4:30 i paused the video and just sat there, wondering why it was never taught like this. That moment you wrote out 1+ m^2, the reason behind the formula "clicked" instantly, far better than the "memorize this it's on the exam" approach my classes took. Excellent job!
Adding up tiny bits is the way to go with integrals. It is sad how many come away thinking areas under curves is the most fundamental form of integration. And that is what leads directly into this argument. There isn't even really any cleverness to it. You just have to understand integration the correct way.
One could also write dL = sqrt(dx² + dy²) and then pull the dx out of the square root. Mathematically questionable, but gives the right result dL = sqrt(1 + (dy/dx)²) dx. If one wants to do that in a more rigorous way, one has to use Delta instead of d, and at the end, take the limit of Delta x going to zero.
I remember getting a similar problem in class long ago about a rope around the Earth. The question was "If you start with a rope wrapped around the equator of the Earth, how much extra rope would you need to raise it 1 m above the surface?" Finding out it was only 6.28 m was pretty mind blowing at first (Just found out this question is such a classic it has its own Wiki page)
I asked pals this about 50 years ago only using a sphere the size of the sun. When they would not believe the answer I said imagine taking that 6.28m and making 4 pieces 1.57m long and adding into 4 points along the rope at 90° intervals and so move each quarter of the rope 754mm away in one direction then another 754mm away at right angles to the first move. This is a move that totals 754*1.414mm (square root of 2 from Pythagoras) or over a metre away from the surface at the centre of the quarters averaging 1m off the surface in total.
Fundamentally, it's the exact same problem, it's just that your curve is closed (connects to itself). If instead there were 2 posts set 50 meters apart, and you were hold the rope from above somewhere in the middle (ie. when taunt, it forms an isosceles triangle), and if you were to give it slack, you get the same problem. Likewise, there's nothing special about 2 dimensions. It also takes surprisingly little extra volume (compared to the mass of the earth) to make the earth 1km wider in radius (510000000 km^3 compared to the 1083000000000 km^3 of the earth).
Here's a real world example for you- I work in telecommunications, and our data will often run on a fiber optic cable which is strung between poles above ground. If a garbage truck (it's always the garbage trucks!) takes down a fiber line, we can measure the length to the fiber break by sending a light pulse which reflects off of the broken end of the fiber and returns to the source. This time delay (with the speed of light in glass) gives us the length of fiber- this is called Optical Time Domain Reflectometry (OTDR) if you want to read more about it. The time measurements have a precision on the order of tens of nanoseconds, giving break locations with a precision on the order of several meters. Unfortunately, that measured distance is more like the arc length, but what we really need is the ground position where we need to send the technician with the splice equipment. Since the fiber cables hang in a catenary-like curve, we have to solve this problem every time a break takes place above ground... or get lucky, and hear from the police or a traffic report the intersection where someone has crashed and downed a pole.
Wow! That is fascinating! So if I have this correct, the distance you get out from the light pulse is how far down the line it is, but the line distance and the ground distance is not the same, because the sag makes it so more cable length is used than the ground length. So the OTDR would be an overestimate of the length to the break, correct?
If the poles are Always 50 apart and have 51 of Fiber...and you meassure 102 then Just 102/51= 2 * 50 = 100 Thats my quick guess to that Problem Greatz from Germany and have a nice Day opo
similar thing for high voltage power lines. The temperature has a huge effect on the length of the cables and even on the hottest day of the year you will want a safe distance to the ground for the high voltage running trough them. But at the same time you don't want the cable to not be to short at the coldest days. Which is why global worming is a problem for those power lines. If the increase in spread for the once in a thousand years high and low temperature is to big, the lines will have to be redesigned. I am not sure if this has already happend for this reason anywhere on the world, but if the patern holds it will.
Q1 breaks my brain. You're only extending a single meter. So how can someone move more than a meter? And you're letting the rope out at an angle to the direction of travel, which I intuit from vectors ought to have less of an effect than if the rope-letting and travelling were aligned. And yet I can just do Δ = √(((50+α)÷2)²-(50÷2)²) where 'Δ' is the distance travelled and 'α' is how much rope I let out, and set α = 1 and, yep, Δ = 5.02 m, and I'm *still* making a 'wat' face
I was also surprised the first time when I saw the result, and I repeated the calculation again, thinking I was wrong.just one meter of additional rope leads to a vertical displacement of more than five metres. It is like an amplifying effect with a gain of approximately five to one
@@alexandersmith6140 Not sure if this will be intuitive but ill try: imagine the plane with a bunch of tightly packed evenly spaced concentric circles, and you are standing on the edge of a large one. if you walk away from the center, you will pass through many circles quickly. but if you walk perpindicular to that direction instead, at first you will hardly pass any circles, because you are almost just walking along your starting circle. More mathematically: its because if the derivative of a vector valued function is perpendicular to the function, then the derivative of its length is zero.
@@JushuaAbraham-sj2xl It's a tangent thing. As an angle approaches 90° the tangent value approaches infinity. That is why no matter how hard you pull one end of the rope there will always be some sag (It may be too small to measure for a light cord). The angle in this case is that between the horizontal rope and the vertical pull of gravity. So the tangent of this angle for a central drop of 3.4m and a length of 25m is 25/3.4 giving an angle of 82.25° which is getting close to 90°. The anchor point for fall arrest systems requires that they can withstand a large pull, well in excess of the weight of a human body, the effect of this tangent multiplier can be huge. A use of this is the Spanish windlass where twisting rope works the same way to multiply the effort of twisting a rope into a large pulling effort. Given the angle of 82° and the estimated weight of an adult human with kit may be 80kg the person pulling on the rope would probably get pulled off the cliff face before they managed to brake the rope that was slipping with the friend on it since the effective weight they were trying to pull up would probably only have to about 600kg when it reached that 1m of slip.
The large amount of horizontal motion needed to take up a tiny bit of slack on the rope also relates to the mechanical advantage that pulling the rope horizontally has on the tension of the rope, how taut rope needs to be to not deflect by that much.
Ships have a piece of small rope called a "tattle tale" secured to 2 points on a mooring line. When there is no tension on the mooring line, the tattle tale has a lot of slack in it. As the mooring line stretches, the 2 points move apart and the tattle tale slack is removed. This way, the ship's crew can tell when there is too much strain on the mooring line. The length of the tattle tale and how far apart to secure the ends depends on the specifcations of the mooring line.
Math was just numbers to me, but after coming across your channel. I see the beauty in it. The art. Every video you've posted has been blowing my mind.
What he assumes is that a parabola is "close enough" -- he mentioned that the real curve is cosh(x), which has the e^x and e^-x terms you're thinking of, but that a parabola is fairly close and it's much easier to do the arc-length calculation for it. It's kind of handwavey.
The athletics example about slack reminded me of the high jump - with a relatively stiff bar, there is still over an inch in height difference at the stanchions and in the middle of the high jump.
for Q1 I actually guessed 5m. I didn't expect to be so close, but I knew it had to be much bigger than the difference in length. I'd always been surprised by the 5-12-13 Pythagorean triangle which shows that a tiny difference in length between two lines creates a big distance between them at an angle, so I remembered it well.
People like you that knows how to explain maths well, they deserve a place in Heaven. It is always a strugle for me to understand a book math, but then I see videos like yours and it makes everthing so easy.
As an amateur in sewing and pattern design, this has caused issues for me. A 1mm discrepancy in a 30cm seam can result in a 1cm arc! That's why I think half the battle is cutting fabric pieces as accurately as possible.
"The simplist way is to ask your friend" "Hey!! How much did you fall off!!" "Four point threAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHH!!!!!!" "Approximately 3 seconds"
I work on sea going tugboats. We tow barges with a cable. Knowing the sag in the cable is important so it doesn’t touch the bottom. We call it shining the wire. The company policy is add 10 feet of depth for ever layer off the drum. The difficulties arise because every layer is a different diameter, every drum is a different size,and the hight of the barge is always different. The cables themselves are usually 1500 feet long ,but you never use the last layer on the drum because that is how it holds on. The diameter of the cable is mostly 1 31/32”. The math is daunting. We rely on experience.
