Amount of Time, Wire Length

[Guest]

Wire .02 inches in diameter is wound on a spool 5.5 inches in length and .75 inches in diameter, turning at 25 revolutions per second. Determine amount of time (seconds) to wind the first layer and length of first layer.

[Guest]

5.5 / 0.02 = 275 revolutions for 1 layer

At 25 revs. per second....... time for 1 layer is....275 / 25 = 11 Seconds.

Length of 1 revolution..... 3.14 x 0.75 = 2.355 inches

Length of 275 revolutions.... 275 x 2.355 = 647.625 inches

[Guest]

A question:

How would the total time to fill the spool and total capacity be determined ?

[Guest]

You know 275 turns for layer 1. And you know 25 revs per second
Need to know how many layers in filled spool.
Or... how many turns....
Or... the finished diameter of the spool....

Let d = diameter of the wire and r = radius of the wire......

Is each layer wound directly on top of previous layer coil....all wire diameters are they in alignment...same number of turns in each layer... and each layer adds another 2d to the diameter of the spool...for each layer added...

Or is the second layer wound in the groove between the coils of the previous layer....this will mean alternate layers (even layers) will have 1 less coil ......and the diameter will be increased by 2x (sqroot(d^2 - r^2) for each layer added. All odd layers will have same number of turns as layer 1.

You should be able to work it out from this when you have got the facts....

......Post your working and I will see how you get on.........Consider the problem logically..........

"You must be doers of the word and not only hearers who mislead themselves. Those who hear but don't do the word are like those who look at their faces in a mirror. They look at themselves, walk away, and immediately forget what they were like."

[Guest]

I am not sure how to solve. I don't think volume, proportion, or rate- time- distance will work.

[Guest]

How you go about solving the problem depends on what the finished component is.......
Is it simply just a spool of wire, say 100 feet of wire wound on a spool just to contain it.?.?.? So need to know what length of wire you want on the spool.....

Is it a spool for a solenoid, insulated copper wire wrapped on a spool, The pulling force or the voltage rating will depend on the number of turns..... ampere turns. So the coil size is usually rated as a number of turns....????.....Example 3000 turns of 0.02 inch wire......So need to be given the number of turns required.

Is it a coil for electronics....Example needs to be a certain resistance.....The resistance total resistance will be based on the length of wire in the coil. So need to be given the length of wire required.

You need to identify the final dimensions of the coil either, number of turns, length of wire or outer diameter of finished spool and then work out what dimension you want to find based on the methods in the previous post.

.......and you need to have a go at it yourself, then post it........ then I may be able to help......

[Guest]

Work Logically......a step at a time.....

If it takes 20 bites to eat an apple......then after the first bite I have eaten 1/20 part of the apple.
After 10 bites I have eaten 10/20 or half of the apple.
After 20 bites I have eaten 20/20 of the apple or all of the apple......
But....then I am left with the core.......
So I didn't eat all of the apple in 20 bites......only all of the bit that could be eaten.......
So there are errors and roundings in most calculations.....but we can work out an answer that is near enough correct to satisfy the conditions of the problem in hand.......

So for you wire/spool problem......If you don't know how precisely the wire will be guided and wound on the spool, it doesn't really matter but it may leave a slight error, but at least you should get an answer and a method used on how you calculated it........and you can refine it when you have worked through it. Consider the wire wound directly aligned on the wires of the layers below, this is the simplest solution. Then as a second method consider the wire wound on the "groove" between the wires on the layer below......There will not be that much difference.....will it matter to the problem in hand..? ?. Also is the wire wound as an exact spiral round the spool or do we consider it as complete loops fitting perfectly...??? No matter how you consider it you should be able to show logically your working and come up with and answer......

[Guest]

length of wire- 1 mile - 5280 feet
simple wire

5280 * 12 = 63,360 inches

63360 / 2.355 = 26,904.46 = 26,904 revolutions

26904 * 11 = 295,944 seconds

I think I miscalculated.

[Guest]

Are you sure you are thinking hard enough......
you divided by 2.355 which is the length of 1 revolution on the first layer. There are only 275 revs on the first layer.

Depending on how it is wound the next layer will be on top of the first and will be longer...
And the third layer will be longer still and so on until 1 mile of wire is wound on.....

[Guest]

63,360 * 2.355 = 149,212.80 = 149,213 revolutions

149,213 * 11 = 1,641,343 seconds

If that is not correct I don't know.

[Guest]

You are not explaining your method of working.......

You have multiplied the number of inches in 1 mile by the number of inches in one revolution and gave the units of your answer for this part as revolutions.......
Why did you think multiplying inches by inches per revolution would give revolutions..?

If you divide the inches in 1 mile (63360) by the inches per revolution you will get the number of revolutions....
because....inches divided by (inches / revolution) is same as using "turn divisor upside down and multiply" so we get....
inches x (revolutions / inches) ....the inches cancel and we are left with the units "revolutions"

So 63,360 / 2.355 = 26904 revolutions.

