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How funny... worked that time! Kinda wish I had a CAD program now...

Thanks so much! Will try this today.

I have the screen shot (I'm on a Mac), but I can't get it to post here in my reply. I did as you suggest, but it won't show...

I do appreciate the help SO much! Hey if you need any design work like business cards or invitations to something or whatever, let me know! THAT I can do!

PS

Tried again to upload screen shot jpg and it's not showing in preview.

Bob - I thought I was smart - must be in other subjects. *sigh* I barely even made it through Algebra in high school, but #1 that was about 35 years ago, and #2 I thought it was because I didn't do my homework. Math for the most part is completely over my head. I did try writing down the formula on paper and made notes, to no avail. I hate to ask you to just tell me what it is, but at this point I have to. I'm kind of under the gun on this project, and they're in a hurry, so I need to get a sample of at least one completed today. If you could give me the numbers on this according to my measurements, I'm sure I can follow suit and determine the rest of pots of different sizes. Would you be willing? I'm an artist and make things look pretty. The synapses on the other side of my brain misfire when it comes to math! LOL

I think I got as far as determining R1=5.0625 (since the diameter of the large circle is 10.125) and R2=4.25 (since the diameter of the bottom of the pot is 8.5). Good so far, right? Then I divide the two? I get .84 (rounded off). What do I then do with that figure? I'm not sure exactly how to arrive at L. I know, I know, I didn't get very far.

Oh - the total height of the pot is 9.125, but the area the label will wrap around is 7.375

My other problem I just noticed is in order to create this "warp" or "arc" of the label in Illustrator, it only allows me to punch in a percentage, not a degree. Is that a problem?

Thanks for your responses! Bob, I'm not sure I understand any of that since I never took trig, but I have a fairly decent IQ so perhaps I can figure it out - I'd just need a bit of time. But yes, the diagram you show is what I have but flipped over. When the wrap is flat, it will have a curve and the outer end points will have a slant, which is what I'm trying to find. In Illustrator, I can use the warp tool to create and arc of a specified percentage, so I figure I can use that once I determine the correct formula to do so.

bobbym: The height of the pot itself is 9.125 but the height of the label will be 7.375 inches.

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Let me start by saying I know absolutely nothing about trig, which I assume I would have to know to figure this out. I'm designing labels that wrap around a flower pot, but cannot figure out how to arrive at what percentage my warp or arc has to be. The pot is 31.5 inches at the top, 27.75 inches at the bottom (circumferences), and the label area will be 7.375 inches high. I have several other pots of varying sizes I'll need to create label wraps for. Is there a particular formula I could use to figure this out? Not sure if you need this, but the opening at the top is 10 inches, and the width across the bottom is 8.5 inches. Thanks so much!

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