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| shreksbro Quote | Reply | This message was updated on 2/7/2007 9:13:14 PM by shreksbro | Pyramid geometry posted on: 2/7/2007 7:20:36 AM Sorry if you read an earlier version of this, I have been working on the subject for many hours now, I've solved some of my earlier problems and I'm a bit more clear on what I need to achieve, so I have been editing like a fiend. I am dealing with square-based pyramids. I am trying to create a spreadsheet (in Excel) which will allow me to enter a base length and height for a pyramid, and from just those two numbers generate the following figures: 1. The angle of the apex, when viewed as a cross-section through opposite sides; 2. The angle of the apex, when viewed as a cross-section through opposite corners; 3. The dimensions and angles of the triangular side panels when laid flat. Sadly I never was able to grasp trigonometry. I know soh cah toa comes into it somewhere but I can't figure out how to manipulate it into giving me the reults I need. Any help very gratefully received - and don't be afraid to dumb it down for me  |
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cramwit
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Pyramid geometry
replied on: 5/8/2007 3:51:30 PM a = altitude n = side length The diagonal base of the pyramid is sqrt(2) * n perpendicular cross section is simply n since you want the apex angle these values have to be treated as the 'rise' of your triangle & you use only 1/2 of each of them a is the run of the triangles you are looking at. the apex angle is two of these trigonometric constructions combined. (i hope this is clear enough) you have rise & run so you have the tangent requirements. Since you want the angle you use the arctangent function. Diagonal apex angle: 2 * arctan( ( 0.5 * sqrt(2) * n ) / a ) not sure if that can be simplified Perpendicular cross section apex angle: 2 * arctan( ( 0.5 * n ) / a ) you have n as the base of pyramid panel triangle. the sides are the hypotenuse of Diagonal apex triangles or h^2 = a^2 + ( 0.5 * sqrt(2) * n )^2 h^2 = a^2 + 0.25 * 2 * n^2 h^2 = a^2 + 0.5 * n^2 h = sqrt( a^2 + 0.5 * n^2 ) using h from above two equal base angles are ( i think ) arccos( (n/2) / h ) = arccos( n /(2h) ) apex angle is: 180 degrees - 2 * arccos( n / (2h) ) or (very iffy) in radians pi - 2 * arccos( n / (2h) ) You should have someone sharp in trig to double check all this, my trig is not just rusty, it is ancient |
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