@@ian_simbotin the barge is 75’ wide , 450’ long. When it is empty ,it draws 4’, max load 27’. The point of attachment for the tow bridle is about 12’ above the waterline light. The last point of contact on the stern of the tug is about 7’ above the waterline. When everything is up to speed we are moving at 7knts. The max load for the barge is 14,000 short tons. At sea in normal conditions we put out 7 layers .
It should be actually quite intuitive for anyone who has been playing with stretched string or rope. It's so easy to make large sag even if string is really tightly stretched. If you think about it for a minute, you'd realize that according to Hooke law, the elongation of the string is small, as well as the applied force.
I like this point. I wonder if it is becoming less intuitive because people are on screens so much and don't interact with physical objects. I had an engineering professor tell me a few years ago that the students are great at computers and coding, but none of them know how to use a wrench!
I had to imagine 50 meters on a sports field, one end of the rope at the goal, and how hard it would be to pull all the visible sag out of the rope. Then the amount of calculated sag started to make sense.
I just used Heron’s formula after summing 50, plus 2 times 25, plus 1 and used the base of 50m for the lower part of the triangle and the area I got of approximately 125.6m for a 5.02m drop of your friend when you add 1 extra meter of rope.
Not placing him above or below anyone else, but there are a lot of good teachers in various disciplines on TH-cam. I’m glad you found one who works for you.
In structural engineering applications, reminds us of calculating the sag of a hanging power cable under its own weight. The shape the cable takes is known as a catenary. According to the problem posed here, the hyperbolic cosh function closely approximates this sag calculation. I didn’t realize this during my undergraduate years studying civil engineering. It was a difficult problem to solve with few computer programs those days.
I posted a comment about the similarity of this problem with the one where a (very, very long) rope is girdling the earth around the equator. Add one meter of slack, and then pinch and raise the rope so it's taut again. How high is it gonna go? A fellow commenter asked for the answer, so here goes: There's gonna be a baseline between the points where the rope goes tangent to the surface and starts to gain altitude. But we cannot use the Pythagorean theorem because the excess length (slack) is relative to the slightly curved equator, not the straight chord through the earth. We still use sine and cosine (actually, tangent and cosine) in a right triangle with corners at the Earth's center, apex of the raised rope, and tangent point. As you can guess, or you can see if you try it, one has to make use of approximations. But they are excellent and yield fairly accurate results. Speaking of, the simple version (triangle case) of the chasm problem can be tackled in an approximate fashion, even though the exact answer is ridiculously simple. The approximate answer for the sag, s, in terms of x=0.5m (half the excess length, or slack) and L=25m (half the chasm gap, or baseline) is: sag = s = sqrt(2xL) I mean, sqrt(2*x*L) = sqrt(25) = 5m The exact value is s=5.0249... Now, to better appreciate the effectiveness of the approximation, and to bridge the gap (so to say) between the original problem here and the other one with the Earth's equator, let's consider a huge chasm (Grand canyon type of chasm) of 5km, but the same amount of slack, 2x=1m The approximate formula yields: sag = s = 50 meters, while the exact value is s = 50.0025 So, two things: (i) The approximation gets better when L/x gets larger, and (ii) The sag is much greater for the same amount of slack! Now, back to the equator. The answer for the height (instead of sag) is h = 121.5 meters, or so, which is about 133 yards, or almost 400 feet. And, by the way, the baseline between the tangent points is 78.7 km, or 48.9 miles
6:54 Dos cosh(x) assume no bending stiffness (tension only), evenly distributed mass and gravity, and parabolic assumes no tension stiffness (bending only)? And if so would the "truth" not be somewhere between them?
Here is a good website that talks about the different assumptions that end up with the parabola vs the catenary. In practice, the catenary assumptions are much more reasonable than the ones that lead to a parabola, but in a long shallow curve, there isn't too much difference between the two. lbcc.pressbooks.pub/structuraldesign/chapter/chapter-4-catenary-cables-and-arches-2/
This reminds me of the "string girdling earth" problem where an additional 6.3m of extra string allows a meter of additional height all around the earth. Very counterintuitive
And it works for a circle of any radius! If you add 6.3 meters of string around a ball, it would add 1 meter of additional radius. This fact is even more counterintuitive for me
3:43 Just a “In case it helps comment” This slope section has different scales for the X and Y axis, this may confuse some how the values are found. Thanks for the videos, nice job!
If instead of the parabolic approximation we use a circlar segment and trunkated Taylor series sin(x)=x-x^3/6, cos(x)=1-x^2/2, we get h=sqrt(153/8)=4.3732 approximately. It is almost the same as the parabolic approximation, however we can easily compute all the numbers without Wolfam Alpha.
The exact drop depends on the properties of the rope, and the weight of both the rope and the man. To simplify the problem I assumed the maximum drop possible, where the rope sections are kept straight by the weight of the man. (Rope weight is assumed to be insignificant) Assuming a right angle triangle is created , the triangle sides are hypotenuse, 25.5m, the new length, the original length 25m, and the drop in m. This gives a maximum drop of 5m.
Without any calculations i say its less than 1m cuz the 1m is spread across the gap other than that i could try to collapse the right side on the left side to get the same arch to stright diffrence across the rope except the meddle and use that to get the spread of the middle compared to the rest(aka a percentage)×1=the middle gap acrch to stright deference (i have no mathmaticla proof but it fell like the right way to do this with the least amount of calculations) Edit:welp i was waayy off
Another place you see the underestimating of Arc Length is in sailing yacht sizes. They are classified by length. For example a 42ft yacht being much, much bigger internally than a 40ft yacht. That extra 2ft resulting in enough increased volume for additional cabins!
Too often they don't know how, because the "why" is already obvious to them, and they don't have time to figure out the most intuitive way to explain it to someone who doesn't already get it And there's usually a lot of people in the class, and different people will find different explanations make sense. I guess it's not easy to find ways that work for everyone, not when they _also_ have to make time to drill everyone on the precise details I think maths is one of the hardest subjects to teach well, especially in large groups, because it usually relies so much on other simpler maths that the student needs to know already. If the students don't already have a good intuition for the simpler maths, it's even harder to give them a good intuition for the later stuff. So the problem just compounds on itself as time goes by (And I expect half the students have already given up on the idea that maths is interesting or makes any sense, and they just want to be told how to pass the exam anyway)
Assuming the weight is in the middle of the rope, and it is perfectly inelastic, the absolute maximum drop should be the height of a right triangle winh hypotenuse af 25.5m and length of 25m, which is just about 5 meters. The slack variation is much harder to calculate intuitively, but given that it's a hyperbola, without appealing to major math tools (as you asked us not to) my intuition says it would be between 2 and 4 meters drop (about half the weighted distance at least)
On a related note I was on site and had to record the dimensions of an existing arched bridge. From afar it looked like a segment of a circle. I got a mate to I ball a spirit level end to end and measured the bow in the middle to where a string line would have been. Then measured along the curve thinking I had enough information to define the circle and the rise at one metre intervals. I spent hours trying to figure out the trig to get the actual resulting complete circle. I ended up doing some sort of iterative calc in excel to get the answers. Im
The string-girdling-the-earth problem is just as enjoyable, or maybe a lot more. But it isn't the one mentioned in the other comments here; in fact, it's similar to this one, because the question is the same; namely, given a certain (small) amount of slack, how far can it get you? For the circumference of the earth, you add one meter of length to the rope encircling earth, and then you pluck the rope and pull up until it's taut again. Find how high above the earth does it reach. Like some giant plucking the rope above the chasm here, instead of hanging down. And the straight 50-meter-chasm is replaced by the round earth with a circumference of about 40 thousand km. So, it's gonna be geometry again, like the triangle case, but a little more nasty.
