# A new computation of G from the Cavendish experiment

21 views

### pioneer1

Aug 27, 2007, 9:47:27 PM8/27/07
to
Hi,

I used Cavendish's observations (Experiment IV of August 12, 1797) to
compute a value for G. I believe that this was not done in modern
times. G is usually computed from the mean density of the earth
computed by Cavendish using the formula N^2/(a constant) B. But I am
getting a value about 2.5 times greater than CODATA value of G.

I would be greateful to anyone who could take a look and let me know
why the discrepency.

The computations are here:

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_and_G

Thanks again.

### Edward Ruden

Aug 28, 2007, 8:34:33 PM8/28/07
to
On Aug 27, 7:47 pm, pioneer1 <1pione...@gmail.com> wrote:
..
> http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_and_G

You need to make your calcuations more clear and self-contained. It
requires too much detective work to even figure out what your
variables are defined as in terms of measurements. Like, how do you
define theta? Deflection from equilibrium without gravity from either
big ball would be appropriate for it's occurance in the equations, but
is that how you define it, or is it deflection between going from M to
M' attraction (which should be 2*theta)? Such a confusion could be the
key to the discrepancy. It's not clear what *you* think the variables
should represent, so we'll never know.

### Uncle Al

Aug 29, 2007, 6:00:22 AM8/29/07
to

More recent data,

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

### Dr J R Stockton

Aug 30, 2007, 1:37:18 AM8/30/07
to
In sci.physics.research message <1188268491....@19g2000hsx.goog
legroups.com>, Tue, 28 Aug 2007 01:47:27, pioneer1 <1pio...@gmail.com>
posted:

>I used Cavendish's observations (Experiment IV of August 12, 1797) to
>compute a value for G. I believe that this was not done in modern
>times. G is usually computed from the mean density of the earth
>computed by Cavendish using the formula N^2/(a constant) B. But I am
>getting a value about 2.5 times greater than CODATA value of G.

The ratio is suspiciously close to 8/3. I don't find the description of
the experiment easy enough to follow; and it seems not to agree with
another description.

--
(c) John Stockton, Surrey, UK. *@merlyn.demon.co.uk / ??.Stoc...@physics.org
Web <URL:http://www.merlyn.demon.co.uk/> - FAQish topics, acronyms, & links.
Correct <= 4-line sig. separator as above, a line precisely "-- " (SoRFC1036)
Do not Mail News to me. Before a reply, quote with ">" or "> " (SoRFC1036)

### pioneer1

Aug 30, 2007, 1:37:20 AM8/30/07
to
On Aug 29, 6:00 am, Uncle Al <Uncle...@hate.spam.net> wrote:
> pioneer1 wrote:

If it is fine with the moderators I would like to discuss in this
thread only the computations related to the original Cavendish
experiment. I would appreciate if you could take a look at the
computations. Thanks.

### pioneer1

Aug 30, 2007, 1:37:19 AM8/30/07
to

Ok. I would like to make the calculations self-contained. Let me know
what is not clear. For theta, you might want to check Figure 1 here:

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_and_G

In Cavendish's pendulum, as shown in the figure, the scale division 20
was where the pendulum arm was at rest. At 20 divisions theta equals
zero. In this particular experiment he moved the weights from M to M'.
At M the rest point of the arm was at 18.01 divisions. When he moved
the weights to M' the rest point of the arm moved (as calculated by
Cavendish) to 24.04. I used r = 24.04 - 20 as the angle of
displacement to compute restoring torque tau = k theta. I computed
theta as theta = gyration arm / r = 0.0547 radians.

I don't understand why I should be using 2 theta. Can you explain?

Thanks.

### DRLunsford

Aug 30, 2007, 10:55:51 AM8/30/07
to

Well the arccos of (sqrt(1/2.67)) is 52 deg 16 min, which happens to
be the latitude of Cambridge, more or less.

