The Old VCR Problem

When videocassette recorders first came out, the only way of knowing where the tape was was a counter, usually connected to the takeup side of cassette.  When the tape first started the empty takeup side would turn rapidly and the counter would advance quickly as well.  As more tape played or was recorded the takeup side would get larger, turning (and counting) more slowly.

If given the exact diameter of the spindle, speed and thickness of the tape, and gear ratio between the takeup reel and counter, it would be straightforward to come up with a way to relate the counter to time the tape has been playing and/or recording.  The problem is how to derive this factor without it. All of these come out in the final conversion, in fact, it is not necessary to even use pi in this problem except to factor it out in the first step.

Let's assume a constant tape speed and that the tapes will always be rewound (and counter zeroed) before using.  If we know the counter value after recording for exactly 1 hour and after exactly 2 hours, would that be enough information?  Certainly.  There is a direct ratio between the cross sectional area of tape on the takeup side and time as well as between the counter and change in radius of any "donut".

Since the counter starts with a zero value, the ultimate equation will be
               1)          Time = [(x + R)2 - R2] / K
where the K is a conversion factor, x is a counter value, and R is the radius of the spindle in counter units.  In the diagram, the red area represents the tape for the 1st hour while the yellow area does the same for the 2nd hour.  Both have the same area--the yellow donut is bigger but also thinner.  We can equate these donuts and solve for R, or assume two red donuts equal one red AND yellow donut.  Everything in the equation would be divided by the conversion factor K so it isn't even shown (yet!).

2 * ( [x1 + R]2 - R2)              = (x2 + R)2 - R2
2 * ([x12 + 2x1R + R2] - R2) = x22 + 2x2R + R2 - R2
                   2 * (x12 + 2x1R) = x22 + 2x2R
                         2x12 + 4x1R = x22 + 2x2R
                         4x1R - 2x2R = x22 - 2x12
                         R(4x1 - 2x2) = x22 - 2x12
                                         R = (x22 - 2x12) / (4x1 - 2x2)

Since we have R, we can plug it into Time = [(x + R)2 - R2] / K as well as x1 which is the counter reading for one hour.  We already know Time = 1 hour so we can rearrange: K = [(x + R)2 - R2] / 1 hour.  The number we get is our conversion factor and includes everything mentioned in the 2nd paragraph above.  Dividing [(x + R)2 - R2] by our K will give the hours for any counter value--further dividing K by 60 will yield the time in minutes in equation 1).  To determine the counter for a given time, subtract the minutes from both sides of the equation and solve the quadratic for a positive x.

Note this diagram is an isomorph of the first one.  The areas are defined by subtracting the area of one triangle (instead of circle) from another and it is still necessary to subtract R2 from (R + x)2.  Once * slope * conversion factor is divided out, it is the same problem.

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In another forum, Fetika created this diagram for R1 = 500 and R2 = 950 (or x1 = 500 and x2 = 950).