OpenVSP

$ c_i \equiv \mbox{Chord length at span station} $
$ b_i \equiv \mbox{Section span} $
$ S_i \equiv \mbox{Section area} $
$ S_{total} \equiv \mbox{Total Area} $
$ \bar{c} \equiv \mbox{Weighted chord sum} $
$ L_{ref} \equiv \mbox{Reference length} $
$ D \equiv \mbox{Diameter} $
$ X_{area} \equiv \mbox{Max cross sectional area} $
$ F \equiv \mbox{Fineness ratio} $
$ Re \equiv \mbox{Reynolds number} $
$ V_{inf} \equiv \mbox{freestream velocity} $
$ \nu \equiv \mbox{kinematic viscosity} $
$ C_{f} \equiv \mbox{friction coefficient} $
$ f \equiv \mbox{flat plate drag} $
$ S_{wet} \equiv \mbox{wetted area} $
$ S_{ref} \equiv \mbox{reference area} $
$ C_D \equiv \mbox{coefficient of drag} $

Wetted areas for all components are taken from comp geom and as such as are appropriately trimmed.

Reference length is calculated according to it's geometry type internal to VSP; either a “Wing” type or a “Body” type. If the result of these calculations are zero, due to rotation or abnormal geometry shape the other equation is attempted as a fail safe. If neither of these methods yield a result greater than zero, a default value of 1.0 is used as to prevent division by zero when calculating fineness ratio.

Wing reference length is found by taking the area weighted chord.

First find the width of the wing section. $$ \Delta_x = |x_{le(1)} - x_{le(end)}| $$ $$ \Delta_y = |y_{le(1)} - y_{le(end)}| $$ $$ \Delta_z = |z_{le(1)} - z_{le(end)}| $$ $$ b_i = \sqrt{ {\Delta_x}^2 + {\Delta_y}^2 + {\Delta_z}^2} $$ $$ S_i = b_i * \frac{c_i + c_{\left(i+1\right)}}{2} $$ $$ S_{total} = \sum\left(S_i\right) $$ $$ \bar{c} = \sum\left(\frac{c_i + c_{\left(i+1\right)}}{2}\right) $$ $$ L_{ref} = \frac{\bar{c}}{S_{total}} $$

Body geometry reference lengths are calculated by taking the distance between to the front and back ends of the degenerated stick. $$ \Delta_x = |x_{le(1)} - x_{le(end)}| $$ $$ \Delta_y = |y_{le(1)} - y_{le(end)}| $$ $$ \Delta_z = |z_{le(1)} - z_{le(end)}| $$ $$ L_{ref} = \sqrt{ {\Delta_x}^2 + {\Delta_y}^2 + {\Delta_z}^2} $$

Thickness to chord takes the max thickness to chord from the degenerate stick created from Degen Geom.

Fineness Ratio is calculated by taking the diameter calculated using the max cross sectional area taken from the degenerate stick from Degen Geom.

$$ D = 2 * \sqrt{\frac{X_{area}}{\pi}} $$ $$ F = \frac{D}{L_{ref}} $$

Depending on the atmosphere input type, the kinematic viscosity is calculated accordingly and used to find the Reynolds number for the geometry.

$$ Re = \frac{V_{inf} * L_{ref}}{\nu} $$

Laminar Percent takes a 0 to 100 value and at default is set to 0% for a fully turbulent flow.

The laminar friction coefficient is calculated using the following equation: $$ C_f = C_{f (100\%Turb)} - \left(C_{f (\% Partial Turb)} * \% Lam\right) + \left(C_{f (\% Partial Lam)} * \% Lam\right) $$

where (using a chosen friction coefficient and it's corresponding inputs) $$ C_{f (\% Partial Turb)} = f\left(Re_{Lam}\right) $$ $$ C_{f (\% Partial Lam)} = f\left(Re_{Lam}\right) $$ where $$ Re_{Lam} = Re * \% Lam $$

$ Q \equiv \mbox{Scale factor applied to drag coefficient} $

$$ f = S_{wet} * Q * C_f * FF $$

$$ C_D = \frac{f}{S_{ref}} $$

Utilizing this feature the user is able to combine the wetted area of any geometry with that of another. Currently, geometries can only be grouped with their ancestors and geometries of the same shape type.

