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Hi all,
I have to calculate the surface area of a domed structure. Basic dimensions are 62m diameter x 18m high.
In my case the dome is not a true spherical shape, but formed by 36nr radial curved beams with flat sheeting curved on the vertical only. My own approximate calculations show curved/actual = 10% extra which looks excessive. Any thoughts on where I’m going wrong?
A radius of 32 and a height of only ~16.2 is not a hemisphere. If the beams form a true arc above the vertical portion then there is math somewhere to figure out the area of 16.2m of a 32m radius sphere.
https://en.wikipedia.org/wiki/Spherical_cap: If the radius of the base of the cap is https://upload.wikimedia.org/math/0/...e269772661.png, and the height of the cap is https://upload.wikimedia.org/math/2/...3e3a695e91.png, then the volume of the spherical cap is[1] https://upload.wikimedia.org/math/b/...d1c6bb5a1a.png
and the curved surface area of the spherical cap is[1]
https://upload.wikimedia.org/math/2/...d081b02c51.png
I get 3257.2M^2
add the vertical area = circumference * 1.8 = 362M^2
Do you get 3619M^2 total surface area?
Hi Sam
I never said it was a hemisphere (see the image); a diameter and arc height only as one solution which i determined using this tool, as per the spreadsheet. The cap area formula gives me 4037m2, It's really the difference between the segmented and true curve I can't believe.
Thanks for the assistance
Regards
MD
Sorry, I get in the habit of speaking to the world, here. I didn't mean to imply that you didn't instantly know that.
I can't see that SS on this laptop, so I was only using the numbers in your post..
Hi Sam
Here's a screenshot
Just curious, by "segmented" area do you mean the area of a triangle with a base of Circum/n and a height of 16.2
(or radius*x)
Both the area you came up with and the one derived by me with the Wiki formula are in the same ballpark. (+/- 5%). Practically slipstick accuracy. Unless it's critical, I would use the Mean.
What does this mean? I didn't understand '3nr' and 'curved on the vertical'Quote:
but formed by 36nr radial curved beams with flat sheeting curved on the vertical only.
Hi Paul
The "dome" comprises 36 triangles laid on straight purlins between the arched beams, so each triangle is only curved in one plane.
Only an approximation using flat triangular segments, I get 3491.4 m2
1. 18 high x 31 base = 35.85 m length to peak
2. each of 36 panels has 62m x pi / 36 as base = 5.41 m as base
3. 1/2 of each panel = 5.41/2 m = 2.70 m as base (to make right triangle)
4. area of 1/2 panel = .5 x B x H = .5 x [2.70 x 35.85] = 48.49 m2
5. area of panel = 2 x 48.49 = 96.98 m2
6. area of 36 panels = 96.8 x 36 = 3491.4 m2
Attachment 16300
Attachment 16301
According to this calculator:
62 is not the diameter. According to that site, the Radius of the sphere that the Dome is the Cap of is 35.69444.
The panel base width is pi*ArcD/36 = 5.41
The height of a (circle) arc 5.41W with Radius 31 is 0.11794
That is the approximate difference in radii of a R:=31 circle touching the center of a panel and a circle touching the edges of that panel. A difference of ~ .38%.
The difference in the lengths of arcs one panel wide of those Radii is 2*pi*.11794 or 0.741%. The difference in the widths of a flat panel and a radiused one is in the ballpark of half that because that difference is a sine function.
Now we can use this calculator to compare the areas of a dome 62 wide and one 62.11794 wide
R:=31: A = 7056.0171
R:=31.05897: A = 7078.6326
Average = 7067.32485
Approximate error factor of flat vs arced panels = +- 0.1597%
Is it correct to assume that it's a segment of a perfect sphere?
In this calculator picture, the yellow seems to be a slice from the top of a perfect sphere
Attachment 16306
I could imagine squishing the top down to get something like the dome picture
I believe so.Quote:
Is it correct to assume that it's a segment of a perfect sphere?
At least that is what I assumed, since the manufacturing costs of a parabola are more then the gains in strength over a simple arc.
'Squishing' the top really compromises the strength of even an arc, plus the added manufacturing costs of building in the ever changing radius of the compound curve.
I was a builder in a previous life :)
Thanks both
Paul, There is an error here
1. 18 high x 31 base = 35.85 m length to peak
The height of the triangle has to be measured on the curve.
Sam
re 62 is not the diameter. 62m is the measured diameter of the dome, I can't measure the sphere, obviously.
I thought I would try another approach.
The roof comprises 36 equal panels, each with a centre angle of 10 degrees. This gave me a base length of 6.54; obviously wrong.
Using this calculator, the inscribed 36 sided shape has a side of 5.4037 which with a hypotenuse of 37.555 gives a "flat" centre angle of 4.125 degrees rather than the expected 5, and a total area of 3643m². I guess that the 10% difference is correct despite the visual similarity.
Yes, but like I said it was only an approximation using 36 flat trianglesQuote:
Paul, There is an error here
1. 18 high x 31 base = 35.85 m length to peak
The height of the triangle has to be measured on the curve.
The 35.85 m is the length of the panel starting on the outer edge of the bottom of the dome (0 m height)to the point at the middle of the top of the dome (18m height)
2 * 25.69444Quote:
re 62 is not the diameter. 62m is the measured diameter of the dome, I can't measure the sphere, obviously.
Then Arc calculator provided the radius/diameter of the sphere. (2 * 25.69444) However I used the circumference of the Dome when calculating the width of the panels and the difference in radii at their centers and their edges.
Q: all of these approaches seem to assume that the Dome is a Spherel Cap, that is a slice of the sphere = 4037 m^2
http://www.had2know.com/academics/sp...alculator.html
Looking at Mac's picture, I still think it looks like the sphere's cap was squished a little and therefore the area would be less than that
Thanks both for the help.
I know that the spherical cap area is a maximum and must exceed that of the flat sectional construction. I can use the former for my budget costing, and the roof sections can be physically measured when access is available on site. I'll let you know how it turns out.
Regards
Malcolm
I think that the min would be the flat triangles in #9, or 3491 m2
Split the difference? :devil2:
The apex of a panel is 1/2 the length of the arc as shown in the image in Paul's #11.
Remember, they overlap or have seams. BTDT. :banghead:
thanks sir
thanks sir
regards
ashish