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s S --- = --- d DWhere: s = Scale Model Size of Sunspot d = Scale Model Diameter of Sun S = Actual Size of Sunspot D = Actual Diameter of the Sun (1,400,000 km) Substitute, cross-multiply, do dimensional analysis, solve for "S": s (mm) S (km) -------- = -------------- d (mm) 1,400,000 km s (mm) x 1,400,000 km S (km) = ----------------------- d (mm)
If the scale model size of the largest sunspot = 3mm, then its actual size would be approximately 24,706 km wide.
If the scale model size of the largest sunspot = 3mm, then its actual size would be approximately 24,706 km -- nearly double the Earth's diameter! Approximately 110 Earth diameters would fit inside the Sun's diameter (1,400,000/12,735 = ~110).
The scale of the image = 1/8,240,000,000 (1,400,000,000,000 mm/170 mm = ~8,240,000,000). Features on the Sun are nearly 8.25 billion times larger than the image!
(largest) km hm dam m dm cm mm (smallest)
2--1--0
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For example, convert 1,100 cm to m by moving the decimal point 2 places to
the left: 1,100 cm = 11 m (approximately 36 feet).