That would be the case if the spool diameter didn't get any bigger as we wound on more wire. As we wind on each layer the diameter of the spool will get bigger by 2 diameters of the wire (2 x 0.02 inches) assuming the wire is wound directly on top of the wire on the layer below with the centres of the wires aligned.

So the diameter of the spool where we are winding the second layer will be 0.75 + 0.04 = 0.79 inches

So 1 revolution on the second layer will be 3.14 x 0.79 = 2.4806 inches and 275 revs will use up 682.165 inches and will take 11 seconds revolving at 25 revs per second.

So Layer 1 uses up 647.625 inches .....we had calculated this in previous post...
and Layer 2 uses up 682.165 inches .....

We could keep on doing this for each layer until the mile (63360 inches) of wire was all used up.....
But.....you should be able to work out a formula that will help do it.

It will be in the form of a series where the winding diameter of each layer will be the (spool diameter) for the first layer, then the (spool diameter) + (2 x wire diameter) for the second layer, then the (spool diameter) + 2(2 x wire diameter) for the third layer, then the (spool diameter) + 3(2 x wire diameter) for the fourth layer.....and so on.....

You should see a pattern emerging.......Let L = layer number.....
The diameter then becomes the (spool diameter) + (L-1)(2xwire diameter) for the first layer, then the (spool diameter) + (L-1)(2 x wire diameter) for the second layer, then the (spool diameter) + (L-1)(2 x wire diameter) for the third layer, then the (spool diameter) + (L-1)(2 x wire diameter) for the fourth layer...and so on...

This becomes the (spool diameter) + (1-1)(2xwire diameter) for the first layer, then the (spool diameter) + (2-1)(2 x wire diameter) for the second layer, then the (spool diameter) + (3-1)(2 x wire diameter) for the third layer, then the (spool diameter) + (4-1)(2 x wire diameter) for the fourth layer...and so on....

This becomes the (spool diameter) + (0)(2xwire diameter) for the first layer, then the (spool diameter) + (1)(2 x wire diameter) for the second layer, then the (spool diameter) + (2)(2 x wire diameter) for the third layer, then the (spool diameter) + (3)(2 x wire diameter) for the fourth layer....and so on

The diameters will need to be multiplied by 3.14 to get the length of wire per revolution and multiplied by 275 to get length of wire per layer. You will then need to get the sum of "n" terms in the series when this equals 63360 and all the wire is used up.

[Guest]

I still don't understand.

[Guest]

You must understand something about problem....Do you understand what the question is asking...never mind the solution...
We have already worked out the details of the first layer of wire on the spool.
We know the spool length, spool diameter, the wire diameter, the speed of winding, number of turns per layer, length of wire on the first layer, also the new diameter on which will be wound the second layer (assuming the layers fit neatly directly over the tops of the wire on the previous layer).
Some of these details were given in the question and some others were calculated.....
From what we have done so far see if you can put numbers to each item in the list...
That will let me know if you understand the question this far....and I can help with what items you are not able to put numbers to...

The next bit of the problem was your question......
You mentioned 1 mile of wire....

So you want 1 mile (63360 inches) of wire wrapped on the spool......OK

Assume simple wrap each layer is same wrapped directly on top of the wires on the previous layer.
That is a right hand spiral, left hand spiral, right hand and so on for each layer. Diameter will increase by 2 wire diameters each layer....

Following on from previous post you should end up with something like this......
In general terms......Length of wire per layer.....

And for each revolution....This is an arithmetic series of diameters having a starting value (0.75), a common difference (2 wire dia.) and "n" terms or in this case "L" terms....So...

Length of wire per layer = (Pi ) x (turns per layer)((starting spool diameter) + ((Layer number minus 1) x (Two wire diameters)))

3.14 x 275(0.75+(L-1)(0.04))

3.14 x 275(0.75 + 0.04L - 0.04)

3.14 x 275(0.75 + 0.04(L - 1)

= 647.625 + 34.54(L - 1)

or 3.14 x 275(0.71 + 0.04L)
= 613.085 + 34.54L

when L=1 = 647.625 as calc before......
when L=2 = 682.165

You will need the Formula for Sum of "n" terms in the series....."n" terms or in this case "L" terms....
....and solve for "L"....to get number of layers for 1 mile of wire.....

[Guest]

I can't do it. Never mind.

[Guest]

Thanks anyway.

[Guest]

You were the one who posed the question......I was only trying to help and let you have a go at it yourself....
Here is the remainder of my solution......Hopefully you will read it and try to understand it......
It is a practical problem with a reasonable solution......and my answer seems reasonable.....I think it is correct until someone proves me wrong.
Its like eating the apple.....I took it in stages.....or bites and worked my way through it.....
I may have made a few assumptions and there may be some minor errors but solution generally good enough for practical purposes........
So following on from my previous post......................
You will need the Formula for Sum of "n" terms in the series....."n" terms or in this case "L" terms....
....and solve for "L"....to get number of layers for 1 mile of wire.....

So... For a series....(a), (a + d), (a + 2d), (a + 3d),.....up to the "n" term (a + d(n-1))......