This can be used for example if the gear pod is modeled seperately from the fuselage but the wetted area of the gear pod should be applied with the drag qualities (e.g. length, form factor, etc.) of the fuselage.

By default, subsurfaces are incorporated as a part of the geometry as a whole. In other words, the surfaces do not subtract any wetted area from the geometry or have any of their own unique properties. However, the parasite drag tool let's the user choose these as options if they desire.

The three options are:

• Treat as Parent
• Separate Treatment
• Zero Drag

Treat as Parent: The default option, incorporates the wetted area of the subsurface as part of a continuous geometry.

Separate Treatment: Allows the user, to some extent, control the qualities of the subsurface. However, due to limitations of the methodology used, the geometry based qualities (e.g. L_{ref}, Re, etc.) are derived from the parent geometry. The user is allowed to change the form factor equation type, manually set the laminar percentage, and manually set the interference factor for the subsurface.

Zero Drag:The subsurface wetted area is subtracted from the total wetted area of the geometry and no longer contributes to the drag of the component.

$ x \equiv \mbox{distance along chord} $
$ k \equiv \mbox{roughness height} $
$ \gamma \equiv \mbox{specific heat ratio} $
$ r \equiv \mbox{recovery factor} = 0.89 $
$ n \equiv \mbox{viscosity power-law exponent} = 0.67 $

$ l \equiv \mbox{length of component} $
$ d \equiv \mbox{diameter of component} $
$ w \equiv \mbox{width at maximum cross sectional area} $
$ h \equiv \mbox{height at maximum cross sectional area} $
$ FR \equiv \mbox{Covert Fineness Ratio} = \frac{l}{\sqrt{wh}} $
$ l_{r} \equiv \mbox{Length of Fuselage} $
$ A_{x} \equiv \mbox{Cross Sectional Area of Fuselage} $
$ Q \equiv \mbox{Interference Factor} $

$$ C_f = \frac{1.32824}{Re^{1/2}}\ $$

$$ C_f = \frac{0.455}{\ln^2(0.06\ Re)}\ $$

$$ C_f = \frac{0.225}{\log(Re)^{2.32}}\ $$

$$ \frac{1}{\sqrt{C_f}}\ = 3.46\log\left(Re\right) - 5.6 $$

$$ \log\left(Re\ C_f\right) = \frac{0.242}{\sqrt{C_f}}\ $$

$$ \frac{1}{\sqrt{C_f}}\ = 4.15\log\left(Re\ C_f\right) + 1.70 $$

$$ \frac{1}{\sqrt{C_f}}\ = 4.13\log\left(Re\ C_f\right) $$

$$ C_f = \frac{0.0592}{Re^{1/5}}\ $$

$$ C_f = \frac{0.074}{Re^{1/5}}\ $$

$$ C_f = \frac{0.027}{Re^{1/7}}\ $$

$$ C_f = \frac{0.058}{Re^{1/5}}\ $$

$$ C_f = \frac{0.455}{\log\left(Re\right)^{2.58}}\ $$

$$ C_f = \frac{0.472}{\log\left(Re\right)^{2.5}}\ $$

$$ C_f = \frac{1}{\left(2\ \log\left(Re\right) - 0.65\right)^{2.3}}\ $$

$$ C_f = \frac{0.370}{\log\left(Re\right)^{2.584}}\ $$

$$ C_f = \frac{0.427}{\left(\log\left(Re\right) - 0.407\right)^{2.64}}\ $$

$$ C_f = \frac{0.42}{\ln^2\left(0.056\ Re\right)}\ $$

$$ C_f = \left(1.89 + 1.62\ \log\left(\frac{l}{k}\right)\right)^{-2.5} $$

$$ C_f = \left(2.87 + 1.58\ \log\left(\frac{x}{k}\right)\right)^{-2.5} $$

$$ C_f = \left(1.4 + 3.7\ \log\left(\frac{x}{k}\right)\right)^{-2} $$

$$ C_f = \frac{\left(1.89 + 1.62\ \log\left(\frac{l}{k}\right)\right)^{-2.5}}{\left(1 + \frac{\gamma - 1}{2}\ M_\infty\right)^{0.467}} $$

