If anyone else takes a look here, note that these are not necessarily correct; there seems to be something wrong with my derivations, which is why I'm poking around here.

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${\displaystyle Rc^{2}=2GM\;R={2G\,M \over c^{2}}\;M={c^{2}R \over 2G}}$

${\displaystyle E=M\,c^{2}={c^{4}\,R \over 2G}}$

${\displaystyle S={k\,A \over 4\ell _{P}^{2}}={kc^{3}\,A \over 4\hbar G}={\pi kc^{3}\,A \over 2hG}}$

${\displaystyle T={E \over S}}$ - no guarantee over this, or that the ${\displaystyle E}$ here is the one equal to ${\displaystyle Mc^{2}}$.

${\displaystyle \sigma ={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}={2\pi ^{5}k^{4} \over 15c^{2}2^{3}\pi ^{3}\hbar ^{3}}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}}$

${\displaystyle q=e_{b}A\,}$

${\displaystyle A=4\pi \,R^{2}=4\pi \left({2G\,M \over c^{2}}\right)^{2}=4\pi {4G^{2}\,M^{2} \over c^{4}}={16\pi G^{2}\,M^{2} \over c^{4}}}$ - but see Weisstein's section below and New A section

${\displaystyle S={\pi kc^{3}\,R^{2} \over \hbar G}={2\pi ^{2}kc^{3}\,R^{2} \over hG}}$

${\displaystyle S={4\pi kG\,M^{2} \over \hbar c}={8\pi ^{2}kG\,M^{2} \over hc}}$

${\displaystyle T={E \over S}=M\,c^{2}{\hbar c \over 4\pi kG\,M^{2}}={c^{4}\,R \over 2G}{\hbar G \over \pi kc^{3}\,R^{2}}}$

${\displaystyle T={hc^{3} \over 8\pi ^{2}kG\,M}={\hbar c^{3} \over 4\pi kG\,M}}$

${\displaystyle T={hc \over 4\pi ^{2}k\,R}={\hbar c \over 2\pi k\,R}}$

${\displaystyle e_{b}=\sigma \,T^{4}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}\left({\hbar c^{3} \over 4\pi kG\,M}\right)^{4}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}\left({hc^{3} \over 8\pi ^{2}kG\,M}\right)^{4}}$

${\displaystyle e_{b}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}{\frac {\hbar ^{4}c^{12}}{256\pi ^{4}k^{4}G^{4}\,M^{4}}}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}{\frac {h^{4}c^{12}}{4096\pi ^{8}k^{4}G^{4}\,M^{4}}}}$

${\displaystyle e_{b}={\hbar c^{10} \over 15,360\pi ^{2}G^{4}\,M^{4}}={hc^{10} \over 30,720\pi ^{3}G^{4}\,M^{4}}}$

${\displaystyle q=\sigma \,T^{4}A={\sigma h^{4}c^{4} \over 64\pi ^{7}k^{4}\,R^{2}}={hc^{2} \over 480\pi ^{2}\,R^{2}}}$

${\displaystyle q={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}\left({\hbar c^{3} \over 4\pi kG\,M}\right)^{4}{16\pi G^{2}\,M^{2} \over c^{4}}}$

${\displaystyle q={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}{\frac {\hbar ^{4}c^{12}}{256\pi ^{4}k^{4}G^{4}\,M^{4}}}{16\pi G^{2}\,M^{2} \over c^{4}}}$

${\displaystyle q={\frac {c^{6}\hbar }{960\pi G^{2}\,M^{2}}}}$

${\displaystyle Rc^{2}=2GM\;R={2G\,M \over c^{2}}\;M={c^{2}R \over 2G}}$

${\displaystyle E=M\,c^{2}={c^{4}\,R \over 2G}}$

${\displaystyle S={k\,A \over 4\ell _{P}^{2}}={kc^{3}\,A \over 4\hbar G}={\pi kc^{3}\,A \over 2hG}}$

${\displaystyle T={E \over S}}$ - no guarantee over this, or that the ${\displaystyle E}$ here is the one equal to ${\displaystyle Mc^{2}}$.

