Form Factor:Pyramid

Equations

For pyramid of base edge-length 2R, and height H. The angle of the pyramid walls is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha} . If $H<R/\tan \alpha$ then the pyramid is truncated (flat top).

• Volume $V_{pyr}={\frac {4}{3}}\tan(\alpha )\left[R^{3}-\left(R-{\frac {H}{\tan(\alpha )}}\right)^{3}\right]$
• Projected (xy) surface area $Sp_{pyr}=4R^{2}$

Form Factor Amplitude

$F_{pyr}(\mathbf {q} )={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)$

where

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{iq_{1}H}+\,\,{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\\K_{3}&=\,\,+{\text{sinc}}(q_{3}H)e^{iq_{3}H}+\,\,{\text{sinc}}(q_{4}H)e^{-iq_{4}H}\\K_{4}&=-i{\text{sinc}}(q_{3}H)e^{iq_{3}H}+i{\text{sinc}}(q_{4}H)e^{-iq_{4}H}\end{alignedat}}$

${\begin{alignedat}{2}q_{1}={\frac {1}{2}}\left[{\frac {q_{x}-q_{y}}{\tan \alpha }}+q_{z}\right]&\,\,,\,\,\,\,&q_{2}={\frac {1}{2}}\left[{\frac {q_{x}-q_{y}}{\tan \alpha }}-q_{z}\right]\\q_{3}={\frac {1}{2}}\left[{\frac {q_{x}+q_{y}}{\tan \alpha }}+q_{z}\right]&\,\,,\,\,\,\,&q_{4}={\frac {1}{2}}\left[{\frac {q_{x}+q_{y}}{\tan \alpha }}-q_{z}\right]\\\end{alignedat}}$

Isotropic Form Factor Intensity

$P_{pyr}(q)=\left\{{\begin{array}{c l}{\frac {\Delta \rho ^{2}V_{pyr}^{2}}{(qR)^{6}}}???&\mathrm {when} \,\,q\neq 0\\4\pi \Delta \rho ^{2}V_{pyr}^{2}&\mathrm {when} \,\,q=0\\\end{array}}\right.$

Derivations

Form Factor

For a pyramid of base-edge-length 2R, side-angle $\alpha$, truncated at H (along z axis), we note that the in-plane size of the pyramid at height z is:

$R_{z}=R-{\frac {z}{\tan \alpha }}$

Integrating with Cartesian coordinates:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&=\int \limits _{V}e^{i\mathbf {q} \cdot \mathbf {r} }\mathrm {d} \mathbf {r} \\&=\int \limits _{z=0}^{H}\int \limits _{y=-R_{z}}^{+R_{z}}\int \limits _{x=-R_{z}}^{+R_{z}}e^{i(q_{x}x+q_{y}y+q_{z}z)}\mathrm {d} x\mathrm {d} y\mathrm {d} z\\&=\int \limits _{0}^{H}\left(\int \limits _{-R_{z}}^{+R_{z}}e^{iq_{x}x}\mathrm {d} x\right)\left(\int \limits _{-R_{z}}^{+R_{z}}e^{iq_{y}y}\mathrm {d} y\right)e^{iq_{z}z}\mathrm {d} z\end{alignedat}}$

A recurring integral is (c.f. cube form factor):

${\begin{alignedat}{2}f_{x}(q_{x})&=\int _{-R_{z}}^{R_{z}}e^{iq_{x}x}\mathrm {d} x\\&=\int _{-R_{z}}^{R_{z}}\left[\cos(q_{x}x)+i\sin(q_{x}x)\right]\mathrm {d} x\\&=-{\frac {2}{q_{x}}}\sin(q_{x}R_{z})\\&=-2R_{z}\mathrm {sinc} (q_{x}R_{z})\\\end{alignedat}}$

Which gives:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&=\int \limits _{0}^{H}\left(-2R_{z}\mathrm {sinc} (q_{x}R_{z})\right)\left(-2R_{z}\mathrm {sinc} (q_{y}R_{z})\right)e^{iq_{z}z}\mathrm {d} z\\&=4\int \limits _{0}^{H}R_{z}^{2}\mathrm {sinc} (q_{x}R_{z})\mathrm {sinc} (q_{y}R_{z})e^{iq_{z}z}\mathrm {d} z\end{alignedat}}$