-drl

### Edward Ruden

Aug 30, 2007, 10:55:52 AM8/30/07
to
I've found your mistakes and corrected them below. Incidentally, I
myself have personnally measured G to within 1% in 1980 in Junior Lab
when I was a physics major at Case Western Reserve University. I made
improvements to the Cavendish balance and analysis techniques that
resulted in a more accurate measurement than any previous student had
ever obtained on that instrument, according to the profession. So, I
can personnally assure you that G is within 1% of its moderrn quoted
value without any need for conformity with the hide-bound reactionary
physics establishment :-)

I will scan and upload my final report to my reprints page at
http://home.comcast.net/~rudenbiz/physics/pubs/index.htm
Give me a week or so.

On Aug 29, 11:37 pm, pioneer1 <1pione...@gmail.com> wrote:

> theta as theta = gyration arm / r = 0.0547 radians.

The above is an example of exactly what I mean when I say you don't
define things clearly. You define r in your Fig. 1 as the distance on
the scale of the arm with the mass in the M' location to the center of
the scale (which you assume to correspond to theta=0). But what is
"gyration arm"? The arm is a physical object, not a measurement.
Furthermore, r should be in the numerator. All your variables should
be defined in terms of elementary measurements of distance and mass
for people to understand you.

Now, as for theta, you can't assume that theta=0 when the scale reads
20 divisions of 1/20'th of an inch. You'd have to rotate the weights
to the neutral positiion half-way between the two extremes and measure
the scale to determine that. You said, however, only that measurements
where taken with M' and M at the near positions, at which point the
equilibrium position on the scale read S = 24.04 divisions and S' =
18.01 divisions, respectively. Since the apparatus is symmetric, that
means the gravity-free equilibrium position is half-way between
(21.025 div). r, then should be defined as r=(S-S')/2= 3.015 div =
0.383508 cm. Theta, then is r divided by the radius of the arm. For
this, you have an inconsistency in the definition of L. In the linked
calculation of moment of inertia it's defined as the length of the arm
L=186.18 cm, but on the main page, its numerical value is half that
(ie, the radius). I'll take the former definition, so theta = 2r/
L=0.00412 cm. This itself only provides a minor to you G calculation,
reducing it to 1.341E-7.

A larger error apparently results from you confusion about L and/or
what the torque is. The torque due to Gravity expression in you main
page is correct, but only if you assume that L is the *length*. Recall
that the wire is torqued by *two* masses on each end of the arm,
resulting in a torque of (L/2)*2GMm/s^2. Since you, however, use the
radius definition, you need to divide your G estimate by two to give
G=6.708E-8. This is about 1% from it's modern value, just as Cavendish
obtained.

### pioneer1

Sep 2, 2007, 2:41:49 PM9/2/07
to
On Aug 30, 10:55 am, Edward Ruden <rudenbz...@yahoo.com> wrote:

> I've found your mistakes and corrected them below.

This is great! Many thanks for your help with this. I agree with
everything you wrote and I corrected the computations in the wiki and
I gave you credit for the corrections (I hope this is ok).

> Now, as for theta, you can't assume that theta=0 when the scale reads
> 20 divisions of 1/20'th of an inch. You'd have to rotate the weights
> to the neutral positiion half-way between the two extremes and measure
> the scale to determine that.

Ok. But Cavendish neglected this step. He computed the rest points
from 3 successive measurements and he included them in his chart.

>You said, however, only that measurements
> where taken with M' and M at the near positions, at which point the
> equilibrium position on the scale read S = 24.04 divisions and S' =
> 18.01 divisions, respectively.

I made a couple of mistakes with numbers here. Now I corrected the
figure 1 in the wiki. Correct numbers are:

S = 24.02 and S' = 18.10

>...r, then should be defined as r=(S-S')/2= 3.015 div =
> 0.383508 cm.

