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Aug 27, 2007, 9:47:27â€¯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.

Aug 28, 2007, 8:34:33â€¯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

..

> 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.

Aug 29, 2007, 6:00:22â€¯AM8/29/07

to

More recent data,

<http://www.npl.washington.edu/eotwash/publications/pdf/prl85-2869.pdf>

<http://www.npl.washington.edu/eotwash/publications/pdf/mst10-454.pdf>

<http://www.npl.washington.edu/eotwash/publications/pdf/prd54-1256R.pdf>

--

Uncle Al

http://www.mazepath.com/uncleal/

(Toxic URL! Unsafe for children and most mammals)

http://www.mazepath.com/uncleal/lajos.htm#a2

Aug 30, 2007, 1:37:18â€¯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.

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)

Aug 30, 2007, 1:37:20â€¯AM8/30/07

to

On Aug 29, 6:00 am, Uncle Al <Uncle...@hate.spam.net> wrote:

> pioneer1 wrote:

> pioneer1 wrote:

> More recent data,

>

> <http://www.npl.washington.edu/eotwash/publications/pdf/prl85-2869.pdf>

> <http://www.npl.washington.edu/eotwash/publications/pdf/mst10-454.pdf>

> <http://www.npl.washington.edu/eotwash/publications/pdf/prd54-1256R.pdf>

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.

Aug 30, 2007, 1:37:19â€¯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.

Aug 30, 2007, 10:55:51â€¯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

Aug 30, 2007, 10:55:52â€¯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 :-)

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.

Sep 2, 2007, 2:41:49â€¯PM9/2/07

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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.

Sep 5, 2007, 8:02:32â€¯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.

>

> ... 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.

Sep 8, 2007, 9:25:42â€¯PM9/8/07

to

On Sep 5, 8:02 am, Edward Ruden <rudenbz...@yahoo.com> wrote:

> I've uploaded my old final report to

> http://home.comcast.net/~rudenbiz2/physics/pubs/Ruden_80_Cavendish_Experiment.pdf

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.

Recently I came across this page

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.

Sep 9, 2007, 10:15:26â€¯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

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.

Sep 11, 2007, 12:31:35â€¯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

experiment. It can be downloaded at

English in my last post. Bailey 1884 performs a detailed damped

harmonic oscillator analysis of his own version of the Cavendish

experiment. It can be downloaded at

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

Sep 15, 2007, 4:16:15â€¯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.

I would appreciate your comments on this. Thanks

>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

experiment please let me know.

>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.

Sep 17, 2007, 7:32:48â€¯AM9/17/07

to

> .... 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.

>

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.

Sep 19, 2007, 6:27:59â€¯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:

> 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.

I will read the rest of your post more carefully and reply later. But

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.

I would appreciate your comments on this. Thanks again.

Sep 19, 2007, 10:57:00â€¯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

Sep 25, 2007, 7:30:45â€¯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:

..> 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.

But your solution is

y(0) = A + B/w

so it is not 0 when t = 0.

Any thoughts on this?

Sep 26, 2007, 2:09:24â€¯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

enough to ask for that.

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

enough to ask for that.

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

you're a bit over your head at the moment.

Oct 15, 2007, 10:32:33â€¯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.

Thanks again for your help.

>

> 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

> you're a bit over your head at the moment.

Good advice. Thanks.

Oct 19, 2007, 10:11:17â€¯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.

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