This can be simplified automated solving. For a regular pyramid, we obtain:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {4{\sqrt {2}}}{q_{x}q_{y}}}{\frac {\left({\begin{array}{l}-q_{y}\left(-q_{x}^{2}+q_{y}^{2}-2q_{z}^{2}\right)\cos(q_{y}R)\sin(q_{x}R)\\\,\,\,\,-q_{x}\cos(q_{x}R)\left(2i{\sqrt {2}}q_{y}q_{z}\cos(q_{y}R)+\left(q_{x}^{2}-q_{y}^{2}-2q_{z}^{2}\right)\sin(q_{y}R)\right)\\\,\,\,\,+i{\sqrt {2}}q_{z}\left(2e^{i{\sqrt {2}}q_{z}R}q_{x}q_{y}-\left(q_{x}^{2}+q_{y}^{2}-2q_{z}^{2}\right)\sin(q_{x}R)\sin(q_{y}R)\right)\end{array}}\right)}{q_{x}^{4}+(q_{y}^{2}-2q_{z}^{2})^{2}-2q_{x}^{2}(q_{y}^{2}+2q_{z}^{2})}}\end{alignedat}}$

Form Factor near q=0

qy

When $q_{y}=0$:

${\begin{alignedat}{2}q_{1}&=q_{3}\\q_{2}&=q_{4}\\K_{1}&=K_{3}\\K_{2}&=K_{4}\\\end{alignedat}}$

So:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{3}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{4}\end{array}}\right)\\&={\frac {H}{q_{x}0}}\left({\begin{array}{l}\cos \left[q_{x}R\right]K_{1}\\\,\,\,\,+\sin \left[q_{x}R\right]K_{2}\\\,\,\,\,-\cos \left[q_{x}R\right]K_{1}\\\,\,\,\,-\sin \left[q_{x}R\right]K_{2}\end{array}}\right)\\&={\frac {H}{q_{x}}}{\frac {0}{0}}\\\end{alignedat}}$

qx

When $q_{x}=0$:

${\begin{alignedat}{2}q_{1}&=-q_{4}\\q_{2}&=-q_{3}\\K_{1}&=\,\,+{\text{sinc}}(+q_{1}H)e^{+iq_{1}H}+\,\,{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}\\K_{2}&=-i{\text{sinc}}(+q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}\\K_{3}&=\,\,+{\text{sinc}}(-q_{2}H)e^{-iq_{2}H}+\,\,{\text{sinc}}(-q_{1}H)e^{+iq_{1}H}\\K_{4}&=-i{\text{sinc}}(+q_{2}H)e^{-iq_{2}H}+i{\text{sinc}}(-q_{1}H)e^{+iq_{1}H}\end{alignedat}}$

Since sinc is an even function:

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+\,\,{\text{sinc}}(q_{2}H)e^{-iq_{2}H}=K_{3}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}=K_{4}\\K_{3}&=\,\,+{\text{sinc}}(q_{2}H)e^{-iq_{2}H}+\,\,{\text{sinc}}(q_{1}H)e^{+iq_{1}H}=K_{1}\\K_{4}&=-i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}+i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}=K_{2}\end{alignedat}}$

And:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {H}{0q_{y}}}\left({\begin{array}{l}\cos \left[-q_{y}R\right]K_{1}\\\,\,\,\,+\sin \left[-q_{y}R\right]K_{2}\\\,\,\,\,-\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\end{array}}\right)\\&={\frac {H}{0q_{y}}}\left({\begin{array}{l}\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\\\,\,\,\,-\cos \left[+q_{y}R\right]K_{1}\\\,\,\,\,-\sin \left[+q_{y}R\right]K_{2}\end{array}}\right)\\&={\frac {-2H}{0q_{y}}}\sin \left(q_{y}R\right)\left[-i{\text{sinc}}(q_{1}H)e^{+iq_{1}H}+i{\text{sinc}}(q_{2}H)e^{-iq_{2}H}\right]\\&={\frac {2iH\sin(q_{y}R)}{0q_{y}}}\left[{\text{sinc}}(q_{1}H)\left(\cos(+iq_{1}H)-i\sin(+iq_{1}H)\right)-{\text{sinc}}(q_{2}H)\left(\cos(-iq_{2}H)-i\sin(-iq_{2}H)\right)\right]\\\end{alignedat}}$

qz

When $q_{z}=0$:

${\begin{alignedat}{2}q_{1}&=q_{2}\\q_{3}&=q_{4}\\\end{alignedat}}$

So:

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(q_{1}H)e^{iq_{1}H}+\,\,{\text{sinc}}(q_{1}H)e^{-iq_{1}H}\\K_{2}&=-i{\text{sinc}}(q_{1}H)e^{iq_{1}H}+i{\text{sinc}}(q_{1}H)e^{-iq_{1}H}\\K_{3}&=\,\,+{\text{sinc}}(q_{3}H)e^{iq_{3}H}+\,\,{\text{sinc}}(q_{3}H)e^{-iq_{3}H}\\K_{4}&=-i{\text{sinc}}(q_{3}H)e^{iq_{3}H}+i{\text{sinc}}(q_{3}H)e^{-iq_{3}H}\end{alignedat}}$

q

When $q=0$:

${\begin{alignedat}{2}q_{1}&=q_{2}=q_{3}=q_{4}=0\\\end{alignedat}}$

So:

${\begin{alignedat}{3}K_{1}&=+1+1&=2\\K_{2}&=-i+i&=0\\K_{3}&=+1+1&=2\\K_{4}&=-i+i&=0\end{alignedat}}$

And:

$F_{pyr}(\mathbf {q} )={\frac {H}{0\times 0}}\left({\begin{array}{l}\cos \left[(0)R\right]2\\\,\,\,\,+\sin \left[(0)R\right]0\\\,\,\,\,-\cos \left[(0)R\right]2\\\,\,\,\,-\sin \left[(0)R\right]0\end{array}}\right)$

qx and qy

When $q_{x}=q_{y}=0$:

${\begin{alignedat}{3}q_{1}&=q_{3}&=+{\frac {q_{z}}{2}}\\q_{2}&=q_{4}&=-{\frac {q_{z}}{2}}\\\end{alignedat}}$

${\begin{alignedat}{2}K_{1}&=\,\,+{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+\,\,{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}\\K_{2}&=-i{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+i{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}\\K_{3}&=\,\,+{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+\,\,{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}=K_{1}\\K_{4}&=-i{\text{sinc}}(+q_{z}H/2)e^{+iq_{z}H/2}+i{\text{sinc}}(-q_{z}H/2)e^{+iq_{z}H/2}=K_{2}\end{alignedat}}$

So:

${\begin{alignedat}{2}F_{pyr}(\mathbf {q} )&={\frac {H}{q_{x}q_{y}}}\left({\begin{array}{l}\cos \left[(q_{x}-q_{y})R\right]K_{1}\\\,\,\,\,+\sin \left[(q_{x}-q_{y})R\right]K_{2}\\\,\,\,\,-\cos \left[(q_{x}+q_{y})R\right]K_{1}\\\,\,\,\,-\sin \left[(q_{x}+q_{y})R\right]K_{2}\end{array}}\right)\\\end{alignedat}}$

To analyze the behavior in the limit of small Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_x} and $q_{y}$, we consider the limit of $q\to 0$ where $q_{x}=q_{y}=q$. We replace the trigonometric functions by their expansions near zero (keeping only the first two terms):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \lim_{q\to0} F_{pyr}(\mathbf{q}) & = \frac{H}{q q} \left( \begin{array}{l} \cos\left[ (q-q)R \right] K_1 \\ \,\,\,\, + \sin\left[ (q-q)R \right] K_2 \\ \,\,\,\, - \cos\left[ (q+q)R \right] K_1 \\ \,\,\,\, - \sin\left[ (q+q)R \right] K_2 \end{array} \right) \\ & = \frac{H}{q^2} \left( \begin{array}{l} \left[ 1 - \frac{ ((q-q)R)^2 }{2!} + \cdots \right] K_1 \\ \,\,\,\, + \left[ (q-q)R - \frac{((q-q)R)^3}{3!} + \cdots \right] K_2 \\ \,\,\,\, - \left[ 1 - \frac{ ((q+q)R)^2}{2!} + \cdots \right] K_1 \\ \,\,\,\, - \left[ (q+q)R - \frac{((q-q)R)^3}{3!} + \cdots \right] K_2 \end{array} \right) \\ & = \frac{H}{q^2} \left( \begin{array}{l} \left[ 1 - \frac{ ((q-q)R)^2 }{2!} - 1 + \frac{ ((q+q)R)^2}{2!} \right] K_1 \\ \,\,\,\, + \left[ (q-q)R - \frac{((q-q)R)^3}{3!} - (q+q)R + \frac{((q-q)R)^3}{3!}\right] K_2 \\ \end{array} \right) \\ & = \frac{H}{q^2} \left( \begin{array}{l} \left[ \frac{ ((2q)R)^2}{2!} - \frac{ ((q-q)R)^2 }{2!} \right] K_1 \\ \,\,\,\, + \left[ (q-q)R - (2q)R \right] K_2 \\ \end{array} \right) \\ & = \frac{ (2qR)^2}{2!}\frac{H K_1}{q^2} - \frac{ ((q-q)R)^2 }{2!}\frac{H K_1}{q^2} + (q-q)R \frac{H K_2}{q^2} - 2qR \frac{H K_2}{q^2} \\ & = \frac{ 4R^2 H K_1}{2} - \frac{ R^2 H K_1}{2}\frac{(q-q)^2}{q^2} + R H K_2\frac{(q-q)}{q^2} - \frac{2 R H K_2}{q} \\ & = 2R^2 H K_1 \end{alignat} }