Ok. With the corrected numbers I got:

r = (S-S')/2=2.96 div = 1.48" = 3.759 cm

>Theta, then is r divided by the radius of the arm.

Right, my mistake. Corrected:

theta = 3.759/93.09 = 0.004038

> ... the wire is torqued by *two* masses on each end of the arm,

> resulting in a torque of (L/2)*2GMm/s^2.

Ok. I corrected this, and I get

6.67294*10^-8

This is 0.9998 of the recommended value. So, very close. But I think
that 5 digit accuracy is not justified here. Cavendish gave the
density of the earth to only two digits, so maybe the result of this
computation should be

6.67*10^-8.

Thanks again for your help with this. I'll check your computation when
it is up on your site. I am glad that someone else is interested in
the original Cavendish experiment. I have other questions about the
experiment that I hope to post soon.

### Edward Ruden

Sep 5, 2007, 8:02:32 AM9/5/07
to
On Sep 2, 12:41 pm, pioneer1 <1pione...@gmail.com> wrote:
> ... so maybe the result of this
> computation should be
>
> 6.67*10^-8.
>

Yes.

I've uploaded my old final report to
http://home.comcast.net/~rudenbiz2/physics/pubs/Ruden_80_Cavendish_Experiment.pdf
What we've done so far here is straightening out the most basic
analysis. Once you get the error in G down to the percent level,
you'll want to perform a more accurate analysis. For example, in your
calculation of the torsion constant, you have negected that fact that
both damping and the gravitational field *gradient* perturb the
oscillation period. Also, you neglect the gravitational attraction of
the more distance stationary ball on each pendulum ball. I don't know
what Cavendish himself considered, but these three effects are
considered in my final report. It turns out that the field gradient
effect actually cancels out if you take the equilibrium theta
difference between M and M' tests as we have - but you need to prove
this for a valid analysis.

### pioneer1

Sep 8, 2007, 9:25:42 PM9/8/07
to
On Sep 5, 8:02 am, Edward Ruden <rudenbz...@yahoo.com> wrote:

Many thanks for posting your Cavendish lab paper. I have been looking
to find the equation of motion for the experiment for a while now.

http://www.sas.org/tcs/weeklyIssues_2005/2005-07-01/feature1/index.html

which has the same differential equation but I couldn't figure out the
solution. It was a nice surprise to find that you have the solution as
well.

I have many questions. But first I would like to find the solution
without the simplifying assumption that you make

x = theta d << a

and compare it with the solution you actually used in your experiment.
Do you know that solution?

How did you obtain w_0^2? In pendulum motion w^2 = K/I, I understand
that part. But I couldn't figure out the algebra for the second term
4GMmd^2/Ia^3.

I also tried to simplify the equation of motion by ignoring damping
due to air. I assumed that R=0 and that eliminated the exponential. I
also eliminated the phase angle phi and also substituted B and w_0^2
and w_0 to see the constant terms explicitly. Since it is difficult to
write those equations in the newsgroup I put them in my wiki.

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_%28equation_of_motion%29

As far as I understand, using this equation of motion we add a
constant factor to the trigonometric simple harmonic motion of the
pendulum. Did you notice in your measurements the effects of the force
in the pendulum motion? How did the the pendulum arm vary because of
the force term?

Thanks again, this was very helpful.

### Edward Ruden

Sep 9, 2007, 10:15:26 PM9/9/07
to
On Sep 8, 7:25 pm, pioneer1 <1pione...@gmail.com> wrote:
well.
> ... I would like to find the solution

> without the simplifying assumption that you make
>
> x = theta d << a

This assumption is necessary to linearize the differential equation.
That means it consists of a sum of zero'th through second derivatives
of theta with constant coefficients equal to a constant. The reason
for this is that such a dif eq has a simple analytic solution
consisting of an exponentially decaying sine wave. Without
linearization, your stuck you'll need to solve it numerically. You you
want to take that route, get a book on numerical methods.

