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For a neutrally buoyant horizontal turbulent round jet, a segment's horizontal momentum flux changes only if an external horizontal force or ambient momentum input exists. Neither is present, soPositive buoyancy adds a vertical force, not a horizontal one. Therefore $d{M}_x/ds=0$ remains true for an inclined forced plume even when $F_0\ne 0$; in general $M_x=M_0\cos {\theta }_0$.

The net upward force on a segment is $g({\rho }_0-\rho )\pi b^{2}ds$. Since $F=g({\rho }_0-\rho )\pi b^{2}V$ and $V=|\mathbf{M}|/Q$, vertical momentum balance givesIn kinematic plume fluxes, the equations are $m_x^{\prime }=0$, $m_z^{\prime }=fq/m$, where $m=(m_x^2+m_z^2{)}^{1/2}$. The centreline geometry satisfiesFor $F_0>0$, $Q>0$ beyond the source, and an initially nonvertical turbulent round jet pointing right, $M_z$ increases and the centreline turns toward the vertical. The magnitude of momentum flux is not conserved once the tangent has a vertical component.

Put $m_0=M_0/{\rho }_0$ and $f_0=F_0/{\rho }_0>0$ under the literal mass convention. Their dimensions are $[m_0]=L^4{T}^{-2}$ and $[f_0]=L^4{T}^{-3}$. Dimensional analysis givesThe jet length compares source momentum with buoyancy. The region near the source behaves as a turbulent round jet; sufficiently far away, the forced plume behaves like a pure plume. With the constant entrainment coefficient retained explicitly, substantial turning occurs over a length of order ${\ell }_J=L_J/\sqrt{E}$, where $E=2\alpha \sqrt{\pi }$. Statements involving $s\ll L_J$ or $s\gg L_J$ normally hold this dimensionless coefficient fixed.

Near the point source, $m\sim m_0$ and the Batchelor entrainment hypothesis gives $q\sim E\sqrt{m_0}s$. ThereforeIntegrating the centreline tangent givesTo leading order the centreline is a straight turbulent round jet at its source inclination. For the horizontal source, the first curvature is the horizontal forced-plume trajectoryFor a vertical source, $\cos {\theta }_0=0$, the trajectory stays vertical rather than developing this cubic transverse displacement. If $F_0=0$, there is no finite buoyancy-induced jet length and the turbulent round jet remains straight.

For a horizontal source, $m_x=m_0$ throughout. Far downstream $m_z\gg m_0$, so the kinematic plume fluxes obey the vertical pure plume equations to leading order,Set $q=q_{\ast }{f}_0^{1/3}{\hat{s}}^{5/3}$ and $m_z=m_{\ast }{f}_0^{2/3}{\hat{s}}^{4/3}$ with $\hat{s}=s+s_0$. Matching exponents and coefficients gives $(5/3)q_{\ast }=E\sqrt{m_{\ast }}$ and $(4/3)m_{\ast }^2=q_{\ast }$. Consequentlyand the requested literal mass and density-weighted buoyancy fluxes areto leading order. The radius is $b\sim (6\alpha /5)\hat{s}$ and the axial velocity decreases as ${\hat{s}}^{-1/3}$.

The centreline becomes nearly vertical but retains finite horizontal drift:Since $dz/ds=1-O({\hat{s}}^{-8/3})$, there is a finite positive geometric deficit $D={\lim }_{s\to \infty }(s-z)$, and $z=s-D+o(1)$. The equivalent vertical plume virtual origin is therefore located atFor the horizontal point-source forced plume, the virtual source is downstream and below the actual source. The horizontal turbulent round jet first entrains substantial ambient fluid while rising only a little. Thus at the height where it turns upward it already has a finite radius and volume flux; an equivalent pure plume must have begun rising from below to acquire these. The turning displacement, entrained radius divided by its far-field spreading angle, and virtual vertical depth all scale asAt fixed entrainment coefficient this is simply $O(L_J)$. The arclength origin $s=-s_0$ should not be interpreted as a physical height: the curved near field supplies the additional shift $D$. The numerical constants depend on the chosen top-hat plume model and entrainment coefficient; scaling does not fix them.

The inclined forced plume bends from a cubic near-field trajectory toward a vertical asymptote. Dashed far-field spreading lines extrapolate to the equivalent plume virtual origin; the marked location is obtained from the integral model for this illustration, not a universal numerical prediction.

A circular turbulent round jet supplied with kinematic momentum flux $m_0$ but negligible buoyancy flux conserves $m=m_0$. A top-hat plume model and constant Batchelor entrainment give $dq/ds=2\alpha \sqrt{\pi m_0}$. From a point source with zero volume flux, $q=2\alpha \sqrt{\pi m_0}s$, radius $b=2\alpha s$, and axial velocity $V=m_0/q\propto s^{-1}$.