Take a unit square piece of paper with vertices ABCD. Join edges AB and AC to make a cone-like new three dimensional structure which is durable and water-tight.
a) What volume of water will the paper cone hold, if it is held with the axis of the cone oriented vertically?
b) What volume of water will the paper cone hold, if it is held however is necessary to maximize the amount of water it can hold?
Part a) is equivalent to a quarter circle shaped into a cone.
The curved portion has length 2pi/4 = pi/2 and becomes the circumference of the base. So the radius of the base is 1/4. The height of the cone can be found by Pyth. as sqrt(15)/4.
V=(1/3)pi(1/4)^2*(sqrt(15)/4) = pi*sqrt(15)/192
I think part b) is tougher than it looked at first (and it looked tough). Basically once you tilt the cone-like shape the water takes the shape of an oblique cone with an elliptical base. The axial angle of the cone is arctan(1/sqrt(15)) or about 14.5 degrees.
From inspection of my paper model it seems the optimum tilt is probably more than this (*see below) but I thought this angle might be a good place to start since the seam of the come become vertical so at least the height of the cone is simply 1.
The best I could do so far with this simplification is the ellipse has a major axis of tan(2*arctan(1/sqrt(15))= sqrt(15)/7. The length up the side opposite the seam of the cone is 8/7. I'm stuck on how to find the minor axis of the ellipse.
(*I made a paper model with an 8.5 inch sheet of paper. Holding up the cone-like shape I placed a round pencil across the gap at the top. Holding the cone at various angles the pencil shows how the high-water-mark changes. I can imagine adding water and tipping it a bit further than 14.5 degrees until at some point it starts to pour out the sides.)
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Posted by Jer
on 2020-09-12 21:57:12 |
Solution to part a) & progress so far on b)