> How did you obtain w_0^2? In pendulum motion w^2 = K/I, I understand
> that part. But I couldn't figure out the algebra for the second term
> 4GMmd^2/Ia^3.

Remember, it's not just pendulum motion. There is a graviational
gradient and damping the effect the period too. Linearization also
requires approximating the torque due to gravity by a + b*theta, where
a and b are constants. a is the torque at theta=0 and b = dN/dtheta at
theta = 0. This a correction due to the gravitational gradient. b,
then gets added to the theta coefficient K coefficient and has the
effect of decreasing omega_0 in the general solution to the dif eq. To
understand this, and why the actual frequency omega_f is shifted by
damping, read an introductory book on differential equation.

> As far as I understand, using this equation of motion we add a
> constant factor to the trigonometric simple harmonic motion of the
> pendulum. Did you notice in your measurements the effects of the force
> in the pendulum motion? How did the the pendulum arm vary because of
> the force term?

If I understand you correctly, the "constant" we add to the sinusoidal
motion is the primary effect we're looking for. That is the
equilibrium theta the damped oscillator will theoretically settle to
due to the gravitational torque on the wire if we could wait all day.
Unfortunately, very slow drifts in the apparatus due to diurnal
temperature variation and such will cause significant error if we wait
too long. That's why Cavendish (an everyone else) simply tracks the
oscillation and interpolates the equilibrium theta.

Incidentally, I'm am working my way through Cavendish's original
report. I'm not sure yet, but I don't think he takes into account
damping and gravitional gradients fully as I do. It would be
interesting to reanalyse his oscillation extremum data with these
things taken into account. It would require solving for the
equilibrium theta of a damped harmonic oscillator from the times and
amplitudes of the extrema. Does anyone know if this has been done
before? If not, why don't you do that.

### Edward Ruden

Sep 11, 2007, 12:31:35 PM9/11/07
to
Here's followup on linearizing the Torque. Sorry about mangling the
English in my last post. Bailey 1884 performs a detailed damped
harmonic oscillator analysis of his own version of the Cavendish

http://www.alphysics.com/cavendishexperiment/cavendishexperiment.html

(thanks for the links!). The rhs of the first equation on p 218 is the
linearized torque (he later goes on to treat damping). One significant
source of error is the fact that, due to the inverse square law, the
gravitational field is not uniform. This complicates the trajectory of
the pendulum, making inferences about its oscillation center more
difficult. Letting y=theta and taking prime to mean time derivative,
the equation of motion with linear viscous damping can be written:

ay''+by'+cy+d=N(y)

where a ,b, c, and d are constants, and N(y) is the torque as a
function of y. The y dependence results from the inverse square law.
Equations of the form shown, but with the rhs equal to zero have the
simple analytic general solution - an offset decaying harmonic
oscillator where the frequency is a function of a, b, and c. An
analytic solution cannot be found given the exact form of N(y) but,
fortunately, any continuous function may be represented by a line for
sufficiently small displacements about a given point. We approximate
N(y), then, by

N(y) = e+fy

where e is N at y=0 (which we take to be at the center of the
apparatus, not the oscillation), and f is dN/dy at y=0. Our equation
of motion then becomes

ay''+by'+(c-f)y+(d-e)=0

This equation, therefore has a offset decaying sine wave general
solution, but where the coefficients and, therefore, the frequency
have been altered. You will find this approximation perfectly adequate
for the quality of the data Cavendish took. However, Cavendish only
records three extremum points of the oscillation. This is sufficient,
however, (after a little calculus max/min analysis and some algebra)
to fit a unique offset decaying sine wave to the data. We then can
find the oscillation center. This is something worth doing for the
original Cavendish data, assuming it has not already been done by
*someone* in the last 200 years. In other words, how good is
Cavendish's data proper (independent of his primative analysis)?