$$ f = \frac{\left(1 + 0.22\ r\ \frac{\gamma - 1}{2}\ {M_e}^2\ \frac{Te}{Tw}\right)}{\left(1 + 0.3\ \left(\frac{Taw}{Tw} - 1\right)\right)} $$ $$ C_f = \frac{0.451\ f^2\ \frac{Te}{Tw}}{\ln^2\left(0.056\ f\ \frac{Te}{Tw}^{1+n}\ Re\right)} $$

$$ FF = 1 + \frac{t}{c}\ \left(2.94206 + \frac{t}{c}\ \left(7.16974 + \frac{t}{c}\ \left(48.8876 + \frac{t}{c}\ \left(-1403.02 + \frac{t}{c}\ \left(8598.76 + \frac{t}{c}\ \left(-15834.3\right)\right)\right)\right)\right)\right) $$

$$ FF = 1 + 4.275\ \frac{t}{c} $$

Recreated Data from DATCOM is shown in the Figure and is used to find the Appropriate Scale Factor for use in the DATCOM Equation through interpolation. $$ FF = \left[1 + L\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4\right] * R_{L.S.} $$

$$ FF = 1 + 2\ \left(\frac{t}{c}\right) + 60\ \left(\frac{t}{c}\right)^4 $$

$$ FF = 1 + Z\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4 $$ $$ Z = \frac{\left(2-M^2\right) \cos\left(\Lambda_{\frac{c}{4}}\right)} {\sqrt{1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)}} $$

$$ FF = 1 + \frac{2.2 \cos^2\left(\Lambda_{\frac{c}{4}}\right)} {\sqrt{1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)}} \left(\frac{t}{c}\right) + \frac{4.84 \cos^2\left(\Lambda_{\frac{c}{4}}\right) \left(1 + 5\cos^2\left(\Lambda_{\frac{c}{4}}\right)\right)} {2\left(1-M^2\cos^2\left(\Lambda_{\frac{c}{4}}\right)\right)} \left(\frac{t}{c}\right)^2 $$

$$ FF = 1 + 2.7\ \left(\frac{t}{c}\right) + 100\ \left(\frac{t}{c}\right)^4 $$

$$ FF = 1 + 1.8\ \left(\frac{t}{c}\right) + 50\ \left(\frac{t}{c}\right)^4 $$

$$ FF = 1 + 1.44\left(\frac{t}{c}\right) + 2\left(\frac{t}{c}\right)^2 $$

$$ FF = 1 + 1.68\left(\frac{t}{c}\right) + 3\left(\frac{t}{c}\right)^2 $$

$$ F^* = 1 + 3.3\left(\frac{t}{c}\right) - 0.008\left(\frac{t}{c}\right)^2 + 27.0\left(\frac{t}{c}\right)^3 $$ $$ FF = \left(F^* - 1\right)\left(cos^2\left(\Lambda_{\frac{c}{2}}\right)\right) + 1 $$

$$ F^* = 1 + 3.52\left(\frac{t}{c}\right) $$ $$ FF = \left(F^* - 1\right)\left(cos^2\left(\Lambda_{\frac{c}{2}}\right)\right) + 1 $$ $$ Q = 1.2 $$

$$ FF = 1 + \frac{60}{FR^3} + 0.0025\ FR $$

$$ FF = 1 + \frac{0.35}{FR} $$

$$ FF = 1 + \frac{1.5}{\left(\frac{l}{d}\right)^{1.5}} + \frac{7}{\left(\frac{l}{d}\right)^3} $$

