In this video, I calculate the height covered by a coil of cable of known length around a spool, counting only the portion of the height that includes a full loop. I first note that as a cable is w...
Transcript
The solution to problem 6, and this is a cable wound around a spool. So a cable has radius R and length L and is wound around a spool with radius capital R without overlapping. What is the shortest length along the spool that is covered by the cable? Yeah. Now, this question here took me quite a while to wrap my head around the spool, and mainly because, well, the question doesn't explain exactly what it's, what it wants properly. And the solution to many was absolutely horrible in explaining, so I had to resort to GROC AI. converse with it to explain what in the world's going on. But anyways, let's take a look at the solution. So as the cables wrapped around the spool, think of the top or bottom of the cable, forming a helix of radius capital R plus R measured from the center line of the cable. I added this because the solution many did not add it. That's very, very, very important. Otherwise, you'll be lost. So let H be the vertical distance between coils, and again, ignoring, well, first of all, ignoring the initial non-slanted coil winding. It's a number at the beginning you have this, uh, flat part that's not slanted afterwards you're gonna get a slant like that and so on and also measured from the central line of the cable so again these bold ones I included these because it's very very important and the solution does not include that or the question the question doesn't even include it so we have this coil right here so here's just the coil but again ours won't have any overlapping we're not gonna have a double like that but but nonetheless this portion here, and this portion here is what we're interested in. Notice how there's a slant here. As you're going up, there's a slant. It's a constant slant. All right, and this one is constant slant here. And then at the beginning, the first one here, that's a flat one. We're going to ignore that. So, yeah, so what we're going to have is, well, here, first of all, I'm going to do right here, is I'm going to unwind. If I unwind this, if we unwind, and just fix up the spelling. So we unwind, we're going to have initial portion that's going to be like that, and that's going to be flat, and then afterward it's all going to go at a slant. It's going to go to a slant like that the whole time. And again, this, now this is a cable, so if we have it, we're going to have a full slant once we unwind this all, and this is going to be the diameters 2R. So the number that's 2R. But now this, the height difference, or this H right here, again, what we're asked to find is what is the total height, essentially? That's all we're essentially asked. It says, what is the shortest length along the spool that is covered by the cable. So that's going to be basically a height like that. That's what we're doing, like the full height around that. And this would correspond to somewhere around here. So like, yeah, something like that. And again, it depends, because we have a finite length, we have a length L there. So it's going to be, reach it a point there. So, yeah, assuming, in this case, it's going to be super long, I got, so that's, etc. But anyways, you get the idea, that's going to be the full height. But now if we look at this part here, what I keep emphasizing, the center line, that's where this capital uppers are. So if we take a look at that, so let's say we had this portion here, and then zoom out there, or zoom in, I'll just do zoom in. Yeah, zoom in. So zoom in. We're going to get something like this. And also I'm going to draw here. Let's say this is, see, this part right here would be, I'll put a dot for coming out of the page. And then this one is going into the page with an X. So just the usual electromagnetic, the circuits design. So we're going to have it like here. So make it even bigger. So it's, so there is coming out of the page. and then we're going to have a slant. It's going to be a slant like this, and this is going to be going into the page. And now this, and this is, and then it goes all the way wrap, wraps around, and so on, so we're going to get another wrap, and then this goes another one, just give you an idea, like that. So that's how the coil one looks like here, so that's this part here. All right, and now the radius, yeah, so there's a spool. there's the, I'll say the center of the right of the spools like this, and then it goes all the way down here. So then the radius is going to be, and it's up to the center line. So it's going to go, there's the full distance across where this is. Now, this is the ending of the spool, that's just wherever the spool is, and then that's all the way around it. So this is going to be R, and then we're going to go to the center line, lower case R. So that's the radius of the cable. Alright, so that's what that is. And now, this height difference, H, that