E [ 1/tanθ_A + 1/tan(θ_C/2) ]
   C_i = C - C_A - C_B
       = C - E [1/sinθ_B + 1/sinθ_A]

To find the area A₁ (triangle area of ABC) and A₂ (triangle area of cut out A_iB_iC_i), use Heron's Formula,
   A₁ = [ P₁ (P₁ - A) (P₁ - B) (P₁ - C) ]^0.5
         where P₁ = (A + B + C) / 2
   A₂ = [ P₂ (P₂ - A_i) (P₂ - B_i) (P₂ - C_i) ]^0.5
         where P₂ = (A_i + B_i + C_i) / 2

Finally, volume, inside surface area and outside surface area are,
   V = (A₁ - A₂) D
   A_outer = (B + A) D + (C_A + C_B) D + 2 (A₁ - A₂)
   A_inner = (B_i + A_i) D

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

27 Nov 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used arcsin which gave incorrect angle when any angle was over 90^o. The correct method uses the arccos formulation for the triangle angles. Derivation details are given above.
2) Old equations had division by zero, incorrect alpha angle. Corrected but old equations not used (see previous item number 1).

The basic Prism object is a three sided extruded shape. For the "Hollow Prism B Open V Shape " shape, there is a cutout along the length of the prism (D). The cutout shape is also a simple three sided prism scaled the same as the full prism. One side of the cutout touches side C so the final shape looks like a 'V". The wall thickness is the same for all sides, including the two ends.

Technical Calculations
Before actual surfaces and volumes are calculated, various dimensions and angles are required. The diagram shows new constants that are needed.

Using the Law of Cosines,
   θ_A = cos^-1[ (B² + C² - A²) / (2 B C) ]
   θ_B = cos^-1[ (A² + C² - B²) / (2 A C) ]
   θ_C = cos^-1[ (A² + B² - C²) / (2 A B) ]

Using trigonometry,
   C_A = E / sinθ_B
   C_B = E / sinθ_A

Inside triangle edge dimensions,
   A_i = A - E [ 1/tanθ_B + 1/tan(θ_C/2) ]
   B_i = B - E [ 1/tanθ_A + 1/tan(θ_C/2) ]
   C_i = C - C_A - C_B
       = C - E [1/sinθ_B + 1/sinθ_A]

To find the area A₁ (triangle area of ABC) and A₂ (triangle area of cut out A_iB_iC_i), use Heron's Formula,
   A₁ = [ P₁ (P₁ - A) (P₁ - B) (P₁ - C) ]^0.5
         where P₁ = (A + B + C) / 2
   A₂ = [ P₂ (P₂ - A_i) (P₂ - B_i) (P₂ - C_i) ]^0.5
         where P₂ = (A_i + B_i + C_i) / 2

Finally, volume, inside surface area and outside surface area are,
   V = (A₁ - A₂) D
   A_outer = (B + A) D + (C_A + C_B) D + 2 (A₁ - A₂)
   A_inner = (B_i + A_i) D

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

28 Nov 2020 - Kurt Gramoll
1) Equations were modified in the program to match the above equations. Previously, the calculations used arcsin which gave incorrect angle when any angle was over 90^o. The correct method uses the arccos formulation for the triangle angles. Derivation details are given above.

The basic Prism object is a simple three sided extruded shape. For the "Solid Prism" shape, there center is solid.

Technical Calculations
The area of a triangle, A_tri, can be calculated if the length of each side is known. This is called Heron's Formula. Using A, B, and C as the length of each side, the triangle area is

    
        where p = (A + B + C) / 2

Outer Surface Area, A_outer, is calculated by using the area of each of the two triangular ends, A_tri, and adding the three rectangular surfaces between the two ends.

    A_outer = 2 A_tri + (A + B + C) D

Volume, V, of the solid prism is the area of the triangular end, A_tri, times its length.

    V = D A_tri

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

The Projectile Body shape is only one of two shapes that are over defined, caused an over constraint system (the other is the Upper Body). It can be noted, that I, J with E and F can be used to determine the location of the radius arc for both the top and bottom curves. But, this arc curve center and length can also be determined by the two points on the curve and a tangent line at one of the points. Thus, it is over constraint.

This shape has 5 redundant dimensions, I, J, C, D and G. G is defined with E and F. The inside dimensions C and D are defined by the outside dimensions and the thickness set at L (vertical distance).

If I, J, C, D and G are not used, then the shape can be determined. The actual weight equations do use I, J, C, D and G, so it has not been deleted from the Weight Estimator input page.

This over constraint condition can be seen by noticing the inside arc, top or bottom half, can be defined by two points on the curve and one tangent line. The circle diagram shows that two points and a tangent line through one of the points can define a curve without needing the radius I or J, or the location of the center point. The two points on the shape are defined by A and F (also the tangent line), and by B and F for the lower portion. The upper portion is similar.

