Huygens and Bernoulli's brachistochrone - Rijksuniversiteit Groningen

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H&B – p.16/27Detailed formulation1. K(ϕ) is tangent to E and perpendicular to C ;2. |K(ϕ)| is radius of curvature of C in C(ϕ)Comments:- Pendulum attached to cusp of Echord evolves along ‘cheeks’ E=⇒ mass follows C- K⊥C ∼ instantaneous rotation in rolling-wheelconstruction of cycloid;NB: K⊥C ⇒ dynamics of friction-less beadH.W. Broer, Huygens’ isochrone slinger. Euclides 70(4) (1995) 110-117
Check with high school computationsLemma: K(ϕ)//E ′ (ϕ) andK(ϕ)⊥C(ϕ)Proof: On the one handK(ϕ) = C(ϕ) − E(ϕ) =0@2 sin ϕ−2 − 2cos ϕ1A =0101@ 2 sin ϕ 2 cos ϕ 2−2 cos 2 ϕ2A ‖@ sin ϕ 2−cos ϕ 2A;On the other handE ′ (ϕ) =0@ 1 − cos ϕ−sin ϕ1A =0@2 sin 2 ϕ2−2sin ϕ 2 cos ϕ 21A ‖0@ sin ϕ 2−cos ϕ 21Aetc., etc., etc.NB: Check that k(ϕ) = arclength along E from E(ϕ) till extremal:On the one hand k(ϕ) = 2 √ 2 √ 1 + cos ϕ = 4 cos ϕ 2 ,On the other hand the arclength equals 4 sin ϕ+π = 4sin( ϕ 2 2 + π 2 ) = 4 cos ϕ 2 ;“figures !”QEDH&B – p.17/27
Page 1 and 2: H&B - p.1/27Huygens and Bernoulli
Page 3 and 4: H&B - p.3/27Huygens and BernoulliCh
Page 6 and 7: H&B - p.6/27Commentary1. Potential
Page 8 and 9: H&B - p.8/27Arclength cycloidarclen
Page 10: H&B - p.10/27Evolute and evolventSo
Page 14 and 15: The master himself . . .H&B - p.14/
Page 18 and 19: H&B - p.18/27Johann Bernoulli’s s
Page 20 and 21: H&B - p.20/27DiscretisationHorizont
Page 22 and 23: H&B - p.22/27OpticsPn 1 = 1/v 1α 1
Page 24 and 25: H&B - p.24/27Deduction cycloidSnell
Page 26 and 27: H&B - p.26/27Commentary1. Challenge

Huygens and Bernoulli's brachistochrone - Rijksuniversiteit Groningen

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