In simple terms the Formula for the Sum of "n" terms is found from the following... Let "S" = Sum of "n" terms, "a" = first term, "d" = the common difference and "l" is the last term.....The terms can be described as either the (first + difference and so on) or the (last - difference and so on)

Sum= (a) + (a + d) + (a + 2d) + (a + 3d).....+ (l - 3d) + (l - 2d) + (l - d) + (l) .......Eqn for Sum (1)

The same Written in reverse order gives.....

Sum = (l) + (l - d) + (l - 2d) + (l - 3d)....+ (a + 3d) + (a + 2d) + (a + d) + (a) .......Eqn for Sum (2)

Now adding Eqn(1) and Eqn(2) we see most of the terms cancel and leaving.....

(2 x Sum) = Sumof(a + l) up to "n" terms = n(a + l)

So the Twice the Sum... 2S = n(a + l) or S = (n/2)(a + l)

If We don't know the last term so "l" needs to be described in terms of the series.... The last will be the "n"th term...

So "l" = a + d(n - 1)

So... 2S = n(a + a + d(n-1)

So... 2S = n(2a + d(n-1))

Or... S = (n/2)(2a + d(n-1))

and in terms of this question Sum will be......where "n" in the above is "L" for layers in this question.

S = (L/2)(2a + d(L - 1))


S = (L/2)((2 x 647.625)+(34.54(L-1)))

63360 = (L/2)((1295.25 + 34.54L - 34.54))

63360 = 647.625L +17.27L^2 - 17.27L

63360 = 630.355L + 17.27L^2

solve this quadratic for "L" gives.......

L = 45 or -82 .....45 is the +ve answer....take the +ve answer...

So 45 layers... For 1 mile of wire.....

45 x 275 = 12375 revolutions required to wind the wire on

45 x 11 = 495 seconds = 8.25 minutes to wind it on.

Finished Wound coil outer diameter = ((45 x 0.04) + 0.75) = 2.55 inches

Seems reasonable................Simple.......................

[Guest]

I posted the problem and understand it- I understand now how the first layer was calculated. I know you were trying to help and I appreciate that. The problem only asked for the first layer and did not ask for the capacity and time to fill the spool - I was just wanting to know. As you were explaining the solution, I was getting more confused. I understand you said it is a series but I still did not know how to solve. When you explained the series and the quadratic I understand the solution more than I did but am not totally clear. Thanks again.

[Guest]

Cheers......I am glad I was able to help.....
Yes we made the problem much harder by working out how to fill the spool with a mile of wire......and then I included deriving the formula for the sum of the layers......and we needed a quadratic to find the number of layers......
Best Regards........

[Guest]

A question:

I chose a mile as the length of wire. If I had used a smaller number would the procedure be basically the same just not as many steps involved ? Thanks again.

[Guest]

If less of the same wire was to be wound on the same size of spool and the same information is required to be found, the method would be the same, just the number of layers would be different. We did the previous work in alot of detail this can be condensed now that we know all the formula. Treating it as an arithmetic series can deal with all layers even the first layer.

So for say half a mile (31680 inches) of wire wound on the spool this would be the sort of minimum calculations.....

(Spool length) / (wire dia.) = 5.5 /0.02 = 275 revolution per layer.
(275 revs) / (25 revs per second) = 11 seconds to wind 1 layer (any layer).
Length of first layer = 3.14 x 0.75 x 275 = 647.625 inches (this is the first term of the series)
The increase in diameter of spool+wire will still be (2 x 0.04) = 0.04 inches per layer added.
And when multiplied by 3.14 x 375 gives 0.04 x 3.14 x 275 = 34.54 inches of wire added per layer (this is the common difference of the series).
The series terms will still be the same and the formula for the Sum of the layers will still be the same....but the Sum now equals 31680...

S = (L/2)((2 x 647.625)+(34.54(L-1))) ......Standard formula for Sum of n terms in a series.....

31680 = (L/2)((1295.25 + 34.54L - 34.54))

31680 = 647.625L +17.27L^2 - 17.27L

31680 = 630.355L + 17.27L^2

solve this quadratic for "L" gives.......we are wanting to find the number of layers "L" in half a mile...

L = 28.3 or -64.8 .....28.3 is the +ve answer....take the +ve answer...

So we will take 29 layers to complete the last layer started ... For 1/2 mile of wire.....

29 x 275 = 7975 revolutions required to wind the wire on

29 x 11 = 319 seconds = 5.32 minutes to wind it on.

Finished Wound coil outer diameter = ((29 x 0.04) + 0.75) = 1.91 inches.

So you can see that because the diameter is getting bigger each layer added it takes approx. 30 layers to wind the first 1/2 a mile of wire and only approx 15 more layers to wind the next 1/2 mile of wire.

The diameter is 1.91 inches for 1/2 a mile and only approx 0.6 inches bigger when the next 1/2 mile of wire is added.

We only need the quadratic because we don't know the number of layers and need to find it.

If we knew the number of layers or the number of turns and wanted to find the total wire used we would just simply have to work out from the Standard formula for the Sum and get a value for "S".