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Deschele Schilder
Deschele Schilder
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Suppose the water consists of microscopic non-interacting [apart from temperature dependent [but T<100particles (C)]apart from elastic collisions), say microscopic sand particles. All these particles come out of the pipe with an average horizontal velocity will follow (in the mean) a parabolic trajectory downwards. The greater the diameter of the pipe the longer it takes for the part of the water (modeled by the particles) closer to the upper part of the pipe to reach the ground than the part of the water closer to the lower part. This is a constant though and has no influence on the shape of the falling water mass. Of course,The particles on the diameterupper side of the mass will decreasepipe follow the closer it getssame parabola as the particles on the lower side of the pipe. It's easy to see that this results in de deformation of the initial circular cross-section of the water when leaving the pipe, into an ellipse form for the cross-section when hitting the ground. And because the particles don't interact, there are no tidal forces so the diameter (long axis) of the water mass when arrivingellipse on the ground is minimal andhas the velocity maximalsame value as the diameter of the initial circle. So the cross-section of the beam isn't diminishing but stays the same because the particle density diminishes (w.r.t.in contrast to what followsreal water). And of course, particlesParticles on the outer side of the water mass will manage to escape, but this has a negligible effect. At last, when the water mass falls the particles get separated more and more (instead of forming dropletsSo in real water after falling some time, which is due to the viscosity and accompanying tidal forces). In this case, the form of the falling (particle modeled) water beam is parabolic and the form the beam has when arriving on the ground is an ellipse.

Now suppose the opposite: Suppose the water consists of microscopic particles that are all bound to each other by small strings. In this case, there is also a mean horizontal for the particle, but aside from the collisions (which don't affect the mean parabolic form of the falling water mass), the microscopic particles all pull on each other by means of the little springs, which means that tidal forces working on the water mass appear. The particles don't move further from each other the longer they fall, so no droplets will be formed. Because of the tidal forces, the lower parts of the water mass experience a force upwards and consequently arrive at the floor later than in the former case and with a broader diameterthe same cross-section and velocity as the initial cross-section and velocity. It's like shooting a long ironbicycle chain in a horizontal direction by letting it come off from a high building. The first part of the chain will arrive later on the groundbig, then in the caseconstantly rotating, the chain parts are unchainedgear blade in a horizontal direction from a high building. It's obvious that the form of the second kind of water coming out horizontally from a pipe doesn't have the form of a parabola (just like the form of the chain taking of horizontally after which it falls to the ground).

So tidal forces, which in your case of a real falling mass of water are present, due to the water's viscosity. Though the Reynolds number of water is very high the form is almost parabolic, which is to say not parabolic. The cross-section gets smaller when the beam is closer to the floor (the water density isn't changing), but has a minimal ellipse form. Also, because of the tidal forces, the velocity at the ground is a little smaller than would be the case if there were no tidal forces (or in other words, viscosity).

So the from is almost a parabole, which is to say the form is not a parabole, though the exact form I can't tell you. In the case of the chain, it could be a catenary. A simple experiment to see thisthat a fluid with a low Reynolds number doesn't create a parabolic form when you launch a beam of the stuff vertically is to make a beam of honey (low Reynolds number) and a beam of water (high Reynolds number) come out at the same type of horizontal pipe with the same velocity and compare the two ensuing forms ofspots on the falling waterfloor (before the fluid with the high Reynolds number breaks up) where they arrive. The "more" parabolic (high Reynolds number), the further the beam will arrive on the ground (measured as the horizontal distance between the pipe's exit and honeythe impact spot on the floor). It will be more difficult to compare the final cross-sections of the beam.

Suppose the water consists of microscopic non-interacting [apart from temperature dependent [but T<100(C)] collisions) particles. All these particles come out of the pipe with an average horizontal velocity will follow (in the mean) a parabolic trajectory downwards. The greater the diameter of the pipe the longer it takes for the part of the water closer to the upper part of the pipe to reach the ground than the part of the water closer to the lower part. This is a constant though and has no influence on the shape of the falling water mass. Of course, the diameter of the mass will decrease the closer it gets to the ground. And because the particles don't interact, there are no tidal forces so the diameter of the water mass when arriving on the ground is minimal and the velocity maximal (w.r.t. to what follows). And of course, particles on the outer side of the water mass will manage to escape, but this has a negligible effect. At last, when the water mass falls the particles get separated more and more (instead of forming droplets in real water after falling some time, which is due to the viscosity and accompanying tidal forces). In this case, the form of the falling water beam is parabolic.

Now suppose the opposite: Suppose the water consists of microscopic particles that are all bound to each other by small strings. In this case, there is also a mean horizontal for the particle, but aside from the collisions (which don't affect the mean parabolic form of the falling water mass), the microscopic particles all pull on each other by means of the little springs, which means that tidal forces working on the water mass appear. The particles don't move further from each other the longer they fall, so no droplets will be formed. Because of the tidal forces, the lower parts of the water mass experience a force upwards and consequently arrive at the floor later than in the former case and with a broader diameter. It's like shooting a long iron chain in a horizontal direction from a high building. The first part of the chain will arrive later on the ground, then in the case, the chain parts are unchained. It's obvious that the form of the second kind of water coming out horizontally from a pipe doesn't have the form of a parabola (just like the form of the chain taking of horizontally after which it falls to the ground).

So tidal forces, which in your case of a real falling mass of water are present, due to the water's viscosity. Though the Reynolds number of water is very high the form is almost parabolic, which is to say not parabolic.

So the from is almost a parabole, which is to say the form is not a parabole. A simple experiment to see this is to make a beam of honey (low Reynolds number) and a beam of water (high Reynolds number) come out at the same type of horizontal pipe with the same velocity and compare the two ensuing forms of the falling water and honey.

