Dr. Ecco Solution
Solution to "Treasure Arrow," DDJ, July 2005.
Let's first solve the problem symbolically. The pole has length L, mass Mp, and the arrowhead has weight Ma. We represent the pole's mass Mp as a point mass at the center.
Now, let's say we have five elastics including ones at the ends and at every quarter point. We know that the lengths of the bands must form a straight line because the pole is stiff. This observation (due to my colleague Alan Siegel) gives us a constraint in addition to balanced torque and balanced vertical force. Here, s1 is the stretch of the leftmost band in centimeters. is the difference from one band to the next one to the right, again in centimeters.
s1+=s2
s2+=s3
s3+=s4
s4+=s5
The balance of vertical forces gives us:
5s1+((1+2+3+4))=Mp+Ma
The balance of torques gives us:
LMp/2 from the pole.
Countertorques:
L(s1+(3/4)(s1+)+(1/2)(s1+2)+(1/4)(s1+3))
=L((s1(1+3/4+1/2+1/4))+
(3/4+1+3/4))
=L(2.5s1+2.5)
=2.5L(s1+)
So (1) Mp/2=2.5(s1+), from torque balance; (2) Mp+Ma=5(s1+2), from vertical balance. From (2), ((Mp + Ma)/5)- 2=s1. Therefore, using the torque balance:
Mp/2=2.5(s1+)
=2.5((((Mp+Ma)/5)-2)+)
= 2.5((((Mp+Ma)/5)-))
=(2.5((Mp+Ma)/5))-(2.5)
=(((Mp+Ma)/2))-(2.5)
So:
2.5=Ma/2
=Ma/5
Therefore, s1=Mp/5+Ma/5-2Ma/5= Mp/5-Ma/5. This implies that if Mp= Ma, then s1=0. Since these two quantities are in fact equal, s1=0, so the left band is at its rest length of 1 meter, the right band is down 60 centimeters, and the arrow is pointing to a point that is 100 cm+260 cm=2.2 meters from the top. Further, is 20 centimeters. So, Mp=Ma=100 kilograms.
Optimal Farming: Errata and Reader Improvements
Alan Dragoo was the first to point out the bug in my solution to problem 1. He suggested a design where one circle would cover a central square and four smaller rectangles would cover side rectangles. Denis Birnie independently arrived at the same solution, as well as a very interesting solution for open problem three. You can find those at http://cs.nyu.edu/cs/faculty/shasha/papers/birniedobbssol.doc.
DDJ