Editing 2509: Useful Geometry Formulas

Jump to: navigation, search

Warning: You are not logged in. Your IP address will be publicly visible if you make any edits. If you log in or create an account, your edits will be attributed to your username, along with other benefits.

The edit can be undone. Please check the comparison below to verify that this is what you want to do, and then save the changes below to finish undoing the edit.
Latest revision Your text
Line 30: Line 30:
 
Randall approximates the area of the 2D shape on the page as the sum of the area of the triangle formed by the major axis of the ellipse and the two lines, and half of the area of the ellipse (<sup>π</sup>/<sub>2</sub>&nbsp;ab) since most of the upper half of the ellipse overlaps the triangle.  The equation for this area is A = 1/2 πab + bh.  This is captioned below the figure.
 
Randall approximates the area of the 2D shape on the page as the sum of the area of the triangle formed by the major axis of the ellipse and the two lines, and half of the area of the ellipse (<sup>π</sup>/<sub>2</sub>&nbsp;ab) since most of the upper half of the ellipse overlaps the triangle.  The equation for this area is A = 1/2 πab + bh.  This is captioned below the figure.
  
The actual area of a picture of a cone is not Randall's approximation, because the sides connect at the points on the ellipse where they can spread widest and form tangents to the ellipse, and such points are a little higher than those which define the major axis. This is most obvious in cases when h is only a little larger than a. The area can be computed to be exactly A = b (a arccos(-a/h)) + √(h<sup>2</sup>-a<sup>2</sup>)).
+
The actual area of a picture of a cone is not Randall's approximation, because the sides connect at the points on the ellipse where they can spread widest and form tangents to the ellipse, and such points are a little higher than those which define the major axis. This is most obvious in cases when h is only a little larger than a. The area can be computed to be exactly A = (π - arccos(a/h))ab + b √(h<sup>2</sup>-a<sup>2</sup>).
  
 
The 3D right circular cone commonly depicted by this drawing would have a volume of πr<sup>2</sup>h/3 where r=a=b.  The area of the "lower" surface would be πr<sup>2</sup>, while the surface area of the upper conical surface would be πr√(h<sup>2</sup> + r<sup>2</sup>).  Neither of these areas can correspond with the caption in the comic, nor does the total surface area (the sum of these two).
 
The 3D right circular cone commonly depicted by this drawing would have a volume of πr<sup>2</sup>h/3 where r=a=b.  The area of the "lower" surface would be πr<sup>2</sup>, while the surface area of the upper conical surface would be πr√(h<sup>2</sup> + r<sup>2</sup>).  Neither of these areas can correspond with the caption in the comic, nor does the total surface area (the sum of these two).

Please note that all contributions to explain xkcd may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see explain xkcd:Copyrights for details). Do not submit copyrighted work without permission!

To protect the wiki against automated edit spam, we kindly ask you to solve the following CAPTCHA:

Cancel | Editing help (opens in new window)