Calculating Dimensions for Constructing Super Adobe (Earth Bag) Domes
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The document explains how to calculate the distance from the centerline to the inner wall of a dome at any given height above the ground using the Pythagorean theorem. It discusses the properties of circles and demonstrates the calculation through various examples, including a specific dome with defined dimensions. The relationship between heights, radii, and distances from the centerline is outlined with detailed steps for practical application.
Calculating Dimensions for Constructing Super Adobe (Earth Bag) Domes
- 1. 1 Calculation for Domes: Distance from Centerline to Inner Wall for Any Given Height above Ground This document describes the basis for the calculation, and provides examples. Basis of the calculation The vertical cross-sections of the domes considered here are circular arcs. The properties of circles allow us to calculate the distance of interest via the Pythagorean Theorem. Relevant properties of circles All points on a circle are at the same distance from its center. That distance is called the radius of the circle. Confusingly, the same word (radius) is also used for any segment from the center to a point along the circle: Now, we’ll add to our diagram a horizontal line passing through the circle’s center. Please note that this horizontal line represents the springline of the dome. If we now draw a vertical segment, as shown below, then a right triangle is formed by the vertical segment, the radius, and the segment from the center of the circle to the point where the vertical segment intersects the springline: Center Radius Center Radius Center Radius
- 2. 2 Notice that the hypotenuse of our right triangle is a radius drawn to the point where the vertical segment touches the circle. Of course, we could also form a right triangle by first drawing a horizontal line at some height H above the springline. If the height H is small enough, then line we just drew will cut our circle at two points. 1 We’ll choose one of those points, and call it P. In a dome, point P would be on the inside of the wall, at height H above the springline: Now, we’ll draw a vertical segment from P to the springline: 1 If H is equal to the radius of the circle, then the dashed line will be tangent to the circle. In other words, the dashed line will cut the circle at only one point, which is at the top of the circle. H P H Center Radius
- 3. 3 Next, we form the right angle by drawing a horizontal segment from the center of the circle to the point at which our vertical segment intersects the springline. We’ve now formed our right angle: Finally, we add the hypotenuse (the radius drawn to P). Our right triangle is now complete. We would follow the same procedure to construct the right triangle for any height that interests us. Here are some examples: The length of the hypotenuse is the same for all such triangles, since it’s equal to the radius of the circle. Notice that we needn’t use the full circle to construct our right triangles. We could use only a semicircle, H P H P H P
- 4. 4 or even just the part of the circle that corresponds to the dome we wish to construct: To continue, we’ll examine one such triangle in greater detail. We’ve added the point Q, which is directly over the center of the circle, and at the same height as P. Note that the length of the base of the triangle is equal to the distance from Q to P. The angle at Q is a right angle, so we now have two different right triangles, as shown below: Both of the right triangles have the same hypotenuse (whose length is equal to the radius of the circle):Therefore, the two red triangles (above) are identical. We can use either to calculate the distances we’ll need for constructing a dome. Since we’re interested in finding distances from the axis of the dome to the inside of the wall, we’ll use Triangle II in the sections that follow. The Pythagorean Theorem For convenience, we’ll re-draw Triangle II from the previous section, labeling the length of each side of the triangle to help us apply the Pythagorean Theorem: Q P Q P I II Q P Q P
- 5. 5 According to the Pythagorean Theorem, the relationship between X, H, and R is 𝑋2 + 𝐻2 = 𝑅2 , from which 𝑋 = √𝑅2 − 𝐻2 . Calculation of Distances from the Centerline of the Dome to the Inside of the Wall, for Any Height above the Springline In a hemispherical dome, the line L coincides with the centerline of the dome. Therefore, the distance from the dome’s centerline to its inside wall, at any height H that interests us, is just X. In such a case, the equation we just saw, above, is exactly the one we need: In a hemispherical dome, 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑒𝑛𝑡𝑒𝑟𝑙𝑖𝑛𝑒 𝑡𝑜 𝑖𝑛𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑎𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 𝐻 = 𝑋 = √𝑅2 − 𝐻2 . RH L X X RH R is the radius of the circle (and therefore of the curve of the wall). H is whatever height above the springline that might interest us. L Is the (imaginary) vertical line that passes through the center of the circle. X is the distance that there is between L and the inside wall of the dome, at height H. L
- 6. 6 Many domes for earthbag buildings are not hemispherical; instead, their vertical cross-section has the following shape: Each side is an arc of a circle whose center is offset from the centerline (CL) of the dome: How can we find the distance between the centerline of the dome and the inside wall in this case? To simplify the explanation, we’ll make use of the dome’s symmetry about its centerline. Because of that symmetry, we need consider only one side: From our previous work, we know how to find the distance X for any given height H: L CL R CL R R The radius R is the same for both sides.
