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Turning Needle Cars. Francis Hunt [email protected] A Vision of the Future. Area swept out = π (½) 2 = π /4. Turning round in π /8. Turning round in π /8. Turning round in π /8. Turning round in π /8. What is the least possible area?. Turning round in ε. Idea 1

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A vision of the future
A Vision of the Future

Area swept out = π(½)2 = π/4





Turning round in 83
Turning round in π/8

What isthe least possible area?


Turning round in
Turning round in ε

Idea 1

  • The area swept out in parallel parking can be made arbitrarily small

Besicovitch’s Solution



Turning round in2
Turning round in ε

Divide base into 2m equal segments (here m = 3)


Turning round in3
Turning round in ε

Divide base into 2m equal segments (here m = 3)


Turning round in4
Turning round in ε

Divide height into m + 2 equal bands




Turning round in7
Turning round in ε

area of shape has been reduced by a factor of 2/(m+2)

By choosing m large enough we can make this arbitrarily small


Conclusion
Conclusion

  • the area of the spikagon can be made arbitrarily small (by choosing m large enough)

  • the area of the 2m parallel parks can be made arbitrarily small (by driving far enough)

  • the total area needed to turn our needle car round can therefore be made arbitrarily small


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