Looking for a specific calculation.

Being that I'm in need of some penetrating oil on the old noggin to crank out computations of old, I think this is a great opportunity for one or more of the engineering students in the CR4 community to step up and give some input, for a change.

I believe the simplest formula lies in the realm of calculus, but I am wide open to be corrected on that.

You know you may be right as I remember the calculation was quite a long string.

The material may stretch and may not be wound uniformly the calculated length therefore may not be exact.

The length can be estimated

The number of layers can be calculated as - (n-a)/b

the average length per layer can be calculated as = (a + (n-a)/2) * Pi.

The total length then is = avg length * number.

Correction : the number of layers should be (n-a)/2b (working with diameters and not radius)

Averaging is good and the formula does work but i know the radius changes by the thickness of the material each time the roll is wrapped

As there can be stretching / bending issues: what about determining the weight of 1 meter (or any other unit) of this material?

Total weight / weight of the unit you've measured = total number of units you have in the roll.

Should the roll have a plastic / cardboard core, it's weight must be discounted from the total.

Simple... ain't it?

for the number of feet in a belt roll [tightly wound on core]

A= diameter of coil in inches + diameter of coil hole in inches

belt length in feet = (A) (# of coils)(0.131)

CR3

This is a straight area calculation. The area of the roll is equal to the area of the sheet.

Where R is the roll outer radius, r is the roll inner radius and t is the sheet thickness, solve for L total length.

PI(R^2-r^2) = t x L

L= PI(R^2 - r^2)/t

OK i think we're geting close hereb but there is still something not right, so lets say the problem goes as such:

The core diameter of a roll of material is 4" OD, the od of the finished roll is 30" OD

the thickness of the material wrapping from the core OD to the roll OD is .012" per layer.

Solve to find the linear footage of material wrapped on the roll.

Am I beginning to sound like a math teacher? ........... Geeze I hope not!!

Ouch, makes my head hurt.

It's back to school for many of us. Start here.

Found that thru this site which I now believe holds the formula. (I'm too lazy to search)

In any case, I am sure that it will be very, very long.

4819.0277 feet.

please elaborate and show the actual calulation, as I am not coming up with the same number for some reason.

ok whats missing her is the answer this formula provides leaves you with the measurement in inches. What I was missing is the conversion of inches to feet.

Got it now

Least I think I do

The formula I gave you is accurate it is from machinists ref.

let D=30 d=4 T=.012

Formula [((D-d)÷2)+d] * Pi * [(D-d)÷T] = total length

[((30-4)÷2)+4] * Pi * [(30-4)÷.012] = 115715.3294

I'm taking the average dia. Calculaing the circumference and multiplying by the number of coils in the roll. (30 -4 ÷ .012)

It worked on the examples I used prior to this posting.

The problem I keep having using the averaging method is that it is off of the known length of a given roll by approxomately 500 ft.

Accuracy is the key on making it work every time. I have to be able to account for as close to the exact footage as possible as the material is very expensive.

The formula will work and be accurate, assuming the material is wraped tight and does not stretch. average is a poor choice of terminology.

Perhaps the approach of weighing the material would be best.

Good luck

so what is it you want? Linear feet or feet² ?

Linear feet. What I am going for is to be able to know the length of material left on a roll of a given diameter based on the above known figures.

By doing this i will know in feet how much of this material i have in stock.

It's for our material handler to take inventory of stock on hand.

I am making a spreadshet for him to use as a quick refrence.

Hi, Its about 50+ years ago now, but I make your
example roll to be 1,606 yards. (4820 ft. 57843 inch)

Please correct me if I'm wrong.
(the calc. can be amended to suit your spread sheet)

From 4"d to 30"D the material radius is 13 inches (30 - 4)/2.
with a thickness 0.012 - equals 1083 layers. (13 / 0.012)

From 4"d the first layer is 12.56 inches (pi x d)
and at 30" the last layer is 94.26 inches. (pi x D)
The average layer is 53.41 in. (94.26 + 12.56 / 2)

Therefore: the total length is 53.41 x 1083 layers = 57,843 inches.

Approximately 1,606 yards.

From this one can create a spread sheet to calc. most rolls.

Hope this helps.

jt.

http://www.fastfreeads.com

Thanks jt, it looks like your agreeing with hendrik on using an averaging setup. I was hoping to gain a bit more accurate by creating a formula which will factor in each layer as the calculation is done. But this is definately a start and getting very close to actual measurements provided on the rolls i am currently receiving.