${\displaystyle \sigma ={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}={2\pi ^{5}k^{4} \over 15c^{2}2^{3}\pi ^{3}\hbar ^{3}}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}}$

${\displaystyle q=e_{b}A\,}$

${\displaystyle A={8\pi G^{2}\,M^{2} \over c^{4}}={8\pi G^{2} \over c^{4}}\left({c^{2}\,R \over 2G}\right)^{2}={8\pi G^{2} \over c^{4}}{c^{4}\,R^{2} \over 4G^{2}}}$ - see Weisstein's section below

${\displaystyle A={2\pi R^{2}}\,}$

${\displaystyle S={kc^{3} \over 4\hbar G}{2\pi R^{2}}={\pi kc^{3} \over 2hG}{2\pi R^{2}}={kc^{3} \over 4\hbar G}{8\pi G^{2}\,M^{2} \over c^{4}}={\pi kc^{3} \over 2hG}{8\pi G^{2}\,M^{2} \over c^{4}}}$

${\displaystyle S={\pi kc^{3}\,R^{2} \over 2\hbar G}={\pi ^{2}kc^{3}\,R^{2} \over hG}={2\pi kG\,M^{2} \over \hbar c}={4\pi ^{2}kG\,M^{2} \over hc}}$

${\displaystyle T={E \over S}}$

${\displaystyle T={c^{4}\,R \over 2G}{2\hbar G \over \pi kc^{3}\,R^{2}}={c^{4}\,R \over 2G}{hG \over \pi ^{2}kc^{3}\,R^{2}}=M\,c^{2}{\hbar c \over 2\pi kG\,M^{2}}=M\,c^{2}{hc \over 4\pi ^{2}kG\,M^{2}}}$

${\displaystyle T={\hbar c \over \pi k\,R}={hc \over 2\pi ^{2}k\,R}={\hbar c^{3} \over 2\pi kG\,M}={hc^{3} \over 4\pi ^{2}kG\,M}}$

Damn, the different area formula is making things even worse?!? Maybe it's the definition of ${\displaystyle E}$ that's not what I'd expect? It needn't be the mass-energy equivalence of the entire mass of the black hole.

${\displaystyle T={E \over S}={E \over {2\pi kG\,M^{2} \over \hbar c}}={E\hbar c \over 2\pi kG\,M^{2}}={\hbar c^{3} \over 8\pi kG\,M}}$

${\displaystyle E={\hbar c^{3} \over 8\pi kG\,M}{2\pi kG\,M^{2} \over \hbar c}={c^{2}\,M \over 4}}$ - Why this apparent equivalence? I have no idea. It could also be that ${\displaystyle T={E \over 4S}}$.

Let's try a comparison with the gravitational binding energy formula. I doubt it has any relevance, but I'll see what it generates.

${\displaystyle U={3G\,M^{2} \over 5\,r}={3G\,M^{2} \over 5{2G\,M \over c^{2}}}={3Gc^{2}\,M^{2} \over 10G\,M}={3c^{2}\,M \over 10}}$ - Well, it's in the ballpark, and I'm pretty sure that binding energy formula doesn't allow for relativistic effects. But I still don't think it has anything to do with it.

${\displaystyle e_{b}=\sigma \,T^{4}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}\left({hc^{3} \over 4\pi ^{2}kG\,M}\right)^{4}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}\left({\hbar c^{3} \over 2\pi kG\,M}\right)^{4}}$

${\displaystyle e_{b}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}{\frac {h^{4}c^{12}}{256\pi ^{8}k^{4}G^{4}\,M^{4}}}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}{\frac {\hbar ^{4}c^{12}}{16\pi ^{4}k^{4}G^{4}\,M^{4}}}}$

${\displaystyle e_{b}={hc^{10} \over 1920\pi ^{3}G^{4}\,M^{4}}={\hbar c^{10} \over 960\pi ^{2}G^{4}\,M^{4}}}$

${\displaystyle q={hc^{10} \over 1920\pi ^{3}G^{4}\,M^{4}}A={\hbar c^{10} \over 960\pi ^{2}G^{4}\,M^{4}}A}$

${\displaystyle q={hc^{10} \over 1920\pi ^{3}G^{4}\,M^{4}}{8\pi G^{2}\,M^{2} \over c^{4}}={\hbar c^{10} \over 960\pi ^{2}G^{4}\,M^{4}}{8\pi G^{2}\,M^{2} \over c^{4}}}$

${\displaystyle q={hc^{6} \over 240\pi ^{2}G^{2}\,M^{2}}={\hbar c^{6} \over 120\pi G^{2}\,M^{2}}}$

${\displaystyle {\hbar c^{3} \over 8\pi kG\,M}={hc^{3} \over 16\pi ^{2}kG\,M}}$

${\displaystyle e_{b}=\sigma \,T^{4}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}\left({hc^{3} \over 16\pi ^{2}kG\,M}\right)^{4}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}\left({\hbar c^{3} \over 8\pi kG\,M}\right)^{4}}$