@@xenmaifirebringer552, I said I'll get back with the answer soon, but it's been more than a week now; sorry for the delay. Anyway, I'm gonna post now a regular comment with the answer, so anyone can see it; if I post it here as a reply, or a reply-to-a-reply, etc, it would just get buried and lost into oblivion, which will happen anyway, but, hey, I'm trying.....
0:51 I guessed roughly 5 meters. I did Pythagorean with the hypotenuse 25.5 being half of the actual rope and 25 being one of the legs as the half the distance between the tree. It’s probably wrong but I don’t know what else to apply to this situation
Analogous to the problem cited in this video, there is a practical example of "A cylindrical tank laid horizontal, length of the cylinder along the ground". Calculate the volume of Diesel Fuel while filling this tank at any instant. I have an engineering background but too old to figure this out. Perhaps youngster can suggest a formula for this problem.
It is a lot more complicated there, though, since you need to allow for the stretch in the line. They usually use fairly static webbing that doesn't stretch much, but as we've seen here a small amount of stretch can drop the midpoint by a lot.
Wonder where my errors are adding up from. I just ran the number on my catenary model and I'm getting about 4.36m, which is awfully close but not exact. Without going into too many details, let L be the length of half of the rope and D be the distance from the end to the middle. Then, you can find your solution catenary as h(x) = cosh(ax)/a + C where h is the height, x is distance from the centre, and C doesn't matter since we can move "ground level" to wherever we would like. The parameter a winds up being a solution to sinh(aD) = aL Finding the appropriate a = 0.0138152, you can plug in L = 25.5 and D = 25 to obtain h(D) - h(0) = 4.360337.
Made me think of bernoullis brachistochrone problem that shows up in physics mechanics courses, the shape ends up being a cycloid. I’d love to see a vid on that
I like to explain this to my students as "there's not a lot of length difference between a tight rope and a saggy one, and that's why it's pretty much impossible to load at long rope and keep it straight". Think high voltage lines, or laundry...
The slack plays a bit role in a ship's anchor ⚓ . Because the chain slacks it is hard to un slack it by increasing the horizontal distance if for insane there is a wave it is stable
Why don't you use the p value associated to the parabola to try to guess the ecuation considering 4p for the horizontal distance? Is it a bad guess? Just suppose the horizontal distance is latus rectum.
Interesting and my thanks. I ran into this problem when we were discussing the length of cable or wire between telephone poles in a High School Physics class or two. Turned out that there had to be some drop because of gravity. Which related to the amount of tension applied to the wire. A big deal if you think about it, and relates the the reliability of the then telephones and electricity coming out of your light sockets.
Having seen tbe answer to how far the rope drops or sags when 1m is added to 50m, here's another "surprising" answer: suppose there was a rope running around the equator of the earth. Assuming the earth was a smooth solid sphere of radius 6,378,000,000 m how much extra rope would be needed to raise the rope 1m off the ground all the way round the equator?
My idea was to divide the graph into two equal triangles with the following properties ( adjacent side = x, opposite = 25 m, hypotenuse = y). In addition, I thought that the sum of both hypotenuses must be 51 m and since both are equal, I derived a formula from this. 51 m = 2root(25 m² + x²) This results in two answers: Answer 1: x = Root(101)/2 Answer 2: x = -(root(101)/2) I hope my thoughts where correct
How TH-cam got me here today I don't know. This really blew my mind. My intuition was wrong. After watching, what I was imagining was how any slack, would instantly create a arc from a huge circle. as x (delta of the chord)->0, y I wasn't doing the integration, but just some geometry. Looking at this from the arc and coord ratio, 4.3431 was my answer, between your two examples. 78.5 would be the max amount of rope could go before it no longer fits that limit of a circle, and then it would droop down.
MOre interesting is the question how strong the Rope pulls ON the fixing Points left and right depending ON the deepness of the sagging. (Antenna Problem)
My intuitive guess is that the first problem is so simple to calculate, we should calculate. The answer to the second problem is, "less", unless I want to do the not so simple calculation for a catenary curve.
It depends on the stretchyness of the rope. If it has no real give to it, they would drop about 5 meters (using the Pythagorean Theorem), so I would guess that it would be less if there was no weight pushing down on the rope. Then it depends on the weight of the rope and whatever the formula for a catenary is 😀.
It would be interesting to iterate between the two questions. Start with the triangle, then have it be a quadrilateral, and so on. I have to admit that the use of a calculator that does the integration for you is a bit dissatisfying, because, at that point, the intuition disappears that connects that part. I think the above might help keep it all connected.
If this is the method of solution you expect, make sure to specify to your students that you're seeking an _approximation_, not an exact solution. Most students delving into advanced calculus will assume they're expected to give exact solutions. To them, these questions feel more like 'invent a method for deriving the distance between the apex of an arc of length x with unknown shape, and the midpoint of the arc's terminations.' To these students; the problem is the unknown shape of the arc. This is why it's important to specify that you expect an approximation, rather than a solution.
The first question is easily splved by just using the pitagorean theorem, it's about ~5.025m The second question I know hanging ropes use some kind of hyperbolic function but as an engineering in training approximating it with a parabola should be fine hahaha But you asked for intuition and not me calculating the arc length of the parabola, so I'd say it should be less than 5 cuz the load is pretty evenly distributed across the entire rope and it's the same total length so I'd say it's like 4m :3
The first answer being 5 meters blew my mind the math checks out but how logically does that make sense just breaks my brain thinking about it but I guess it must come from the rest of the line as well not just the 1 meter of slack added in like you would intuitively think.
The way the first puzzle is drawn the rope is straight. The pythagorean theorem tells us the rope falls just over 5 meters. Real world it will be less because the eope follows a catenary curve.
1. The shape of the curve strongly depends on the properties of the rope. Assuming that it is a parabola is not valid assumption. 2. The basketball ball trajectory did not count with the perspective. The ball was also moving away from the camera, therefore the used curve was not correct.
The most unintuitive question is how much longer the earth circumference compare at the ground level vs 1 meter above above it . just around 6.28 meters
i got carried away and did calculatings, the first one (should be) about 5.02 m (i may be stupid??) with the classic sqrt((51/2)^2-25^2) i dont know the second one
preliminary answer, haven't watched the video yet. will check later the initial and final point of the chord describe an angle theta at the center of the circle. start by finding this angle, and the radius. by geometry, r*theta = 51 m and, by trig, r*sin(theta/2) - r*sin(-theta/2) = 50 m. by simmetry, the last one is just r*sin(theta/2) = 25 m. solve for r and theta to get r ≈ 74.124998 m and theta ≈ 0.688027 rad. the height lays in correspondence with the midpoint of the angle, thus its corresponding value will be given by r - r*cos(theta/2) ≈ 4.34 m EDIT oh, it was a problem on a parabolic curve. thought it was a problem on a circle from the thumbnail. still close though.
The reason I forgot about the arc length formula and only derive it when I can use it.... is that... there are infinitely more possible arcs than can be described by simple functions. Even with functions which we can easily integrate... the formula makes it hard to integrate... and sometimes impossible to do so.
0:50 Woah, intuition is so lacking here. Intuition kind of told me between 0,5 and 1m and that is just so wrong. If you approximate the problem with 2 triangles, you get more than 5m
how is it possible that the lenght of x in the pythagoreon theorem is 5m, when the rope is only increasing in 1 meter lenght? Shoudn't x be max. 1 meter long ?
That is a good question and I think it's a common deception to think that the max it can increase will be 1m. However, that x is not physically part of the rope. Perhaps thinking about the opposite situation could help. Say you have a pole 5m tall with a rope that is 25m long and placed at the bottom of the pole and pulled tight at some other point 25m away. If you extended the rope by 1m, you could only go 1m up the pole if you made sure the rope lay on the ground the whole way except for the 1m up the pole. However you could reach the top of the pole using the shortest route (the hypotenuse).
@@ZacharyBlue Thank you! I guess the problem with the understanding does not lie in the specific problem, but rather the fundamental undestanding of the pythagoreon theorem. I will try to understand it more
Great example! And important as too much tension can cause them to break with high wind and too loose can cause them to bump into things and also break
For the first one, using the Pythagorean theorem, I got that the middle of the rope should drop by sqrt(25.25) meters. I have NO clue how to do the second part!