Hi again doublej_b,

Weight is a good way forward, in that if you can pre-weight
a then known length (after use) this can be a good estimate
for all future rolls. Providing:
a) the protective wrapping is the same, and
b) the material does not vary with moisture. eg. paper rolls.

Weight can be a very good indication; not so accurate as
measurement; with a calc. you can be surprisingly accurate
- to within a few feet, or even inches; which is not a bad guess
in 1,000's of yards.

I have always admired the ancients for achieving feats of
measurement, even for the planets, with remarkable accuracy.

Best regards

jt.

http://www.fastfreeads.com

Thanks for the input JT.

Weighing is not an oprion here although it would make it very close every time.

Problem is i don't have a scale capable of weighing something that heavy.

The rolls are over 300 lbs and the best i can do is a bathroom scale and it is way to small.

I know there is a proper formula that will take into acount each and ever lap in the roll and give a length based on that but I can't for the life of me remember what it is.

Post #7 got it right, but changed the measurements - and all omitted some possibly helpful additional information.

To get decent accuracy you will need to measure effective thickness of the material while it is on the roll, rather than relying on nominal thickness or measuring it when it is off the roll - suggestion below. Aside from that we just need to calculate the cross sectional area and divide it by the material thickness:

Just to be certain I'm not causing confusion, I'll redefine the terms given in the question as I understood them:
n = outside diameter of material on roll, a = inside diameter of material on roll (or outside diameter of core), b = material thickness when wound on the roll.

The cross-sectional area of the part of the roll where you have the material is
pi . (n/2)² - pi . (a/2)²

So the length of the material is ∏ /4 . (n² - a²) / b

To get reasonable accuracy, it is best to measure the thickness of the material as wound on the roll. Just (?) count the number of layers = M. The wound thickness is then (n-a)/2/M. [This theory as presented here doesn't account for partial layers properly, because it assumes uniform radii; however, you will still get the best practical results by using the average measured diameters in different directions and accounting for part-filled layers.]

So, if you can manage to count the number of layers, the length of the material will be very close to:
∏ . M . (n + a) / 2

You may find it surprising that the length seems to increase when you increase the diameter of the core - but this is correct because the number of layers will decrease faster than (n+a) increases.

Hope this is what you needed.

Fyz

Double JB:

Have you checked with your material supplier? Surely this is a problem they have encountered before with other customers. And for many materials stretch is a huge issue, so the supplier would know how much tension is normally applied and what the stretch factor would be.

I suggest you start there and not mess with theoretical stuff that could be way off.

I agree - if that solution is available.

Fyz

Unfortunately I have checked with several material suppliers and none of them have the answer. They have counters on their machines and that is the number they go by.

I guess the easiest way to be accurate would be to have our machine operators subtract the used feet from the rolls and write it on the roll when they are done.

But just try to get people to do that everytime. Or any time for that matter.

Oh and don't forget the deer in the headlights look you get when you give them one more new thing to do.

It's so hard to get good help now days.

I'm afraid you'll probably just have to go with 23, then. It may be worth additionally sample-checking the stretch/shrinkage as you unwind at different points of a typical roll if accuracy is important.

Fyz

Thanks I appreciate the help.

I will give it a shot as it seems more complete and comes closest to give me the best number or closest number.

"I guess the easiest way to be accurate would be to have our machine operators subtract the used feet from the rolls and write it on the roll when they are done."

Last time I bought curtain material, this was how they kept tabs on their rolls, so it can be done...but I know what you mean about the DITH look.

Good luck

Double JB,

You never told us what's on the rolls!

I had a similar problem with woodgrain material we were slitting, with accounting wanting to know how much of the material remained on each roll when a portion was removed and slit into narrower strips and rewound. But wait it gets worse! They also wanted to tie this into the TIME it took to do the slitting so they could apply a labor cost to the slit material! And, not all materials would run at the same speed!

The machine was old and antiquated, a real home-brew job. But we were able to add a fixed diameter pinch roller with a counter that would tell us exactly how many linear feet (revolutions counted x circumference of roller) were removed. Keeping good records we got good at correlating linear feet to weight of the original rolls and converting $/pound to $/ft. Then it was a simple matter of noting the count and speed each time they changed speed and/or materials, and marking the reels with the amounts removed (purchased material) or wound (slit material). Then runtime could be calculated for each run and set-up time was added on a standard basis, with occasional time study to recheck the standard, which relieved the operator of the chore and onus of recording their own set-up time.

