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Roe-Hulmn Intitute of Technology Roe-Hulmn Scholr Mthemticl Science Technicl Report (MSTR) Mthemtic 8-15-9 Flttening Cone Sen A. Broughton Roe-Hulmn Intitute of Technology, brought@roe-hulmn.

1 Roe-Hulmn Intitute of Technology Roe-Hulmn Scholr Mthemticl Science Technicl Report (MSTR) Mthemtic Flttening Cone Sen A. Broughton Roe-Hulmn Intitute of Technology, Follow thi nd dditionl work t: Prt of the Geometry nd Topology Common Recommended Cittion Broughton, Sen A., "Flttening Cone" (9). Mthemticl Science Technicl Report (MSTR). Pper Thi Article i brought to you for free nd open cce by the Mthemtic t Roe-Hulmn Scholr. It h been ccepted for incluion in Mthemticl Science Technicl Report (MSTR) by n uthorized dminitrtor of Roe-Hulmn Scholr. For more informtion, plee contct bernier@roehulmn.edu.

2 Flttening Cone S. Allen Broughton Mthemticl Science Technicl Report Serie MSTR 9-1 Augut 15, 9 Deprtment of Mthemtic Roe-Hulmn Intitute of Technology Fx (81) Phone (81)

3 Flttening Cone S. Allen Broughton 15 Aug 9 Content 1 How thi problem got trted 1 Problem ttement nd olution 3 Verifiction tht AB i locl iometry 5 4 The rel exmple 6 5 Picture nd ttchment 7 1 How thi problem got trted A locl mnufcturing deign compny clled with the following problem. We wnt to mnufcture cut off lnted cone from flt heet of metl. If the cone w norml right cone we know tht we would imply cut out ector of circle nd roll it up. However the cone i lnted. We wnt to know wht the flttened hpe look like o tht we cn cut it out nd roll it up to cloely pproximte correct finl hpe. We lo wnt to minimize the mount of wted metl fter the hpe i cut out. I replied tht No, I don t know of ny formul for thi, let me think bout it. Cn you end me picture. The compny ent CAD drwing. The drwing i the firt of the picture in the picture nd ttchment ection (Section 5). The problem, nd it generliztion my be olved nlyticlly but the nlyticl olution i given in term of indefinite integrl which rrely cn be evluted in cloed form. The olution my be found numericlly which re good enough the crete picture of the flttened out cone. In Section we decribe the problem of flttening out 1

4 cone over curve prmetriztion problem in the differentil geometry of urfce. In Section 3 we verify tht the cone urfce relly my be flttened out by howing tht the flttening mp (or i invere the roll up mp) i n iometry. In Section 4 we crry out the computtion in the motivting exmple, coming up with formul to decribe the outline of the flttened out hpe. Finlly Section 5 how picture of the rolled up cone, the flttened out region, nd the Mple workheet tht compute the flttened out region. The workheet lo contin everl view of the cone to upply viul verifiction. Problem ttement nd olution Let P (), L, be pce curve of length L prmeterized by rclength nd let Q be ny point in R 3. Conider the cone over Q determined by P (), we give the detiled definition hortly. The cone i flt in the differentil geometry ene nd therefore my be flttened to region S in the plne. The gol of thi note i to decribe the region S in term of polr coordinte. The olution to thi problem i importnt in mnufcturing where the flttened region i rolled into cone or truncted cone hpe. An importnt function in our nlyi will be ditnce from Q to P () l() = P () Q. (1) Define the cone mp A from the rectngle R = [, L] [, 1] by A(, t) = (1 t)q + tp () The imge C of A, i clled the cone over P () bed t Q. We would like to fltten C into ector S. Nmely, find ector S in the plne decribed in polr coordinte by S = {(r, θ) : θ Θ, r ρ(θ)} () for ome Θ nd ρ(θ) to be determined, nd mp B : S R for σ to be determined, B(r, θ) = ( ) r σ(θ), ρ(θ) uch tht the compoite mp A B given by ( AB(r, θ) = 1 r ) Q + ρ(θ) ( r ρ(θ) i locl iometry from S onto C, except t the cone point. ) P (σ(θ)) (3)

