Need formula for Spiral Orange Juice Can

[bsmith6356]

Hi there- I'm a graphic artist and work on a lot of "Spiral" cans. A good examples would be a Pillsbury Crescent roll can (after you whack it on the counter and unroll it). Or, a frozen Concentrate orange juice can unrolled. Anyway, I've created formulas in Excel to setup the page size that the template and graphics will be built on, but there is one measurement I can't come up with without manually measuring. I've attached an illustration.
Thank you!
Brent
Charlotte, NC

 

[BvU]

? Is two parts: 7.125 plus 12.45 times the cosine of 31.738 degrees. Be careful when entering this on a calculator. Excel works in radians, so use

= cos(31.738 / 180 * pi() )​

 

[bsmith6356]

Hi BvU-

I can’t thank you enough for your help on this! I didn’t realize you’d responded (and so quickly), the forum email got filtered into a different inbox. My numbers aren't coming out the same way as yours, and trying to figure out where I'm going wrong. I've attached 2 files, one that Illustrates the problem and also an Excel doc that shows the formula you suggested. Would you mind taking a look?

Thanks again,
Brent

 

[BvU]

Can't say I&Q is very legible; can distinguish question on two parts, though:
in mathematics (also in excel and on most calculators) , multiplication goes before addition and subtraction, so in

7.125 plus 12.45 times the cosine of 31.738 degrees

you calculate the cosine of 31.738 degrees: 0.85046 (*)
you multiply with 12.45

to get 10.588​

then you add 7.125

to get 17.713 thingies (inches, I believe).​

That the 7.125 is one term should be very obvious from your drawing ! No multiplying with anything, clearly.

(*) the cosine of 30 degrees is ##{1\over 2}\sqrt 3 = 0.8666, so you want to be close to that -- your 0.948 is the cosine of 31.738 radians = 1818.45 degrees; you don't want that...​

All this based on your diagram, designated 'Same template as image on the left, but cleaned up unnecessary information. This is how I have to layout the graphics on template'.

Time to scrutinize your 'cleaning up': your cleaned up picture shows 12.450 where the original picture on the left has a red line that is a lot longer.

Do you really mean to leave out the 'unvarnished width' and the 'Adj label bleed' ?

Scrutinizing the original picture is worth doing also: the 12.45 and the 6.861 give an angle of 28.858 degrees, not 31.738. So I wonder how that 'wind angle' is obtained.

Red lines and blue lines do not begin and end at the same points

And so on...

 

[bsmith6356]

Ugh, you’re right, I screwed up when creating the Illustrations. And yes, you’re also right that I shouldn’t be eliminating the unvarnished width and the Adjacent Label Bleed. I’d need to get the full “Sheet Height,” the magic number for which I need a formula.

I use the 31.738 wind angle the provided to calculate the rotation angle which orients the template the direction it will run through the press. In this case the rotation angle would be

-58.262 or (Wind Angle – 90)

The additional amount that needs to be added to the “Crossmatch” of 12.45 inches are the areas you pointed out (in orange on new PDF) and shown below, but I know they can’t be used as is…since they are horizontal measurements not taking into account the angle.

.188 (Eyetrack) + .062 (Adj Label Bleed LEFT) + .062 (Adj Label Bleed RIGHT)

Which throws a wrench in for me because I don’t have those measurements (angled).

I’ve been creating automated workflows to step files and add Printers’ marks, I’ll just have to put in a stopping point to manually rotate and measure the height. If you think of any ideas, let me know. Thanks again for all of your help!
Brent