$$ FF = 1 + \frac{2.2}{\left(\frac{l}{d}\right)^{1.5}} + \frac{3.8}{\left(\frac{l}{d}\right)^3} $$

$$ FF = 1 + \frac{2.8}{\left(\frac{l}{d}\right)^{1.5}} + \frac{3.8}{\left(\frac{l}{d}\right)^3} $$

$$ FF = 1.02\left(1.0 + \frac{1.5}{\left(\frac{l}{d}\right)^{1.5}} + \frac{7.0}{\left(\frac{l}{d}\right)^3 \left(1.0 - M^3\right)^{0.6}}\right) $$

$$ \Lambda = \left(\frac{l_{r}}{\frac{4}{\pi}A_{x}}\right)^{0.5} $$ $$ FF = 1 + \frac{2.2}{\Lambda^{1.5}} - \frac{0.9}{\Lambda^3} $$

Jenkinson suggests a constant Form Factor for typical nacelles on wings. $$ FF = 1.25 $$

Jenkinson suggests a constant Form Factor for typical nacelles on aft fuselages. $$ FF = 1.50 $$

The parasite drag tool includes several options for atmosphere models as well as the option for the user to have control over 2 atmospheric qualities between Temperature, Pressure, and Density.

Upper Limit: 82,021 feet

Upper Limit: 84,852 meters

$ \frac{t}{c} \equiv \mbox{thickness to chord ratio of selected geometry} $
$ M \equiv \mbox{freestream Mach number for flight condition} $
$ M_{cr} \equiv \mbox{Critical Mach number, the point at which drag creep begins to occur} $
$ M_{DD} \equiv \mbox{Drag Divergence Mach number, the point at which drag significantly begins to rise} $
$ M_{DD,eff} \equiv \mbox{Effective Drag Divergence Mach number, Drag Divergence Mach number with consideration fro wing sweep} $
$ \Delta_{CD} \equiv \mbox{Additional CD due to Transonic Drag effects} $
$ \phi_{25} \equiv \mbox{Average quarter chord sweep of selected geometry} $
$ \gamma \equiv \mbox{Specific heat ratio; typically 1.4} $
$ C_{L} \equiv \mbox{Design Lift Coefficient} $

Example plots are given showing Drag Divergence Mach number using each equation with contours of quarter chord sweep angle at 0°, 15°, 30°, 45°, 60°, and 75° utilizing the following inputs: $$ C_{L} = 1.0 $$ $$ Airfoil Type = Conventional $$ $$ \gamma = 1.4 $$ $$ A_{F} = 0.8 $$ $$ K_{A} = 0.8 $$

$$ \Delta_{CD} = 20 * \left(M - M_{cr}\right)^4 $$

$ M^{*} \equiv \mbox{1.0, conventional airfoils; maximum t/c at about 0.30c} $
$ M^{*} \equiv \mbox{1.05, high-speed (peaky) airfoils, 1960-1970 technology} $
$ M^{*} \equiv \mbox{1.12 to 1.15, supercritical airfoils [Conservative = 1.12; Optimistic = 1.15]} $
$$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ \frac{t}{c} = 0.30\cos{\phi_{25}}\left(\left(1 - \left( \frac{5 + {M_{DD,eff}}^2}{5 + {M^*}^2}\right)^{3.5}\right)\frac{\sqrt{1 - {M_{DD,eff}}^2}}{{M_{DD,eff}}^2}\right)^{\frac{2}{3}} $$

$ M^{*} \equiv \mbox{1.0, conventional airfoils; maximum t/c at about 0.30c} $
$ M^{*} \equiv \mbox{1.05, high-speed (peaky) airfoils, 1960-1970 technology} $
$ M^{*} \equiv \mbox{1.12 to 1.15, supercritical airfoils [Conservative = 1.12; Optimistic = 1.15]} $
$$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ \frac{t}{c} = 0.30\cos{\phi_{25}}\left(\left(1 - \left( \frac{5 + {M_{DD,eff}}^2}{5 + \left(k_{M} - 0.25c_{l}\right)^2}\right)^{3.5}\right)\frac{\sqrt{1 - {M_{DD,eff}}^2}}{{M_{DD,eff}}^2}\right)^{\frac{2}{3}} $$