Why is the frequency altered by linearizing N in this way, you ask?
Well, it's because it subtracts a contribution to acceleration
proportional to y, effectively lowering the spring constant! Damping
lowers the frequency too; the viscous drag slows the pendulum causing
the period to lenghten. Wikipedia has a nice article on this at

http://en.wikipedia.org/wiki/Harmonic_oscillator

### pioneer1

Sep 15, 2007, 4:16:15 PM9/15/07
to
On Sep 9, 10:15 pm, Edward Ruden <rudenbz...@yahoo.com> wrote:

> > x = theta d << a
>
> This assumption is necessary to linearize the differential equation.

I have a couple of problems with this assumption.

1. I am not sure that it is justified. We are trying to measure a
small quantity which depends on the change in a, i.e., delta a = a -
x. The change in a is relevant to this experiment. (A mathematical
argument for this is needed.)

2. The differential equation is non-linear. This is a non-linear
problem. If we make it linear the equation of motion no longer
represents the motion of the pendulum. If we make the differential
equation linear then differential equation and its solution answer two
fundamentally different questions.

3. Linearization removes theta. Let's write the force term as

GMmd/(a-theta d)^2

This way I see that all terms acting on the pendulum arm (torsion,
damping and force) have theta in them. If we remove theta d the force
term no longer acts on the arm:

GMmd/a^2

does not act on the pendulum arm because it is independent of theta.

Computing G from a term that does not act on the attracted weight does
not make sense to me. If we remove theta from the force term we
effectively remove the force as an agent altering the motion of the
pendulum.

>Without
> linearization, your stuck you'll need to solve it numerically. You you
> want to take that route, get a book on numerical methods.
>

I am working on this. I asked the question on PlanetMath
http://planetmath.org/?op=getmsg&id=16676 and some people are trying
to help.

> Incidentally, I'm am working my way through Cavendish's original
> report. I'm not sure yet, but I don't think he takes into account
> damping and gravitional gradients fully as I do.

I think you are right. If you have any results regarding the original

>It would be
> interesting to reanalyse his oscillation extremum data with these
> things taken into account.

I would try this but I want to understand the issue about the equation
of motion first.

### Edward Ruden

Sep 17, 2007, 7:32:48 AM9/17/07
to
On Sep 15, 2:16 pm, pioneer1 <1pione...@gmail.com> wrote:
>
> > > x = theta d << a
> .... The change in a is relevant to this experiment.
>
Let's clear this up first. There is no change in a. On my page 11, a
is defined as the distance from the center of the big ball to the
center of the box holding the pendulum. This approximation lets x
represent theta directly without trig functions (aka the small angle
approximation), and it lets us approximate the gravitational field's x-
dependence by a straight line tangent to the real field at x=0 (good
for small x)
>
> GMmd/(a-theta d)^2
>
Yes, multiplied by two, this is an accurate estimate of the torque on
the wire due to gravity. I approximate this by A + B*x, where
x=theta*d, A is the torque at x=0, and B is the torque's derivative
w.r.t. x at x=0 This is a first order Taylor expansion.

> If we remove theta d the force term no longer acts on the arm:
>
> GMmd/a^2
>
> does not act on the pendulum arm because it is independent of theta.
>

No. Remember, this is half the torque due to gravity. This
contribution to not having a theta dependence simply means it is
constant. It's still acting to shift the equilibrium theta. When the
big balls are rotated to the opposite side, the *sign* of the torque
changes, causing the equilibrium theta to shift in the other
direction. The goal of the analysis is to determine that shift
accurately. Note, however, as explained, I do not make such a crude
approximation. The Taylor expansion accounts for its dependence on
theta, but only to first order. This is not absolutely necessary, but
it improves the accuracy of the analysis to better than one percent.

### pioneer1

Sep 19, 2007, 6:27:59 AM9/19/07
to
On Sep 17, 7:32 am, Edward Ruden <rudenbz...@yahoo.com> wrote:
> On Sep 15, 2:16 pm, pioneer1 <1pione...@gmail.com> wrote:

> Let's clear this up first. There is no change in a.

