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Subject: "Toroidal Helix Problems"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #640
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Jeremy
guest
Aug-04-07, 07:53 PM (EST)
 
"Toroidal Helix Problems"
 
   It is an unfortunate fact for me that I am far better at posing interesting problems than I am at solving them. So I hope that one of you whiz-bang people out there can help me out.

Let us imagine that we have a torus with a large radius R (distance from center of hole to ring center) and a small radius r (radius of the ring cross-section). Now let us imagine a circular disk of radius r that is inside the torus just touching the outer surface. We have a point on the disk that is at the innermost part of the surface at radius R-r from the torus center. We move the disk along the inside of the torus at velocity V while rotating the disk with rim velocity v until the point on the disk is 180 degrees from where it'started on the outmost part of the torus at radius R+r.

Question 1:

What is the helical distance D traveled given V,v,R and R? For a bonus point what is the distance traveled for an arbitrary amount of angle?

Questions 2 & 3:

Imagine that we have a series of "latitude" lines going around the torus that the described toroidal helix is cutting across. Does the helix cut the lines at a constant angle or does this angle vary? How do we calculate the angle?


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  Subject     Author     Message Date     ID  
Toroidal Helix Problems Jeremy Aug-04-07 TOP
  RE: Toroidal Helix Problems alexb Aug-05-07 1
     RE: Toroidal Helix Problems Jeremy Aug-05-07 2
         RE: Toroidal Helix Problems alexb Aug-05-07 3
  RE: Toroidal Helix Problems mr_homm Aug-05-07 4
  RE: Toroidal Helix Problems mpdlc Aug-08-07 5
     RE: Toroidal Helix Problems mr_homm Aug-08-07 6
         RE: Toroidal Helix Problems mpdlc Aug-08-07 7
         RE: Toroidal Helix Problems mpdlc Aug-08-07 8

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alexb
Charter Member
2065 posts
Aug-05-07, 03:57 PM (EST)
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1. "RE: Toroidal Helix Problems"
In response to message #0
 
   >Let us imagine that we have a torus with a large radius R
>(distance from center of hole to ring center) and a small
>radius r (radius of the ring cross-section). Now let us
>imagine a circular disk of radius r that is inside the torus
>just touching the outer surface. We have a point on the disk
>that is at the innermost part of the surface at radius R-r
>from the torus center. We move the disk along the inside of
>the torus at velocity V while rotating the disk with rim
>velocity v until the point on the disk is 180 degrees from
>where it'started on the outmost part of the torus at radius
>R+r.

Question -1:

Are you sure this ever happens?

Question 0:

Do you assume there is no friction between the disk and the torus? For, if there is no friction, the disk may not travel from the innermost circle to the outermost one.


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Jeremy
guest
Aug-05-07, 06:52 PM (EST)
 
2. "RE: Toroidal Helix Problems"
In response to message #1
 
   I'll try explaining again in a different way, a picture would certainly help. The circular disk is inside the torus and on a plane that is aligned parallel and coincident with the axis going through the torus hole i.e the disk is perpendicular to the torus surface and is rotating inside the torus on that plane. Now imagine that this plane is itself rotating around the torus with velocity V alogn the circular path that is the center of the torus ring. The path of a point on the rim of the disk will trace out a helix on the surface of the torus. The disk is assumed to be flush with the interior of the torus and frictionless. Is that clearer?


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alexb
Charter Member
2065 posts
Aug-05-07, 10:09 PM (EST)
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3. "RE: Toroidal Helix Problems"
In response to message #2
 
   Aha. I first imagined the disk inside the proverbial "doughnut's hole" apparently moving on the outside of the surface. (Which is to explain the background for my questions.)

So you have a point on the surface of a torus originally on the innermost circle. The point participates in two simultaneous rotations: one in the plane through the axis of the torus, the other along with this plane around the axis. Of interest is the moment in time when the point gets onto the outer rim of the torus and the distance covered till then.

The time is easy to find. Think of rotating the torus while keeping the plane through the axis fixed. The motion of the point inside this plane is independent of the rotation of the torus. The angular distance to be covered is Pi while on the surface the distance is r·Pi with the velocity of v: T = r·Pi/v.

Now, I was about to remark that the definition of V is still ambiguous and I rather it was an angular velocity. But at this point an icon popped up in the task bar alerting me to a new message. The message appeared to be related to the current thread. Thus I post the above as is and see what transpires next.


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mr_homm
Member since May-22-05
Aug-05-07, 10:10 PM (EST)
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4. "RE: Toroidal Helix Problems"
In response to message #0
 
   Hi Jeremy,

I'll have a go at your proposed problem.

The velocity V you mention must be assumed to be the velocity of the center point of the disk. (Since a rotating extended body does not have a single unique velocity, a representative point must be selected, and I assume you mean the default choice, i.e. the centroid.)

In that case, the motion of the point has two velocity components perpendicular to each other, and so its net velocity is the vector sum of these. The speed v is indeed constant by assumption, but the velocity of the point in the direction around the hole of the torus is not constant, since the plane of the disk turns as it moves, so as to always fill the torus cross section.

In that case, the angular velocity of the motion around the torus is w = V/R, and the linear velocity in the direction around the torus is therefore wx, where x = R-r·cos(a), assuming a is the angular position of the point on the disk and that a = 0 when the point is on the inside surface of the torus.

Therefore, the speed of the point through space is sqrt(v^2 + (wx_^2) = sqrt(v^2 + ((V/R)·(R-r·cos(a))^2) = V·sqrt((v/V)^2 + (1 -(r/R)cos(a))^2). This must be integrated across the time required for the half rotation around the disk. Because of the constant velocity v of the point around the disk, a = vt/r by the definition of radian measure; hence dt = r/v·da, and so the integration can be done in the angular variable a.

