Programming Challenge 24-1
I would assume that in real life [let's say] 5 layers of material would measure more than 100 mils -- there's got to be some little bit of air layer or non-sticking-together dusting or surface irregularities or other don't-quite-lie-precisely aspect. I assume that should be ignored, but maybe an extra mil or two per layer should be assumed.
My $0.05, ignores stuff like spiral step increase.
(defun JOHN:CH124 ( / ans rod tod thk rad num tot)
(if (not AH:getvalsm)(load "Multi Getvals.lsp"))
(setq ans (AH:getvalsm (list "Enter values" "Roll OD" 5 4 "10" "Thickness mil" 5 4 "20" "Tube OD" 5 4 "2")))
(setq rod (atof (car ans))
Thk (/ (atof (cadr ans)) 1000.)
tod (atof (caddr ans))
)
(cond
((> tod (- rod 0.125))
(progn (alert "Roll is to small for calulation \n\nWill now exit please run again")(exit))
)
((> tod rod)
(progn (alert "Inside roll tube is set too large bigger than roll \n\nWill now exit please run again")(exit))
)
((= thk 0.016)(princ))
((= thk 0.02)(princ))
((= thk 0.027)(princ))
(alert "You have entered a incorrect thickness 16 20 or 27 are valid ")
)
(setq num (fix (/ (/ (- rod tod) 2.0) thk)))
(setq rad (+ (/ tod 2.0) (/ thk 2.0)))
(setq tot (* pi (* 2.0 rad)))
(repeat (- num 1)
(setq rad (+ rad thk))
(setq tot (+ tot (* pi (* 2.0 rad))))
)
(alert (strcat "You have " (rtos (/ tot 12) 2 1) " feet left"))
(princ)
)
(JOHN:CH124)
11" dia 3.5" core 20 mil is this close ? You question was the length of the roll available.
@john.uhden wrote:
....
BTW #2, I doubt that you will develop the formula on your own. I couldn't.
Oh, I know the formula(s); it's just a matter of encoding them. The number of layers [assuming you can realistically measure the outside diameter of the roll to the nearest fiftieth of an inch with any precision, and the sheet material doesn't compress in the measurement, and so on] is easy enough, and the amount of additional sheet length for each successive layer. There's a quite short formula for how much additional roll length is to be added beyond just the tube circumference x the number of layers, based on the number of layers but not involving (repeat)ing to add some for each layer, successively more for each, one at a time. But I'll keep that to myself until I get to putting it all together.
To be even more unrealistically realistic to a greater precision than would ever be possible, the length of the first layer should be measured at* the mid-way line of the sheet thickness -- there would be slight compression inboard of that, and slight stretching outboard. [* The actual no-compression-or-stretch plane is really not exactly at the mid-thickness line, but that's at least much closer than the inside face, i.e. than the tube surface for the first layer. Getting it exactly I think would require specifics about the physical properties of the material.] And I would think a general calculation has to be based on the assumption that all layers go all the way around, and the last layer ends right over where the first one begins.
@john.uhden wrote:
If there isn't at least 3 feet more than what's needed for the next liner, then they'll send the roll back ....
Your BTW #2 implies there's something hard-to-derive about the calculations, which implies a certain level of precision that I assumed, which 3' of "slop" argues against, but overthinking or not, here's my approach.
The formula I alluded to before is in line 49. Think of the incremental increase in circumference per layer as billiard balls -- the cumulative total over the layers [beyond the first one] of one bit of extra to add for the second layer, two for the next, etc. The formula for the total of a "pyramidal" stack like that, with X rows, is X times {half-of-X-plus-one-half}. Just plug in X [X=5 gives you the 15 billiard balls], and no repeating X times, and adding up along the way, is needed. One line of code sums up the whole thing.
The code assumes the 3 standard thicknesses you list are all that are allowed, but it could be made to allow additional values instead of limiting it to those 3. It offers 20-mil as initial default. It offers 3.5" as the tube initial default diameter, though for that it allows any positive size input. It forbids invalid [including negative] inputs for tube diameter and outside roll diameter in relation to it.
And in case one enters a roll diameter that is not a multiple of sheet thickness beyond the tube diameter [as is the case with your 11" vs. 3.5", which at 20 mils comes to 187.5 layers], it (fix)es the result, since you can't have a half-layer in terms of thickness. It could instead round it to the nearest value up or down, instead of (fix)ing [rounding downward only].
Using an online archimedies spiral calculator, this is what I get:
Of course, the O & I diameters would need to be measured with a micrometer for the answer to be precise and I would assume that given that, there would be a minimum error of 2*pi*11/12 feet. Also, stretch couldn't be calcuated without more information such as the initial stretch from the factory, the stress in place, and the modulus of elasticity I believe. That doesn't even consider the fact that if it is stretched from the factory, then each layer would be slightly less thick than nominal.
@dbroad wrote:
Using an online archimedies spiral calculator, this is what I get: .... the O & I diameters would need to be measured with a micrometer for the answer to be precise....
Oh, so much stuff we can get over-picky about, but I find it fun to consider the details, however unrealistic. This is why @Sea-Haven in Message 4 said "ignores stuff like spiral step increase."
The spiral length can't be precise because of course the wrapping doesn't really follow a spiral path:
The true shape of the stepping would be affected by the flexibility of the material, and by how tight it's pulled in the winding. And it would gradually broaden and smooth out as the number of layers increases.
If one actually had a need to account for such things in detail, it's hard to imagine how they could be calculated.