5 In the compoite mp A B the origin i mpped to the cone point Q,the curve given r = ρ(θ) i mpped to the curve P (), nd the rdil line egment r ρ(θ) i mpped iometriclly to egment determined by Q nd point on the pth. Hence AB(r(θ), θ) = P (σ(θ)), for ome function σ nd the complete formul for the mp AB i completely determined from the geometry. If the mp of rdil line egment i to be iometric then we mut hve ρ(θ) = l(σ(θ)), (4) thu ρ(θ) i determined once σ(θ) i known. Since the curve r = ρ(θ) i mpped iometriclly to the curve AB(r(θ), θ) = P (σ(θ)) Then we mut hve: θ d du P (σ(u)) θ du = θ θ σ (u) P (σ(u)) du = (ρ(u)) + (ρ (u)) du (ρ(u)) + (ρ (u)) du θ θ σ (u) du = (ρ(u)) + (ρ (u)) du ince the left nd right hnd ide re the rclength in three pce nd polr coordinte. It follow tht σ(θ) i imple rclength long the curve r = ρ(θ) nd (σ (θ)) = (ρ(θ)) + (ρ (θ)) (5) From thi eqution nd eqution 4 we get (σ (θ)) = (ρ(θ)) + (ρ (θ)) (1 (l (σ(θ))) ) (σ (θ)) = (l(σ(θ))) = (l(σ(θ))) + (l (σ(θ))σ (θ)) (σ (θ)) (l(σ(θ))) = (1 (l (σ(θ))) ) σ l(σ(θ)) (θ) = ± (1 (l (σ(θ))) ) Simplifying by etting = σ(θ), d dθ = σ (θ) we get the differentil eqution d dθ = l() 1 (l ()). Thi differentil eqution cn be olved by eprting nd integrting θ = σ(θ) 1 (l ()) d (6) l() 3

6 In prticulr Defining T (σ) by we my define σ by Θ = L T (σ) = 1 (l ()) d (7) l() σ 1 (l ()) d (8) l() σ(θ) = T 1 (θ) (9) Propoition 1 Let P (), l(), T (σ), σ(θ), ρ(θ) = l(σ(θ)), A, B, Θ be defined bove. Then the ector S, which i the domin of the complete cone mp ( AB(r, θ) = 1 r ) ( ) r Q + P (σ(θ)), (1) ρ(θ) ρ(θ) i defined by S = {(r, θ) : θ Θ, r ρ(θ)} (11) Exmple Let P () = ( co ( ), in( ), c), π, nd Q = then l() = + c nd Since π then θ = θ = σ(θ) d + c σ(θ) + c σ(θ) = θ + c σ(θ) π θ + c π θ π + c nd o Θ = π + c ρ(θ) = + c 4

7 3 Verifiction tht AB i locl iometry We need to how tht orthonorml frme on S i tken to n orthonorml frme on C. On S we tke the orthonorml frme r nd 1 r θ. We compute ( ) dab = d (( 1 r ) ( ) ) r Q + P (σ(θ)) r dt t= ρ(θ) ρ(θ) = P (σ(θ)) Q ρ(θ) = P (σ(θ)) Q l(σ(θ)) ( ) 1 dab r θ We clculte = 1 r = 1 r d dt t= ( rρ (θ) ρ (θ) Q (( r 1 ρ(θ + t) ) + 1 r ) ( Q + r ρ(θ + t) ) ) P (σ(θ + t)) ) ( rρ (θ) ρ (θ) P (σ(θ)) + rσ (θ) ρ(θ) P (σ(θ)) = ρ (θ) ρ (θ) Q ρ (θ) ρ (θ) P (σ(θ)) + σ (θ) ρ(θ) P (σ(θ)) = ρ (θ) ρ (θ) (P (σ(θ)) Q) + σ (θ) ρ(θ) P (σ(θ)) dab ( ) ( ) 1 dab dab r r θ ( ) ( ) dab = r r = P (σ(θ)) Q ρ(θ) = ρ (θ) ρ(θ) P (σ(θ)) Q l(σ(θ)) ( ρ (θ) P (σ(θ)) Q ρ(θ) = 1 ρ (θ) (P (σ(θ)) Q) + σ (θ) = ρ (θ) ρ(θ) + σ (θ) ρ(θ) (P (σ(θ)) Q) P (σ(θ)) ) ρ(θ) P (σ(θ)) + σ (θ) ρ(θ) (P (σ(θ)) Q) P (σ(θ)) Now nd (P (σ(θ))) Q) (P (σ(θ))) Q) = ρ (θ) (P (σ(θ))) Q) P (σ(θ))σ (θ) = ρ (θ)ρ(θ) (P (σ(θ))) Q) P (σ(θ)) = ρ (θ)ρ(θ) σ (θ) 5