${\displaystyle e_{b}={2\pi ^{5}k^{4} \over 15c^{2}h^{3}}{h^{4}c^{12} \over 65,536\pi ^{8}k^{4}G^{4}\,M^{4}}={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}{\hbar ^{4}c^{12} \over 4,096\pi ^{4}k^{4}G^{4}\,M^{4}}}$

${\displaystyle e_{b}={hc^{10} \over 491,520\pi ^{3}G^{4}\,M^{4}}={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}}$

${\displaystyle P={hc^{10} \over 491,520\pi ^{3}G^{4}\,M^{4}}A={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}A}$

${\displaystyle P={hc^{10} \over 491,520\pi ^{3}G^{4}\,M^{4}}{8\pi G^{2}\,M^{2} \over c^{4}}={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}{8\pi G^{2}\,M^{2} \over c^{4}}}$

${\displaystyle P={hc^{6} \over 61,440\pi ^{2}G^{2}\,M^{2}}={\hbar c^{6} \over 30,720\pi G^{2}\,M^{2}}}$

${\displaystyle T_{H}={\kappa \over 2\pi }}$ ( natural units)

${\displaystyle T={\hbar \,c^{3} \over 8\pi GMk}}$

${\displaystyle P={\hbar \,c^{6} \over 15360\,\pi \,G^{2}M^{2}}}$

${\displaystyle t_{\operatorname {ev} }={5120\,\pi \,G^{2}M_{0}^{\,3} \over \hbar \,c^{4}}}$

${\displaystyle T_{BH_{sol}}=60\mathrm {nK} =6\times 10^{-8}\mathrm {K} }$

${\displaystyle T_{BH}(4.5\times 10^{22}\mathrm {kg} )=2.7\mathrm {K} }$

${\displaystyle P_{sol}=10^{-28}\mathrm {W} \,}$

${\displaystyle t_{\mathrm {ev} _{sol}}=10^{67}\mathrm {year} =3\times 10^{74}\mathrm {s} }$

${\displaystyle t_{\mathrm {ev} }(10^{11}\mathrm {kg} )=3\times 10^{9}\mathrm {year} =9\times 10^{16}\mathrm {s} }$

${\displaystyle P=3.563\,45\times 10^{32}\left[{\frac {\mathrm {kg} }{M}}\right]^{2}\mathrm {W} }$

${\displaystyle t_{\mathrm {ev} }=8.407\,16\times 10^{-17}\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\mathrm {s} \ \ \approx \ 2.66\times 10^{-24}\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\mathrm {yr} }$

${\displaystyle M_{0}=2.282\,71\times 10^{5}\left[{\frac {t_{\mathrm {ev} }}{\mathrm {s} }}\right]^{1/3}\mathrm {kg} \ \ \approx \ 7.2\times 10^{7}\left[{\frac {t_{\mathrm {ev} }}{\mathrm {yr} }}\right]^{1/3}\mathrm {kg} }$

${\displaystyle t_{\mathrm {ev} }(2.28\times 10^{5}\mathrm {kg} )=1\mathrm {s} }$

${\displaystyle E(2.28\times 10^{5}\mathrm {kg} )=2.05\times 10^{22}\mathrm {J} }$

${\displaystyle P(2.28\times 10^{5}\mathrm {kg} )=6.84\times 10^{21}\mathrm {W} }$

${\displaystyle P=\sigma T^{4}A={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}\left({\hbar \,c^{3} \over 8\pi GMk}\right)^{4}A}$

${\displaystyle P={\pi ^{2}k^{4} \over 60c^{2}\hbar ^{3}}{\frac {\hbar ^{4}c^{12}}{4096\pi ^{4}k^{4}G^{4}M^{4}}}A}$

${\displaystyle P={c^{10}\hbar \over 245,760\pi ^{2}G^{4}M^{4}}A}$

${\displaystyle A{c^{10}\hbar \over 245,760\pi ^{2}G^{4}M^{4}}={\hbar \,c^{6} \over 15360\,\pi \,G^{2}M^{2}}}$

${\displaystyle A={\frac {\frac {\hbar \,c^{6}}{15,360\,\pi \,G^{2}M^{2}}}{\frac {c^{10}\hbar }{245,760\pi ^{2}G^{4}M^{4}}}}}$

${\displaystyle A={\frac {\hbar \,c^{6}}{15,360\,\pi \,G^{2}M^{2}}}{\frac {245,760\pi ^{2}G^{4}M^{4}}{c^{10}\hbar }}}$

${\displaystyle A={\frac {16\pi G^{2}M^{2}}{c^{4}}}}$

Odd, it does come out the same as the usual formula as surface area of a sphere. My mistake must lie elsewhere.

