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From: Robert Tiismus on 13 Mar 2010 10:37
Today I calculated, for my own amusement, the correlation between my old
VHS device counter value and playing time from the beginning of the
tape. I went ahead and used my HP50g in the process to integrate the
necessary '\v/(X^2+1)' with INTVX (\v/ being the trigraph for square
root sign).
Resulting in

'X/2*\v/(X^2+1)+-1/2*LN(ABS(\v/(X^2+1)-X))'

Trying to simplify the result was not entirely succsessful. The result
in itself is right of course, but doing the same thing by hand resulted in

'1/2*(X*\v/(X^2+X)+ASINH(X))'

which has better looks :) Anyway, simplifying the last equation on HP50G
expands the ASINH and makes the equation ugly again. Maybe I am
overlooking something, but shouldn't SIMPLIFY make the equations more
simple? Furthermore, it's not entirely intuitive that
'-LN(ABS(\v/(X^2+1)-X))' (as in the INTVX answer) is equal to
'LN(ABS(\v/(X^2+1)+X))', the later being the calculation method of
ASINH. This is not to bash the integration routines of HP50G, they work
surprisingly well to get rid of the integration tables, but maybe to
have suggestions how to polish the output of the routine.

Robert Tiismus
From: John H Meyers on 13 Mar 2010 14:20
On 3/13/2010 9:37 AM, Robert Tiismus wrote:

> Today I calculated, for my own amusement,
> the correlation between my old VHS device counter value
> and playing time from the beginning of the tape.

I'm wondering how a hyperbolic function got into it.

Isn't the length of tape wound or unwound in N revolutions
just N times the mean of the starting and ending circumferences
of the wound or unwound section of the tape,
on the reel whose movement the tape counter indicates?

The ending circumference, in turn, is 2pi times N tape thicknesses
more or less than the starting circumference.

If the tape counter is proportional to the number of revolutions
of one of the reels, and if the playing time is proportional
to the length of tape that has been wound or unwound,
then unless I've made a gross mistake
(which hasn't happened in the entire previous two or three hours :)
shouldn't the playing time be a quadratic polynomial of N?

Another way to look at it is that the tape length wound or unwound
is proportional to the gain or loss in area of the windings,
which is sort of proportional to the difference between two squares
of a revolution counter having an "absolute zero" at some offset to N.

I'll come back in a few days to find out where I may have gone wrong.

[r->] [OFF]
From: Veli-Pekka.Nousiainen on 13 Mar 2010 22:11
Robert Tiismus wrote:
> Today I calculated, for my own amusement, the correlation between my old
> VHS device counter value and playing time from the beginning of the
> tape. I went ahead and used my HP50g in the process to integrate the
> necessary '\v/(X^2+1)' with INTVX (\v/ being the trigraph for square
> root sign).
> Resulting in
>
> 'X/2*\v/(X^2+1)+-1/2*LN(ABS(\v/(X^2+1)-X))'
>
> Trying to simplify the result was not entirely succsessful. The result
> in itself is right of course, but doing the same thing by hand resulted in
>
> '1/2*(X*\v/(X^2+X)+ASINH(X))'
>
> which has better looks :) Anyway, simplifying the last equation on HP50G
> expands the ASINH and makes the equation ugly again. Maybe I am
> overlooking something, but shouldn't SIMPLIFY make the equations more
> simple? Furthermore, it's not entirely intuitive that
> '-LN(ABS(\v/(X^2+1)-X))' (as in the INTVX answer) is equal to
> 'LN(ABS(\v/(X^2+1)+X))', the later being the calculation method of
> ASINH. This is not to bash the integration routines of HP50G, they work
> surprisingly well to get rid of the integration tables, but maybe to
> have suggestions how to polish the output of the routine.
>
> Robert Tiismus

