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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and Engineering Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . Chapter 8 Special cases 8.1 Pyramid volume I have been promising to explain the factor of onethird in the volume of a pyramid: V = 1 hb 2 . 3 Although the method of special cases mostly cannot explain a dimensionless constant, the volume of a pyramid provides a rare counterexample. I first explain the key idea in fewer dimensions. So, instead of immediately b h = b explaining the onethird in the volume of a pyramid, which is a difficult three dimensional problem, first find the corresponding constant in a twodimensional problem. That problem is the area of a triangle with base b and height h : The area is A bh . What is the constant? Choose a convenient triangle, perhaps a 45degree right triangle where h = b . Two of those triangles form a square with area b 2 , so A = b 2 / 2 when h = b . The constant in A bh is therefore 1 / 2 no matter what con b and h are , so A = bh / 2 ....
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This note was uploaded on 02/24/2012 for the course MECHANICAL 6.055J taught by Professor Sanjoymahajan during the Spring '08 term at MIT.
 Spring '08
 SanjoyMahajan

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