View Factor Calculator: Analytical and Monte Carlo Method

In this web page, radiation view factors (also know as configuration factors) for different geometrical configurations can be calculated by the analytical formula and Monte Carlo method. The Monte Carlo calculation uses WebGL 2.0. Depending on the browser and hardware configuration, this functionality might not work. The number of rays for the view factor calculation is 100,000. The seeds of the random number generator are fixed, so the results are reproducible.

Differential surface to circular disk

r ( r > 0 ) :
h ( h > 0 ) :
θ [°] ( 0 ≤ θ ≤ 180 ) :

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\begin{gather} F_{d1-2} = \frac{r^2}{r^2+h^2} \cos \theta, ~~ \mathrm{where}~~\theta \le \arctan \frac{h}{r} \\ F_{d1-2} = 0, ~~ \mathrm{where}~~\theta > \arctan \frac{r}{h} + \frac{\pi}{2} \end{gather}

Differential surface to circular disk in parallel plane

r ( r > 0 ) :
h ( h > 0 ) :
a ( a ≥ 0 ) :

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\begin{gather} F_{d1-2} = \frac{1}{2} \left[ 1 - \frac{Z-2R^2}{\sqrt{Z^2 - 4R^2}} \right] \\ \mathrm{where}~~H = \frac{h}{a}, ~~ R = \frac{r}{a}, ~~ Z = 1 + H^2 + R^2 \\ \end{gather}

Differential surface to rectangular surface in parallel plane

a ( a > 0 ) :
b ( b > 0 ) :
c ( c > 0 ) :

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\begin{gather} F_{d1-2} = \frac{1}{2\pi} \left( \frac{A}{\sqrt{1+A^2}} \arctan \frac{B}{\sqrt{1+A^2}} + \frac{B}{\sqrt{1+B^2}} \arctan \frac{A}{\sqrt{1+B^2}} \right) \\ \mathrm{where}~~A = \frac{a}{c}, ~~ B = \frac{b}{c} \end{gather}

Differential surface to rectangular surface in 90° angle

a ( a > 0 ) :
b ( b > 0 ) :
c ( c > 0 ) :

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\begin{gather} F_{d1-2} = \frac{1}{2\pi} \left( \arctan \frac{1}{Y} - \frac{Y}{\sqrt{X^2+Y^2}} \arctan \frac{1}{\sqrt{X^2+Y^2}} \right) \\ \mathrm{where}~~X = \frac{a}{b}, ~~Y = \frac{c}{b} \end{gather}

Differential surface to sphere

r ( 0 < r < h ) :
h ( 0 < r < h ) :
θ [°] ( 0 ≤ θ ≤ 180 ) :

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\begin{gather} F_{d1-2} = \left( \frac{r}{h} \right)^2 \cos \theta, ~~ \mathrm{where}~~\theta \le \arccos \frac{r}{h} \\ F_{d1-2} = 0, ~~ \mathrm{where}~~\theta > \arcsin \frac{r}{h} + \frac{\pi}{2} \end{gather}

Differential surface to cylinder

r ( 0 < r < h ) :
h ( 0 < r < h ) :
l ( 0 < l ) :

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\begin{gather} F_{d1-2} = \frac{L}{\pi H} \left[ \frac{1}{L} \arctan \frac{L}{\sqrt{H^2-1}} + \frac{X-2H}{\sqrt{XY}} \arctan \sqrt{\frac{X(H-1)}{Y(H+1)}} - \arctan \sqrt{\frac{H-1}{H+1}} \right] \\ \mathrm{where}~~L = \frac{l}{r}, ~~ H = \frac{h}{r}, \\ X = (1+H)^2 + L^2, ~~ Y = (1-H)^2 + L^2 \end{gather}

Differential surface to right triangle in parallel plane

h ( h > 0 ) :
l ( l > 0 ) :
θ [°] ( 0 < θ < 90 ) :

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\begin{gather} F_{d1-2} = \frac{D}{2\pi A} \arctan \left( \frac{D \tan \theta}{A} \right), ~~ \mathrm{where}~~D = \frac{d}{h}, ~~ A = \sqrt{1+D^2} \end{gather}

Disk to parallel coaxial disk

h ( h > 0 ) :
r₁ ( r₁ > 0 ) :
r₂ ( r₂ > 0 ) :

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\begin{gather} F_{1-2} = \frac{1}{2} \left\{ X - \sqrt{X^2 - 4\left( \frac{R_2}{R_1} \right)^2} \right\} \\ \mathrm{where} ~~ X = 1 + \frac{1 + R_2^2}{R_1^2}, ~~ R_1 = \frac{r_1}{a}, ~~ R_2 = \frac{r_2}{a} \end{gather}