This gives rV/v·(integral from 0 to pi (sqrt((v/V)^2 + (1-(r/R)cos(a))^2)·da). I don't think this integral has a closed-form antiderivative. It looks to me like it will turn into an elliptic integral of some sort. It is a definite integral, so there may be a solution for its numerical value in a table if you can process it into the appropriate form.

Hope that helps!

--Stuart Anderson


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mpdlc
guest
Aug-08-07, 06:48 AM (EST)
 
5. "RE: Toroidal Helix Problems"
In response to message #0
 
   If I understood correctly your post the point of contact M of the disk D with the toroidal surface T has an speed with two components always perpendicular to each other. Given the position of point M in any instant component v will be in direction of the tangent to meridian of the toroidal surface and V will be tangent to the parallel of M. Similarly to the helix on a cylinder.

Subsequently the modulus of the total combined speed W= SQRT( v^2 + V^2)

One stated the above the solution is immediate

First the time inverted by M to go from the inner equator of the torus to the outer equator it will be obviously T= r*pi/v. Then the total distance traveled by M from will be D= T* SQRT( v^2 + V^2).

Second the angle that the trajectory form with the tangent parallel remain constant. The value of the tangent of the said angle it is the quotient of the two components of the speed. Tan = v/V.

If you need to know the poison of M in any given time below is the parametric equation of the path of M on the toroidal surface

X = (R-r*cos(a))cos(h)
Y = (R-r*cos(a))sin(h)
Z = R*sin(a)

Where a = v*t/r ; h = V*t/R being t the time. So by substitution of a and h above you get the time equation of M trajectory.


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mr_homm
Member since May-22-05
Aug-08-07, 10:04 AM (EST)
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6. "RE: Toroidal Helix Problems"
In response to message #5
 
   @mpdic:

The equations you give for X, Y, and Z are correct (except in the last one, you need "r" not "R"), but together they contradict your earlier statement that V is the velocity along a parallel of M. Taking the derivatives, you get

X' = (r*sin(a)*a')*cos(h) - (R-r*cos(a))*sin(h)*h'
Y' = (r*sin(a)*a')*sin(h) + (R-r*cos(a))*cos(h)*h'
Z' = r*cos(a)*a'.

X'^2 + Y'^2 = (r*sin(a)*a')^2 + ((R-r*cos(a))*h')^2 after canceling terms, and so

X'^2 + Y'2 + Z'^2 = (r*a')^2 + ((R-r*cos(a))*h')^2.

Since a' = v/r and h' = V/R, this simplifies to

X'^2 + Y'2 + Z'^2 = v^2 + (V-Vr/R*cos(a))^2 =

v^2 + V^2(1-r/R*cos(vt/r))^2, which is obviously not v^2 + V^2.

This is because, as I mentioned in my earlier post, the velocity component tangent to a parallel of M, for a point on the rim of the disk, will not be V, because of the normal vector to the disk rotates as the disk moves around the torus. This gives an additional velocity parallel to V, which you can see in the formula above. By the way, this formula for the total speed now agrees with the one I gave earlier, although this time, the derivation proceeds by differentiating your paramatrized curve, instead of directly calculating the velocity components. Unfortunately, this results in a hard integral.

Your solution would be correct in the limit of large R/r, i.e. for a true helix wrapped around a cylinder.

Hope this helps!

--Stuart Anderson


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mpdlc
guest
Aug-08-07, 12:17 PM (EST)
 
7. "RE: Toroidal Helix Problems"
In response to message #6
 
   Thank you Stuart for your observation. After reviewing my draft paper I messed up with r, R and also with the greek letter rho . I fully agree that the expression for Z should be Z = r*sin(a) .

Regarding to the parametric equation my intention was to write h = V*t/rho meaning rho = R-rcos(a) Maybe I am wrong, but my interpretation is that the component of velocity always in following the parallel remains the same, what means the rotating speed of point M around the torus main axis will vary because of the term r*cos(a) added to R in the expression for h

I do agree with you that the disk has to have a yawing component like it happen in the regular helix however this yawing component varies it is not constant like in the regular helix and in both cases it is responsible for the twist of the curve, but being its axis of yawing passing through, the center of the disk, the contact point M and the center of the circumference of the meridian passing on M will not yield any component at the velocity for M.

Having said that I do not contest your interpretation since the enunciate of the whole subject it is not well defined ( to define a vector you need the modulus, the direction and the application point what it is not the case for V in the post) so it can be interpreted in a different ways. So is licit for you and as I unintentionally posted in my equation for h = V/R to consider the rotation as constant.


However since the author was apparently was looking for a sort of helix I tried to make sense and get a loxodromic route so the angle with parallel being always constant, then given v in order to get this route, the horizontal component of M Vh has to remain constant too and shortcut Vh as V. I do agree with you that otherwise to solve analytically the integral it will be at least a nasty if not impossible task.

Thank you for your always useful remarks.


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mpdlc
guest
Aug-08-07, 09:39 PM (EST)
 
8. "RE: Toroidal Helix Problems"
In response to message #6
 
   After my last post I was rethinking in the equation of the trajectory and I concluded that the first assumption I made of h = V*t/(R-r*cos(a)) it is wrong too since the term V/(R-r*cos(a)) represent the rotation at precisely the instant t but not for all times from the initial instant so the correct term for h it is much more complicated and its value is

h = Integral (V/(R-r*cos(a)))dt between 0 and t

I found the analytical solution for the integral on a Table of Integral book for values of R > r but the expression if so cumbersome that it just too complicated to write in a normal word processor, then by substitution of h we will get if not the solution of the problem, since V has not been well defined, a least the loxodromic rhumb line on a toroidal surface.


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