4 Jan 2021 - Kurt Gramoll
1) Equations not correct. Cannot update since the shape is not physically possible. Input variables need to be reduced.

The Hollow Projectile A Closed shape has an ogive (a partial circular arc curve) top section and a cylindrical bottom section. Both sections are hollow. The rate of curvature of the top section can be changed by modifying the radius dimension, D. The location of arc cure ('+' symbol) is determined by the program to match the other dimensions and D.

It is possible to have a non-tangent connection between the top and bottom sections at their common joint.

The thickness is defined by the difference of A and E (and divided by 2), and is constant in both sections.

The bottom is enclosed. The thickness of the bottom is assumed to be the same as the upper shell wall.

The horizontal line between the ogive section and cylinder, the top vertical line at the joint, and vertical line at the bottom center do not technically exist. They are artifacts from the 3D modeling program.

The location of the arc center, the '+' symbol on diagram, is determined from the lower point of C dimension (adjacent to B dimension) and the tip. Both these points are known.

As illustrated in the circle diagram, two points and the radius can define an arc curve without needing other dimensions. There is a second possible circular arc (dashed line), but it is not realistic so is not considered for the shape.

Technical Calculations
There are two sub-shapes that make up the Hollow Projectile with a closed end, an ogive solid of revolution and cylinder. Cylinder surface and volume is well known, but the ogive solid surface and solid equations need to be derived and explained. This section will concentrate on the ogive solid (curved top section).

First, the ogive curve needs to be established. Then it can be revolved to create the solid. Recall, an ogive curve is a section of a circular arc as mentioned in introduction section above. The ogive needs to be described as an equation in an x-y cartesian coordinate system so it can be integrated about the x-axis to generate a solid.

The diagram at the right shows the Ogive Curve (red line) that describes the edge of the upper section of the projectile. That curve is rotated about the x-axis giving the upper solid projectile shape. The lower shape is the cylindrical section.

   y = [D² - (x - D cosα)²]^0.5 - D sinα

This form is important so that the curve can be rotated about the x-axis to give a solid. The parameters A, B, C and D are the input variables in the Weight Estimator program.

Using the ogive curve, the volume can be calculated as (see Solid Projectile for details),

V = π [ (D² + b₁² - a₁²) C + a₁ C² - C³/3]
   - π b₁ { (C-a₁) [D² - (C-a₁)²]^0.5 + D² sin^-1[ (C-a₁)/D] }
   + π b₁ { (-a₁) [D² - a₁²]^0.5 + D² sin^-1[ (-a₁)/D] }

where a₁ = D cosα and b₁ = D sinα

The surface is,

S = 2πD [ C + b₁ sin^-1[(a₁ - C)/D] + b₁ sin^-1(C/D) ]

The V and S equations work for both solid and hollow sections of the shape.

The vertical wall thickness at the point, t_p, is not the same as the wall thickness at the tangent point, t_t. nor the perpendicular thickness, t,at any point on the curve. This can be seen in the diagram at the right. Therefore, new dimensions D_in, C_in are needed to calculate the inside hollow volume and surface using the above equations. If the known t_t is used to find D_in and C_in, there error will be 3-4% for thick walled projectiles.

The inside radius is E/2 where E is a known input variable. The new inside radius of curvature, D_in, can be found using the angle α and Pythagoras' theorem (triangle 1),

   D_in = [ (D sinα + E/2)² + (D cosα - C)² ]^0.5

Note, E, D, and C are the input variables and are not calculated variables. The angle α is calculated from the equation derived in the solid projectile section.

The distance C_in is also needed. It can also be found angle α and pythagoras' theorem (triangle 2) along with the new dimension, D_in,

   D_in² = [ (D sinα)² + (D cosα - C + C_in)² ]^0.5

   C_in = [ D_in²+ (D sinα)²]^0.5 - (D cosα - C)

The new variables, D_in and C_in are used for interior hollow volume and inner surface.

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

5 Dec 2020
1) The equations in WE were changed to the equations derived above. The old equations did not adjust D for smaller inside hollow shape and they incorrectly used wall thickness at tangent for vertical wall thickness at the point (top). The wall thickness perpendicular to the surface is constant. Also, the 'offset' variable (calculated in subroutine) did not adjust A, C or D for the smaller shape. Error was 3-4% for thick projectile shapes.
2) LEFT OUT weight of the bottom disk, missing volume = π A² (A-E)/4. Fixed so this volume is included.

The Hollow Projectile A Open shape has an ogive (a partial circular arc curve) top section and a cylindrical bottom section. Both sections are hollow. The rate of curvature of the top section can be changed by modifying the radius dimension, D. The location of arc cure ('+' symbol) is determined by the program to match the other dimensions and D.

It is possible to have a non-tangent connection between the top and bottom sections at their common joint.