Right. I meant change in (a - x). Thanks for correcting.

Robert Israel kindly ran a numerical calculation in Maple and posted
the result at sci.math

Can you take a look at that thread? I see that in your paper you
computed damping by way of your curve fitting program. Do you think
there is a way to compute the damping coefficient for the original
Cavendish experiment data? It would be great to compare linear
equation you used and the non-linear solution and see if there is a
difference.

Judging from the Maple graph, it looks very close to linear motion but
there may be a difference if we can compare the numbers for each
solution.

There is also a discrepency in the result. My calculation from the
original experiment shows a value for theta 200 times greater than
what Maple gave as computed by Robert Israel.

### Edward Ruden

Sep 19, 2007, 10:57:00 PM9/19/07
to
On Sep 19, 4:27 am, pioneer1 <1pione...@gmail.com> wrote:

> Do you think
> there is a way to compute the damping coefficient for the original
> Cavendish experiment data?

Yes. as I said on Sep 11, "Cavendish only

records three extremum points of the oscillation. This is sufficient,
however, (after a little calculus max/min analysis and some algebra)
to fit a unique offset decaying sine wave to the data"

What you want to do fit the function

x(t)=A*exp(-B*t)*sin(C*t+D)+E

to the measure extrema with coordinates (t1,x1), (t2,x2), (t3,x3)
where the first and last are assumed to be extrema of one sign and
(t2,x2) is the intemediate extremum of the other sign. The t2
measurement actually overspecifies the problem since it can be shown
that t2=(t1+t3)/2 for our fitting function (the extrema are equally
spaced for the decaying sine wave). To avoid this, we simply define
t2=(t1+t3)/2. That leaves 5 measurements and 5 fitting parameters
(A...E), subject to the constraint that the measured times correspond
to extrema. That means we need five equation. They are:

xi=x(ti), where i=1,2,3 and 0=x'(ti) where i=1,3

Prime, here, means the derivative wrt t. Setting the derivatives = 0
means t1 and t3 are extrema. Normally 5 equations and five unknowns
means a lot of work, but there are some tricks with equation ratios
and differences that makes it relatively easy. Once you find A...B
for the big balls at both extremes, you can determine G, as shown in
my final report.

Based on our exchanges, I suspect you'll find the math difficult. If
you can't manage it and if I have time in the near future I can trudge
through it and post the results. Reader contributions are welcome to,
of course. Again, the question we are trying to answer is, "How good

### pioneer1

Sep 25, 2007, 7:30:45 AM9/25/07
to
On Sep 19, 10:57 pm, Edward Ruden <rudenbz...@yahoo.com> wrote:
> On Sep 19, 4:27 am, pioneer1 <1pione...@gmail.com> wrote:
..

I am trying to compare the linear solution as stated in your paper
with the numerical solution as solved by Robert Israel. He kindly
posted

the y(t) for one period with 10 seconds intervals.

But I ran into a problem regarding the phase, or it seems like it.

Without the damping term your solution is

y(t) = A cos w t + B/w

w == omega_0^2 == omega_f (when there is no damping)
B == 2GMmd/Ia^2

The numerical solution has

y(0) = 0

I would think that this should be the case since initially the angle
is zero.

y(0) = A + B/w

so it is not 0 when t = 0.

Any thoughts on this?