I must admit to always taking the lazy path of writing ds^2 = dx^2 + dy^2 and just pretend what follows is rigorous, where I think of ds as "just a bit of arc length" and dx and dy as "a tiny change in x and y"; if I zoom in enough each step is "approximately a right triangle". "Divide" by dx^2 ds^2/dx^2 = 1 + (dy/dx)^2. Take the square root multiply over by dx and integrate. Also works for different geometries like in special relativity where you have ds^2 = c^2 dt^2 - dx^2. The dx * sqrt(1+m^2) notion is a bit more rigorous of a take (without sacrificing intuition) that I might use the next time I explain this formula to someone.
For the first one, it has to be less than 1/2 meters. My guess is 1/4 m And for the case without friend it should be even less maybe 10cm So apparently I have no intuition and mistook the upper bound for the lower bound 😂 I actually derived the formula for the chain line by setting up the differential equations when I was in ~10th grade. Unfortunately I no longer have the post-it I calculated it on but as far as I remember I also calculated it by setting up the length of the line for a short distance to get the weight per small segment in x-direction
At first for both questions I guessed about a metre at most? Then I calculated the drop for the first (as in the video) and realised I *massively* underestimated the drop. So for the second... my gut went to 3 metres, but after last time I decided to guess 4.5 metres. I'm glad i continued to ignore my gut 😆
No weight other than the weight of the rope? That would be catenary curve. Time to pull out my Calculus of Variations textbook from 45 years ago? Definitely not an easy problem.
So I solved this problem, I am going to give the answer, I dont know if is correct. I based my answer on Greco-Babylonian geometry and the basis of developing Acrs and Spans. The basis of the technique is that if you bisect a circle or inscribe a hexagon into a circle of radius r you create two known answers, In the first case you have a 2 suoerpositioned spans of two, in the second case you have 6 end on end spans of length one. In the first case you have have two superposition spans of length 2, so 2 x 2 = 4. this is the coversge of 4/2pi. We can bisect either span creating a tangent outward. Doing this we have an ouward facing triangle of 1/2 span (2 * 1/2 = 1) and the cosine of 1 is zero, so the ouward facing triangle is 1 -0 in the outer part of the busecting radial. The new span created is this 1^2 + 1^2 = s^2 of the square root. We can repeat this process for generating sides of polygons until at some point Lspan * sides = 2 pi. This is roughly 13 steps for most computers. We can do the same with the hexagon. So the ratio you give is 50/51 = 0.980392 which is close to a span 39.375 degrees 0.980438. This was derive from a series of back tables generating by using the 180° based and 120°[60°] based span tables. In these tables you can combine variables of two angles to come up with precise information for a third angle and then go about busecting the soans and creating outer triangles to creat new smaller spans. I try to keep the babylonian (360 = 2 pi() whole angles. This particular angle comes out of a theoretical 315° that splits 3 times to 39.375° As a consequence its half span is 19.6875° [Yes, indeed I have this on an Excel worksheet an no I did not make this for the puzzle, I was testing to see how far babylonians and ancients greeks could fill out a span table just using the pythagorean theorem, right now the table has the largest gap of 1.5°. I could just continue the process until the perimeter of the polygon did not change, then easily get every span to 2° by making the arguemnt that cosine of trivially small angles is indistinuishable from 1, so that if I have an angle like say 1 degree, and I divide it say 7 times to get 1/(2^7) degrees, if I just multiple the sin of 1° x 2 and reverse the division process, I can return all the parameters for 2°, but since the ancients dont have 64 bit processors that would defeat the effort]? While i could just get the cosine of 19.6875 that would be cheating in a greek, persian or babylonian scribal school. So from this we can see that the span of 39.375 is 0.67378 and the bisect to the span is 0.94158. Ok this is the answer but we (the scribe needs to fix it) need to fix it so that the construction crew understands it. You span is 50. 50 ~= 0.67378 x r. We dont know r. R = 74.2. R * 0.94158 = 68.875. Thus the dustance between the arc and the span at the bisect is 74.2 - 68.9 or 4.37 units. Or 4 and 3/8ths units. And so, 2000+ years before calculus and integrals, this problem can be solved (crossing fingers and hoping I didnt make any errors😎) Now of you are a stickler for ancient history, you should note I am using the decimal system instead of fractions,
Why would you go trough all the cool math and then just say 'nah, the real cosh is too hard' ? Don't pick the rope problem if you will not solve it. You could have sticked to the basketball case ! That was already cool !
You lay down a rope to cover the equator, it will be roughly 40,000 kilometers, if you increase its length by ONE meter, guess how high the rope can go above the ground??
Great lesson. my only comment would be to "call it as it is" i.e. it's a Catenary or Hyperbolic curve, not a parabola. you do mention it and compare the two, but it's math and it should be more exact. I would rather teach someone to recognize, this curve is special. I enjoy this stuff, thank you.
4:30 i paused the video and just sat there, wondering why it was never taught like this. That moment you wrote out 1+ m^2, the reason behind the formula "clicked" instantly, far better than the "memorize this it's on the exam" approach my classes took. Excellent job!
This reply falls in the 2:58 statement on math conceptual education in the video.
So glad this has helped!!
Adding up tiny bits is the way to go with integrals. It is sad how many come away thinking areas under curves is the most fundamental form of integration. And that is what leads directly into this argument. There isn't even really any cleverness to it. You just have to understand integration the correct way.
One could also write dL = sqrt(dx² + dy²) and then pull the dx out of the square root. Mathematically questionable, but gives the right result dL = sqrt(1 + (dy/dx)²) dx.
If one wants to do that in a more rigorous way, one has to use Delta instead of d, and at the end, take the limit of Delta x going to zero.
@@bjornfeuerbacher5514 There is so much of integral calculus that is mathematically questionable, yet gives the right answer. Especially in physics.
I remember getting a similar problem in class long ago about a rope around the Earth. The question was "If you start with a rope wrapped around the equator of the Earth, how much extra rope would you need to raise it 1 m above the surface?"
Finding out it was only 6.28 m was pretty mind blowing at first
(Just found out this question is such a classic it has its own Wiki page)
I mean that's a prettt diferent question, all you need is 2(pi)R
I asked pals this about 50 years ago only using a sphere the size of the sun. When they would not believe the answer I said imagine taking that 6.28m and making 4 pieces 1.57m long and adding into 4 points along the rope at 90° intervals and so move each quarter of the rope 754mm away in one direction then another 754mm away at right angles to the first move. This is a move that totals 754*1.414mm (square root of 2 from Pythagoras) or over a metre away from the surface at the centre of the quarters averaging 1m off the surface in total.
Classic, but different. Math is fun.
Fundamentally, it's the exact same problem, it's just that your curve is closed (connects to itself). If instead there were 2 posts set 50 meters apart, and you were hold the rope from above somewhere in the middle (ie. when taunt, it forms an isosceles triangle), and if you were to give it slack, you get the same problem. Likewise, there's nothing special about 2 dimensions. It also takes surprisingly little extra volume (compared to the mass of the earth) to make the earth 1km wider in radius (510000000 km^3 compared to the 1083000000000 km^3 of the earth).
Here's a real world example for you- I work in telecommunications, and our data will often run on a fiber optic cable which is strung between poles above ground. If a garbage truck (it's always the garbage trucks!) takes down a fiber line, we can measure the length to the fiber break by sending a light pulse which reflects off of the broken end of the fiber and returns to the source. This time delay (with the speed of light in glass) gives us the length of fiber- this is called Optical Time Domain Reflectometry (OTDR) if you want to read more about it. The time measurements have a precision on the order of tens of nanoseconds, giving break locations with a precision on the order of several meters.
Unfortunately, that measured distance is more like the arc length, but what we really need is the ground position where we need to send the technician with the splice equipment. Since the fiber cables hang in a catenary-like curve, we have to solve this problem every time a break takes place above ground... or get lucky, and hear from the police or a traffic report the intersection where someone has crashed and downed a pole.