Note that since $\mathrm {sinc}$ is symmetric $K_{2}=K_{4}=0$. When Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q_y} are small (but not zero and not necessarily equal), many of the above arguments still apply. It remains that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_2 \approx K_4 \approx 0} , and:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \lim_{(q_x,q_y)\to0} F_{pyr}(\mathbf{q}) & = \frac{H K_1}{q_x q_y} \left( \cos\left[ (q_x-q_y)R \right] - \cos\left[ (q_x+q_y)R \right] \right) \\ & = \frac{H K_1}{q_x q_y} \left( \left[ 1 - \frac{ ((q_x-q_y)R)^2}{2!} + \cdots \right] - \left[ 1 - \frac{((q_x+q_y)R)^2}{2!} + \cdots \right] \right) \\ & = \frac{H K_1}{q_x q_y} \left( \frac{(q_x+q_y)^2 R^2}{2!} - \frac{(q_x-q_y)^2 R^2}{2!} \right) \\ & = \frac{H R^2 K_1}{2 q_x q_y} \left( (q_x+q_y)^2 - (q_x-q_y)^2 \right) \\ \end{alignat} }

Isotropic Form Factor Intensity

To average over all possible orientations, we note:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{q}=(q_x,q_y,q_z)=(-q\sin\theta\cos\phi,q\sin\theta\sin\phi,q\cos\theta)}

and use:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} P_{pyr}(q) & = \int\limits_{S} | F_{pyr}(\mathbf{q}) |^2 \mathrm{d}\mathbf{s} \\ & = \int_{\phi=0}^{2\pi}\int_{\theta=0}^{\pi} \left| \frac{H}{q_x q_y} \left( \begin{array}{l} \cos\left[ (q_x-q_y)R \right] K_1 \\ \,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\ \,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\ \,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4 \end{array} \right) \right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\ & = \frac{H^2}{q^2} \int_{0}^{2\pi}\int_{0}^{\pi} \frac{1}{\sin^4\theta \sin^2\phi\cos^2\phi} \left| \left( \begin{array}{l} \cos\left[ (q_x-q_y)R \right] K_1 \\ \,\,\,\, + \sin\left[ (q_x-q_y)R \right] K_2 \\ \,\,\,\, - \cos\left[ (q_x+q_y)R \right] K_3 \\ \,\,\,\, - \sin\left[ (q_x+q_y)R \right] K_4 \end{array} \right) \right|^2 \sin\theta\mathrm{d}\theta\mathrm{d}\phi \\ \end{alignat} }

Regular Pyramid

A regular pyramid (half of an octahedron) has faces that are equilateral triangles (each vertex is 60°). The 'corner-to-edge' distance along each triangular face is then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{face,c-e} = R \tan(60^{\circ}) = \sqrt{3} R}

This makes the height:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} (d_{face,c-e})^2 & = (H)^2 + (R)^2 \\ H^2 & = (d_{face,c-e})^2 - (R)^2\\ H & = \sqrt{ (\sqrt{3} R)^2 - (R)^2 }\\ & = \sqrt{ 3 R^2 - R^2 }\\ & = \sqrt{ 2 } R \\ \end{alignat} }

So that the pyramid face angle, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha} is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \tan(\alpha) & = \frac{ H }{ R } \\ \alpha & = \arctan \left( \frac{\sqrt{ 2 } R}{R} \right) \\ & = \arctan( \sqrt{2} ) \\ & \approx 0.9553 \\ & \approx 54.75^{\circ} \end{alignat} }

The square base of the pyramid has edges of length 2R. The distance from the center of the square to any corner is H, such that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} \cos(45^{\circ}) & = \frac{R}{H} \\ H & = \frac{R}{ 1/\sqrt{2} } \\ & = \sqrt{2} R \end{alignat} }

Surface Area

For a non-truncated, regular pyramid, each face is an equilateral triangle (each vertex is 60°). So each face:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} S_{face} & = 2 \times \left( \frac{ R R \tan(60^{\circ}) }{2} \right) \\ & = R^2 \sqrt{3} \end{alignat} }

The base is simply:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} S_{base} & = 2 R \times 2 R \\ & = 4 R^2 \end{alignat} }

Total:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} S_{pyr} & = 4 \times R^2 \sqrt{3} + 4 R^2 \\ & = 4(1 + \sqrt{3}) R^2 \end{alignat} }

Volume

For a regular pyramid, the height Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H=\sqrt{2}R} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tan(\alpha)=H/R = \sqrt{2}} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} V_{pyr} & = \frac{4}{3} \tan (\alpha) \left[ R^3 - \left( R - \frac{H}{ \tan (\alpha)} \right)^3 \right] \\ & = \frac{4}{3} \sqrt{2} \left[ R^3 - \left( R - \frac{ \sqrt{2} R }{ \sqrt{2}} \right)^3 \right] \\ & = \frac{4\sqrt{2}}{3} R^3 \\ \end{alignat} }