Base disk to inside surface of cylincer

h ( h > 0 ) :
r₁ ( r > 0 ) :

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\begin{gather} F_{1-2} = 2H \left[ \sqrt{1+H^2} - H \right] \\ \mathrm{where} ~~ H = \frac{h}{2r} \end{gather}

Identical, parallel, directly opposed rectangles

a ( a > 0 ) :
b ( b > 0 ) :
c ( c > 0 ) :

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\begin{align} F_{1-2} = &\frac{2}{\pi XY} \left\{ \ln \left[ \frac{(1+X^2)(1+Y^2)}{1+X^2+Y^2} \right]^{1/2} + X \sqrt{1+Y^2} \arctan \frac{X}{\sqrt{1+Y^2}} \right. \\ &\left. + Y \sqrt{1+X^2} \arctan \frac{Y}{\sqrt{1+X^2}} - X \arctan X - Y \arctan Y \right\} \\ &\hspace{70pt}\mathrm{where} ~~ X = \frac{a}{c}, ~~ Y = \frac{b}{c} \end{align}

Two rectangles with one common edge and 90° angle

h ( h > 0 ) :
w ( w > 0 ) :
l ( l > 0 ) :

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\begin{align} &F_{1-2} = \frac{1}{\pi W} \left( W \arctan \frac{1}{W} + H \arctan \frac{1}{H} - \sqrt{H^2 + W^2} \arctan \frac{1}{\sqrt{H^2 + W^2}} \right. \\ &\left. + \frac{1}{4} \ln \left\{ \frac{(1+W^2)(1+H^2)}{1+W^2+H^2} \left[ \frac{W^2(1+W^2+H^2)}{(1+W^2)(W^2+H^2)} \right]^{W^2} \left[ \frac{H^2(1+W^2+H^2)}{(1+H^2)(W^2+H^2)} \right]^{H^2} \right\} \right) \\ &\hspace{110pt}\mathrm{where} ~~ H = \frac{h}{l}, ~~ W = \frac{w}{l} \end{align}

Unit sphere to rectangle

h ( h > 1 ) :
l₁ ( l₁ > 0 ) :
l₂ ( l₂ > 0 ) :

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\begin{align} F_{1-2} &= \frac{1}{4\pi} \arctan \sqrt{\frac{1}{H_1^2+H_2^2+H_1^2H_2^2}} \\ &\mathrm{where} ~~ H_1 = \frac{h}{l_1}, ~~H_2 = \frac{h}{l_2} \end{align}

Unit sphere to coaxial disk

h ( h > 1 ) :
r ( r > 0 ) :

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\begin{align} F_{1-2} &= \frac{1}{2} \left[ 1 - \frac{1}{\sqrt{1+R^2}} \right] \\ &\mathrm{where} ~~ R = \frac{r}{h} \end{align}

Sphere to coaxial cone

h ( h > 0 ) :
r₁ ( r₁ > 0 ) :
r₂ ( r₂ > 0 ) :
θ [°] ( 0 < θ < 90 ) :

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\begin{align} F_{1-2} &= \frac{1}{2} \left[ 1 - \frac{1+H+R/\tan\theta}{\sqrt{(1+H+R/\tan\theta)^2 + R^2}} \right] \\ &\mathrm{where} ~~ H = \frac{h}{r_1}, ~~ R = \frac{r_2}{r_1}, ~~ \theta \ge \arcsin \frac{1}{1+H} \end{align}

Interior of outer cylinder to exterior of coaxial inner cylinder

h ( h > 0 ) :
r₁ ( r₁ > 0 ) :
r₂ ( r₂ > 0 ) :

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\begin{align} &F_{2-1} = \frac{1}{R} \left( 1 - \frac{H^2+R^2-1}{4H} - \frac{1}{\pi} \left\{ \arccos \frac{H^2-R^2+1}{H^2+R^2-1} \right. \right. \\ &\left. \left. - \frac{\sqrt{(H^2+R^2+1)^2-4R^2}}{2H} \arccos \frac{H^2-R^2+1}{R(H^2+R^2-1)} - \frac{H^2-R^2+1}{2H} \arcsin \frac{1}{R} \right\} \right) \\ &\hspace{50pt} \mathrm{where} ~~ R_1 = \frac{r_1}{h}, ~~R_2 = \frac{r_2}{h}, ~~A = R_2 + R_1, ~~B = R_2 - R_1 \end{align}

Interior of cone to base disk

h ( h > 0 ) :
r ( r > 0 ) :

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\begin{align} &F_{1-2} = \frac{1}{\sqrt{1+H^2}}, ~~ \mathrm{where} ~~ H = \frac{h}{r} \end{align}