${\displaystyle T_{article}={T_{mine} \over 2}}$

${\displaystyle P_{article}={q_{mine} \over 16}}$

For some reason, ${\displaystyle T_{mine}}$ is coming out twice as large as ${\displaystyle T_{article}}$. It naturally follows that this would lead to ${\displaystyle q_{mine}}$ being 16 times larger than ${\displaystyle P_{article}}$. I thought event horizons might have a surface area that was not equal to that of a conventional sphere, but that doesn't seem to explain it.

${\displaystyle S={\frac {Akc^{3}}{4\hbar G}}}$

${\displaystyle T={\frac {\hbar c^{3}}{8\pi kGM}}}$

(was Black hole entropy)

${\displaystyle S_{BH}={\frac {kA}{4l_{\mathrm {P} }^{2}}}}$

${\displaystyle l_{\mathrm {P} }={\sqrt {G\hbar \over c^{3}}}}$

(was Laws of black hole mechanics)

${\displaystyle T_{H}={\frac {\kappa }{2\pi }}}$ ( natural units)

${\displaystyle S_{BH}={\frac {A}{4}}}$ ( natural units)

I noticed this in the table of equations with their nondimensionalized equivalents:

Thermal energy per particle per degree of freedom

${\displaystyle E={\frac {1}{2}}kT}$

This may be the key, if it's half of ${\displaystyle kT}$, much like ${\displaystyle E={\frac {1}{2}}mv^{2}}$ threw me off with that half, when I was very young.

${\displaystyle 2E=kT\,}$

${\displaystyle E={\frac {kT}{2}}}$, ${\displaystyle T={\frac {2E}{k}}}$, ${\displaystyle k={\frac {2E}{T}}}$

${\displaystyle r_{s}={\frac {2Gm}{c^{2}}}}$

${\displaystyle r_{s}=m\times 1.48\times 10^{-27}}$

${\displaystyle S={Akc^{3} \over 4G\hbar }}$

${\displaystyle A={8\pi M^{2}G^{2} \over c^{4}}}$ - Here's where it's different! That's not the formula for an undistorted sphere.

${\displaystyle S={2\pi kGM^{2} \over \hbar c}}$

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${\displaystyle P_{SH}={G\rho \hbar \over 90}}$

${\displaystyle P_{SH}={\hbar c^{6} \over 960\pi G^{2}\,M^{2}}}$

${\displaystyle \rho ={M \over ({4\pi \over 3})R_{H}^{3}}={M \over ({4\pi \over 3})({2G\,M \over c^{2}})^{3}}={3c^{6} \over 32\pi G^{3}\,M^{2}}}$

${\displaystyle S_{bh}={kAc^{3} \over 12\hbar G}}$

${\displaystyle S_{bh}={\frac {1}{3}}S_{Bek}={\frac {1}{3}}\left[{kAc^{3} \over 4\hbar G}\right]}$

${\displaystyle M=1.9889\times 10^{30}\mathrm {kg} }$

${\displaystyle R=2953.3\mathrm {m} \,}$

${\displaystyle E=1.7876\times 10^{47}\mathrm {J} }$

${\displaystyle A=2\pi R^{2}=5.4800\times 10^{7}\mathrm {m^{2}} }$ or ${\displaystyle A=4\pi R^{2}=1.0960\times 10^{8}\mathrm {m^{2}} }$

${\displaystyle S={2\pi kG\,M^{2} \over \hbar c}=7.2427\times 10^{53}\mathrm {J/K} }$

${\displaystyle T={\hbar c^{3} \over 8\pi kG\,M}=6.1702\times 10^{-8}\mathrm {K} }$

${\displaystyle P={\hbar c^{6} \over 15,360\pi G^{2}\,M^{2}}=9.0081\times 10^{-29}\mathrm {W} }$

${\displaystyle e_{b}={P \over A}}$

${\displaystyle e_{b}={P \over 2\pi R^{2}}=1.6438\times 10^{-36}\mathrm {W/m^{2}} }$ or ${\displaystyle e_{b}={P \over 4\pi R^{2}}=8.2191\times 10^{-37}\mathrm {W/m^{2}} }$

${\displaystyle e_{b}=\sigma T^{4}=5.6705\times 10^{-8}\mathrm {W \over m^{2}K^{4}} \,(6.1702\times 10^{-8}\mathrm {K} )^{4}=8.2191\times 10^{-37}\mathrm {W/m^{2}} }$

