Newton at Niagara Falls

After a visit to Niagara Falls, I was puzzled by one observation. Shouldn’t the path of the water make a significantly noticeable arch? Especially at the Canadian side, the water looks more like a hanging curtain. Maybe it’s simply an illusion?

Considering that the river has significant velocity, I would expect the “arch” path to be more prominent. Of course, later in the free-fall, rocks interfere. But, just passed the brink, shouldn’t significant arching be detected?

In short, I would expect the water to fall further out (away) from the brink. Even at lesser velocities, arching is observed. For example, using a simple garden hose placed horizontally against the top of the railing of a deck, with end aligned with edge of railing and pointed out to the yard, a noticeable arch is observed.

I’ve played with some numbers, guessing at reasonable river speeds, and the displacement in the y-direction should be substantial considering the distance of the fall.

I know…I’m standing on the brink myself, right? I should take a vacation when on vacation! :o

This may not answer your question, but have you considered that there may be an arching that you simply can’t see? The river is between 3 and 7 feet deep at the edge of the precipice. Are you sure that when you look at what you perceive to be the edge, it isn’t actually several feet behind and below the point where you’re looking?

Another difference between the river and your analagous hose: the river is running purely on gravity, which means that the layer of water that is physically on the edge is being forced downward by all the water above the bottom layer. In a hose, the “shallow” depth of the hose stream means that there is insignifiant downward pressure on the stream as it exits the nozzle.

(I suspect that the very nature of the dynamics of water exiting a hose my also play a part, but I don’t have any hydraulic theory to back that up.)

Tom~

Tom, I have considered what you say, and I can only WAG that it must be an illusion of perception. I can only WAG that it is very hard to compare the arch against any reference behind the falls.

The second part of your answer addresses hydraulic pressure within the river due to its depth. But, once over the brink, the water is in free-fall and there would be no hydraulic pressure (static head) no matter how deep the river at the brink.

Strictly, the initial velocity in the x-direction governs. (I’m assuming all motion to be in the x-direction despite eddies, etc., etc. I think one would agree the motion is primarily in the x-direction up to the brink.) Also, from fluids, one would expect the top surface of the river to have a velocity vector of greatest magnitude.

Likewise, the garden hose pressure does not enter into a free-fall problem; water pressure simply grants some initial velocity in the x-direction.

Jinx,
Your WAG is correct. The scale is large, and the arching takes place on a relatively small scale. The speed of the river isn’t that great, the falling water quickly reaches a high veloctity in the Y direction

(Vy = -32ft*sec)), hence a steep angle (angle is ARCTAN(Vy/Vx)). We’re used to seeing similar motion on a small scale, and we expect the path shape of the falling object to scale up, but it doesn’t. The same faulty reasoning leads people to believe they can throw a ball incredible distances off the top of a building. The movies don’t help because they will frequently film a scale model and try to pass it off as full size, after seeing enough of these scenes we think they’re real.

Thanks, Frolix8, but did you mean to say Vy is -32 ft/sec? Are you confusing velocity with acceleration due to gravity of 32 ft/sec^2? I believe you meant to mention that the water quickly reaches a terminal velocity, perhaps, in the y-direction? I haven’t cranked the numbers real hard, but maybe it’s just a coincidence? (Still, a terminal velocity in the “y” wouldn’t impact motion in the “x” direction.)

You are correct, also, that it is hard to judge the speed other than “swift”! There are ways to find it, but the park authorities don’t take too highly to that sort of thing!

I’m afraid we’re over-complicating the answer to the OP’s question. References to movement in the x- and y-axes are on the right track, and the answer lies there.

For our purposes, the hydraulic properties of water are best ignored, since they’re essentially irrelevant to the degree of a waterfall’s arch. Instead, what we have is a pretty simple exterior ballistics problem.

First of all, what Jinx refers to as the river’s “significant velocity” in the x-axis is less significant to the shape of the arch than one might think. Using average flows and depths, my calculator tells me that the water’s average speed in the x-axis ranges between 2 and 6 fps at the brink. Given that, past the brink, the water accelerates in the y-axis in pretty much the same manner as any other falling body (i.e., at approximately 32 f/s/s), the x-axis speed loses its significance very quickly.

