CalMeachem’s solution is correct, although there is a slight typo. Because the first wrap is just around the spindle, the sum over 2pit should be from 0 to M-1. (For M wraps, you have M-1 2pit terms.)
Hence,
L = 2piRM + pitM(M-1)
t = 0.5 mm
R = 5 mm
L = 10,000 mm
Solving this gives me 132.24 wraps.
Whoops – I can’t do math. I actually get 70.85 wraps.
I get 70.852. I suspect the people who got 79 thought that the diameter increased by .5 mm with each turn, but it increases by 1 mm since there are two thicknesses of tape, one each side. Then you get the equation in mm
pi(10+11+12+…+(n+9)) = 10000
n being the number of turns, which becomes
pi(n^2+19n)/2 = 10000
and solves to n = 70.852. There is probably a very tiny error since the equation was derived on the assumption that n was an integer. On the other hand, the equation is an approximation since, (1) the tape has to stretch to allow its top surface to be longer than its undersurface and (2) I am not sure I really understand what happens when you change layers, which introduces a discontinuity in the tape’s direction. So 70+ is probably as good an answer as you can get.
I can’t believe you people are still with this. I already gave you the answer. I have done it by three different methods and they crosscheck so if there’s soemthing wrong I must be misunderstanding the problem. I have done quite a few calculations of transformer windings in the past so this is quite an easy problem.
The radius of the first turn is the radius of the core plus half the thickness of the tape and each turn the radius increases by the thickness of the tape.
Turn Radius Length Accum
1 5.25 32.99 32.99
2 5.75 36.13 69.12
3 6.25 39.27 108.38
4 6.75 42.41 150.80
5 7.25 45.55 196.35
6 7.75 48.69 245.04
7 8.25 51.84 296.88
8 8.75 54.98 351.86
9 9.25 58.12 409.98
10 9.75 61.26 471.24
. . . .
66 37.75 237.19 8915.84
67 38.25 240.33 9156.17
68 38.75 243.47 9399.65
69 39.25 246.62 9646.26
70 39.75 249.76 9896.02
71 40.25 252.90 10148.92
Proportional part: 0.4112
So my answer remains 70.41
Can someone prove it wrong or show me what I am misunderstanding?
>> Sailor’s solution neglects the fact that the radius of each layer is increasing
CalMeacham’s post indicates he did not understand my solution.
Using calculus the integral length of the spiral is
L = 10 * PI() * n + PI()/2 * n^2
where L is length (10,000) and n is the number of turns. Resolving the second degree equation you get 70.41
can anyone think of another method of doing it? I can’t but probably others exist.
ahhh…
i forgot that the thickness is actually 1mm…
doh!
The answer to the first OP is in fact:
sailor!
That was who, as the OP asked, can answer the question fastest.
sailor showed the perfect attitude towards seemingly hard problems with a nice bit of insight and a quick solution. It reminds me of the two flies between the two trains problem: no need for infinite or converging series.
I have, however, a slightly different answer from a stricly numerical solution. It only arises from assuming that the tape winds up perfectly on the spool with deforming. I treated it as a beam under load and with that there is a difference a few decimal places down the line.
Again. sailor’s right. If you disagree, show your work…
>> It reminds me of the two flies between the two trains problem: no need for infinite or converging series.
MonkeyMensch, evidently great mathematical minds think alike because that is exactly what came to my mind too. 
For those who may not know what we are talking about: the problem is this: On a railway, at t=0, two trains are 100 miles apart and traveling towards each other. Train A travels East at 35 MPH and train B travels west at 15 MPH. A fly departs from train A flying at 50 MPH towards the other train and, when it reaches it instantly turns around and flies in the opposite direction and keeps doing this between the two trains until they crash.
Q. a) What is the total distance flown by the fly from t=0 until the trains crash?
Q. b) In what direction was the fly going when the trains crash?
The answers can be found without even using pen and paper.
