Bertrand Curves











Bertrand curves and its properties

A pair of space curves C and C1 and are called Bertrand curves if principal normal to C is also principal normal to C1 , i.e, n1=n




Principal Noemals are Same
o+
az = 1.00
el = 0.30
P
P1
Principal Noemals are Same



Properties of Bertrand curves

Let C and C1 are Bertrand curves then

  1. Distance between corresponding points of Bertrand curves is constant
  2. Tangent at corresponding points of Bertrand curves inclined at constant angle
  3. Curvature and torsion of either curves are connected by linear relation
  4. Torsion of Bertrand curves have same sign and their product is constant

Let C and C1 are Bertrand curves then mathematics of above statements are

  1. If λ is distance between corresponding points of Bertrand curves, then λ=0
  2. if t is tangent to C and t1 is tangent to C1, dds(t.t1)=0
  3. if κ and τ are curvature and torsion of C then
    aκ+bτ+c=0 ; κ and τ are connected by linear relation
    if κ1 and τ1 are curvature and torsion of C1 then
    a1κ1+b1τ1+c1=0 ; κ1 and τ1 are connected by linear relation
  4. if τ and τ1 are torsion of C and C1 then τ and τ1 have same sign, means τ.τ1=positive and τ.τ1=constant


Proof
  1. Distance between corresponding points of Bertrand curves is constant
    o+
    O
    P
    P1
    r
    r1
    n=n1
    , 7+2*Math.sin(t)];}; function (t) { return [ 7+5*Math.cos(t), 7+5*Math.sin(t)];}; board.create('curve',[r, 0, Math.PI],{strokeColor:'red'}); board.create('functiongraph',[r1, 0, Math.PI],{strokeColor:'red'}); Let C and C1 are Bertrand curves then
    n1=n(i)
    Let P be a point on C and P1 be its corresponding on C1 with
    OP1=r1
    OP=r
    Then,
    PP1 is principal normal to C
    or PP1=λn
    where λ is distance between corresponding points P and P1
    or OP1OP=λn
    or r1r=λn
    or r1=r+λn (ii)
    Differentiating (ii) w. r. to. s , we get
    t1ds1ds=t+λn+λ(τbκt)
    or t1ds1ds=t+λn+λτbλκt
    or t1ds1ds=(1λκ)t+λn+λτb
    Taking dot product by n1=n, we get
    (t1ds1ds)n1=[(1λκ)t+λn+λτb]n
    or 0=[0+λ1+0] t1.n1=0,t.n=0,b.n=0
    or 0=λ
    It shows that, distance between corresponding points of Bertrand curves is constant
    The first property established.

  2. Tangent at corresponding points of Bertrand curves inclined at constant angle
    Let t is tangent to C and t1 is tangent to C1,
    and also let, α is angle between t and t1, then
    cosα=t.t1|t|.|t1|
    or cosα=t.t1
    Differentiating w. r. to s, we get
    dds(cosα)=dds(t.t1
    or dds(cosα)=(ddst).t1+(ddst1)t
    or dds(cosα)=(κn).t1+(κ1n1ds1)ds)t
    In Bertrand curves, n1=n , so
    Replacing n by n1 and also replacing n1 by n in the right, we get
    dds(cosα)=(κn1).t1+(κ1nds1)ds)t
    or dds(cosα)=0+0
    or dds(cosα)=0
    It shows that,
    derivative of cosα is zero
    so
    cosα is constant
    or α is constant
    Hence, yangent at corresponding points of Bertrand curves inclined at constant angle
    The second property established.

  3. Curvature and torsion of either curves are connected by linear relation
    Let C and C1 are Bertrand curves then using (A), we write
    t1ds1ds=t+λn+λτbλκt
    t1ds1ds=(1λκ)t+0×n+λτb
    or t1ds1ds=(1λκ)t+λτb (iii)
    Also, we know that,
    o+
    az = 1.00
    el = 0.30
    t_1
    b_1
    t
    b
    n=n_1
    • principal normals to C and C1 are same
    • tangents to C and C1 are inclined at constant angle α
    • binormals to C and C1 also inclined at same constant angle α

    Hence,
    t1=cosαt+cos(90α)b
    or t1=cosαt+sinαb(iv)
    Comparing (iii) and (iv), we get
    ds1ds=1λκcosα=λτsinα
    We have to show relation between κ and τ, so
    Taking, second and third part, we get
    1λκcosα=λτsinα
    or (1λκ)sinα=λτcosα
    or (λsinα)κ+(λcosα)τ+sinα=0
    This shows that κ and τ are connected by linear relation in the form
    aκ+bτ+c=0 where a,b,c are constants given by a=(λsinα),b=(λcosα),c=sinα
    Next,
    Since relation between C and C1 are reverse in terms of α and λ we have
    λ will be λ and α will be α for C1
    Thus, while writting expression for κ1 and τ1 , we get
    (λsin(α))κ1+(λcos(α))τ1+sin(α)=0
    or (λsinα)κ1+(λcosα)τ1+(sinα)=0
    This shows that κ1 and τ1 are connected by linear relation in the form
    a1κ1+b1τ1+c1=0
    where a1,b1,c1 are constants given by
    a1=(λsinα),b1=(λcosα),c1=(sinα)
    The third property established.



  4. Torsion of Bertrand curves have same sign and their product is constant
    Let C and C1 are Bertrand curves then using (B), we get
    ds1ds=1λκcosα=λτsinα
    We have to show relation between τ and τ1, so
    Taking, first and third part, we get
    ds1ds=λτsinα (v)
    Since relation between C and C1 are reverse in terms of α and λ . Thus, while writing expression for τ1 , we get
    dsds1=λτ1sin(α) (vi)
    Multiplying (v) and (vi), we get
    ds1dsdsds1=λτsinαλτsin(α)
    or1=λτsinαλτ1sinα
    or1=λ2ττ1sin2α
    orττ1=(sinαλ)2
    Torsion of Bertrand curves have same sign and their product is constant
    The forth property established.

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