is, well, it's going to be just this right here. The difference between the center lines, that's H. And again, it's the same thing as putting it here. It's just exactly the same thing, H. Yeah. Yeah, so these are the equivalent there. So that is what it means. So basically, as a full loop there, that's the difference height, H. So you're going to move up by H. and then again, and then 2R is going to be like that, if that we'll put it H here, but then the radius is going to be at an angle like that, so we have a 2R, like that. All right, all right, and then lastly, another thing that the solution method doesn't explain is, well, we're going to have from here to here, this is the full diameter, so let's say this is going to be the full diameter is like this. But then if we had it in looking at it from the top view, and I'll get to the top view soon, but here's the diameter, and this is going to be just two times capital R plus lowercase R. And then again, this measure from the center line, that's the key part here, and it's also measure center line, all the way across here. And also, number, this height difference measure from the center line, just to give you an idea, but it's the same thing as going from the other part. Yeah, all the way across center line, and so on. Yes, but for this case here, for this, to end up going here, then has to loop all the way around here. So that has to do a full 360 degree. So in other words, it goes from here, all the way across like that. Let's go back in. So the full length of that, yeah, let's do it like that. So this full length like this, going in, and then going back out here, coming back out. So it's going to be inward out. It's going to be like that. This full circumference. I'll draw it again better. And here quickly, just a pause video and just find an image on, just a brave search set of Google search. So, yeah, we have it here. So there's the full overlapping here. So you can see how it goes all the way across there. Yeah, so one full loop before it gets this from center line to center line is going to be, it's going to be radius. Again, it's going to be radius. R plus R. I'll make it actually in, and maybe, yeah, this is from the center line from center line. This is going to be, I'll put R plus R, yes, it's good enough. All right here, I just quickly just fix that up. You just move it over like that, it's a bit more visible, and now the circumference here is going to be, well to see. I'll just draw it all the way across like that, or let's try it one more time. So yeah, the circumference is going to be roughly, yeah, it's going to be equal to see the circumference. And all right here, I just quickly just added this setup there, just this thin circle line. So now this C here, I'm going to draw this right here, and I just fix it up. So C, the scrumference equal to, again, a recall just pi times the diameter, this is just D, capital D, this equals to now, well, 2 pi r, capital R plus R. So we're going to keep that in mind. Yeah, because now we can finally get to the next part here. I want to go through all this context, just to help make sense of this, because the Solutions Manual did not do that. All right, so now we're going further, so then from similar triangle, so we're going to draw this, basically zoom into this, and then we also know that the full circumference is that. So what we're going to do is, we're going to have a setup, it looks like this. So there is, I'll draw it like this, a bit more vertical. So we have a setup like this. and then it goes, we're just exaggerating a bit. And then these are, again, these are parallel. I'll just draw it here, parallel, so it's parallel lines. And now we have, this is the centerline height there from each, uh, uh, overlapped, I mean, each, uh, wire, uh, circumference. This is gonna be our H. And now the, um, diameter of, this can be right angle, diameter of the wires to R, these radius is just lowercase R, and now the full circumference, this is C, this equals to 2 pi, and then R plus R. So again, that's just this whole setup here, going back up here, so it goes from here, all the way straight line until this matches up with that. So that's what it is. It's one full 360 there. This is if you were to unwrap it all. So unwrap it, this would, actually, this would go like, yeah, would go all the way there. Be a bit lower, yeah, be like this, and then it keeps going until it matches up. Something like that, yeah, but it's not too scale. This should be a bit higher up. Yeah, this would be a bit higher up, but yeah, it's going to be like this all the way, and then you're going to have this matching up, and so on. So again, it's very, very hard to see, because it's a full 360, it's going to be a bit higher. Alright, yeah, yeah, so I could have made that a bit better, I just realized, yeah, this should be a bit higher somehow, but again, you're trying to flatten 3D there, it's pretty hard, but it's gonna be like this, and then again, the wire keeps going, because as a full length, L, et cetera, all the way to L. And so, okay, so we just have that, so now we're gonna use similar triangles. So we know this part right here, well, we know H squared using Pythagorean theorem. 