Wow! That is fascinating! So if I have this correct, the distance you get out from the light pulse is how far down the line it is, but the line distance and the ground distance is not the same, because the sag makes it so more cable length is used than the ground length. So the OTDR would be an overestimate of the length to the break, correct?
@@MathTheWorld You got it! So in that sense, the OTDR shot is precise, but not accurate. :D
If the poles are Always 50 apart and have 51 of Fiber...and you meassure 102 then Just 102/51= 2 * 50 = 100
Thats my quick guess to that Problem
Greatz from Germany
and have a nice Day
opo
similar thing for high voltage power lines. The temperature has a huge effect on the length of the cables and even on the hottest day of the year you will want a safe distance to the ground for the high voltage running trough them. But at the same time you don't want the cable to not be to short at the coldest days. Which is why global worming is a problem for those power lines. If the increase in spread for the once in a thousand years high and low temperature is to big, the lines will have to be redesigned. I am not sure if this has already happend for this reason anywhere on the world, but if the patern holds it will.
Wait you're telling me you don't have cable metrage between intersections recorded from installation and have to go by road lengths and maps?
For Q 1 Simple pythagorean formula 5.02 meters. For Q 2 I think it involves catenary calculations
Q1 breaks my brain. You're only extending a single meter. So how can someone move more than a meter? And you're letting the rope out at an angle to the direction of travel, which I intuit from vectors ought to have less of an effect than if the rope-letting and travelling were aligned. And yet I can just do Δ = √(((50+α)÷2)²-(50÷2)²) where 'Δ' is the distance travelled and 'α' is how much rope I let out, and set α = 1 and, yep, Δ = 5.02 m, and I'm *still* making a 'wat' face
I was also surprised the first time when I saw the result, and I repeated the calculation again, thinking I was wrong.just one meter of additional rope leads to a vertical displacement of more than five metres. It is like an amplifying effect with a gain of approximately five to one
@@alexandersmith6140 Not sure if this will be intuitive but ill try: imagine the plane with a bunch of tightly packed evenly spaced concentric circles, and you are standing on the edge of a large one. if you walk away from the center, you will pass through many circles quickly. but if you walk perpindicular to that direction instead, at first you will hardly pass any circles, because you are almost just walking along your starting circle. More mathematically: its because if the derivative of a vector valued function is perpendicular to the function, then the derivative of its length is zero.
@@JushuaAbraham-sj2xl It's a tangent thing. As an angle approaches 90° the tangent value approaches infinity. That is why no matter how hard you pull one end of the rope there will always be some sag (It may be too small to measure for a light cord). The angle in this case is that between the horizontal rope and the vertical pull of gravity. So the tangent of this angle for a central drop of 3.4m and a length of 25m is 25/3.4 giving an angle of 82.25° which is getting close to 90°. The anchor point for fall arrest systems requires that they can withstand a large pull, well in excess of the weight of a human body, the effect of this tangent multiplier can be huge. A use of this is the Spanish windlass where twisting rope works the same way to multiply the effort of twisting a rope into a large pulling effort. Given the angle of 82° and the estimated weight of an adult human with kit may be 80kg the person pulling on the rope would probably get pulled off the cliff face before they managed to brake the rope that was slipping with the friend on it since the effective weight they were trying to pull up would probably only have to about 600kg when it reached that 1m of slip.
for question 2 my initial intuition is to solve it with potential energy
The large amount of horizontal motion needed to take up a tiny bit of slack on the rope also relates to the mechanical advantage that pulling the rope horizontally has on the tension of the rope, how taut rope needs to be to not deflect by that much.
You can also use this mechanical advatage - for example to tow a car stuck in mud (i remember this example from a very old math book for children :) )
Ships have a piece of small rope called a "tattle tale" secured to 2 points on a mooring line. When there is no tension on the mooring line, the tattle tale has a lot of slack in it. As the mooring line stretches, the 2 points move apart and the tattle tale slack is removed. This way, the ship's crew can tell when there is too much strain on the mooring line. The length of the tattle tale and how far apart to secure the ends depends on the specifcations of the mooring line.
Math was just numbers to me, but after coming across your channel. I see the beauty in it. The art. Every video you've posted has been blowing my mind.
Wow thank you! This is such a huge compliment and part of our mission goal for our channel
The struggles trying to compute for it, encourage me more specially when i made a blunder.
I dont understand why we assumed the rope makes a parabola. I always thought a uniform density line used catenaey shapesninvolving e^x and e^-x terms
What he assumes is that a parabola is "close enough" -- he mentioned that the real curve is cosh(x), which has the e^x and e^-x terms you're thinking of, but that a parabola is fairly close and it's much easier to do the arc-length calculation for it. It's kind of handwavey.
@@BrooksMoses Yes, thank you for clarifying this!
That is the definition of hyperbolic cosine. 2coshx=exp(x)+exp(-x)
…and in the thumbnail it looks like the arc of a circle.
Yeah I thought it was an arc of a circle when I saw the thumbnail, but got confused when he didn’t say it was a catenary arch.
The athletics example about slack reminded me of the high jump - with a relatively stiff bar, there is still over an inch in height difference at the stanchions and in the middle of the high jump.
That's a great example of how a small amount of curve can lead to big differences!
for Q1 I actually guessed 5m. I didn't expect to be so close, but I knew it had to be much bigger than the difference in length. I'd always been surprised by the 5-12-13 Pythagorean triangle which shows that a tiny difference in length between two lines creates a big distance between them at an angle, so I remembered it well.
People like you that knows how to explain maths well, they deserve a place in Heaven. It is always a strugle for me to understand a book math, but then I see videos like yours and it makes everthing so easy.
Thank you so much. You made our day!
As an amateur in sewing and pattern design, this has caused issues for me. A 1mm discrepancy in a 30cm seam can result in a 1cm arc! That's why I think half the battle is cutting fabric pieces as accurately as possible.
"The simplist way is to ask your friend"
"Hey!! How much did you fall off!!"
"Four point threAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHH!!!!!!"
"Approximately 3 seconds"
We got the same answer
I work on sea going tugboats. We tow barges with a cable. Knowing the sag in the cable is important so it doesn’t touch the bottom. We call it shining the wire. The company policy is add 10 feet of depth for ever layer off the drum. The difficulties arise because every layer is a different diameter, every drum is a different size,and the hight of the barge is always different. The cables themselves are usually 1500 feet long ,but you never use the last layer on the drum because that is how it holds on. The diameter of the cable is mostly 1 31/32”. The math is daunting. We rely on experience.
Oh, man, I love this question and I'm gonna try and work on it to see if I can help at all. C'mon, guys, we gotta help our man here!
@@ian_simbotin the barge is 75’ wide , 450’ long. When it is empty ,it draws 4’, max load 27’. The point of attachment for the tow bridle is about 12’ above the waterline light. The last point of contact on the stern of the tug is about 7’ above the waterline. When everything is up to speed we are moving at 7knts. The max load for the barge is 14,000 short tons. At sea in normal conditions we put out 7 layers .
Oh, boy, lots of jargon; I hope I can manage, but if I need translation or clarification, I'll ask...
@@ian_simbotin no problem . I thought I was going kind of light on nautical terms. Sorry.
Aye aye, Skipper; it's just that I'm just about illiterate when it comes to nautical things. Anyway, it won't hurt to learn a few things
It should be actually quite intuitive for anyone who has been playing with stretched string or rope. It's so easy to make large sag even if string is really tightly stretched. If you think about it for a minute, you'd realize that according to Hooke law, the elongation of the string is small, as well as the applied force.
I like this point. I wonder if it is becoming less intuitive because people are on screens so much and don't interact with physical objects. I had an engineering professor tell me a few years ago that the students are great at computers and coding, but none of them know how to use a wrench!
I had to imagine 50 meters on a sports field, one end of the rope at the goal, and how hard it would be to pull all the visible sag out of the rope. Then the amount of calculated sag started to make sense.
