Re: ATM 5 hour error-free straight bolt tracking

[Anthony Stillman <atmer@flash.net>]

What a great idea.  After I read Doug's post, I blew through the math on a
scrape of paper.  This is what I came up with.  (Those who find derivations
boring, please skip to the last paragraph)

For circular motion at a constant angular velocity we have

theta = omega * t

where
  theta     the time dependent angular position
  omega     the angular velocity
  t         time

Again for constant circular motion the rate of change in position angle
with respect to time is

d/dt (theta)  = d/dt (omega * t)
                    = omega

Further the second derivative, that is the angular acceleration is

d/dt ( d/dt (theta) ) = d/dt ( d/dt ( omega * t) )
                      = d/dt ( omega )
                      = 0

these are the conditions to be matched by the Sliding Tangent Drive ;-)

>From the description, the angle theta of the sliding board is defined by a
translation and board length.  Specifically

theta = arcsin ( c * t  / h )

where
  c    translation constant in length per unit time
  h    board length from sliding contact to swivel nut afixed to board
  t    time

The first derivative of theta with respect to time is

d/dt (theta) = d/dt ( arcsin (c*t/h))
             = c / (h^2 - c^2 * t^2)^0.5

valid for -pi/2 <= theta <= pi/2, i.e. straight down and straight up

Clearly this isn't a constant with respect to time.  However for large h
and small t its not bad.  How not bad?  The second derivative of theta is

d/dt ( d/dt ( theta ) ) = d/dt(c / (h^2 - c^2 * t^2)^0.5 )

                        = c^3 * t / (h^2 - c^2 * t^2)^1.5

We would like this to be zero.  To approximate that c should be small.


Roughly omega above is 0.0044 radians per minute.  Ideally the first
derivative of the sliding tangent drive should be this as well.  Hence

  0.0044 = c / (h^2 - c^2 * t^2)^0.5

We want to keep c small so I chose a 1 RPM motor driving a 36 threads per
inch rod for a c value of

  c = 1 RPM / 36 RPI
    = 0.0278 IPM  (inches per minute)

Hence if
  t = 0
then
  h = c / 0.0044
    = 0.0278 / 0.0044
    = 6.31 inches

As noted above we would like the second derivative to be zero, and at t=0
it is.  In fact the deviation from zero acceleration is very small.
However as the drive proceeds error accumulates.  Specifically for time in
minutes and theta in radians

Time     Desired theta   Actual theta    Desired omega    Actual omega
t = 10      0.0436          0.0436        0.004363323     0.004363323
t = 30      0.1309          0.1313                        0.004363324
t = 60      0.2618          0.2649                        0.004363326
t = 120     0.5236          0.5512                        0.004363335
t = 229     0.9992          1.5308                        0.004363365

At thirty minutes the accumulated error amounts to 1.3 arc seconds.  At one
hour the accumulated error is 10.6 arc seconds.  After two hours the error
is 1.6 degrees.  However the drive rate is still nearly ideal.  Hence there
would be little need to rewind the drive.

So there you have it, the meaning of life, the path to inner peace and the
answer all the great wonders of the universe.  Someone, please check my
work.

Anthony


PS The value for omega is not sidereal.  A slight change in h will adjust
for this.  A slight change in h will also phase shift the error and entend
the drives useful range.  Thankyou Doug for posting your idea.