8 squared plus, yeah, 8 squared is equal 2, and this part here, we'll just call that, yeah, adjacent there. This is just going to be, I'll put a blank there. Yeah, I'll just call it adjacent. Yeah, let's put it adjacent there, AD. Like that, AD squared plus 2R squared. So then basically move this, yeah, move this one to the other side, and then square it all, move this over to there, and then square that, we're going to get a AD, this part right here. This is AD is equal to, or put it adjacent like this, maybe. So it doesn't look like it's two variables. A and D, yeah, is equal to. square root, and then we're only taking the positive size, plus minus, we'll just take the positive, because we want a physical value, not a negative. It's going to be now, h squared, minus the, this is square that 4 r squared, like that. Alright, now we'll see the similar triangles set up, though. Well, this is this angle's theta. Well, there's two triangles there. This theta, but we also have this one here. So, what we get is, well, in this case, we're going to have, you know, we're going to have cosine. I'm going to put this as, yeah, this is a hypotenuse there. I'm just going to put that as cosine theta. This equals to, actually, not cosine, we're going to do sine theta. Yeah, sign theta right here. Yeah, actually it's not sign theta at all, it's going to be tan theta, yeah, so tan theta, number tan theta is equal to opposite, so opi over adjacent, so that equals to 2R, so we're going to have a 2R for the small triangle over square root h squared minus 4r squared. all right, so that equals that, but it also equals to, well, this is the, for the big triangle, there's, there's the new opposite, and there's the new adjacent, yes, it's the same exact setup there. So what this means is this equals to, well, C over H, which equals to 2 pi r. plus capital R over H, like that. All right, so now that we have this. What we want to eventually solve for is H, that's what we want. So what I'll do is, well, the next setup is, let's just write this down. All right, let's just, yeah, well, we have this, and we have this. What I'll do is, we want to get rid of the square root there. I'm going to, I'm going to basically square everything. So square this whole setup there. So I'll just put it here first, as yes, what we get is 2R. Just get rid of that square root there, H. squared minus 4r squared and then this equals to 2 pi r plus capital r over h we're going to square this all now square this so we're not changing anything is multiply by itself so we're going to get a yeah this top part's going to be now we're going to have a 4 r squared over now and also move this over do that and also move it over to the other side So we don't have this denominator there and also move this one over to this side after we squared. Let's do it all in one go. So we're going to get, you know, we're going to get as well. Also, we also have this 2 and 2. We're going to cancel that out. So that's going to have this now. We're going to have this H squared, R squared, R squared. So we're R squared, H squared equals to, we're going to get a R plus R. squared and then times it by h squared minus 4r squared because that's this part here we're getting rid of the square root move it over to this side and yeah we also have this pie squared so we forgot the pi squared there so let's put that in here I'll move this over All right here, it's moved over, so pi squared like that. Alright, so now we can do some more cancellations. Actually, we're not going to do that much cancellations, what we'll do instead is, well, we want to multiply this stuff out, so we want to get this H squared out of there. So we're going to, this is going to equal to 2, and then also move the R squared to the other side. Yeah, so I'm just going to start doing some algebra right here. Is this going to equal to pi squared? I'll keep this R plus R squared. out like that, and then this is going to be multiplied inside H squared minus now we're going to have a, yeah, that we're going to have a 4r squared, and then we're going to have this next setup is going to be, yeah, 4R squared, we have pi squared, r plus r squared like that, so I have all that. set up there, all right, just erase this. So now what we have is, yeah, I'll keep this here, R squared, H squared like that. So the next setup is, well, this one will move it over to the other side and move this one into this side and factor out the H squared, and then factor out this and this. So we're going to get, then move that as the first equation, so we're going to get, h squared, and then we're going to have this, that's going to be the positive one now, pi r plus capital r squared minus r squared like that. All right, that's what we're doing. And then the next one's going to be positive, now we've moved it over