Mr B you are partially correct : I think I'm FULLY correct....

Posted by Denis Borris on October 09, 2002 at 12:53:27:

In Reply to: I got "unlost" .. Mr B you are partially correct posted by Subhotosh Khan on October 09, 2002 at 11:21:17:

: radius of the base of small cone = R * T/(2*pi)

I have : R * T / 360 (T is in DEGREES)

So if original circle's radius is 5, and I cut at 75 degrees:
base radius = 5 * 75/360 = 1.0416666....

: height of the small cone = sqrt[R^2 - {R * T/(2*pi)}^2]
: =R/(2pi)*sqrt[(2pi)^2 - T^2]

Similarly to radius, I have : R * sqrt(360^2 - T^2) / 360

: radius of the base of large cone = R * (2pi-T)/(2*pi)

I have : R * (360 - T) / 360

: height of the large cone = sqrt[R^2 - {R * (2pi-T)/(2*pi)}^2]

I have : R * sqrt[T(720 - T)] / 360

I'm SURE you'll find mine equals same as yours; BUT:
due to "nature" of problem, mine is simpler, isn't it?
There is no need to deal with radians et al.... CORRECT??

I ran a "brute strength" using R=5 (and MY formulas!) :
MAX Volume = 57.0800738... when T = 116.6449866... degrees.

AGREE? If you agree, you're correct :))

Follow Ups:

• Hokay.... Denis Borris 16:26:44 10/09/02 (3)
  • Please check your equation... Subhotosh Khan 16:53:07 10/09/02 (2)
    • NO! Denis Borris 18:14:54 10/09/02 (1)
      • Yes .... Subhotosh Khan 08:54:59 10/10/02 (0)

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