I just used Heron’s formula after summing 50, plus 2 times 25, plus 1 and used the base of 50m for the lower part of the triangle and the area I got of approximately 125.6m for a 5.02m drop of your friend when you add 1 extra meter of rope.
I don’t know who you are but you are surely one of the best teachers for the modern generation.
Thank you this will make Doug so happy to hear! Doug Corey is a mathematics/mathematics education professor at BYU
Not placing him above or below anyone else, but there are a lot of good teachers in various disciplines on TH-cam. I’m glad you found one who works for you.
Such a well done video. Thank you !
In structural engineering applications, reminds us of calculating the sag of a hanging power cable under its own weight. The shape the cable takes is known as a catenary. According to the problem posed here, the hyperbolic cosh function closely approximates this sag calculation. I didn’t realize this during my undergraduate years studying civil engineering. It was a difficult problem to solve with few computer programs those days.
I posted a comment about the similarity of this problem with the one where a (very, very long) rope is girdling the earth around the equator. Add one meter of slack, and then pinch and raise the rope so it's taut again. How high is it gonna go?
A fellow commenter asked for the answer, so here goes:
There's gonna be a baseline between the points where the rope goes tangent to the surface and starts to gain altitude. But we cannot use the Pythagorean theorem because the excess length (slack) is relative to the slightly curved equator, not the straight chord through the earth. We still use sine and cosine (actually, tangent and cosine) in a right triangle with corners at the Earth's center, apex of the raised rope, and tangent point.
As you can guess, or you can see if you try it, one has to make use of approximations. But they are excellent and yield fairly accurate results. Speaking of, the simple version (triangle case) of the chasm problem can be tackled in an approximate fashion, even though the exact answer is ridiculously simple.
The approximate answer for the sag, s, in terms of x=0.5m (half the excess length, or slack) and L=25m (half the chasm gap, or baseline) is:
sag = s = sqrt(2xL)
I mean, sqrt(2*x*L) = sqrt(25) = 5m
The exact value is s=5.0249...
Now, to better appreciate the effectiveness of the approximation, and to bridge the gap (so to say) between the original problem here and the other one with the Earth's equator, let's consider a huge chasm (Grand canyon type of chasm) of 5km, but the same amount of slack, 2x=1m
The approximate formula yields:
sag = s = 50 meters,
while the exact value is
s = 50.0025
So, two things:
(i) The approximation gets better when L/x gets larger, and
(ii) The sag is much greater for the same amount of slack!
Now, back to the equator.
The answer for the height (instead of sag) is
h = 121.5 meters, or so,
which is about 133 yards, or almost 400 feet.
And, by the way, the baseline between the tangent points is
78.7 km, or 48.9 miles
Thank you Professor Math The World, this will do me good justice.
I really liked this video, the way you explained how we get to the arc length formula was so clear.. it really blew my mind!
Glad you enjoyed it!
6:54
Dos cosh(x) assume no bending stiffness (tension only), evenly distributed mass and gravity, and parabolic assumes no tension stiffness (bending only)?
And if so would the "truth" not be somewhere between them?
Here is a good website that talks about the different assumptions that end up with the parabola vs the catenary. In practice, the catenary assumptions are much more reasonable than the ones that lead to a parabola, but in a long shallow curve, there isn't too much difference between the two.
lbcc.pressbooks.pub/structuraldesign/chapter/chapter-4-catenary-cables-and-arches-2/
@@MathTheWorld Honestly a super link what a grate read.
This reminds me of the "string girdling earth" problem where an additional 6.3m of extra string allows a meter of additional height all around the earth. Very counterintuitive
And it works for a circle of any radius! If you add 6.3 meters of string around a ball, it would add 1 meter of additional radius. This fact is even more counterintuitive for me
Yes, you add 2*pi(*1units)
3:43 Just a “In case it helps comment”
This slope section has different scales for the X and Y axis, this may confuse some how the values are found.
Thanks for the videos, nice job!
If instead of the parabolic approximation we use a circlar segment and trunkated Taylor series
sin(x)=x-x^3/6, cos(x)=1-x^2/2,
we get h=sqrt(153/8)=4.3732 approximately. It is almost the same as the parabolic approximation, however we can easily compute all the numbers without Wolfam Alpha.
The exact drop depends on the properties of the rope, and the weight of both the rope and the man.
To simplify the problem I assumed the maximum drop possible, where the rope sections are kept straight by the weight of the man. (Rope weight is assumed to be insignificant)
Assuming a right angle triangle is created , the triangle sides are hypotenuse, 25.5m, the new length, the original length 25m, and the drop in m. This gives a maximum drop of 5m.
Without any calculations i say its less than 1m cuz the 1m is spread across the gap
other than that i could try to collapse the right side on the left side to get the same arch to stright diffrence across the rope except the meddle and use that to get the spread of the middle compared to the rest(aka a percentage)×1=the middle gap acrch to stright deference (i have no mathmaticla proof but it fell like the right way to do this with the least amount of calculations)
Edit:welp i was waayy off
O the slope formula explanation to derive arc length formula makes so much sense
Considering a circunference I got the closed approx ==> DROP = sqrt(6·SLIP·LENGTH)/4
for any SLIP
Another place you see the underestimating of Arc Length is in sailing yacht sizes. They are classified by length. For example a 42ft yacht being much, much bigger internally than a 40ft yacht. That extra 2ft resulting in enough increased volume for additional cabins!
professors never seem to explain why any of this is possible. This video makes it incredibly straight forward to understand why
Then you have really bad professors. In math, explaining _why_ something works should always be the most crucial part.
Too often they don't know how, because the "why" is already obvious to them, and they don't have time to figure out the most intuitive way to explain it to someone who doesn't already get it
And there's usually a lot of people in the class, and different people will find different explanations make sense. I guess it's not easy to find ways that work for everyone, not when they _also_ have to make time to drill everyone on the precise details
I think maths is one of the hardest subjects to teach well, especially in large groups, because it usually relies so much on other simpler maths that the student needs to know already. If the students don't already have a good intuition for the simpler maths, it's even harder to give them a good intuition for the later stuff. So the problem just compounds on itself as time goes by
(And I expect half the students have already given up on the idea that maths is interesting or makes any sense, and they just want to be told how to pass the exam anyway)
Assuming the weight is in the middle of the rope, and it is perfectly inelastic, the absolute maximum drop should be the height of a right triangle winh hypotenuse af 25.5m and length of 25m, which is just about 5 meters. The slack variation is much harder to calculate intuitively, but given that it's a hyperbola, without appealing to major math tools (as you asked us not to) my intuition says it would be between 2 and 4 meters drop (about half the weighted distance at least)
On a related note I was on site and had to record the dimensions of an existing arched bridge. From afar it looked like a segment of a circle. I got a mate to I ball a spirit level end to end and measured the bow in the middle to where a string line would have been. Then measured along the curve thinking I had enough information to define the circle and the rise at one metre intervals. I spent hours trying to figure out the trig to get the actual resulting complete circle. I ended up doing some sort of iterative calc in excel to get the answers. Im
For those that play with ropes and overhead conductors, a huge change in tension can be caused by a small change in the arc length!
The string-girdling-the-earth problem is just as enjoyable, or maybe a lot more. But it isn't the one mentioned in the other comments here; in fact, it's similar to this one, because the question is the same; namely, given a certain (small) amount of slack, how far can it get you?
For the circumference of the earth, you add one meter of length to the rope encircling earth, and then you pluck the rope and pull up until it's taut again. Find how high above the earth does it reach. Like some giant plucking the rope above the chasm here, instead of hanging down. And the straight 50-meter-chasm is replaced by the round earth with a circumference of about 40 thousand km. So, it's gonna be geometry again, like the triangle case, but a little more nasty.
Now give us the answer please, how high does the pinched rope raise? I need to know...
@@xenmaifirebringer552, I said I'll get back with the answer soon, but it's been more than a week now; sorry for the delay. Anyway, I'm gonna post now a regular comment with the answer, so anyone can see it; if I post it here as a reply, or a reply-to-a-reply, etc, it would just get buried and lost into oblivion, which will happen anyway, but, hey, I'm trying.....