Hope this helps.

What we produce here is decals.

The materials we use vary from polyester mylars to vinyls, included in that are some paper rolls. Some have adhesives and some do not. Thicknesses range from .006 - .015 on rolls. All are laminated to a removable paper liner which makes for a pretty stable product as far as stretching goes.

When multiple rolls are measures for diameter and footage as listed by the manufacturer, they all are within about .010 diameter and footage is listed as being the same.

The numbers I have been giving previously are just base numbers to calculate from.

When measured with calipers the thickness of the material is .012 and the core outer dimension is 3.75. The outside roll diameter is 26.5 inches and the footage is 4200 ft.

I know these numbers are different from the original given numbers but I figured they would be easier to work with.

The problem occuring when using an averaging program is when the OD changes to half that of the original with the id remaining the end result becomes a bit more skewed.

Now I know that there will not be half the amount of material on half the roll because of the ever increasing diameter. I also know that about 2/3 of the roll makes up for about half the footage. (I have tested footage calcs by using a slitter that can measure footage) Problem is we can't use this for every roll as it is a production machine.

So for the bean counters to get what they want and for inventories to be accurate. I need a calc to be as accurate as possible. I remember from my math days in HS that there is a calculation to figure the length of a spiral and looked online for that, but it is still an averaging program.

What would like is something would work best in an excel program so the computer can make the necessary adjustments to come up with an accurate length. So i have the calculating power of a computer to do this with but not the formulation to make it work.

After all is said and done it will be put on the server and available for the use of both accounting and warehousing to maintain an extremely accurate inventory of roll materials.

I'm still trying to figure out the best means for sheet stock but measuring is pretty acurate for it right now.

So there is my delimma in full. Leave it to the mechanic to find the right way to fix everyone's problems.

my local hardware store has a counter fixed to the spool with the material being fed through a counter. It is not after market. It is manufactured specifically for apps such as yours. It is along the lines of what our compadre from St. Louie was describing.

Thanks but unfortunately we don't have the means to recount rolls as there ar many sitting on shelves right now that have not bee acounted for and no means of running them and rewinding them except by using production equipment.

In other words no offline machines to use for this.

I know......I can sure make things difficult.

LOL that what I'm here for.

I would seriously weigh the benefits of purchasing a scale. If they are so big they are fork trucked there are scales for that too. Simply rig a scale (like a fish scale) and voila.

If the cost is an issue they can be purchased surplus. I see the problem as the mat is so thin. You might talk to the manufacture about edge marking. They probably have machin counters. perhaps you could pay a dollar or two per roll to get an edge mark.?.

CR3

Now there is an idea might be just the solution. Well future solution at least.

I will have to give them a call and see if they have that capability.

Still have the rolls in stock to contend with though and it looks like STL might be on to something a bit different.

OK, you don't like the direct measure approach. How about this:

It seems like a simple calculus problem.

Let's say X is the variable length you want measured, C is the current circumference (a variable), D is the current diameter (also a variable), and T is the thickness of the material (a constant), and L is the number of layers on the roll (assuming for the moment a hubless role).

These formulas apply: C = pi x D , L = 1/2 x D/T, and ΔX = _LΣ⁰ C

By algebra we get ΔX = _LΣ⁰ pi x D AND D=2TL SO ΔX = _LΣ⁰ pi x 2TL

Since pi, 2, and T are all constants, ΔX = pi x 2T x _LΣ⁰ L

Putting this into the infinitesimal(?) form where dX/dL = pi x 2T x L (from L= L to 0) then:

dx = 2T(pi) x L dL and by integration

X = 2T(pi) x L²/2 = pi T L² and since L=D/(2T) and L²= D²/(4T²), then

X = pi D²/(4T), where X is the amount left on the roll for a given D.

but this is for the hubless roll so you must subtract an amount equivalent to the hub diameter.

If all you are interested in is the change then the two hub diameter terms will drop out and simply find the ΔX for the ΔD. To determine the amount left on the roll though you will need to subtract the hub equivalent from X for the final roll diameter.

I hope this make sense. My calculus is a little rusty. If anyone sees where I have gone astray, please correct me!

Can't see anything wrong with the calculus, but can get same formula X = pi D²/(4T) directly by considering the area of the end of the roll, as pointed out by Guest in #7.

Cheers....Codey

"considering the area of the end of the roll"

The "area of the end of the roll" would be the thickness of the sheet multiplied by the width of the roll".

