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!
Welcome to the forum. I am not sure what you need but what is the height?
In mathematics, you don't understand things. You just get used to them.
Welcome to the forum.
It sounds like your pots are an upside down version of my diagram below. (Sorry, couldn't find a picture the right way up. )
The variables in this are:
R1 the radius of the large circle.
R2 the radius of the small circle.
h the height (distance) between the top and the bottom.
s is the slant length.
If you only know h and not s, you will need trig. It will be easier to measure s, and work with this.
To make the label you need a sector of a circle radius L. The top part is cut away, leaving the wrap around area which will fit your pot. Note: if you want some overlap for sticking the wrap around to itself you will need to add a bit to the angle calculated below.
The distance from the centre of this circle to the small arc ( 2πR2 ) will be L - s
Calculate the two radius lengths ( R1 and R2 ) by halving the diameters.
Calculate L as follows:
So that tells you how big the radius of the sector has to be.
Next, what angle should the sector be?
Does this make sense to you? Post back if you want more clarification.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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.
Once you've studied my previous post, you are welcome to post back questions / requests for more explanation etc.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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?
Last edited by vectorized (2012-03-13 02:02:07)
Let's see if I can take you through this one step at a time.
Step 1. What shape do you need to make the label?
Have a look at my diagram below. It shows a large circle with a smaller one inside. And I've drawn two straight lines out from the centre. Two arcs plus the two lines make a sector of the circle. I've coloured the sector using red and pink. Try this with a bit of paper. Don't worry about sizes at the moment. Any size will do.
Cut out the sector (both bits, still joined). Fold it round until the straight lines meet. You should have a cone.
Cut off the pink bit. Re-fold. You should have a cone with the top cut off. (technical name: frustrum)
That's the shape you need.
Step 2. How big must the circle be to make the frustrum?
Now I'm assuming you will know the radius of the top and the bottom. Your values in the last post are correct.
You also need to know how far along the paper it is from the top circle to the bottom circle. This is the slant height along the pot. It is not the same as the vertical height.
I've shown these measurements in the second diagram
Now the distance from the centre to the small circle (L-s) as a fraction of the distance to the big circle is
The small circle is the same fraction of the large circle:
So these fractions are the same.
If that's starting to get confusing, try it with some numbers.
Suppose the small circle is 1/3 of the large circle.
But the distance L-s has to be one third of L. So
Let's say s = 10 inches.
If L-s is one third of L, then s must be two thirds of L. So
check: If L = 15 and s = 10 then L-s = 5 and 5 is one third of 15 tick.
So the formula for calculating L is
Now I've done some algebra to get L = something in this equation. Maybe that's too much for you so here it is with those numbers:
step 3. So what fraction of the whole circle do we need to make this frustrum?
The arc at the bottom is 2 x pi x R1
The whole circle is 2 x pi x L
So the angle is this fraction of a whole circle:
Now I multiplied that fraction by 360 degrees. If your software wants a percentage then I think you need:
Hope that does it.
I think I should be able to work out this formula from here... I may need a bit of assistance from a friend that knows a bit of trig, but this should do it! Thanks so much!
So much for that! We can't figure it out... Can you give me the solution in the numbers that I provided? I think if I have that, I might be able to figure out the next one on my own. Another figure you might need is that the label will be .75 from the top of the pot (which is 9.125 high), and the label is 7.375 tall. I need the percentage of arc and the slant. Sorry to be a pest!
Another figure you might need is that the label will be .75 from the top of the pot
Oh boy that changes everything! The radius for the label isn't 5.0625 at all because your label isn't at the top! And if I start with a pot height of 9.125 and take off 0.75 that leaves 8.375. But you say the label is 7.375. So is there an inch below the label at the bottom?
I can use proportion to calculate the right radiuses (radii) but I need to be certain the label is 1 inch from the bottom.
Once I'm sure of the right radiuses I can work out s, L and the percentage. Do you have Excel or some similar spreadsheet program. Because I could give you the formulas in that format and then you've got it sorted for any other size pot. Just plug in the figures and it'll pop up with the answers.
Yes. The label is 1 inch from the bottom. I noticed, too, that when I create an arc in Illustrator, the ends jut out from the top and it makes it overwrap the pot. I need it to just barely overlap the ends as it wraps. If I try to squeeze it in from the sides, then my artwork distorts. I'm sure I can figure that on my end once I can get the formula down. Yes, I have access to Excel, although I don't know much about it since I work mostly with Adobe Creative Suite programs, being the artsy type.
I'm trying to include a screen shot of the tool I have to work with in Illustrator, but I can't seem to make it work on this forum. My IQ is dropping as we speak. Help?
For screen shots in a Windows operating system I do 'prt sc' for print screen, paste the result into Paint, cut out the bit I want, re-paste into a new Paint screen and save as a .GIF.
When you want to reply with a picture, make sure you click bottom right on 'post reply' not the 'quick reply' option as this doesn't have an Image Upload option.
I'll start work on my Excel sheet.
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!
Tried again to upload screen shot jpg and it's not showing in preview.
Oh I see now! I have to have at least 10 posts to be able to upload an image with my reply. This makes #10. I'll try in my next post!
Never mind. I just tried again. It's just not working. I go to image upload, select one slot for image, select the image on my hard drive, click preview and nothing.
The preview option doesn't show the images. I don't know why? Then it forgets them when you submit.
So try an image upload and submit without a preview.
Anyway, I've been busy and made a spreadsheet. The screen shot of my figures is shown below (hopefully, if it works for me!!!)
The bordered boxes are the ones you would enter values. The rest get worked out by my formulas.
Using your diameters 10.125 and 8.5, the height of the pot 9.125, the distance down from top to label 0.75 and distance up from bottom to label 1.0 the calculated values are L = 46.1337 and percentage = 10.9735
There are plenty of places where I might have slipped up so I cannot guarantee it is bug free. What I suggest is for you to try these values, print off a trial and see if it fits. Let me know and I'll either whoop with amazement or drop to the floor sobbing, depending on your answer.
If it does work then I can do a second screen shot showing the formulas, which you will then have to copy carefully into your own sheet.
It's gone midnight now in the UK so I'm off to bed. I'll check in the morning in about 8 hours.
Thanks so much! Will try this today.
How funny... worked that time! Kinda wish I had a CAD program now...
OK I can see your picture with 12%.
But I wonder what it is 12% of ???
I'll research Illustrator and see if i can find out.
LATER EDIT: As far as I can tell that warp feature does not allow you to control the shape in the way you want. The top and bottom curves must be arcs of a circle and concentric for the label to fit. Adobe don't tell you 'exactly' what warp does, but the curves may not be parts of a circle.
You can draw circles (ie radius L) and position straight lines to cut out a sector. Then you'd have to edit the shape outline to limit it to the sector you want. I've never used Illustrator, but that's how I'd do it.