${\displaystyle e_{b}={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}=8.2191\times 10^{-37}\mathrm {W/m^{2}} }$

${\displaystyle M=2.2827\times 10^{5}\mathrm {kg} }$

${\displaystyle R=3.3895\times 10^{-22}\mathrm {m} }$

${\displaystyle E=2.0516\times 10^{22}\mathrm {J} =5\times 10^{12}\mathrm {ton_{TNT}} }$

${\displaystyle A=2\pi R^{2}=7.2185\times 10^{-43}\mathrm {m^{2}} }$ or ${\displaystyle A=4\pi R^{2}=1.4437\times 10^{-42}\mathrm {m^{2}} }$

${\displaystyle S={2\pi kG\,M^{2} \over \hbar c}=9540.3\mathrm {J/K} }$

${\displaystyle T={\hbar c^{3} \over 8\pi kG\,M}=5.3761\times 10^{17}\mathrm {K} }$

${\displaystyle P={\hbar c^{6} \over 15,360\pi G^{2}\,M^{2}}=6.8386\times 10^{21}\mathrm {W} }$

${\displaystyle e_{b}={P \over 2\pi R^{2}}=9.4738\times 10^{63}\mathrm {W/m^{2}} }$ or ${\displaystyle e_{b}={P \over 4\pi R^{2}}=4.7369\times 10^{63}\mathrm {W/m^{2}} }$

${\displaystyle e_{b}=\sigma T^{4}=5.6705\times 10^{-8}\mathrm {W \over m^{2}K^{4}} \,(5.3761\times 10^{17}\mathrm {K} )^{4}=4.7369\times 10^{63}\mathrm {W/m^{2}} }$

${\displaystyle e_{b}={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}=4.7369\times 10^{63}\mathrm {W/m^{2}} }$

${\displaystyle E_{\mathrm {P} }={m_{\mathrm {P} }\ell _{\mathrm {P} }^{2} \over t_{\mathrm {P} }^{2}}}$

${\displaystyle P_{\mathrm {P} }={m_{\mathrm {P} }\ell _{\mathrm {P} }^{2} \over t_{\mathrm {P} }^{3}}}$

${\displaystyle \sigma ={\pi ^{2} \over 60}=0.16449{E_{\mathrm {P} } \over t_{\mathrm {P} }\ell _{\mathrm {P} }^{2}T_{\mathrm {P} }^{4}}=0.16449{m_{\mathrm {P} } \over t_{\mathrm {P} }^{3}T_{\mathrm {P} }^{4}}}$

${\displaystyle M=9.1373\times 10^{37}m_{\mathrm {P} }}$

${\displaystyle R=1.8275\times 10^{38}\ell _{\mathrm {P} }}$

${\displaystyle E=9.1373\times 10^{37}E_{\mathrm {P} }}$

${\displaystyle A=2\pi \,R^{2}=2.0983\times 10^{77}\ell _{\mathrm {P} }^{2}}$ or ${\displaystyle A=4\pi \,R^{2}=4.1966\times 10^{77}\ell _{\mathrm {P} }^{2}}$

${\displaystyle S={2\pi kG\,M^{2} \over \hbar c}=5.2458\times 10^{76}E_{\mathrm {P} }/T_{\mathrm {P} }}$

${\displaystyle T={\hbar c^{3} \over 8\pi kG\,M}=4.3546\times 10^{-40}T_{\mathrm {P} }}$

${\displaystyle P={\hbar c^{6} \over 15,360\pi G^{2}\,M^{2}}=2.4821\times 10^{-81}P_{\mathrm {P} }}$

${\displaystyle e_{b}={P \over 2\pi \,R^{2}}=1.1829\times 10^{-158}P_{\mathrm {P} }/\ell _{\mathrm {P} }^{2}}$ or ${\displaystyle e_{b}={P \over 4\pi \,R^{2}}=5.9146\times 10^{-159}P_{\mathrm {P} }/\ell _{\mathrm {P} }^{2}}$

${\displaystyle e_{b}=\sigma T^{4}=5.9146\times 10^{-159}P_{\mathrm {P} }/\ell _{\mathrm {P} }^{2}\quad T^{4}=3.5956\times 10^{-158}T_{\mathrm {P} }^{4}}$

${\displaystyle e_{b}={\hbar c^{10} \over 245,760\pi ^{2}G^{4}\,M^{4}}=5.9146\times 10^{-159}P_{\mathrm {P} }/\ell _{\mathrm {P} }^{2}}$