Equation solved for M. $$ A = \frac{M^2\cos^2{\phi_{25}}}{\sqrt{1-M^2\cos^2{\phi_{25}}}}\left(\left(\frac{\gamma + 1}{2}\right)\frac{2.64\frac{t}{c}}{\cos{\phi_{25}}} + \left(\frac{\gamma + 1}{2}\right)\frac{2.64\left(\frac{t}{c}\right)\left(0.34C_{L}\right)}{\cos^3{\phi_{25}}}\right) $$ $$ B = \frac{M^2\cos^2{\phi_{25}}}{1-M^2cos^2{\phi_{25}}}\left(\left(\frac{\gamma + 1}{2}\right)\left(\frac{1.32\frac{t}{c}}{\cos{\phi_{25}}}\right)^2\right) $$ $$ C = M^2cos^2{\phi_{25}}\left(1 + \left(\frac{\gamma + 1}{2}\right)\frac{\left(0.68C_{L}\right)}{\cos^2{\phi_{25}}} + \frac{\gamma + 1}{2}\left(\frac{0.34C_{L}}{\cos^2{\phi_{25}}}\right)^2\right) $$

$$ A + B + C - 1 = 0 $$

If a Peaky Airfoil Type is selected $$ M_{DD} = M + 0.06 $$ otherwise $$ M_{DD} = M $$

$ M_{cc} \equiv \mbox{Crest Critical Mach number} $
$$ a = 0.2 $$ $$ b = 2.131 $$ $$ x = \frac{\frac{t}{c}}{\cos{\phi_{25}}} $$ $$ y = \frac{C_{L}}{{\left(\cos{\phi_{25}}\right)}^2} $$

$$ M_{cc} = \frac{2.8355x^2 - 1.9072x + 0.949 - a\left(1-bx\right)y}{\cos{\phi_{25}}} $$

If Conventional Airfoil Type: $$ \Delta_{CD} = 0.04 $$ If Peaky Airfoil Type: $$ \Delta_{CD} = 0 $$ If Supercritical Conservative: $$ \Delta_{CD} = -0.04 $$ If Supercritical Optimistic: $$ \Delta_{CD} = -0.06 $$

$$ M_{DD} = M_{cc} * \left(1.025 + 0.08\left(1-\cos{\phi_{25}}\right)\right) - \Delta_{CD} $$

$ A_{F} \equiv \mbox{Airfoil Technology Factor, typically between 0.8 and 0.95} $
$$ M_{DD,eff} = M_{DD} * \sqrt{\cos{\phi_{25}}} $$ $$ M_{DD,eff} = A_{F} - 0.1C_{L} - \frac{t}{c} $$

$$ \frac{t}{c} = 0.7185 + 3.107e^{-5}\phi_{25} - 0.1298C_{L} - 0.7210M_{DD} $$

$ K_{A} \equiv \mbox{Airfoil Technology Factor, typically between 0.8 and 0.9} $
$$ M_{DD} = \frac{K_{A}}{cos{\phi_{25}}} - \frac{\frac{t}{c}}{cos^2{\phi_{25}}} - \frac{C_{L}}{10cos^3{\phi_{25}}} $$

$$ a = -1.147 $$ $$ b = 0.2 $$ $$ c = 0.838 $$ $$ d = 4.057 $$ $$ M_{DD} = a (C_{L} - b)d + c + \frac{30}{27}\left(\frac{t}{c} - 0.113\right) + 0.00288\left(\phi_{25} - 29.8\right) $$

$$ M_{cr} = M_{DD} $$

$$ M_{cr} = M_{DD} - 0.02 $$

$$ M_{cr} = M_{DD} - 0.1 $$

$$ M_{cr} = M_{DD} - \left(\frac{1}{80}\right)^{\frac{1}{3}} $$

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