Thus, I’d have to conclude that Jinx’s observation is a matter of perception or appearance, rather than some physical peculiarity of Niagara Falls.

The example of the water hose can be reduced to the exact same principle. The big difference between the water exiting the hose and the water falling at Niagara is simply the initial horizontal (x-axis) velocity. The water from the hose is traveling a LOT faster.

Tom, the water beneath the surface is NOT “pushed down” by the water on top. Hydraulic pressure at any point in a relatively still fluid body is omnidirectional. At the brink of the falls, that pressure disappears, but neither the pressure nor its disappearance has a significant effect on the motion of the water and, thus, the shape of the arch.

I don’t know why fortune smiles on some and lets the rest go free…

T

TBone, some questions: (A) You said that motion in the x-direction is less significant to the shape of the arch than one might think? I don’t follow!

First, in every projectile (ballistics) problem, it is the x-velocity that determines the range. Range is displacement in the x-direction. The “y” direction determines how long the event will last. In this case, the time of the event can be found by relating the “y” displacement to acceleration due to gravity taking Vy=0 at t=0.

The longer the event, the more displacement in the x-direction, and the more arc noticed. (Surely you can recall, in physics, doing academic lab experiments rolling balls off tables and predicting the landing point.) I’m sure you noticed how the parabolic arc will have greater range as the velocity in the x-direction increases. No matter how small the x-velocity, there is displacement in the x-direction which is a parabolic arc, or virtually parabolic, but let’s not split hairs! (Next, someone’ll throw in Coreolis!)

(B) I find it very hard to believe the water from my garden hose, without a nozzle, is moving faster than the Niagara River! How did you come up with 6 ft/s? (~4mph)

<font color="#900000">frolix8: Jinx,
Your WAG is correct. The scale is large, and the arching takes place on a relatively small scale. The speed of the river isn’t that great, the falling water quickly reaches a high veloctity in the Y direction
(Vy = -32ft
sec)), hence a steep angle (angle is ARCTAN(Vy/Vx)). We’re used to seeing similar motion on a small scale, and we expect the path shape of the falling object to scale up, but it doesn’t. The same faulty reasoning leads people to believe they can throw a ball incredible distances off the top of a building. The movies don’t help because they will frequently film a scale model and try to pass it off as full size, after seeing enough of these scenes we think they’re real.</font>

<font color="#009000">Jinx:
Thanks, Frolix8, but did you mean to say Vy is -32 ft/sec? Are you confusing velocity with acceleration due to gravity of 32 ft/sec^2? I believe you meant to mention that the water quickly reaches a terminal velocity, perhaps, in the y-direction? I haven’t cranked the numbers real hard, but maybe it’s just a coincidence? (Still, a terminal velocity in the “y” wouldn’t impact motion in the “x” direction.) </font>*

No, Frolix8 said <code>Vy = -32ftsec</code>. A little miswritten, but the jist is that velocity is linear to the time of freefall; notice the "" symbol is in there, not the “/” as you misquoted.

To be more acurate, the velocity at a given time t is:
<code>V(t) = g * t</code>
g being the acceleration due to Earth’s gravity, 32 ft./sec[sup]2[/sup]. So, after 2 seconds:
<code>
V(2 sec) = 32 ft/sec[sup]2[/sup] * 2 sec = 64 ft/sec.
</code>

Judges 14:9 - So [Samson] scraped the honey into his hands and went on, eating as he went. When he came to his father and mother, he gave some to them and they ate it; but he did not tell them that he had scraped the honey out of the body of the lion.

Jinx, when I said that the x-axis velocity is less significant than one might think, I was referring to the specific problem you presented – the x-axis velocity of the water passing the brink at Niagara Falls. The calculations are quite simple. Using an average depth of 5’ at both falls, an average flow of 150,000 cfs at the American/Bridal Veil Falls and 600,000 cfs at the Horseshoe Falls, and a width of 1060’ for the American and 2200’ for Horseshoe, it’s a matter of pounding a few calculator buttons to find an average horizontal speed of between 2 and 6 fps. Depending on when you were at the falls, flow could have been greater or less, depending on a number of factors, not the least of which is the pair of hydroelectric generating stations that siphon water from American when they’re online.