The thickness is defined by the difference of A and E (and divided by 2), and is constant in both sections.

The horizontal line between the ogive section and cylinder, and the top vertical line at the joint do not technically exist. They are artifacts from the 3D modeling program.

The dimension D is for the arc length from the arc circular center ('+' location) to the outside surface. Dimension D does not represent an angle.

The location of the arc center, the '+' symbol on diagram, is determined from the lower point of C dimension (adjacent to B dimension) and the tip. Both these points are known.

As illustrated in the circle diagram, two points and the radius can define an arc curve without needing other dimensions. There is a second possible circular arc (dashed line), but it is not realistic so is not considered for the shape.

Technical Calculations
There are two sub-shapes that make up the Hollow Projectile with an open end, an ogive solid of revolution and cylinder. Cylinder surface and volume is well known, but the ogive solid surface and solid equations need to be derived and explained. This section will concentrate on the ogive solid (curved top section).

First, the ogive curve needs to be established. Then it can be revolved to create the solid. Recall, an ogive curve is a section of a circular arc as mentioned in introduction section above. The ogive needs to be described as an equation in an x-y cartesian coordinate system so it can be integrated about the x-axis to generate a solid.

The diagram at the right shows the Ogive Curve (red line) that described the edge of the upper section of the projectile. That curve is rotated about the x-axis giving the upper solid projectile shape. The lower shape is the cylindrical section.

   y = [D² - (x - D cosα)²]^0.5 - D sinα

This form is important so that the curve can be rotated about the x-axis to give a solid. The parameters A, B, C and D are the input variables in the Weight Estimator program.

Using the ogive curve, the volume can be calculated as (see Solid Projectile for details),

V = π [ (D² + b₁² - a₁²) C + a₁ C² - C³/3]
   - π b₁ { (C-a₁) [D² - (C-a₁)²]^0.5 + D² sin^-1[ (C-a₁)/D] }
   + π b₁ { (-a₁) [D² - a₁²]^0.5 + D² sin^-1[ (-a₁)/D] }

where a₁ = D cosα and b₁ = D sinα

The surface is,

  S = 2πD [ C + b₁ sin^-1[(a₁ - C)/D] + b₁ sin^-1(C/D) ]

The V and S equations work for both solid and hollow sections of the shape.

The vertical wall thickness at the point, t_p, is not the same as the wall thickness at the tangent point, t_t. nor the perpendicular thickness, t, at any point on the curve. This can be seen in the diagram at the right. Therefore, new dimensions D_in, C_in are needed to calculate the inside hollow volume and surface using the above equations. If the known t_t is used to find D_in and C_in, there error will be 3-4% for thick walled projectiles.

The inside radius is E/2 where E is a known input variable. The new inside radius of curvature, D_in, can be found using the angle α and pythagoras' theorem (triangle 1),

   D_in = [ (D sinα + E/2)² + (D cosα - C)² ]^0.5

Note, E, D, and C are the input variables and are not calculated variables. The angle α is calculated from the equation derived in the solid projectile section.

The distance C_in is also needed. It can also be found angle α and pythagoras' theorem (triangle 2) along with the new dimension, D_in,

   D_in² = [ (D sinα)² + (D cosα - C + C_in)² ]^0.5

   C_in = [ D_in²+ (D sinα)²]^0.5 - (D cosα - C)

The new variables, D_in and C_in are used for interior hollow volume and inner surface.

Conversion Constants to
determine part weight in lb
1 gm/cm³ = 2.2046×10^-6 lb/mm³
= 0.036127 lb/in³

5 Dec 2020
1) The equations in WE were changed to the equations derived above. The old equations did not adjust D for smaller inside hollow shape and they incorrectly used wall thickness at tangent for vertical wall thickness at the point (top). The wall thickness perpendicular to the surface is constant. Also, the 'offset' variable (calculated in subroutine) did not adjust A, C or D for the smaller shape. Error was 3-4% for thick projectile shapes.

The Solid Projectile A Open shape has an ogive (a partial circular arc curve) top section and a cylindrical bottom section. Both sections are solid. The rate of curvature of the top section can be changed by modifying the radius dimension, D. The location of arc cure ('+' symbol) is determined by the program to match the other dimensions and D.

It is possible to have a non-tangent connection between the top and bottom sections at their common joint.

The horizontal line between the ogive section and cylinder section does not technically exist. It is an artifact from the 3D modeling program.

The dimension D is for the arc length from the arc circular center ('+' location) to the outside surface. Dimension D does not represent an angle.

The location of the arc center, the '+' symbol on diagram, is determined from the lower point of C dimension (adjacent to B dimension) and the tip. Both these points are know.

As illustrated in the circle diagram, two points and the radius can define an arc curve without needing other dimensions. There is a second possible circular arc (dashed line), but it is not realistic so is not considered for the shape.