8 Continuing dab Finlly, dab ( 1 r ( r θ ) ( 1 dab r ) ( 1 dab r θ θ ) ) = = ρ (θ) ρ (θ) + σ (θ) ρ(θ) (P (σ(θ)) Q) P (σ(θ)) = ρ (θ) ρ(θ) + σ (θ) ρ (θ)ρ(θ) ρ (θ) σ (θ) = ( ρ (θ) ρ (θ) (P (σ(θ)) Q) + σ (θ) ( ρ (θ) ) ρ(θ) P (σ(θ)) ) ρ (θ) (P (σ(θ)) Q) + σ (θ) ρ(θ) P (σ(θ)) ( ρ ) (θ) ( P (σ(θ)) Q = σ ) (θ) ρ(θ) ρ(θ) + P (σ(θ)) ρ(θ) ρ (θ) σ (θ) ρ (θ) ρ(θ) (P (σ(θ)) Q) P (σ(θ)) ( ρ ) ( (θ) σ ) (θ) = + ρ (θ) σ (θ) ρ (θ)ρ(θ) ρ(θ) ρ(θ) ρ (θ) ρ(θ) σ (θ) ( ρ ) ( (θ) σ ) ( (θ) ρ ) (θ) = + ρ(θ) ρ(θ) ρ(θ) = (σ (θ)) (ρ (θ)) (ρ(θ)) By eqution 5 thi quntity equl 1. 4 The rel exmple Let the pth be circle in the plne tht i offet from the origin, y rdiu nd center (b,, ), nd let Q be the point (,, c) on the z-xi.we my ume tht ( P () = (b + ( co, in( ) ), ) Then ( l() = + b + c + b co ) ( l () = + b + c + b co ) b in = + b + c + b co ( ) 6

9 So tht And 1 (l ()) l() 1 = = = b in +b +c +b co( ) + b + c + b co ( + c + b co + b co + b + c + b co + c + b co + b co + b + c + b co ) T (σ) = σ Θ = π + c + b co + b co + b + c + b co d + c + b co + b co + b + c + b co d There i cloed form for T but it i not helpful. Thu the following numericl pproch i ueful. Select ufficiently lrge N nd for j =,..., N, define σ j = jl N σ j + c + b co θ j = + b co + b + c + b co d ρ j = l(σ j ) where of coure the θ j re computed numericlly for ome numericl election of, b, c. Then Θ = θ N, nd the ector i pproximted by the pir (ρ j, θ j ). 5 Picture nd ttchment The three ttchment to follow re: 1. CAD Picture of the cone.. The region decribed by the flttened out cone. 3. Mple cript tht compute eqution for the flttened cone. Vriou 3D nd D picture re hown. 7

12 Flttening cone Prmetrizing cone et up cone nd function O P d b C$co, $in Q d,, c ; C d 1Kt $Q Ct$P; L d *π* ;, ; P := b C co in Q := c C := t b C co t in 1 Kt c L := π O QP := P-Q; l := qrt(qp[1]^+qp[]^+qp[3]^); l := implify(l); dl := implify(diff(l, )); (1.1.1) QP := b C co in Kc l := b C co C in Cc

