I assume that 230 GSM means grams per square metre? A reel of paper is effectively a spiral, for which there is a mathematical formula: r=R+At, where R is the inner radius of the reel of paper and t is an angle measured in radians. A radian is approximately 57 degrees and 360 degrees is 2(pi) radians. In other words, to convert degrees to radians we multiply degrees by the quantity (pi)/180. The mathematical number pi is about 3.142, the ratio of the circumference of a circle to its diameter.
If we draw a line length R, the inner radius, this is where the paper starts from the middle. This is the starting point and angle t=0. When t=2(pi) we have one wrapping of paper and the length of paper is 2(pi)R. The next wrapping lies over the first wrapping, so the radius of the second wrapping is slightly more than R: it's R+p, so the length of the paper is 2(pi)R+2(pi)(R+p). The angle t after 2 wrappings is 4(pi). The weight of one wrapping is related to the surface area of the cylinder of paper=2(pi)Rw where w is the width of the paper=21". So we need to convert 21" into metres so that we can bring in the given density of 230GSM. 1"=2.54cm=0.0254m, so 21"=0.5334m. We convert the surface area into a weight by multiplying it by the density d=230. Now we have the equation weight=2(pi)Rwd grams.
Because the paper is thin we can consider the reel as being made up of concentric cylinders of paper until we reach the outer radius of 102/2=51"=1.2954m.
Now we have a series: surface area of 1st cylinder + SA of 2nd cylinder + SA of 3rd cylinder +...
This is: 2(pi)Rw+2(pi)(R+p)w+2(pi)(R+2p)w+2(pi)(R+3p)w+...
This can be written: 2(pi)w(R+R+p+R+2p+R+3p+...) which is an arithmetic series. How long is the series? To find out we need to divide the difference between the inner and outer radii by p to get the number of wrappings. So since the outer radius is 1.2954m N, the number of wrappings is (1.2954-R)/p. The series is 2(pi)w(NR+p(1+2+3+...))=2(pi)w(NR+p(N(N-1)/2) and the weight of this is 2(pi)Nwd(R+p(N-1)/2).
To this has to be added the weight of the empty reel, which we can call W: gross weight=W+2(pi)Nwd(R+p(N-1)/2), and N=(1.2954-R)/p.
CONCLUSION: the inner radius R and the paper thickness p in metric units need to be found to find the weight, plus, possibly, the weight of the empty reel.
[In the equation r=R+At, A can be calculated: when t=2(pi), R=R+p, so R+p=R+2(pi)A and A=p/2(pi). r=R+pt/2(pi), and calculus can then be used to solve the problem in a different way. The arithmetic series, however, is probably sufficient to determine the weight to a good approximation.]
