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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

889.0. "Formula for length of a spiral?" by POOL::HALLYB (Lookit all the happy creatures dancin' on the lawn) Mon Jun 20 1988 15:41

On a flat surface there is sitting a thick disk 6 inches in diameter.
There is a notch in the disk at what we will call "12 O'clock", and a
length of half-inch-thick rope coiled flat upon the surface starting
and ending at "12 O'clock".  There are exactly 30 coils.

If you were to view this from above you would see a nice neat spiral,
assuming for simplicity that someone has managed to lay out the rope 
in a smooth fashion.

--> How long is the rope?

Obviously we can estimate the length by pretending we have 30 concentric
circles and summing up their circumferences.  Doing this I get a total
"around" 630 pi.  Is it possible to come up with a better estimate?
For the general "length of a spiral"?  Closed form, perhaps?

No fair peeking at the answers.
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889.1Why not untide the rope and use a yard stick :-)HPSTEK::XIAMon Jun 20 1988 16:287
    I think that is as good as you can get.  I believe any further fine
    tuning of that model will not necessarily improve the accuracy (From
    an engineer's point of view, of course).  The problem is that 
    to further improve the model, the shape of the rope has to be taken
    into consideration.  I mean the "inner side" of the rope covers
    a shorter distance than the "outer side".
    Eugene
889.2The Tape Reel ProblemDRUMS::FEHSKENSMon Jun 20 1988 19:2416
    I once solved this problem sort of in general, trying to find the
    fraction of the total length as a function of the fraction of the
    (maximum radius - minimum radius).  The application was to be able
    to estimate the length (and hence elapsed time) of the partially
    played part of a cassette based on the relative size of the source
    and destination "pancakes".  I believe approaching the problem this
    way is independent of the thickness of the layers, but I may be
    wrong.  Anyway, I used exactly the method you propose, letting the
    layer thickness be dt and integrating suitably.  The answer was
    surprisingly simple, but I've forgotten it anyway, and am too lazy
    to repeat the exercise.  Note, however, that I *assumed* you started
    out knowing the length you are looking for, but maybe you could
    work backwards from this approach.
    
    len.
    
889.3It's not really a spiralAKQJ10::YARBROUGHWhy is computing so labor intensive?Mon Jun 20 1988 20:5011
Near the end of each loop the rope must "reach out" to 1/2" beyond the 
radius of the last loop. Call the arc length of the "reach" s, and the 
radius of the current loop r; call the angle at which the rope leaves the
previous outer edge theta. Then we have

	s = sqrt((r+1/2)^2-r^2)
	theta = arctan(s/r)

then s-r*theta is the difference between the length of the "reaching" rope
and its circular counterpart. The total length of all the loops will be
2*Pi*3+(sum[i/2,i=1..30]) + sum[(s-r*theta), r=3(.5)18].
889.4No, a REAL spiral was intendedPOOL::HALLYBThe smart money was on GoliathTue Jun 21 1988 13:5217
Re: .3 		-< It's not really a spiral >-

.0> If you were to view this from above you would see a nice neat spiral,
.0> assuming for simplicity that someone has managed to lay out the rope 
.0> in a smooth fashion.

    By the above statement back in .0 I meant to imply that in fact
    a perfect spiral was laid out, not a circle-bump-circle as .3 intends.
    
    It actually would be kind of interesting to pursue this from the
    rope's point of view (.1 fails to consider that the rope is a helix
    of strands, therefore has no "inside"/"outside"), but I would really
    prefer to know if there is a general formula for length of a spiral.
    I'm not even sure I know how to parameterize a spiral -- the rope
    notion was just to make sure we all knew what a spiral was.
    
      John
889.5CLT::GILBERTTue Jun 21 1988 19:3133
    It sounds like you want equal distances between successive 'loops'
    of the spiral.  Using polar coordinates, you'd have:

	r(theta+2pi) = w + r(theta)			(1)

    A possible solution to this recursion is:

	r(theta) = theta*w/(2pi)			(2)

    Then, you probably want to compute the *length* of the spiral, from
    theta_0 to theta_1.  That is,

	 theta_1
	integral r(theta) d theta			(3)
	 theta_0

    Using (2), this equals:

	w/(4pi) * ( theta_1^2 - theta_0^2 )		(4)

    Now if we're just interested in N loops 'starting' at radius R,
    we have theta_0 * w/(2pi) = R, and theta_1 = theta_0 + N * 2pi,
    so substituting these into (4) gives the length as:

	w/(4pi) * ( (theta_0+N*2pi)^2 - theta_0^2 )

	= w/(4pi) * ( 4*N*pi * theta_0 + 4*N^2*pi^2 )

	= w/(4pi) * ( 4*N*pi * 2*pi*R/w + 4*N^2*pi^2 )

	= N * pi * ( 2*R + N*w )

    Does this help?
889.6End of Part ICHALK::HALLYBThe smart money was on GoliathWed Jun 22 1988 13:478
    That looks like the right answer, all right.  Polar coordinates
    sure make things easier, don't they.
    
    For the special case N = 1, we have L = 2 * pi * (R + w/2);
    as w --> 0 the spiral approaches a circle and the length approaches
    the formula for circumference of a circle.
    
    So the rope is about 165 feet long.  Thanks.
889.7Arc-length is More DifficultVAXRT::BRIDGEWATERWed Jun 22 1988 15:5447
    Re: .5

    I think the problem of finding the arc-length of a spiral is a
    bit more difficult.  In what follows, I've used a somewhat
    more compact form to make the equations manageable.

    t    = angle in radians
    r(t) = radial length from pole (origin) to point.

    Then the polar equation of a spiral is:

    r(t) = c*t		[ c = w/(2pi) where w is thickness of the rope ]

    Now the arc-length, L, from t1 to t2 is:

           t2
    L = integral sqrt[r(t)^2 + r'(t)^2 ] dt
           t1

    where r'(t) represents the first derivative of r wrt t.  So,

           t2
    L = integral sqrt[c^2*t^2 + c^2] dt         or
           t1


                       t2
    L = [w/(2pi)] * integral sqrt[1 + t^2] dt
                       t1

						       | t=t2
    L = [w/(4pi)] * [t*sqrt(1+t^2) + ln(t+sqrt(1+t^2)) |
						       | t=t1

    
    If you want to measure the length of rope starting on the spiral
    at the pole, then t1=0 and:


    L = [w/(4pi)] * [t2*sqrt(1+t2^2) + ln(t2+sqrt(1+t2^2))]

    If you want to express this in terms of number of revolutions of the
    spiral, n, then make the substitution:

    t2 = 2pi*n

    - Don
889.8CLT::GILBERTWed Jun 22 1988 18:201
    Thanks.  Yes, .5 is wrong (nice try, though).
889.9Solution to Original ProblemVAXRT::BRIDGEWATERWed Jun 22 1988 23:4723
    Re: .0 using equations in .7:

    I didn't reread .0 so some of my assumptions in .7 were wrong.
    Namely, the starting and ending angles of the spiral, t1 and t2,
    should be specified by the following equations:

    w     = 0.5 inches
    r(t1) = w/2pi*t1 = 3 inches
    t2    = t1 + 2pi*30

    Solving for t1 and t2 gives:

    t1 = 12pi
    t2 = 72pi

						       | t=t2
    L = [w/(4pi)] * [t*sqrt(1+t^2) + ln(t+sqrt(1+t^2)) |
						       | t=t1

    Evaluating yields approximately:  1979.27 inches
			         or:   164.94 feet

    - Don