Suppose the water consists of microscopic non-interacting particles (apart from elastic collisions), say microscopic sand particles. All these particles come out of the pipe with an average horizontal velocity will follow (in the mean) a parabolic trajectory downwards. The greater the diameter of the pipe the longer it takes for the part of the water (modeled by the particles) closer to the upper part of the pipe to reach the ground than the part of the water closer to the lower part. This is a constant though and has no influence on the shape of the falling water mass. The particles on the upper side of the pipe follow the same parabola as the particles on the lower side of the pipe. It's easy to see that this results in de deformation of the initial circular cross-section of the water when leaving the pipe, into an ellipse form for the cross-section when hitting the ground. And because the particles don't interact, there are no tidal forces so the diameter (long axis) of the ellipse on the ground has the same value as the diameter of the initial circle. So the cross-section of the beam isn't diminishing but stays the same because the particle density diminishes (in contrast to real water). Particles on the outer side of the water mass will manage to escape, but this has a negligible effect. So in this case, the form of the falling (particle modeled) water beam is parabolic and the form the beam has when arriving on the ground is an ellipse.

Now suppose the opposite: Suppose the water consists of microscopic particles that are all bound to each other by small strings. In this case, there is also a mean horizontal for the particle, but aside from the collisions (which don't affect the mean parabolic form of the falling water mass), the microscopic particles all pull on each other by means of the little springs, which means that tidal forces working on the water mass appear. The particles don't move further from each other the longer they fall, so no droplets will be formed. Because of the tidal forces, the lower parts of the water mass experience a force upwards and consequently arrive at the floor later than in the former case and with the same cross-section and velocity as the initial cross-section and velocity. It's like shooting a long bicycle chain in a horizontal direction by letting it come off from a big, constantly rotating, gear blade in a horizontal direction from a high building. It's obvious that the form of the second kind of water coming out horizontally from a pipe doesn't have the form of a parabola (just like the form of the chain taking of horizontally after which it falls to the ground).

So tidal forces, which in your case of a real falling mass of water are present, due to the water's viscosity. Though the Reynolds number of water is very high the form is almost parabolic, which is to say not parabolic. The cross-section gets smaller when the beam is closer to the floor (the water density isn't changing), but has a minimal ellipse form. Also, because of the tidal forces, the velocity at the ground is a little smaller than would be the case if there were no tidal forces (or in other words, viscosity).

So the from is almost a parabole, which is to say the form is not a parabole, though the exact form I can't tell you. In the case of the chain, it could be a catenary. A simple experiment to see that a fluid with a low Reynolds number doesn't create a parabolic form when you launch a beam of the stuff vertically is to make a beam of honey (low Reynolds number) and a beam of water (high Reynolds number) come out at the same type of horizontal pipe with the same velocity and compare the two spots on the floor (before the fluid with the high Reynolds number breaks up) where they arrive. The "more" parabolic (high Reynolds number), the further the beam will arrive on the ground (measured as the horizontal distance between the pipe's exit and the impact spot on the floor). It will be more difficult to compare the final cross-sections of the beam.

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Deschele Schilder
Deschele Schilder
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You're last "embraced" comment is important. I'll try to answer this question qualitative.

Suppose the water consists of microscopic non-interacting [apart from temperature dependent [but T<100(C)] collisions) particles. All these particles come out of the pipe with an average horizontal velocity will follow (in the mean) a parabolic trajectory downwards. The greater the diameter of the pipe the longer it takes for the part of the water closer to the upper part of the pipe to reach the ground than the part of the water closer to the lower part. This is a constant though and has no influence on the shape of the falling water mass. Of course, the diameter of the mass will decrease the closer it gets to the ground. And because the particles don't interact, there are no tidal forces so the diameter of the water mass when arriving on the ground is minimal and the velocity maximal (w.r.t. to what follows). And of course, particles on the outer side of the water mass will manage to escape, but this has a negligible effect. At last, when the water mass falls the particles get separated more and more (instead of forming droplets in real water after falling some time, which is due to the viscosity and accompanying tidal forces). In this case, the form of the falling water beam is parabolic.

Now suppose the opposite: Suppose the water consists of microscopic particles that are all bound to each other by small strings. In this case, there is also a mean horizontal for the particle, but aside from the collisions (which don't affect the mean parabolic form of the falling water mass), the microscopic particles all pull on each other by means of the little springs, which means that tidal forces working on the water mass appear. The particles don't move further from each other the longer they fall, so no droplets will be formed. Because of the tidal forces, the lower parts of the water mass experience a force upwards and consequently arrive at the floor later than in the former case and with a broader diameter. It's like shooting a long iron chain in a horizontal direction from a high building. The first part of the chain will arrive later on the ground, then in the case, the chain parts are unchained. It's obvious that the form of the second kind of water coming out horizontally from a pipe doesn't have the form of a parabola (just like the form of the chain taking of horizontally after which it falls to the ground).

So tidal forces, which in your case of a real falling mass of water are present, due to the water's viscosity. Though the Reynolds number of water is very high the form is almost parabolic, which is to say not parabolic.

And a second effect comes into play. At a certain velocity, the laminar flow of the water becomes turbulent due to its viscosity which will let the form of the falling water from the pipe, which will make the form of the falling water deviate (due to internal frictions in the water) a bit more from a parabole.

So the from is almost a parabole, which is to say the form is not a parabole. A simple experiment to see this is to make a beam of honey (low Reynolds number) and a beam of water (high Reynolds number) come out at the same type of horizontal pipe with the same velocity and compare the two ensuing forms of the falling water and honey.