The significance of the x-axis speed – which remains relatively constant – and its effect on the apparent “sharpness” of the arc is inversely proportional to the y-axis speed, which increases constantly over time. After only one second, for instance, the y-axis speed exceeds the x-axis speed by a factor of between 5 and 16.

I don’t know what sort of projectiles you may have experimented with, but in any ballistics example I’ve ever seen, the “sharpness” of the parabolic arc is greatest close to the apogee. In the case of a waterfall, we may assume that the brink IS the apogee of the parabolic curve.

As to the speed of the water exiting your garden hose, consider a hose of 5/8" diameter (that’s a big one) discharging just 10 gpm (pretty weak). Approximate speed of the water exiting the hose: 10 fps. Drop the hose diameter to 1/2" (more likely) and the same 10 gpm leaves the hose at nearly 16 fps. Even so, I think if you stood, hose in hand, on the top of an 18-story building (a rough equivalent to the height of Niagara), you’d be flat amazed at how quickly the sream assumes a nearly-vertical orientation.

I don’t know why fortune smiles on some and lets the rest go free…

T

I just performed the exact same experiment with a graphing calculator. If you have access to one I suggest you do the same. Graph y=-x[sup]2[/sup]. This is a good enought approximation to the fall of an object. Focus on the range from y=0 to y=-4. You see a significant arc. Now zoom out to the range y=0 to y=-200. The arc is now almost completely gone. So it’s simply a matter of scale.

TheDude

Tbone, I see what you are saying, but using your data and the formula Q=vA, for the Horseshoe Falls, I get the river moving at 54 ft/s or about 37 mph which is a bit more believeable. Is this the formula you applied? Just curious. (I assumed the water was moving through an imaginary area with cross section of 5’ x 2200’.) Did you lose a decimal place, perhaps, in your calc?

Overall, I can see where the acceleration downward quickly “outranks” the horizontal motion thus giving an almost plumb-vertical appearance. Mystery solved!

They had me over a barrel on this one!

Jinx:

OK, ya nailed me! I didn’t exactly “lose a decimal place.” What I did was quote the wrong units. The numbers – 150,000 and 600,000 – were correct, but the unit should have been gallons per second, NOT cubic feet per second. I apologize!

The conversion factor from cfs to gps is seven-point-something, I believe. (My hydraulics textbooks are in a box somewhere…) Since our calculations aren’t likely to rearrange the world of physics, I think I just rounded it to 8. Dividing your result by eight yields 6.8 fps, a bit higher than the range I quoted. Using the American Falls data, I get about 3.5 fps at average flow – 150,000 gps – but that can be reduced by as much as two-thirds by the hydropower plants I mentioned before.

Incidentally, you’re right there with most people in calling water moving at those speeds – 2-5 mph – as having “significant velocity.” Water seems to be a substance we tend to think of as relatively static – just sitting in a bucket or a puddle or a pond or a lake. Water moving at a walking speed is considered “fast-moving,” simply because we don’t often see water in such a hurry.

And TheDude, BINGO! It is, after all, a matter of scale. In the case of Niagara, if one were able to capture an image of the first, say, 10 feet of the fall, one would see a recognizable section of the beloved parabolic curve that Jinx longs for. But the other 17 stories of fall are going to look an awful lot like a straight, vertical line.

I don’t know why fortune smiles on some and lets the rest go free…

T

I’d say scale but I also want to chuck another variable in. While watching one of those home repair shoes on TV install some of the newest drip edging they show how the curve at the edge will direct the water down into the gutters while shooting the debris over the top of the gutters. Basically the water sticks to the metal drip edge and is thus directed down and back under to fall neatly in the gutter thus keeping debris out. Granted we’re talking a much deeper stream and frankly I don’t know whether there is any type of overhang to Niagara but couldn’t the actual shape of the edge of the falls greatly effect the amount of arc ?

Funnee, it sounds an awful lot like bullshit to me.

H2O is both cohesive and adhesive, when it comes to flow and the ability to carry “debris.” In other words, where the water goes, there goes the debris.