0:51 I guessed roughly 5 meters. I did Pythagorean with the hypotenuse 25.5 being half of the actual rope and 25 being one of the legs as the half the distance between the tree. It’s probably wrong but I don’t know what else to apply to this situation
Analogous to the problem cited in this video, there is a practical example of "A cylindrical tank laid horizontal, length of the cylinder along the ground". Calculate the volume of Diesel Fuel while filling this tank at any instant. I have an engineering background but too old to figure this out. Perhaps youngster can suggest a formula for this problem.
really amazing, explanation, animation and story thank you very much! subbed :)
Welcome aboard!
Are you demanding that the curvature of the rope forms a circular arc or the curve of a slack rope which is neither a circle nor a conic?
Another example where this is *extremely* important: Slack line walking, and canyon jumps as an extension
Yes! Life saving really
It is a lot more complicated there, though, since you need to allow for the stretch in the line. They usually use fairly static webbing that doesn't stretch much, but as we've seen here a small amount of stretch can drop the midpoint by a lot.
Wonder where my errors are adding up from. I just ran the number on my catenary model and I'm getting about 4.36m, which is awfully close but not exact.
Without going into too many details, let L be the length of half of the rope and D be the distance from the end to the middle. Then, you can find your solution catenary as
h(x) = cosh(ax)/a + C
where h is the height, x is distance from the centre, and C doesn't matter since we can move "ground level" to wherever we would like. The parameter a winds up being a solution to
sinh(aD) = aL
Finding the appropriate a = 0.0138152, you can plug in L = 25.5 and D = 25 to obtain h(D) - h(0) = 4.360337.
Made me think of bernoullis brachistochrone problem that shows up in physics mechanics courses, the shape ends up being a cycloid. I’d love to see a vid on that
What is that interactive software for functions?
He used Desmos but Geogebra would also be an option
Yes we used both desmos and geogebra in this video. Desmos for the bit about straight lines and geogebra for the parabola and basketball modeling
I like to explain this to my students as "there's not a lot of length difference between a tight rope and a saggy one, and that's why it's pretty much impossible to load at long rope and keep it straight". Think high voltage lines, or laundry...
The slack plays a bit role in a ship's anchor ⚓ . Because the chain slacks it is hard to un slack it by increasing the horizontal distance if for insane there is a wave it is stable
Why don't you use the p value associated to the parabola to try to guess the ecuation considering 4p for the horizontal distance? Is it a bad guess? Just suppose the horizontal distance is latus rectum.
Interesting and my thanks. I ran into this problem when we were discussing the length of cable or wire between telephone poles in a High School Physics class or two. Turned out that there had to be some drop because of gravity. Which related to the amount of tension applied to the wire. A big deal if you think about it, and relates the the reliability of the then telephones and electricity coming out of your light sockets.
This is a fantastic lesson! (I can adapt this into a class period in about 3 minutes…TEACH LIKE A PIRATE, ARGHH!)
So interesting. Thanks!
Having seen tbe answer to how far the rope drops or sags when 1m is added to 50m, here's another "surprising" answer: suppose there was a rope running around the equator of the earth. Assuming the earth was a smooth solid sphere of radius 6,378,000,000 m how much extra rope would be needed to raise the rope 1m off the ground all the way round the equator?
In what way does the noise in the background make this instruction better? Noisepolution? My professor never had a DJ as assistent!
My idea was to divide the graph into two equal triangles with the following properties ( adjacent side = x, opposite = 25 m, hypotenuse = y).
In addition, I thought that the sum of both hypotenuses must be 51 m and since both are equal, I derived a formula from this.
51 m = 2root(25 m² + x²)
This results in two answers:
Answer 1: x = Root(101)/2
Answer 2: x = -(root(101)/2)
I hope my thoughts where correct
How TH-cam got me here today I don't know.
This really blew my mind. My intuition was wrong. After watching, what I was imagining was how any slack, would instantly create a arc from a huge circle. as x (delta of the chord)->0, y I wasn't doing the integration, but just some geometry.
Looking at this from the arc and coord ratio, 4.3431 was my answer, between your two examples. 78.5 would be the max amount of rope could go before it no longer fits that limit of a circle, and then it would droop down.
MOre interesting is the question how strong the Rope pulls ON the fixing Points left and right depending ON the deepness of the sagging. (Antenna Problem)
My intuitive guess is that the first problem is so simple to calculate, we should calculate. The answer to the second problem is, "less", unless I want to do the not so simple calculation for a catenary curve.
It depends on the stretchyness of the rope. If it has no real give to it, they would drop about 5 meters (using the Pythagorean Theorem), so I would guess that it would be less if there was no weight pushing down on the rope. Then it depends on the weight of the rope and whatever the formula for a catenary is 😀.
It would be interesting to iterate between the two questions. Start with the triangle, then have it be a quadrilateral, and so on.
I have to admit that the use of a calculator that does the integration for you is a bit dissatisfying, because, at that point, the intuition disappears that connects that part. I think the above might help keep it all connected.
If this is the method of solution you expect, make sure to specify to your students that you're seeking an _approximation_, not an exact solution.
Most students delving into advanced calculus will assume they're expected to give exact solutions. To them, these questions feel more like 'invent a method for deriving the distance between the apex of an arc of length x with unknown shape, and the midpoint of the arc's terminations.'
To these students; the problem is the unknown shape of the arc. This is why it's important to specify that you expect an approximation, rather than a solution.
The first question is easily splved by just using the pitagorean theorem, it's about ~5.025m
The second question I know hanging ropes use some kind of hyperbolic function but as an engineering in training approximating it with a parabola should be fine hahaha
But you asked for intuition and not me calculating the arc length of the parabola, so I'd say it should be less than 5 cuz the load is pretty evenly distributed across the entire rope and it's the same total length so I'd say it's like 4m :3
The first answer being 5 meters blew my mind the math checks out but how logically does that make sense just breaks my brain thinking about it but I guess it must come from the rest of the line as well not just the 1 meter of slack added in like you would intuitively think.
The way the first puzzle is drawn the rope is straight. The pythagorean theorem tells us the rope falls just over 5 meters. Real world it will be less because the eope follows a catenary curve.
My intuitive guess is that it's a catenary and we can calculate it with hyperbolic functions (sinh, cosh, etc)
1. The shape of the curve strongly depends on the properties of the rope. Assuming that it is a parabola is not valid assumption.
2. The basketball ball trajectory did not count with the perspective. The ball was also moving away from the camera, therefore the used curve was not correct.
The most unintuitive question is how much longer the earth circumference compare at the ground level vs 1 meter above above it . just around 6.28 meters
i got carried away and did calculatings, the first one (should be) about 5.02 m (i may be stupid??)
with the classic
sqrt((51/2)^2-25^2)
i dont know the second one
preliminary answer, haven't watched the video yet. will check later
the initial and final point of the chord describe an angle theta at the center of the circle. start by finding this angle, and the radius. by geometry, r*theta = 51 m and, by trig, r*sin(theta/2) - r*sin(-theta/2) = 50 m.
by simmetry, the last one is just r*sin(theta/2) = 25 m.
solve for r and theta to get r ≈ 74.124998 m and theta ≈ 0.688027 rad.
the height lays in correspondence with the midpoint of the angle, thus its corresponding value will be given by
r - r*cos(theta/2) ≈ 4.34 m
EDIT oh, it was a problem on a parabolic curve. thought it was a problem on a circle from the thumbnail. still close though.
I guess we're assuming that my friend is infinitely heavy.
The reason I forgot about the arc length formula and only derive it when I can use it.... is that... there are infinitely more possible arcs than can be described by simple functions. Even with functions which we can easily integrate... the formula makes it hard to integrate... and sometimes impossible to do so.
Long bridges during hot days and cool nights. As it heats up and expands, that extra length has to go somewhere.
0:50 Woah, intuition is so lacking here.