### Edward Ruden

Sep 26, 2007, 2:09:24 AM9/26/07
to
You were provided a solution with "initial conditions y(0) = 0 and
y'(0) = 0", as was explained. The solution to a second order
differential equation such as ours requires initial conditions for
y(0) and y'(0). To compare my linearized solution to the more
general one (with full inverse square law) you need to specify the
same initial conditions for the numerical algorithm. You didn't know

Keep in mind too that if you wish to analyze Cavendish's data in terms
of the unlinearized dif eq, the initial conditions won't be so easy to
specify for the numerical solution since you're only given peak y
values and intermediate (t,y) data points. You'll have to interate G
and the damping coefficient, in addition to y(0) and y'(0) to match
the data. That'll be quite a chore. It's still a bit of a chore to fit
parameters A ... E of my linearized solution. My previous reply on
this assumed the times of the extrema were recorded (I see they were
not), but the intermediate data record can cover for that; it's just
more complicate (8 equations and 8 unknown). It gets a bit messy, but
it's still analystically reducible to one equation in one variable,
solved numerically. Maple's good at that.

I admire your persistence, but don't you think you should get some
math and physics courses under your belt? I get the impression that

### pioneer1

Oct 15, 2007, 10:32:33 AM10/15/07
to
On Sep 26, 2:09 am, Edward Ruden <rudenbz...@yahoo.com> wrote:

>To compare my linearized solution to the more
> general one (with full inverse square law) you need to specify the
> same initial conditions for the numerical algorithm. You didn't know
> enough to ask for that.

Right. I am confused about the initial conditions. I think I got them
all wrong. Now it looks like for the experiment I used we don't have
initial conditions of t=0, y=0 and y'=0.

I am using data from Experiment IV. I am starting with the last point
of rest when weights are at negative position, 18.1, and then rest
point 24.02, the first one after he moves weights to positive
position.

Here is the complete data

Point of rest: 18.1
Division 19 at 10h54m45s
Time of mid vibration:
10h55m37s
Division 18 at 10h55m45s

Extreme point: 15.5
Extreme point: 31.3

Division 25 at 11h10m25s
Time of mid vibration:
11h10m40s
Division 23 at 11h11m03s

Point of rest: 24.02

How to translate these into initial conditions?

We may assume 18.1 as the initial point but we don't know the time.
Probably we can compute it. Also now y' is not zero because the arm is
passing 18.1 with a certain velocity. How to find this velocity?

I will have to ask for a new numerical solution for the non-linear
case in the sci.math group. I would appreciate advice.

Also I posted a plot of the comparison of linear and non-linear
solutions for t=0, y=0, y'=0. But there is a big discrepency in
amplitudes and periods.

http://www.densytics.com/wiki/index.php?title=Cavendish_experiment_%28equation_of_motion%29#Comparison_of_linear_and_nonlinear_solutions

>
> I admire your persistence, but don't you think you should get some
> math and physics courses under your belt? I get the impression that

### Edward Ruden

Oct 19, 2007, 10:11:17 AM10/19/07
to
On Oct 15, 8:32 am, pioneer1 <1pione...@gmail.com> wrote:

> How to translate these into initial conditions?

I tried to explain this already. Think about what you have in
mathematical terms. The minimum set of info we need is:
x1, (x2,t2), x3, and (x4,t4) and the fact that x1 and x3 correspond to
positive and negative extrema. The even numbered subscripts are for
intermediate data. The "rest point" doesn't count; that's just an
estimate Cavendish made, and we're trying to do better than Cavendish.
It's annoying that we don't have the times for the extrema, but that's
understandable. The arm is moving so slowly then, it's hard to
estimate that time accurately anyway. Fortunatly, the intermediate
values make up for it.

Now think about what you *want*. If you're solving the "exact" dif eq,
you want x(0), x'(0), G, torsion constant, and damping constant. Those
are *all* needed to solve it numerically. But, since it's solved
numerically, really all you can do it iterate these values until you
converge on a solution where all your knowns are satisfied. You will
find that to be very difficult to code.

However, if you use my linearized solution, you want to find A,
B, ,,,, E from the known parameters. The extremum statements can be
expressed as x'=0 (you need to know how to differentiate an analytic
equation). Other than that, it's "just" algebra and a bit of numerical
root finding in the end. It gets pretty sticky, but it can be done. I
can't make it any simpler than that unless I just went and did it
myself.