Harking back to my house-building days, I hasten to point out that “drip-edge” – actually a molding (usually aluminum) that is fastened to the outside edges of roof sheeting – is designed to (A) be covered by other roofing materials and (B) be exposed to only minute proportions of roof runoff. Its purpose is to protect the edge of the sheeting material (sometimes boards, sometimes plywood, but these days, usually “flakeboard,” “aspenboard,” or some such) from direct exposure to runoff water.

Since “drip-edge” is, by design and application, shielded from direct runoff, I suspect you saw something else. My imagination tells me that there could be some sort of screening or slotted affair that could segregate the water and the crud.

I don’t know why fortune smiles on some and lets the rest go free…

T

You’re right it wasn’t drip edge but here is what it looks like…
http://cgi.pbs.org/cgi-registry/wgbh/thisoldhouse/big_img.pl?is_id=9000235

There may be another factor at work here. All of the calculations here have assumed the underlying rock structure at the edge of the falls is a perfect right-angle ledge. Even if it started out that way, I would think that that centuries of water passing over the edge has eroded it so that it now has an arch shape (or something else more hydrodynamic than a sharp ledge). Though it may look like a sharp ledge from the distance we view the falls at, the region of worn-away rock is probably at least as wide as the 10-foot or so maximum arc length of water from a garden hose.

Billdo, I don’t quite follow. The brink to the NY Falls, for example, is easily observed as being very uneven. No matter how far a piece of uneven “edge” would protrude, the act of falling does not commence until the water has no more support from the riverbed. The brink, regardless of profile, is the very end of the riverbed.

Although the edge, on a macro scale is uneven, I imagine (and this is just imagination – I haven’t been to Niagra Falls in a looooong time) that ends of the edges over which the water falls have been eroded by time, rounding them out. This is in contrast to the sharp metal edge of a garden hose.

Imagine that you have water running off a shelf with a rectangular edge like a bookshelf. That water would arch off of the edge, standing off for a little bit like the water running out of the hose.

However, if the edge of the bookshelf had quater-round molding on the edge (in other words the edge were rounded down) path of the water would be a lot closer to the rounded edge of the shelf. The “support” that the shelf was giving the water would be gradually diminishing over the width of the curve. Therefore the water drop would start not at the “edge” that you perceive, but rather somewhere back where the rounding begins. Another way of looking at this is that the quarter-round molding attached to the edge of the shelf is just filling in with wood the air space under the “arching” water. (Note that I’m using quarter-round molding as a crude approximation of a parabolic arch.)

From a distance, it can be hard to see whether a shelf has a perfectly rectangular edge or a rounded edge, particularly if there is water rushing over that edge. Similarly, I don’t think distance observation of Niagra Falls can clearly identify what the underwater edge looks like, and you’re not getting me out there to dive down and see.

Hmm…this gives a false impression. Take a tank of liquid or the ocean. The hydrostatic pressure increases with depth basically due to the weight of the fluid from above. Hydrostatic pressure will increase in small increments with depth. Hence, it is not equal in all directions.

Jinx, there’s nothing wrong with TBone2’s statement alone. Notice that TBone2 had specifically referred to a given point in water, and it is corerct to say that point of water exerts the same amount of pressure in all directions.

I think it’s clear that you can’t find out how far the warter lands because you’re lacking a variable altogether. You know the flow rate, height of the fall, by you have no idea of it’s speed/depth at the top of the fall. An estimate of 5 feet might sound right, but I wouldn’t bet on it myself.

If you take 55 meters to be the height of the fall (general estimates are between 52 and 57), then the water will reach the bottom in 3.35 seconds. A current of 5 miles per hour would result in the water falling 7.4 meters away from the base. If the actualy current were 30 miles per hour (which isn’t unlikely), the distance would increase to 44.7 meters. Comparing those figures with the height of the falls, I doubt the difference can be neglected.

The rock formation factor can probably be ignored in these calculations though. It’s pretty well known that the edge of the fall looks roughly like a “T”, with the water flowing across the top. The edge is of coruse rounded at the corner of the “T”, but it doesn’t go so far as to become a “D” shaped fall.

Now if one of you could just hike over and stick a yardstick on the top of the falls…