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When I saw that thing I did not understand how that shape is formed? To be ideal, take a vertical smooth plane. Now aim at the wall with a thin water tube. Then the outer layer forms a "parabolic shape" with stagnation point as focus of it. I found it by tracing that shape.enter image description here

How could you explain this observation? Also could you provide the equation in terms of velocity of flow, angle of contact with the wall and gravitational constant?

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    $\begingroup$ When you saw what "thing"? Is this a reference to some popular video or social media post? $\endgroup$ – jpmc26 Jul 3 at 23:49
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    $\begingroup$ @jpmc26 I think it's just a translation problem. $\endgroup$ – Hearth Jul 5 at 15:55
  • $\begingroup$ Please see our FAQ on question titles. $\endgroup$ – DanielSank Jul 11 at 15:21
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Each of the water particles gets pushed to the side by the other particles as the water hits the wall. If we neglect the viscosity of the water, each of these particles follows a throwing-parabola, but under different initial launch angles. If we assume the jet hits the wall horizontally, the water particles are thrown with the same (maximum) initial velocity in every direction. The shape that you observed is then given by the envelope of all possible parabolas.

For all parabolas $$y(x) = x \tan \beta - \frac{g\,x^2}{2\,{v_0}^2 \cos^2\beta} + h_0$$ with initial launch angles $\beta$, the envelope is $$y_\mathrm{H} (x) = \frac{{v_0}^2}{2\,g} - \frac{g\,x^2}{2\,{v_0}^2} + h_0.$$

So it forms indeed a parabola.

Envelope

Edit: The envelope can be derived als follows:

If we define the family of curves implicitly by $$F(x,y,\tan(\beta))=y - x \tan \beta + \frac{g\,x^2}{2\,{v_0}^2 \cos^2\beta}=y - x \tan \beta + \frac{g\,x^2(1+\tan^2\beta)}{2\,{v_0}^2 }=0$$ the envelope of the family is given by (Source) $$F = 0~~\mathsf{and}~~{\partial F \over \partial \tan\beta} = 0$$ We have $${\partial F \over \partial \tan\beta}=-x+\frac{gx^2\tan\beta}{v_0^2}=0 ~~ \Leftrightarrow ~~ \tan\beta=\frac{v_0^2}{gx}$$ Substituting that into $F$ we get $$F=y-\frac{v_0^2}{g}+\frac{g(x^2+v_0^4/g^2)}{2v_0^2}=0 ~~\Leftrightarrow ~~ y_\mathrm{H} (x) = \frac{{v_0}^2}{2\,g} - \frac{g\,x^2}{2\,{v_0}^2}$$

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  • $\begingroup$ Is the maximum initial velocity symmetric if the stream is falling when it hits the wall? $\endgroup$ – Anton Sherwood Jul 4 at 20:49
  • $\begingroup$ @AntonSherwood No, only if it hits the wall horizontally. $\endgroup$ – Azzinoth Jul 5 at 9:31
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    $\begingroup$ I don't think this calculation is convincing. It seems to assume that each particle of water, after it reaches the envelope curve, will continue falling down through the area that's already covered by a rising sheet of water. That's not really what happens in practice; the water instead rolls down along the outside of the "parabola" -- but in order for this combined stream to keep from falling down along your lines it must get some momentum from the upwards water that joins it -- so that that water cannot join the stream tangentially, and the true curve must be lower than you compute. $\endgroup$ – Henning Makholm Jul 5 at 20:56
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    $\begingroup$ @HenningMakholm I agree with that. I think however, that the primary reason for the observed shape is that the water is pushed in all directions when it hits the wall and basically follows gravity after that. All other things like turbulent flow transition, shear viscosity, surface tension, friction etc (see also A.V.S excellent answer) are just real world complications/corrections to that. The OP wanted to understand how such a shape can form in the first place. I think for that purpose it is ok to first consider an idealized version of the problem. $\endgroup$ – Azzinoth Jul 5 at 22:01
  • $\begingroup$ @HenningMakholm I don't even think it would be a too bad approximation, because while the particles that are pushed up from the bottom stay longer at the envelope, the particles coming from the bottom will be pushed down by the same process. So some particles stay up longer and some fall down earlier. If we assume conservation of energy and momentum it will approximately cancel out. $\endgroup$ – Azzinoth Jul 5 at 22:01
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If the water jet impinges on the horizontal surface the liquid flows away in a laminar thin film flow until the flow turns slower and turbulent at some distance from point of impingement zone in a circular hydraulic jump. If the surface which the jet impinges on is vertical this hydraulic jump forms a “rope” which flows circumferentially around the region of radial laminar flow. The physics of the flow is quite complicated, combining laminar to turbulent flow transition, shear viscosity, surface tension, gravity and interaction of fluid with the wall, so there would be no simple solution. However, this physical situation has a practical importance and so it has been studied experimantally:

Image from Wang et al., 2013

So here is a couple of paper dedicated to this topic, the image above is from the first one:

  • Wang, T., Faria, D., Stevens, L. J., Tan, J. S. C., Davidson, J. F., & Wilson, D. I. (2013). Flow patterns and draining films created by horizontal and inclined coherent water jets impinging on vertical walls. Chemical Engineering Science, 102, 585-601, doi:10.1016/j.ces.2013.08.054.

  • Aouad, W., Landel, J. R., Dalziel, S. B., Davidson, J. F., & Wilson, D. I. (2016). Particle image velocimetry and modelling of horizontal coherent liquid jets impinging on and draining down a vertical wall. Experimental Thermal and Fluid Science, 74, 429-443, doi:10.1016/j.expthermflusci.2015.12.010, free pdf.

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  • $\begingroup$ Can you measure the width of the edges and give a relation between edge and depth from the tangent at the vertex. Please... $\endgroup$ – Naga Sandesh Goli Jul 12 at 6:02
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Because if there was no gravity affecting then the water would spread horizontally as well, but due to the effect of gravity the horizontal spread is "dragged" down compared to the initial impact point.

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    $\begingroup$ Its ok. I got it. But what is the equation of it. $\endgroup$ – Naga Sandesh Goli Jul 3 at 9:29
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    $\begingroup$ You asked for an explanation in your question, which I gave. I will let you think about an equation. $\endgroup$ – user207455 Jul 3 at 9:31
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    $\begingroup$ @SolarMike I went to the edits and the first original post said "Then the outer layer forms a parabolic shape with stagnation point as focus of it. I found it by tracing that shape. How could you explain this observation?". This I think supports my original position. However, I admit the asker's original was low effort. Similarly, I see your answer as very low effort and I doubt OP or anyone who would make that question or search it on google doesn't realize that water falls due to gravity, your answer is too obvious. So I still think this answer is insufficient. $\endgroup$ – Santropedro Jul 6 at 6:30

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