I have never seen Part b of this probelm before and from a gut feeling, wouldnt it be impossible to get this answer as the function of displacement of the fly would be rapidly oscillating at this time and would be undiffererentiable?
Shalmanese, your answer is correct. It is an unanswerable question.
Suppose we are calculating an infinite series of alternating positive and negative values the sum of which converges to a given number and we asked, “what is the sign of the last value of the series?” The question has no answer and question b) above is equivalent.
As mentioned by Cabbage, the inner and outer surfaces of the tape are different lengths if it is assumed that the two ends of the tape remain square. So the answer depends on which surface you assume changes length. In reality it is likely that the ends become bevelled due to this, but the question says thickness of the tape remains constant.
If you assume the ends remain square you have to consider what shape the tape conforms to as it is wrapped. That would be concentric arcs for about 335 degrees around the bobbin with a bunch of straight parallel segments tanget to the circle that ramp up to the larger radius due to the tape thickness. If you assumed the inner surface remains the same length (since this provides the maximum number of wraps), there can be 70.04 wraps. It looks like sailor is assuming a bunch of concentric circles instead of accounting for the tape being continuous.
If the ends become bevelled, the number of wraps is around 69.54. That is counting a wrap as including the full thickness of the tape, since the question said tape thickness is constant. If you change the question so the variable thickness at the ends is counted towards being a wrap, then the number would increase to closer to what sailor says.
Since the question asks for the maximum possible and doesn’t specify bobbin length or tape width, the best answer is probably what is implied by Joey P at the beginning. Assume the tape has negligible width and wrap it in a low pitch helix around the bobbin, so the tape doesn’t overlap and radius never increases: 318 wraps (assuming inner surface of tape remains constant length).
Manlob, give me a freaking break. You are making this more complicated than it need be.
Look, in the real world, when you deform a wire or a tape or a simmetrical beam, the central surface remains the same and the top expands unde tension and the bottom contracts under compression. My assumption is that we are in the real world because the problem does not state otherwise. I have calculated wirings for many transformers and this is the way it is done. Any other assumptions are just wrong if we assume we remain in the real world and the way the problem is stated I have no reason to assume otherwise.
>> It looks like sailor is assuming a bunch of concentric circles instead of accounting for the tape being continuous.
Yes because this is the assumption that introduces the least error. I also calculated a perfect spiral using integral calculus but this is less close to reality. The error introduced by the fact that the tape has to jump from one layer to the next is very small and to calculate it we would need a lot more information which is not provided so it is impossible to calculate and can be reasonably neglected.
You are just confounding the issue. If presented with that problem in a test the correct answer if the one I have given. “It depends” would not be considered a correct answer. When presented with a clear problem like that, one assumes the simplest solution within the confines of what is stated. Your assumptions not only complicate the problem unnecessarily but are mostly just plain wrong in the real world.
>> Assume the tape has negligible width and wrap it in a low pitch helix around the bobbin, so the tape doesn’t overlap and radius never increases: 318 wraps (assuming inner surface of tape remains constant length).
Give me a freaking break. Did you read the problem? Did you read the follow up post by the OP? You are assuming things which contradict the problem as stated.
>> the best answer is probably what is implied by Joey P at the beginning.
What the heck are you talking about? Did you read his post? He asked a question and never provided any answer. And his question was if the tape and bobbin were the same width.
Sailor, you specifically asked for other ways of figuring out the number of wraps, and also asked if anyone disagreed with the answer of 70.41 wraps. So please do not get so upset when someone responds to your requests.
When you used integral calculus to get that formula for arc length, what was the equation of the curve? That formula looks like the result of summing the arc lengths of semi-circular arcs arranged in a spiral. I come up with a different formula for arc length (which gives 69.98 wraps when assuming the neutral axis of the tape remains constant length). I used the equation in polar coordinates:
r(theta)= R+ (theta)t/(2pi)
where R is the starting radius and t is the thickness of the tape. I admit that, like assuming concentric circles, this spiral is not a precise description of the geometry of a real tape, only an approximation.