Intuition kind of told me between 0,5 and 1m and that is just so wrong.
If you approximate the problem with 2 triangles, you get more than 5m
funny how common this is!
What are the chances that my next chapter is on this and I see this post!
how is it possible that the lenght of x in the pythagoreon theorem is 5m, when the rope is only increasing in 1 meter lenght? Shoudn't x be max. 1 meter long ?
That is a good question and I think it's a common deception to think that the max it can increase will be 1m. However, that x is not physically part of the rope.
Perhaps thinking about the opposite situation could help. Say you have a pole 5m tall with a rope that is 25m long and placed at the bottom of the pole and pulled tight at some other point 25m away. If you extended the rope by 1m, you could only go 1m up the pole if you made sure the rope lay on the ground the whole way except for the 1m up the pole. However you could reach the top of the pole using the shortest route (the hypotenuse).
@@ZacharyBlue Thank you! I guess the problem with the understanding does not lie in the specific problem, but rather the fundamental undestanding of the pythagoreon theorem. I will try to understand it more
Power lines come to mind first. 😊
Great example! And important as too much tension can cause them to break with high wind and too loose can cause them to bump into things and also break
For the first one, using the Pythagorean theorem, I got that the middle of the rope should drop by sqrt(25.25) meters. I have NO clue how to do the second part!
I must admit to always taking the lazy path of writing ds^2 = dx^2 + dy^2 and just pretend what follows is rigorous, where I think of ds as "just a bit of arc length" and dx and dy as "a tiny change in x and y"; if I zoom in enough each step is "approximately a right triangle". "Divide" by dx^2 ds^2/dx^2 = 1 + (dy/dx)^2. Take the square root multiply over by dx and integrate. Also works for different geometries like in special relativity where you have ds^2 = c^2 dt^2 - dx^2.
The dx * sqrt(1+m^2) notion is a bit more rigorous of a take (without sacrificing intuition) that I might use the next time I explain this formula to someone.
Why not a hyperbolic sin or cos? Sinh or cosh.
For the first one, it has to be less than 1/2 meters. My guess is 1/4 m
And for the case without friend it should be even less maybe 10cm
So apparently I have no intuition and mistook the upper bound for the lower bound 😂
I actually derived the formula for the chain line by setting up the differential equations when I was in ~10th grade.
Unfortunately I no longer have the post-it I calculated it on but as far as I remember I also calculated it by setting up the length of the line for a short distance to get the weight per small segment in x-direction
Make more videos ❤❤
@9:07 33 cm is 13 inches, not 4.
a third of the unit. a third of a meter is (about) 33 cm and a third of a foot is 4 inches
Yes that is what we meant! 1/3 of either unit. Sorry if that wasn’t clear though!
At first for both questions I guessed about a metre at most?
Then I calculated the drop for the first (as in the video) and realised I *massively* underestimated the drop.
So for the second... my gut went to 3 metres, but after last time I decided to guess 4.5 metres. I'm glad i continued to ignore my gut 😆
I don't have time to watch this but the thumbnail's question is kinda easy right😅
It's always embarrassing when the guy half your age just hops up and figures out something like this when you just wanted to go measure it.
the friend (if they're in the middle) drops by a total of root(25.5^2 - 25^2) meters, i think
sqr(50,75) is the solution of the first. When you just drop the lilne, it does not get an arc. That is physics.
No weight other than the weight of the rope?
That would be catenary curve. Time to pull out my Calculus of Variations textbook from 45 years ago? Definitely not an easy problem.
A old rhyme : No force, however great, can pull a thread, however fine, into a horizontal line .
First case (0.5)×√((51) ²-(50) ²)
Second case requires little physics thatim not ready with, i.e tension and gravity and density, soo
My logic says that 50m of the length is "used up" by the width, therefore the extra 1m would comprise the drop, which would make it 0.5m
That'd be true if it sank half a meter from your side, stayed straight and then went up half a meter on the other side.
my guess: with a friend: 8mm, without him 2cm. I tried to tigh a rope straigt several times and it is very hard. And 50m is really long.
So I solved this problem, I am going to give the answer, I dont know if is correct. I based my answer on Greco-Babylonian geometry and the basis of developing Acrs and Spans.
The basis of the technique is that if you bisect a circle or inscribe a hexagon into a circle of radius r you create two known answers, In the first case you have a 2 suoerpositioned spans of two, in the second case you have 6 end on end spans of length one. In the first case you have have two superposition spans of length 2, so 2 x 2 = 4. this is the coversge of 4/2pi. We can bisect either span creating a tangent outward. Doing this we have an ouward facing triangle of 1/2 span (2 * 1/2 = 1) and the cosine of 1 is zero, so the ouward facing triangle is 1 -0 in the outer part of the busecting radial. The new span created is this 1^2 + 1^2 = s^2 of the square root. We can repeat this process for generating sides of polygons until at some point Lspan * sides = 2 pi. This is roughly 13 steps for most computers. We can do the same with the hexagon.
So the ratio you give is 50/51 = 0.980392 which is close to a span 39.375 degrees 0.980438. This was derive from a series of back tables generating by using the 180° based and 120°[60°] based span tables. In these tables you can combine variables of two angles to come up with precise information for a third angle and then go about busecting the soans and creating outer triangles to creat new smaller spans. I try to keep the babylonian (360 = 2 pi() whole angles. This particular angle comes out of a theoretical 315° that splits 3 times to 39.375° As a consequence its half span is 19.6875°
[Yes, indeed I have this on an Excel worksheet an no I did not make this for the puzzle, I was testing to see how far babylonians and ancients greeks could fill out a span table just using the pythagorean theorem, right now the table has the largest gap of 1.5°. I could just continue the process until the perimeter of the polygon did not change, then easily get every span to 2° by making the arguemnt that cosine of trivially small angles is indistinuishable from 1, so that if I have an angle like say 1 degree, and I divide it say 7 times to get 1/(2^7) degrees, if I just multiple the sin of 1° x 2 and reverse the division process, I can return all the parameters for 2°, but since the ancients dont have 64 bit processors that would defeat the effort]?
While i could just get the cosine of 19.6875 that would be cheating in a greek, persian or babylonian scribal school.
So from this we can see that the span of 39.375 is 0.67378 and the bisect to the span is 0.94158. Ok this is the answer but we (the scribe needs to fix it) need to fix it so that the construction crew understands it. You span is 50. 50 ~= 0.67378 x r. We dont know r.
R = 74.2. R * 0.94158 = 68.875. Thus the dustance between the arc and the span at the bisect is 74.2 - 68.9 or 4.37 units. Or 4 and 3/8ths units.
And so, 2000+ years before calculus and integrals, this problem can be solved (crossing fingers and hoping I didnt make any errors😎)
Now of you are a stickler for ancient history, you should note I am using the decimal system instead of fractions,
Why would you go trough all the cool math and then just say 'nah, the real cosh is too hard' ? Don't pick the rope problem if you will not solve it. You could have sticked to the basketball case ! That was already cool !
Surely less than 1m... LOL. Thanks for righting my brain
it still makes no sense for me 😂
You are not alone! So take solace in that I guess haha
I guessed "a meter or two" for the first one, but I was still way off!
For 1st it can't be more than 1m so .5m was my first guess. But now I'm feeling like it could be 1m too
For 2nd it will be much lower so maybe .25m?
5m and 4m. I really had to restrain myself from doing any math, and just go with gut feelings. That's such a huge ask. 😛
Great guess!
You lay down a rope to cover the equator, it will be roughly 40,000 kilometers, if you increase its length by ONE meter, guess how high the rope can go above the ground??
Great lesson. my only comment would be to "call it as it is" i.e. it's a Catenary or Hyperbolic curve, not a parabola. you do mention it and compare the two, but it's math and it should be more exact. I would rather teach someone to recognize, this curve is special. I enjoy this stuff, thank you.
Something like 0.5 m for the first case and like 0.35 m for the second one.
Slack rope without a person on it is not an arc, it is a catenary?