I guess you meant to say "considering the cross-sectional area of the entire roll which is normal to its axis", right?

Well, certainly, if you want to do it the easy way!

That's right, to be quite precise. But this area is only theoretical, of course. To find the length of cloth, there's no need to take any measurements of the roll end, so the end could be cut at an angle without affecting anything - apart that is from the shape of the cloth when unrolled!

"To find the length of cloth, there's no need to take any measurements of the roll end"

Yes, Codey, that is why I was confused by your post #51 where you said that the equation could be found directly by "considering the area of the end of the roll".

What possessed you to say that?

Aaaaah, the light just went one! When you said "roll end", you were thinking of the end of the cylinder formed by the roll (what I would call the "long edge" of the material), but I thought you meant the end (or short edge) of the material as it was wound onto the roll (or the beginning of the material as you unwind it, depending on how you look at it). Thus my confusion.

"Roll end" was a bit ambiguous.

I guess I decided to find the equation by calculus more as an exercise and to show how equations may be derived even if there is no easy geometric solution, as is the case here. Also, 2JB seemed a bit doubtful as to the legitimacy of the equations given in #7, so basically, I decided to seek an alternative approach, albeit one that ended up with basically the same answer.

OK STL Engineer, I see the ambiguity now. Just shows how easily misunderstandings can arise!

Codey

"The core diameter of a roll of material is 4" OD, the od of the finished roll is 30" OD

the thickness of the material wrapping from the core OD to the roll OD is .012" per layer.

Solve to find the linear footage of material wrapped on the roll."

SANITY CHECK for my solution in #36 above:

X = pi D²/(4T), where X is the amount left on the roll for a given D.

So linear footage (inchage?) of material wrapped on the roll is:

X = pi/(4 x .012) x (30² - 4²) = 57,857.66 inches = 4,821.5 ft.

Am I close?

That about as close as i have seen try using these figures and see how close you come to 4200ft.

Roll id=3.75 roll od=26.5 material thickness=.012

three rolls same dimensions same measured footage. As measured and taggen by the material manufacturer.

Well, by my calculations, loosely rolled, no tension/no stretch (not real world of course), the length would be 3,753.5 ft.

Interestingly, if tension was applied and stretch made thickness decrease to .011 inch the total length would be 4094.7 ft.

To make approx. 4200 ft. then the thickness would be 0.0107, not unrealistic I believe. It would also be interesting to know how that 4200 was measured and if they are selling it by the unstretched foot, but measuring it under tension, they are cheating your company because you are paying for more, but actually getting less footage when tension is released, assuming the material is elastic (does not deform permanently).

Now if you tell me that the .012 is the actual thickness while under tension and not after it has been cut off the roll, I must assume there is a problem with one of the other dimensions not being quite right, or else their measurement is inaccurate. How often do you check that you are getting the length they say they are sending you?

I understand that you see at least as much thickness-compression as stretching with fibre-based product - the pressure on the inner layers (due to tension in each successive layer) becomes substantial. Also, depending on the roughness of the surfaces, measurement made with uniform callipers (or micrometers) will give a thickness that is greater than you would see with stacked layers. 11% compression may sound rather high, but is not that implausible.

But I suspect the length of interest is the unwound, non-tensioned length. This could also be slightly smaller (2-3%?) than the manufacturers stated length - which is probably measured at the tension under which the roll is wound. (I imagine that the only practical way to avoid this would be for the manufacturers to measure the difference in lengths between the material under tension and as laid-out sheet. Unless there is firm demand for this, I can't see why they would do this. But you could do it yourself if you can find out the winding tension that the manufacturer uses.

As ever, once you need to include "real-world" effects, you have to go further than pure geometric theory will allow.

Fyz

Ok. Now that I am no longer relying on the information given to me by others, I have learned 2 things.

1. If you want accurate information do not rely on others to give it to you.

2. Listen to yourself when something doesn't seem right.

I must apologize to all of you that have been so helpful in here.

It appears that the information I was receiving from others was wrong and only after deciding that things were just not adding up as they should be i went over to our warehouse location and got the information needed for myself.

First off, the calipers they were using had not been properly zeroed.

Second off the person using them did not know how to read them properly.

Having taken my own pair of certified calipers over there and comparing readings with what he was taking I have discovered that the material in question was only .0105 thick.

Which is why all the discrepancies kept appearing.

