Maybe a trig question

The best thing to begin with is a consideration of end behavior. consider the start of your ribbon to be negative infinity, and the center to be positive infinity. At negative infinity, the value of our function is zero, and at positive infinity it's the width of the wound ribbon, which I'll assign a value of 2 since it'll make things come out nicer.
Since we know it has to be a trigonometric function, we can start narrowing our list of functions from all the trigonometric functions to trigonometric functions that have limits as x goes to positive/negative infinities. this rules out sine, cosine, tangent, secant, cosecant, cotangent, arcsine and arccosine. I'm sure that there are a lot more obscure trig functions for us to eliminate if we wanted to be thorough, but I think it's pretty safe to say that the only trig function that is continuous over a domain from negative to positive infinity, and has an upper and lower limit over its entire domain is arctangent.
In fact, if you take the function arctangent+1, it seems to exhibit the exact behavior we want. It's 0 at negative infinity, the outside of our sphere, and 2 at positive infinity, the center of our sphere.

I'll be working on a calculus proof of this, but that'll take a while longer

Astonishing!
Wish I could say I wish I thought of that but that is monumentaly unlikely what with me all bound in this vocabulary thing. Now all I have to do is try to get Excel to plot an arctangent for me to use in my ribbon cutting.

What I really nead is a form, say a yard long [sorry] a meter long. The bottom edge is straight and the top starts at .25 inches high and gets to 1 inch high by the end. And the top edge follows this arctangent function.

I can just stretch and print a picture of this to the proper scale if I had one, but where is there an example of this curve out there?

I'll keep looking.

Thanks Very much for your help.

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that's arctan right there. I messed up btw, I thought its range was -1 to 1, but the picture reminded me that it's -pi/2 to positive pi/2.

Regardless of what those values are, you'll have to try a couple times to get the scale just right.

A better function for it then would be arctan(x) +pi/2, giving you a width of pi.

Turns out I'm completely wrong. It isn't a trig function, but it certainly involves trig. Three dimensional trig actually.

Imagine a Hemisphere sitting on top of a coordinate grid. Now imagine a spiral cutting outwards so that after each pass, it's the x units farther away from the center than it was before. Imagine the set of points above that line. The distance from the spiral to the hemisphere is our y.
The spiral is defined by the function r=m*theta, where m is the width and theta is the angle measure.
Now the question is though, how high is the function at any point along that spiral? Well, since the distance from the origin is r, we could say that y=sqrt(z^2-r^2) where z is the radius of your hemisphere, but this wouldn't define the function quit the way we want, since I think we're looking to define it in terms of distance along the spiral. This question is very different, and will require some more thinking.
Hopefully somebody else comes along with a more elegant solution to help you out.

The trick is that the periodicity is reduced by the thickness of the ribbon, say eta. As the points move from the outside to the center of the spiral the length of the tape is reducedby pi*r where r is shrinking. Calculus, anyone? Eventually the radius is eta minus the entire radius of the sphere outside.

ah, I was assuming that since you said thin paper, you mean paper of thickness zero. lemme rework it.

Paper or ribbon or whatever does not matter (I think). The mathematical answer will be extended to whatever material is at hand when I get to engineering. For now the infintesimal will do nicely, or the limit could bottom out at a few mils. Just a function to map the hight, y (top edge) of the ribbon as it slopes from the equator to the pole.

Thanks.

Hm. I am not sure there is any trigonometry involved here.

Assuming a tightly wound spiral of thin ribbon, then the length L of ribbon needed to wind up a spiral of radius R should be something like (correct me if I'm wrong)
L=A*R^2 => R^2=L/A,
with A a constant depending on the thickness of the ribbon.
The height h of the hemispherical dome only depends on the distance from the center R as
h=sqrt(r^2-R^2),
with r the radius of the finished dome.
Insert R^2=L/A:

h=sqrt(r^2-L/A).

This is your function h(L). You can determine A in the following way:
1.) Determine r by rolling your ribbon together as tight as you can and measuring the radius.
2.) Measure the unrolled length X of your ribbon.
At the equator, h=0, therefore r^2=X/A
=> A=X/r^2.

(I did not try this out. Please tell me if it works.)