The Ecphorizer
| ( ! ) Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /home/crucl/public_html/EPS/site_page_print.php on line 254 | ||||
|---|---|---|---|---|
| Call Stack | ||||
| # | Time | Memory | Function | Location |
| 1 | 0.0009 | 276376 | {main}( ) | .../site_page_print.php:0 |
| 2 | 3.2155 | 502024 | mysqli_num_rows ( ) | .../site_page_print.php:254 |
| Poor Richard's Almanac XVII |
| R(ichard Bradner Mead |
Issue #14 (October 1982)
I placed a bowl into the storm,
To catch the drops of rain --
A half a globe was just its form,
Two feet across the same;
The storm was o'er, the tempest past,
I to the bowl repaired --
Six inches deep the water stood,
It being measured fair;
Suppose a cylinder, whose base
Two feet across within,
Had stood exactly in that place,
What would the depth have been?
To catch the drops of rain --
A half a globe was just its form,
Two feet across the same;
The storm was o'er, the tempest past,
I to the bowl repaired --
Six inches deep the water stood,
It being measured fair;
Suppose a cylinder, whose base
Two feet across within,
Had stood exactly in that place,
What would the depth have been?
I have not attempted to analyze the solution, as I am not up on computing the cubic contents of globes. (I do remember that the area of a circle is πr2, so I could figure the cubic content of a cylinder. I see that he has taken the diameter squared times 1/4 of π, which produces the same result for the area of the cylinder.) His answer is 2.5 inches, which I am confident is correct.
Contributor Profile
Brad Mead
BRAD MEAD, whose chapters of "Poor Richard's Almanac" brightened many of our early issues, lives in a San Francisco Victorian house with a large collection of porcelain owls.