Arc length of a trochoid

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Wolfram MathWorld gives parametric expressions for a trochoid in terms of a parameter $phi$:
begineqnarray*
x &=& aphi - b sin phi\
y &=& a - b cos phi,
endeqnarray*
but then gives expressions for the arc length, curvature, and tangential angle in terms of a parameter $t$ instead, which is not defined there. Can anyone verify that $t$ is the same as $phi$, or else tell me how they are related?

See http://mathworld.wolfram.com/Trochoid.html.

arc-length

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asked Jul 18 at 18:46

user45160

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Wolfram MathWorld gives parametric expressions for a trochoid in terms of a parameter $phi$:
begineqnarray*
x &=& aphi - b sin phi\
y &=& a - b cos phi,
endeqnarray*
but then gives expressions for the arc length, curvature, and tangential angle in terms of a parameter $t$ instead, which is not defined there. Can anyone verify that $t$ is the same as $phi$, or else tell me how they are related?

See http://mathworld.wolfram.com/Trochoid.html.

arc-length

share|cite|improve this question

asked Jul 18 at 18:46

user45160

1032

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Wolfram MathWorld gives parametric expressions for a trochoid in terms of a parameter $phi$:
begineqnarray*
x &=& aphi - b sin phi\
y &=& a - b cos phi,
endeqnarray*
but then gives expressions for the arc length, curvature, and tangential angle in terms of a parameter $t$ instead, which is not defined there. Can anyone verify that $t$ is the same as $phi$, or else tell me how they are related?

See http://mathworld.wolfram.com/Trochoid.html.

arc-length

share|cite|improve this question

asked Jul 18 at 18:46

user45160

1032

Wolfram MathWorld gives parametric expressions for a trochoid in terms of a parameter $phi$:
begineqnarray*
x &=& aphi - b sin phi\
y &=& a - b cos phi,
endeqnarray*
but then gives expressions for the arc length, curvature, and tangential angle in terms of a parameter $t$ instead, which is not defined there. Can anyone verify that $t$ is the same as $phi$, or else tell me how they are related?

See http://mathworld.wolfram.com/Trochoid.html.

arc-length

share|cite|improve this question

asked Jul 18 at 18:46

user45160

1032

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 18:46

user45160

1032

asked Jul 18 at 18:46

user45160

1032

asked Jul 18 at 18:46

user45160

1032

1032

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

Since this is a first time I hear about trochoids, I just computed the arc length using
$$begineqnarray*
x &=& aphi - b sin (phi) implies x'=a -bcos(phi)\
y &=& a - b cos (phi)implies y'=b sin (phi)
endeqnarray*$$
$$L=int sqrt(x')^2+(y')^2 ,dphi=int sqrta^2+b^2-2 a b cos (phi )=2 |a-b| ,Eleft(fracphi 2|-frac4 a b(a-b)^2right)$$ if $a neq b$.

Looking at the first argument of the elliptic integral, it seems that your are totally correct.

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
  – user45160
  Jul 20 at 4:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, never mind - I see that Wolfram Language defines their parameter differently.
  – user45160
  Jul 20 at 4:55
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Since this is a first time I hear about trochoids, I just computed the arc length using
$$begineqnarray*
x &=& aphi - b sin (phi) implies x'=a -bcos(phi)\
y &=& a - b cos (phi)implies y'=b sin (phi)
endeqnarray*$$
$$L=int sqrt(x')^2+(y')^2 ,dphi=int sqrta^2+b^2-2 a b cos (phi )=2 |a-b| ,Eleft(fracphi 2|-frac4 a b(a-b)^2right)$$ if $a neq b$.

Looking at the first argument of the elliptic integral, it seems that your are totally correct.

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
  – user45160
  Jul 20 at 4:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, never mind - I see that Wolfram Language defines their parameter differently.
  – user45160
  Jul 20 at 4:55
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Since this is a first time I hear about trochoids, I just computed the arc length using
$$begineqnarray*
x &=& aphi - b sin (phi) implies x'=a -bcos(phi)\
y &=& a - b cos (phi)implies y'=b sin (phi)
endeqnarray*$$
$$L=int sqrt(x')^2+(y')^2 ,dphi=int sqrta^2+b^2-2 a b cos (phi )=2 |a-b| ,Eleft(fracphi 2|-frac4 a b(a-b)^2right)$$ if $a neq b$.

Looking at the first argument of the elliptic integral, it seems that your are totally correct.

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
  – user45160
  Jul 20 at 4:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, never mind - I see that Wolfram Language defines their parameter differently.
  – user45160
  Jul 20 at 4:55
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Since this is a first time I hear about trochoids, I just computed the arc length using
$$begineqnarray*
x &=& aphi - b sin (phi) implies x'=a -bcos(phi)\
y &=& a - b cos (phi)implies y'=b sin (phi)
endeqnarray*$$
$$L=int sqrt(x')^2+(y')^2 ,dphi=int sqrta^2+b^2-2 a b cos (phi )=2 |a-b| ,Eleft(fracphi 2|-frac4 a b(a-b)^2right)$$ if $a neq b$.

Looking at the first argument of the elliptic integral, it seems that your are totally correct.

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

Since this is a first time I hear about trochoids, I just computed the arc length using
$$begineqnarray*
x &=& aphi - b sin (phi) implies x'=a -bcos(phi)\
y &=& a - b cos (phi)implies y'=b sin (phi)
endeqnarray*$$
$$L=int sqrt(x')^2+(y')^2 ,dphi=int sqrta^2+b^2-2 a b cos (phi )=2 |a-b| ,Eleft(fracphi 2|-frac4 a b(a-b)^2right)$$ if $a neq b$.

Looking at the first argument of the elliptic integral, it seems that your are totally correct.

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

answered Jul 19 at 4:10

Claude Leibovici

112k1055126

112k1055126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
  – user45160
  Jul 20 at 4:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, never mind - I see that Wolfram Language defines their parameter differently.
  – user45160
  Jul 20 at 4:55
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
  – user45160
  Jul 20 at 4:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, never mind - I see that Wolfram Language defines their parameter differently.
  – user45160
  Jul 20 at 4:55
  

  
  
  
  

Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
– user45160
Jul 20 at 4:46

Thanks for working that out! Your second argument in the elliptic integral is different from theirs. Does it work out to the same thing?
– user45160
Jul 20 at 4:46

Oh, never mind - I see that Wolfram Language defines their parameter differently.
– user45160
Jul 20 at 4:55

Oh, never mind - I see that Wolfram Language defines their parameter differently.
– user45160
Jul 20 at 4:55

add a comment | 

 

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