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OK I can see your picture with 12%.
How funny... worked that time! Kinda wish I had a CAD program now...
Thanks so much! Will try this today.
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.
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!
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'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?
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.
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?
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!
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!
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.