[Manlob], I am quite sure my integral for the spiral is correct. You know why? Because I cross check things… If the equation tallies with a calculation done witha large number of points in Excel then i am reasonably sure it is correct. And you know why I always crosscheck? Because I always make mistakes the first time. So I can assure you that equation is the correct integral of a spiral and I have no idea about half circles (which sounds more complicated to me).
For the heck of it I calculated the number of turns if the tape, on the first turn, would wind tightly around the cylinder most of the way around and then leave in a straight tangent to go over the beginning and start the second turn and then the rest just wind around this. The solution I get is, obviously, a bit lower than 70.4 but is still over 70 so I would say with confidence that under any reasonable and plausible assumptions the number is not under 70.
Whats a bobbin anyway?
Invented by a guy named Bob.
Etymology: origin unknown - - Date: 1530
1 a : a cylinder or spindle on which yarn or thread is wound (as in a sewing machine)
b : any of various small round devices on which threads are wound for working handmade lace
c : a coil of insulated wire or the reel it is wound on
I was asking the equation of the spiral you used to get your formula for arc length. There are variety of types of spirals. The equation for the spiral I gave earlier results in a length different from the formula you give. Your formula results in exactly what is obtained by summing the circumferences of concentric circles. It looks like you used the same equation I used for the spiral, but integrated R*d(theta). That is not how to find arc length. You need to integrate sqrt(R[sup]2[/sup]+ (dR/d(theta))[sup]2[/sup]) d(theta). It turns out that the correct formula for arc length of a spiral gives a less accurate result than the concentric circles value.
I am not saying the simplistic approach gives an inaccurate approximation. It is bound to be very close because it does not account for just a tiny wedge of unfilled space. I will say, however, that accounting for that tiny wedge is the only thing that keeps this question from being completely trivial. Otherwise, it is just a simple exercise in junior high school math. The way the numbers are given makes it look like this may be a question designed for quick mental math. Making a few more simplifying approximations leaves 2*(sqrt(1600)-5), which can be done mentally and equals 70. Done accurately the circle approximation would be 2*(sqrt(5000/pi+25)-5)=70.41.
My head hurts.
>> It turns out that the correct formula for arc length of a spiral gives a less accurate result than the concentric circles value.
>> I am not saying the simplistic approach gives an inaccurate approximation. It is bound to be very close because it does not account for just a tiny wedge of unfilled space.
That’s what I assumed from the very beginning.
>> I will say, however, that accounting for that tiny wedge is the only thing that keeps this question from being completely trivial.
Hmmm, I agree it is pretty simple but read all the posts before mine and tell the posters and the OP they are dumb for not getting the right answer to something so trivial. I do not think “trivial” is the word and what may be trivial for me may not be trivial for others. But if it was trivial for anybody here it was for me and I would not call it trivial. Maybe just easy.
>> The way the numbers are given makes it look like this may be a question designed for quick mental math.
That’s exactly what I did. In that context my first answer, 70.4, is correct and yet you seemed to have a problem with it.
>> I was asking the equation of the spiral you used to get your formula for arc length. There are variety of types of spirals.
The only kind of spiral which meets the OP is one where the radius increases linearly. My spiral starts with r=5 and r increases 0.5 mm/turn. Frankly, all my math was done quick and dirty and transcribing the whole process here is more coding trouble than I am willing to do but because I crosschecked I am certain it is correct.
My initial and simplest answer was 70.4. The spiral model gives the same first decimal. The model with the tape going around in a circle most of the way and then a short tangent to start the next turn gives a number slightly over 70.0 and that is the most accurate model so far. Any answer which yields a result under 70 is based on a bad chosen model.
>> Your formula results in exactly what is obtained by summing the circumferences of concentric circles
It is not exactly the same but pretty close. It seems to confirm I chose a good model, don’t you think?