\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*1000*\cfrac{1}{10}\)
or more clearly,
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*density*\cfrac{1}{particle\,count\,per\,type}\)
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*840*\cfrac{1}{12*6+24*1}\)
we have a diesel boom at \(30.05\,ms^{-1}\) or \(108.22\,kmh^{-1}\) Ooops
And if we apply the expression to gasoline of density \(719.7\,kgm^{-3}\) and on average (\(C_{4}-C_{12}\)), \(C_{8}H_{18}\),
\(\cfrac{1}{2}m_{\rho}v^{ 2 }_{ { max } }=E_{input/proton}=\cfrac{1}{2}c^{ 2 }*3.4354*719.7*\cfrac{1}{8*6+18*1}\)
we have a gasoline boom at \(37.46\,ms^{-1}\) or \(134.86\,kmh^{-1}\) Ooops
These are not cruise speeds. But these speeds suggest that both gasoline and diesel ignite without a spark when pushed at their respective boom speeds. For maximum energy yield, both boom speed should be sustained; at maximum efficiency the energy conversion is nuclear! For less that perfect efficiency (sinusoidal boom speed), the fuel combustion in the heat produced, produces waste gas.
Do we really need the engine to go nuclear?
Note: If we need high speed fuel, the trade off is between low particle count per type of particle and high density fuel. Density decreases with low particle count. Coil up fuel with low particle count but high density comes to mind, but conventional view on hydrocarbons that have many branches is that they burn less effectively. Why should fuel be burning?
