Calculating the surface area of a geodesic dome

I’m no mathematical genius, and Google has failed me. In simple language, doing stuff that can be done with an El Cheapo 24-key pharmacy-bought calculator, how would I go about this? Thanks for your help.

Determine the area of one of the facets, and then multiply it by the number of facets in the dome.

No, the facets are not equal.

How accurate an answer do you require? What I mean is, is this a theoretical math problem, or are you buying paint for your geodesic dome? If the latter, then a section of a sphere would be an adequate approximation.

Work out how many different kinds of facet there are, work out the surface area of each, then do the math on that.

The surface area of a sphere is 4 * Pi * r[sup]2[/sup], so a hemisphere will have a surface area of 2 * Pi * r[sup]2[/sup]. A geodesic dome will be an imperfect approximation of the hemisphere, though, and depending on what value you use for the radius, you will over- or under-estimate the dome’s true surface area.

Adding up the areas of the panels is easier math and more precise. Don’t look for anything more elegant – this is how it’s done.

Here’s a site called Desert Domes that has some useful information, including a “dome calculator”. With this it shouldn’t be too much trouble to calculate the area of the various triangles that make up the surface of your dome.

Here’s the formula for the area of a triangle as a function of the length of its sides:
For a triangle with sides of lengths a, b and c:
Let s = half the perimeter = (a + b + c) / 2
Area = sqrt (s * (s - a) * (s - b) * (s - c))

The area of a regular polygon with n sides and a side length of L is given by A = n L[sup]2[/sup] / (4 tan(pi/n)). This works out to:[ul][li]Triangle: A = 0.433013 L[sup]2[/sup][]Square: A = L[sup]2[/sup] (of course) []Pentagon: A = 1.72048 L[sup]2[/sup]Hexagon: A = 2.59808 L[sup]2[/sup][/ul][/li]Once you have this info, it should be easy to measure the sides of the faces, count how many of each type of polygon you have, and add it all up.

The problem with this formula is that the triangles aren’t equilateral.

What’s kinda interesting about this problem is that if you were trying to estimate the surface area of a hemisphere without knowing the formulas, approximating it by a geodesic dome and adding up the area of each triangle would be an extremely reasonable way to do it.

The area of a triangle is 1/2baseheight. Pick the longest side as the base, and the length from the base to the highest point is the height.

Calculate the area of each type of triangle in the dome. Count the number of each type of triangle in the dome. Add them all up, and that’s the area.

Surface area, not strut lengths…

 Having built polyhedra with sides greater than three,I haven't seen any "Geodesic Domes" with anything other than triangles which are all equal.
 I'm not asking for a cite,but do you have any examples of your statement?

Actually, you should always underestimate it. For a given radius, the dome will be the minimum area you could have.

The Desert Domes site has a nice picture. The red and blue lines are different lengths.

True. But when you have the lengths of the sides it would seem simpler to use the formula based on that, rather than one that requires the height to be determined.

In that picture, I count 30 BBA Isosceles and 10 AAA Equilaterals, so the area is 4.33A^2 + 15A^2*SQRT( (B^2/A^2) - 1/4 )