Having spent about an hour now giving lessons on how to properly zero a set of calipers and explaining how to read them I think I might in the future get better information.

So once again thank you all for your assistance. All solutions involved now will work to some extent but the closest solution to hitting the mark is STL who is only off by 89ft

I would prefer to be at 2-3 ft but this is way better that 500ft

Thank you again everyone.

(Gawd I feel like such an idiot)

Are you 90' short? maybe your supplier is.....nah, nevermind.

First, don't feel foolish - we've all been there.

Based on these new numbers, STL's calcs (same as my formula) give material that is 89.7 feet longer than the manufacturer's figure. There could be all sorts of reasons for this. It may be that this particular material does not pack as tightly as you might like, or that the manufacturer has compensated for pre-stretch values, but the material doesn't fully recover under your conditions (these things can depend on humidity and temperature). It could be that inner layers and outer layers have different thicknesses. If you are receiving 2% longer material than you expect, that should be a useful margin. In the absence of better data, I would work with the manufacturer's figures and base my calculations on an effective thickness of 0.0107" - it is usually better to find you re-ordered a little early than to run out of material. In any event, your error will be smaller for the part-used rolls, which I imagine are the ones that are most "controversial". A final note of scepticism - in case you haven't already, you might check the inner and outer* diameters yourself - maybe there are protective layers included, or the outer diameter is the maximum allowable to allow for batch-to-batch variation? Maybe...

*1% error on outer diameter is more than 2% error in total material...

Fyz

"...but the closest solution to hitting the mark is STL who is only off by 89ft

I would prefer to be at 2-3 ft but this is way better that 500ft"

Are you sure about that? By my calculation, .0105 thickness (with 26.5 material OD and 3.75in. hub diameter) yields about 4289.7ft., so call it 89 or 90ft over. However, it is also common practice for material suppliers to adjust their equipment so that the customer is never shorted and that any error will be on the Plus side, and there is always error, don't kid yourself. Here we are only talking about 2% or so. That may be what is actually happening. Or there may be some other slight error, but I believe the formula is still theoretically correct.

Adjusting the error so that the customer NEVER gets shorted is good business, and the little loss the supplier suffers by sending a little too much is well worth the cost to avoid arguments with good customers, lot rejections, or even losing the business. Sounds like you got yourself a pretty good supplier if you are purchasing 4200 ft and he sends you 4290 ft.!

By the way, 3ft of error out of 4200 is 0.07%! Not 7%, not 0.7%, but .07%! Good luck getting that kind of measuring error out of standard industrial equipment! You are using digital calipers, which I assume are like mine and have an accuracy of only +/- .0005 so if your error is .0005 (I will allow you the other half!) that changes the calculation by 214 ft.! You have got almost 5% error right there!

DON'T PRESS YOUR LUCK!

You are right - I just assumed that 2.jb would be making measurements with a stack that is at least ten layers thick to get the thickness he reported.

In light of my last post here are some suggestions to increase your accuracy.

Switch from digital clipers to an actual micrometer (digital or analog) that will take your accuracy down to +/- .0001 instead of .0005, then take several readings on each reel, at least at the start, middle and end of each reel, and in several points within each section. The more data points you get the more accurate your data will be. Then you can easily find the mean (average) and standard deviation so you may actually have a better idea on the range of your error! You may then find that the calculation gives you very accurate results. If this is critical to your operation, look into using "X-bar and R" or other SPC charts to monitor quality.

And like Fyz said, check if the OD and hubs are being measured with or without any liners that could throw off accuracy. There could even be variation in hub diameter from reel to reel. How accurate is that 3.75 or the 26.5 figure?

"...take several readings on each reel, at least at the start, middle and end of each reel, and in several points within each section. The more data points you get the more accurate your data will be."

Along that same vein, you might consider taking readings from both sides to make sure the cross-sectional thickness is consistent. You may not need to do that all the time, but it would seem to be most useful at the start and middle of the roll. This might even help with quality control of your product.

Blimey..I'm glad I'm not contributing here!...but may I just add a few pennyworth...

If STL's ? Fyz's solution is closer than you have been using...

Use that...but multiply in your empirical real life factor...that way you get the best of both worlds.

Good luck

Del

Well after a nice short busy weekend and a long friday conversation with the powers that bee they are all completely satisfied with getting this close.

Thanks to everyon for all your help.

I plan on checking with the Manufacturer today and see if they do tag on a bit after the fact just for good measure.

Thanks again.