- 7. 7 But the distance we need to know isn’t X; instead, instead, it’s d, in the diagram below: Examining this diagram, we see that if we know X, we can find d by subtracting the distance between the lines L and CL, d = X – (distance between lines L and CL), or, since we know that 𝑿 = √𝑹 𝟐 − 𝑯 𝟐 , d = √𝑹 𝟐 − 𝑯 𝟐– (distance between lines L and CL). Note that the distance between lines L and CL is also the distance between the center of the arc and the center of the base of the dome. This observation allows us to put our answer in the following form: L CL R X H d L CL R X H 𝑿 = √𝑹 𝟐 − 𝑯 𝟐
- 8. 8 Sample Calculations We’ll do our sample calculation for a dome with the characteristics shown in the next figure. The dome is symmetrical. Each side is an arc with a radius of 7.5m, and a center that’s offset 4.5 m from the centerline of the dome: 6m 6m To find d for any height H: Step 1: Find X, from 𝑿 = √𝑹 𝟐 − 𝑯 𝟐. Step 2: Find d from d = X - J. L CL R X H d J (Distance between center of circular arc and center of base of dome.)
- 9. 9 Now, we’ll make three sample calculations. The first two will be done to validate our formulas, and our understanding of them. The first sample calculation: H = 0 If we study our first drawing, We can see that at the springline, the distance between the centerline of the dome and the inside of the wall is 3m. Is this the result that we get from the procedure we developed earlier? We’ll now see. 6m 6m L CL R =7.5m 4.5m
- 10. 10 At the springline, H = 0. We’ll now follow our procedure, Step 1: Find X, from 𝑿 = √𝑹 𝟐 − 𝑯 𝟐. 𝑿 = √𝟕. 𝟓 𝟐 − 𝟎 𝟐 = 7.5m. Step 2: Find d from d = X - J. d = 7.5 – 4.5 = 3m. OK The Second Sample Calculation: H = 6m If we study our first drawing again, We’ll see that at a height of 6m above the springline, the distance between the inner wall and centerline is zero. This observation allows us to make another check of our procedure. 6m 6m To find d for any height H: Step 1: Find X, from 𝑿 = √𝑹 𝟐 − 𝑯 𝟐. Step 2: Find d from d = X - J. L CL R = 7.5m X H d J = 4.5m (Distance between center of circular arc and center of base of dome.)
- 11. 11 Again, R = 7.5m, J = 4.5m, and H = 6m, so according to our procedure, Step 1: Find X, from 𝑿 = √𝑹 𝟐 − 𝑯 𝟐. 𝑿 = √𝟕. 𝟓 𝟐 − 𝟔 𝟐 = 4.5m. Step 2: Find d from d = X - J. d = 4.5 – 4.5 = 0m. OK The Third Sample Calculation: H = 3m Now, we’ll calculate the distance from centerline to inner wall at 3m above the springline. Again, R = 7.5m, J = 4.5m, but his time H = 3m, so according to our procedure, Step 1: Find X, from 𝑿 = √𝑹 𝟐 − 𝑯 𝟐. 𝑿 = √𝟕. 𝟓 𝟐 − 𝟑 𝟐 = 6.874m. Step 2: Find d from d = X - J. d = 6.874 – 4.5 = 2.374m.