All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Cylindrical Cuts (Posted on 2009-01-22) Difficulty: 4 of 5
An object resembling that in the graphic, is carved from a cube of soft material such that its elevations, as seen from the top, front and side, are all that of the drawing at the top.

Note that each curve is a segment of a quadrant circle centred on the opposite corner and having the radius of the side of the square.

While the object may look like a stellated octahedron, each curved stellated extension is actually formed by six surfaces being parts of cylindrical forms (thus there are 48 actual surfaces which form the object).

What percentage of the cube is wasted to create this object?

"Impossible" Solid's Volume was a somewhat related problem.

See The Solution Submitted by brianjn    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts numerical integration | Comment 1 of 5

Let the origin be at the bottom left front corner. Let the bottom be the x-y plane with increasing x going from left to right and y going from near to far, while z goes from bottom to top.

Let's consider the 1/48 of the total volume of the solid that is the wedge formed by

y = 1 - sqrt(1 - x^2)  -- this is based on (y-1)^2 + x^2 = 1, its a front cylindrical surface
y = 1 - z              -- separating this wedge from the one above
x = .5                 -- separating from a wedge on another of the star's arms
                                                  (the one on the top right)
x = 1 - z              -- separating this wedge from the one below

Placing below y = 1 - z and beyond y = 1 - sqrt(1 - x^2) by itself causes this to lie below z = sqrt(1 - x^2).

Integrating is done on ((1-z) - (1 - sqrt(1 - x^2))) dx dz.

The limits of integration on z are from .5 to 1 and on x from 1-z to either .5 or where (1 - sqrt(1 - x^2)) exceeds (1-z) whichever is lower.

The result must be multiplied by 48 to get the volume of the entire object, and then subtracted from 1 to get the portion of the cube that's not in the object.

Private Sub cmdStart_Click()
stp = Val(txtStep.Text)
For z = 0.5 To 1 Step stp
 For x = 1 - z To 0.5 Step stp
  increm = (1 - z) - (1 - Sqr(1 - x * x))
  If increm > 0 Then
   t = t + increm
   Exit For
  End If
Print 48 * t * stp * stp
End Sub

One would expect that the size of the error due to numerical integration would not exceed six times the increment (step size in the above program), as that would be the volume of a cubical shell of that thickness about the cube.

For the integration increments listed at the left, the total volume of the solid comes out to the figure in the right hand column below:

.01       .6906
.001      .6531
.0001     .6491 (the same algorithm in QuickBasic rather than Visual Basic,
                however, produces .6494 to this many places.)

Indeed the difference, in getting more precise approximations, is indeed less than six times the increment used in the less precise approximation.

Therefore it would seem that the volume of the solid itself is 0.649, to three places, and the portion removed from the cube is .351 to that accuracy, or 35.1 % removed.

Edited on January 22, 2009, 4:37 pm
  Posted by Charlie on 2009-01-22 16:35:01

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information