Math Help? Finding Area of Overlap in Venn Diagram

I’ve got two populations: The first is made up of 1,152 people, the second is made up of 2,238 people, and the overlap is made up of 424 people.

Using pi*r[sup]2[/su]=Area, I plug 1,152 and 2,238 into those to get the radius of each circle. Which gives me approximately 19 and 27 for the respective radii. So, if I’m thinking correctly, I can use 19 and 27 units of some measure to create more-or-less proportional circles.

The problem is that I can’t remember (or even solve if I could remember) how to figure out the area of the overlap. That is, how far should the centers of the circles be from one another so that the area of overlap is approximately proportional to 424.

Does that make sense? Can anybody give me a hand on this? (If so, will you?)

Thanks much.

It’s easier to do than to explain how to do. Give me a little time.

Yes, the question does make sense, but the mathematics is a little comples.

First, the radius of a circle with area A would be the square root of (A/pi), so in this case about 19.1492 and 26.6904.

Secondly, the formula for the area of the overlap between two circles is given on this page.

It’s worked out this way. If the centres of the circles are A and B, the radius of the circles are r and R, the distance between the centres of the circles is d, and points where the two circles intersect are C and D, then:

The area you want is equal to the sum of the segments of the circles ACD and BCD, less the sum of the areas of the triangles ABC and ABD. (Those last two triangles are congruent so they have the same area).

The area of the triangles is given in equation 11 of the page I cited above: the first line of the equation is the areas of the two segments, and the second line is the area of the two triangles.

Now to get the area equal to something: I’m not sure how to do that, but I’d be inclined to set up a few values of d in a spreadsheet, see what you get, and interpolate between the closest results…

Once you’ve got that formula for a in terms of R, r, and d, you just need to set a = 424, R = 27, and r = 19 (using whole numbers cause I’m doing this with the Windows calculator), and then solve for d.

So far, I’ve reduced it to d[sup]4[/sup] + 177596d[sup]2[/sup] + 135424 = 0.

You’re much better at algebra than I am.

Thanks guys!

SDMB: Best five bucks ever spent.

There has to be a mistake there: if d is real, the smallest value that d[sup]4[/sup] + 177596d[sup]2[/sup] + 135424 can have is 135424. (The other two terms must be zero or positive). However, I’m not sure that I can offer a better solution.

It’s certainly possible that my arithmetic is off.

I just downloaded a freeware equation solver from Tucows.com and it returned four complex roots. Bummer.

Am I missing something? I’ve never done anything with Venn diagrams that needed areas. The circles were always considered to be symbols, rather than areas.

True, but it’s a nice touch to draw it so that the area of each part is (roughly) proportional to the number of elements in it.

Yep. Our township has 1,152 people working in it and 2,238 people who work living in it, but only 424 people actually live in and work in the township. I decided that I could show that w/ a Venn diagram, but I want the picture to accurately reflect the numbers. I made a crude ball-park estimate of the size of the intersecting area, but I was hoping that there’d be a handy way to just bang it out without any hassle. That is, I was hoping that a math whiz could just say “That’s easy!” and do it in about twenty seconds. I’ve never been good at turning ideas into actual calculations.

:smack:

I was using the formula for chord height, not the area of the circle.

You’ll need to solve numerically using the right formula, cause it has d both inside and outside the arccos calls.

And me without the program I wrote to calculate (trial and error) the roots of cubic and quartic functions. True, it only calculated real roots, but if I had it now I could probably get at imaginary.

Man was that thing ever useful for tests and exams.

Do it by proportion. Calculate the ratio of area to population (pi19^2)/1152 and (pi27^2)/2238. (.9828 and 1.0233, respecitviely)

Average those ratios: 1.003

Multiply by the overlap: 1.003 * 424 = 425.300 square units

maybe it would help if I finished reading the question…d’oh :smack:

Using the above value for the area of overlap, dividing by pi, then taking the square root returns 11.635 as the radius of a circle inscribed in the overlap. Therefore, the radii of the two cirles overlap by 11.635 units. 19+27-11.635=34.364 units.

Oh, nevermind. I took the whole overlap area as a cirlce! I should really stop now.

OK, here’s an idea for a crude calculation:

You have two circles of radius X and Y, Y being the bigger. There is some shared value Z contained partially within X and partially within Y. That’s the basic problem, yes?

ISTM that a proportionally larger amount is in circles of radius X compared to circle Y (19^2 being smaller than 27^2, A units where A is greater than 0 will represent a larger proportional area in circle X than in circle Y) . IOW, the arc length portion of circle X is longer than the arc length portion of circle Y - I had these reversed, then considered a few hypotheticals and thought the way I have it now might be more accurate. Picture a child hugging an adult and I think you’ll see why. ISTM further that the relationship between these two should differ by a factor of Z to some exponent and/or involving pi somewhere, maybe. Alas, that is where my liberal arts education (truthfully that’s more theoretical abstraction than anything else) ends, since I don’t know how to calculate area under an arc not roughly pie-slice shaped.

If someone could derive (or inform us of) some sort of formula regarding the above, I think that might solve it.

Using d = 34, I got an area of about 250. So at least we have a lower bound.

Plowing through the ugly arithmetic expression for the area of the lens, I got:

4A[sup]2[/sup] = - d[sup]4[/sup] + 2158.1386 d[sup]2[/sup] - 119498.5

or d[sup]4[/sup] + 2158.1386 d[sup]2[/sup] - 838602.5 = 0

solving for d[sup]2[/sup] in the quadratic equation gives roots of 508.3 and 1649.8, so d = 22.5 or 40.6. Substituting 22.5 back into the equation gives A = 423.