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 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 < R/ \tan\alpha} then the pyramid is truncated (flat top).

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

Form Factor Amplitude

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 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

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} K_1 & = \,\, +\text{sinc}(q_1 H) e^{i q_1 H} + \,\, \text{sinc}(q_2 H)e^{-iq_2 H} \\ K_2 & = -i\text{sinc}(q_1 H) e^{i q_1 H} + i\text{sinc}(q_2 H)e^{-iq_2 H} \\ K_3 & = \,\, +\text{sinc}(q_3 H) e^{i q_3 H} + \,\, \text{sinc}(q_4 H)e^{-iq_4 H} \\ K_4 & = -i\text{sinc}(q_3 H) e^{i q_3 H} + i\text{sinc}(q_4 H)e^{-iq_4 H} \end{alignat} }

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} 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{alignat} }

Isotropic Form Factor Intensity

This can be computed numerically.

Derivations

Form Factor

For a pyramid of base-edge-length 2R, side-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} , truncated at H (along z axis), we note that the in-plane size of the pyramid at height z 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 R_z = R - \frac{ z }{ \tan \alpha }}

Integrating with Cartesian coordinates:

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} 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^{i q_x x} \mathrm{d}x \right) \left( \int\limits_{-R_z}^{+R_z} e^{i q_y y} \mathrm{d}y \right) e^{i q_z z} \mathrm{d}z \end{alignat} }

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

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} f_{x}(q_x) & = \int_{-R_z}^{R_z} e^{i q_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) \\ & = -2 R_z\mathrm{sinc}(q_x R_z) \\ \end{alignat} }

Which gives:

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} F_{pyr}(\mathbf{q}) & = \int\limits_{0}^{H} \left( -2 R_z\mathrm{sinc}(q_x R_z) \right) \left( -2 R_z\mathrm{sinc}(q_y R_z) \right) e^{i q_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^{i q_z z} \mathrm{d}z \end{alignat} }

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

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} 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^2-2 q_z^2\right) \cos(q_y R) \sin(q_x R) \\ \,\,\,\, -q_x \cos(q_x R) \left(2 i \sqrt{2} q_y q_z \cos(q_y R) +\left(q_x^2-q_y^2-2 q_z^2\right) \sin(q_y R)\right) \\ \,\,\,\, +i \sqrt{2} q_z \left(2 e^{i \sqrt{2} q_z R} q_x q_y-\left(q_x^2+q_y^2-2 q_z^2\right) \sin(q_x R) \sin(q_y R)\right) \end{array} \right) } { q_x^4 + (q_y^2 - 2 q_z^2)^2 - 2 q_x^2 (q_y^2 + 2 q_z^2) } \end{alignat} }

Form Factor near q=0

qy

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_y=0} :

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q_{z}=0} :

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

So:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 $q=0$:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{alignedat}{2}q_{1}&=q_{2}=q_{3}=q_{4}=0\\\end{alignedat}}}

So:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q_{x}=q_{y}=0} :

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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 $q_{x}$ and $q_{y}$, we consider the limit of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q\to 0} where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle q_{x}=q_{y}=q} . We replace the trigonometric functions by their expansions near zero (keeping only the first two terms):

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{alignedat}{2}\lim _{q\to 0}F_{pyr}(\mathbf {q} )&={\frac {H}{qq}}\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 {HK_{1}}{q^{2}}}-{\frac {((q-q)R)^{2}}{2!}}{\frac {HK_{1}}{q^{2}}}+(q-q)R{\frac {HK_{2}}{q^{2}}}-2qR{\frac {HK_{2}}{q^{2}}}\\&={\frac {4R^{2}HK_{1}}{2}}-{\frac {R^{2}HK_{1}}{2}}{\frac {(q-q)^{2}}{q^{2}}}+RHK_{2}{\frac {(q-q)}{q^{2}}}-{\frac {2RHK_{2}}{q}}\\&=2R^{2}HK_{1}\end{alignedat}}}

Note that since Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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:

$\mathbf {q} =(q_{x},q_{y},q_{z})=(-q\sin \theta \cos \phi ,q\sin \theta \sin \phi ,q\cos \theta )$

and use:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{alignedat}{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{alignedat}}}

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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle d_{face,c-e}=R\tan(60^{\circ })={\sqrt {3}}R}

This makes the height:

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{alignedat}{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 {3R^{2}-R^{2}}}\\&={\sqrt {2}}R\\\end{alignedat}}}

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} }