There was a request for estimate of Planetarium dome surface area for painting.
1. Looking at the dome, for eg. in the picture at our website, I estimate 45 4-sided figures visible from the front. Assuming 45 more from the back, and rounding off to 100, I would estimate 100x area of one 4-sided figure.
The 4-sided figure is made of 3 triangles and one trapezium, and its overall shape is like a big equilateral triangle of side approximately 3 metres as measured with a metre scale, an estimate which is likely to be a bit on the low side, would be sqrt(3)/4*3^2 x100,
3.9 sq. metres x 100 = 400 square metres (approx).
1. Looking at the dome, for eg. in the picture at our website, I estimate 45 4-sided figures visible from the front. Assuming 45 more from the back, and rounding off to 100, I would estimate 100x area of one 4-sided figure.
The 4-sided figure is made of 3 triangles and one trapezium, and its overall shape is like a big equilateral triangle of side approximately 3 metres as measured with a metre scale, an estimate which is likely to be a bit on the low side, would be sqrt(3)/4*3^2 x100,
3.9 sq. metres x 100 = 400 square metres (approx).
2. Another estimate is using the distance measured with my feet for radius, 10.75 metres, for the dome. That would make it 2*pi*r^2 = 725 square metres!
This is closer to being correct, because the estimate of 100 4-sided figures by looking at the picture is probably wrong.
But this would be area if the hemisphere was smooth. Actually it is more like a series of pyramids. So, the actual area could be even double this estimate.
(Eg. If a pyramid has 10m x 10m length and breadth, if it has 10m height also, google shows that its surface area would be around 350. Subtracting the base area of 100, that makes it 2.5 times the area of the base alone.)
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