13 pecific vlue O l := L dl := bigd d ; littled d 19.; h d 31.65; d bigd ; b d ; h$bigd c d bigd KlittleD ; t d h c ; L d ub =,b=b,c=c,l ; w d 5; bigd := littled := 19. h := := b := c := t := L := π w := 5 (1.1.) (1..1) plot 3 view O C d ub =,b=b,c=c,c ; CL d C 1, C, C 3 ; O C := t C co t in K t CL := t C co , t in , K t plot3d CL, t = t..1, =..L, orienttion = K85, 5, cling = contrined, xe = boxed ; plot3d CL, t = t..1, =..L, orienttion = K9,, cling = contrined, xe = boxed ; plot3d CL, t = t..1, =..L, orienttion = K85, 87, cling = contrined, xe = boxed ; (1.3.1)

16 Find the ector find function O QP d PKQ; l d qrt QP 1 ^ CQP ^ CQP 3 ^ ; l d implify l ; dl d implify diff l, ; qrt 1 Kdl K d ; l K d implify K ; T d int K, ;

17 QP := b C co in Kc K := l := K := l := dl := K b C co b C b co 1 K b in b C b co C in b in b C b co b C b co b co b C b co b C b co C Cc C Cc Cc C Cc C Cc C Cc C Cc Cb co C Cc T := EllipticF b Kc K C I b c b C b C Cc in K1 Cco, (.1.1) c K b K6 c b Cb 4 Cc 4 C 4 K4 I b 3 c C4 I c 3 b C4 I b c b K b C Cc b C b C Cc Kb EllipticPi b Kc K C I b c b C b C Cc in K1 Cco,

18 K b K b C Cc b Kc K C I b c, K Kb Cc C C I b c b C b C Cc b Kc K C I b c b C b C Cc Cb EllipticPi b Kc K C I b c b C b C Cc in K1 Cco, K b C b C Cc b Kc K C I b c, K Kb Cc C C I b c b C b C Cc b Kc K C I b c b C b C Cc in b co b C b co C Cc C Cc Cb co I b c co K Kb co Kb co b C b C Cc 1Cco Kc K KI b c Kb I b c co Cb co Cb co b C b C Cc 1Cco Cc C KI b c Cb b C b co C Cc b Kc K C I b c b C b C Cc b co 3 Kb co C b co O K b co C co Cc co l d ub =, b = b, c = c, l ; dl d ub =, b = b, c = c, dl ; K Kc

19 K d ub =, b = b, c = c, qrt 1 Kdl l ; plot, l, =..L ; plot, dl, =..L ; plot, K, =..L, ; l := C co dl := K in C co K := 1 K in C co C co

20 K.1 K.

21 find nd plot ector O Θ d evlf Int K, =...L ; N d 1; d d evlf L N ; point d eq j$d, j =..N ; Tvl d eq evlf Int K, =..j, j = point ; outrhovl d eq evlf ub = j, l, j = point ; inrhovl d eq t$outrhovl j, j =1..NC1 ; Θ := N := 1 d := point :=., , , , , , , , , , , , , , , , , , , , ,

22 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , Tvl :=., , ,.49768, ,.81871, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , outrhovl := , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

23 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , inrhovl := , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , O with plot : outpoint d eq outrhovl j $co Tvl j, outrhovl j $in Tvl j, j =1..N C1 ; inpoint d eq inrhovl j $co Tvl j, inrhovl j $in Tvl j, j =1..N C1 ; line1 d inpoint 1, outpoint 1 ; line d inpoint N C1, outpoint N C1 ; diply plot outpoint, cling = contrined, plot inpoint, cling = contrined, plot line1, cling = contrined, plot line, cling = contrined, xe = boxed, ; (..1)

24 outpoint := ,., , 1.537, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,.67346, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , K , , K , , K , , K , , K , , K , , K7.4915, , K , , K , , K , , K , inpoint := ,., , , ,

25 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,.848, , , , , , , , , , , , , , , , ,.93697, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,.81993, , , , , , , , K , , K , , K , , K , , K.56431, , K , , K , , K , , K , , K , , K , line1 := ,., ,. line := K , , K ,

26 K

27 O

4-4 E-field Calculations using Coulomb s Law

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