Hmmm...
If you want to simplify more you could give a try for

STARTEQW.VPN

From: Robert Tiismus on 14 Mar 2010 07:31
You are right. I calculated it as the lenght of the tape spiral, having
linear function r=f(phi)=c*phi in radians. To find the lenght, one must
use integral '\.S(\v/(f^2+df^2)'. If d is thickness of the tape
(~17.5um), then the result is
L(phi)=d/(4*\pi)*(phi*\v/(phi^2+1)+ASINH(phi))
Because phi is large, the result can be simplified to
L(phi)=d*phi^2/(4*\pi)=\pi*d*N^2 .
The starting point is r=0, so the diameter of the drum (~26mm) must be
taken into account.
On the other hand, your approach gives a sum
'\GS(n=0,N,2*\pi*d*n)' The result being, by also simplifying for the
large N:
\pi*d*N^2
which is exactly the same answer as we got before. It would be fun to
calculate the differences of the two approaches, but I speculate that
the difference is smaller than the measurement error :)

My 20 years old Scneider VCR counter increment by one corresponds to
one quarter revolution of the counting drum, so it is indeed counting
linearly.

>
> I'm wondering how a hyperbolic function got into it.
>
> Isn't the length of tape wound or unwound in N revolutions
> just N times the mean of the starting and ending circumferences
> of the wound or unwound section of the tape,
> on the reel whose movement the tape counter indicates?
>
> The ending circumference, in turn, is 2pi times N tape thicknesses
> more or less than the starting circumference.
>
> If the tape counter is proportional to the number of revolutions
> of one of the reels, and if the playing time is proportional
> to the length of tape that has been wound or unwound,
> then unless I've made a gross mistake
> (which hasn't happened in the entire previous two or three hours :)
> shouldn't the playing time be a quadratic polynomial of N?
>
> Another way to look at it is that the tape length wound or unwound
> is proportional to the gain or loss in area of the windings,
> which is sort of proportional to the difference between two squares
> of a revolution counter having an "absolute zero" at some offset to N.
>
> I'll come back in a few days to find out where I may have gone wrong.
>
> [r->] [OFF]

From: Dave Hayden on 14 Mar 2010 09:40
On Mar 13, 11:37 am, Robert Tiismus <exam...(a)example.com> wrote:
> Today I calculated, for my own amusement, the correlation between my old
> VHS device counter value and playing time from the beginning of the
> tape. I went ahead and used my HP50g in the process to integrate the
> necessary '\v/(X^2+1)' with INTVX (\v/ being the trigraph for square
> root sign).
> Resulting in
>
> 'X/2*\v/(X^2+1)+-1/2*LN(ABS(\v/(X^2+1)-X))'
>
> Trying to simplify the result was not entirely succsessful. The result
> in itself is right of course, but doing the same thing by hand resulted in
>
> '1/2*(X*\v/(X^2+X)+ASINH(X))'
>
> which has better looks :) Anyway, simplifying the last equation on HP50G
> expands the ASINH and makes the equation ugly again. Maybe I am
> overlooking something, but shouldn't SIMPLIFY make the equations more
> simple? Furthermore, it's not entirely intuitive that
> '-LN(ABS(\v/(X^2+1)-X))' (as in the INTVX answer) is equal to
> 'LN(ABS(\v/(X^2+1)+X))', the later being the calculation method of
> ASINH. This is not to bash the integration routines of HP50G, they work
> surprisingly well to get rid of the integration tables, but maybe to
> have suggestions how to polish the output of the routine.
>
> Robert Tiismus

Oh this brings back memories. I wrote a program for the 41C to solve
exactly this problem back in 1983. It's # 2418C in the HP Users
Library. Well, it wasn't the exact same problem, it was geared
towards a cassette tape deck not a VCR, but the program will work with
any tape machine where the counter is proportional to the number of
revolutions of the reels.

Once nice insight I had was that you don't actually need to know to
thickness of the tape, diameter of the tape hub, tape speed etc. Once
you know that it's a quadratic function, you can get the constant
values imperically by just running a tape through and noting the
(counter reading, elapsed time) pairs a couple of spots. A little
math then derives the necessary constant terms to the quadratic
equation. The program included a setup program for this purpose.

Dave
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