to the other sides of 4 r squared, pi squared, r plus r squared like that, because we have all that. and yeah this is going to be a bracket like this all right so that's the square bracket so now we could basically divide this out and then square root everything and take the positive of it so what we're gonna get is yeah we will finally get h and then just erase that H equals to, so again plus or minus square root, we just take the positive always, for real purposes, is not, not negative, just because you can always have that for mathematical purposes. Anyway, so I'm going to get is a square root of, well, this top part's going to be, yeah, we're going to divide a bit of all the squared, but this one here, these are all squares. So those all vanish. So, I mean, that's all going to be, well, square root of all them. So, R, so get rid of all those squares there. So 2R, or instead write 2 pi, yes, I'll write 2 pi r. So 2 pi r, and then r plus capital R, divided by, now this whole thing gets square rooted. So we get a pi squared, r plus capital R, and all squared minus r squared lowercase r squared all right yeah so we have all that and we're gonna box this in so that's the height from center line to center line and then with the calculus book the solution value does as well it multiplies by the uh it will it finds the total um the complete cycles there how much multiplies h by the number of complete cycles we can make within the amount of length l we have So, for example, here's the H going to be H times by, let's say, 6, for example. So multiply by an integer amount, and that's the height of that axial distance of the spool. That's pretty much what, again, the question is even asked it. The solution manual doesn't really explain it properly. Yeah, so that is, we found H here. So we found the difference. All right, now, if we parameterize the helix by, and again, this part, the solutions manual in the calculus book does not in any way explain how this is done. Yeah, yeah, basically what we're going to have is we need to find the X, Y, and Z, like, parameters of it, basically, write it in parametric equations. And again, at this part here, I said that this should be a bit higher, actually. It's then it loops around over here. Yeah, this one just loops around. So notice how it loops around and so on. It's pretty good stuff there. But now we need to put them in X, Y, Z components. So what I'm going to do is we can take that center line. So number, it goes to the center line. That's what we're everything is based on, the center line of this cable. So we're going to take the center line. Let's say we have it from here. Here's the one we're going, and then it's going to loop around, and it goes like this, and then there's the bottom part. like this, it's just a rough drawing like that. Again, we're focused on the center line around, and then all the way across there. I'm going to go slow just so we can grasp this, because again, we're dealing with 3D, three dimensions and thickness of lines, which is not something we're used to, so we have that. All right. All right, so we have that one full loop. So in one full loop, we have, this is the height difference, it's going to be our age. So that's the one full loop. It goes like that. And now what I'm going to draw is, I'm going to draw this, I'm going to draw the straight down like here. This is going to be the Z axis, like that. So there's going to be now the Z, there's going to be our Y axis, like that. It's right straight through it. This one is in front of it. That's why I put it like that. also straight, I'll make it down, yeah, that's gonna go like this, and then straight through, this is our X, that's our Y, like that. All right here, and this is just one loop here, so I'm gonna actually keep going here, so again, this is the origin, that's the origin there, and again, I know, this is just now a three-dimensional, yeah, polar coordinates set up here, so what we'll do is just do the usual polar coordinates, so then I'm gonna do is, at this point here, I'm gonna just, erase this, this is going to be like this, extend it out, and then it goes down here. Yeah, so this is going to be like that, this is going to be our x, this is going to be our y here, and this is going to be our angle theta here. But now we're instead of angle, we're going to use it as t. So this is going to be t like that. So that's going to be our angle, and then this is, again, this is all right angles, everything's right angles. and so on, that's the right angle across I got. Yeah, so bear with me, so what we have now, and again, we know that, yeah, cosine T is equal to adjacent over hypotenuse X over, yeah, over, well, this part here is the radius R plus, yeah, I'll do lowercase R plus R, so that's the spool, yeah, the wire plus the capital R, that's this hypotenuse there, it's going to be R plus capital R, and likewise we have, by definition, we have the sign, yeah, the definition, yeah, we also have the sign now is opposite over hypotenuse, sine…